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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 2
512 free, high-quality past-paper-style questions on Equations, formulae and identities for Edexcel IGCSE Maths, each with a full mark scheme. Covers: Algebraic manipulation and factorising; Direct and inverse proportion and rearranging formulae; Linear and simultaneous equations; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Simplify fully (2x3y)4
| Marking point | Mark | Notes |
|---|---|---|
| Raise every factor to the power 4: 24 = 16, (x3)4 = x3×4 = x12, y4. Answer: 16x12y4 | ✓ 1 | Answer D |
Answer: D
Common error: A power of a power means multiply the indices (3 × 4 = 12), not add them. The number 2 must also be raised to the power 4, not multiplied by 4.
Factorise completely 12x2y − 18xy2
| Marking point | Mark | Notes |
|---|---|---|
| The highest common factor of 12x2y and 18xy2 is 6xy. 12x2y ÷ 6xy = 2x, 18xy2 ÷ 6xy = 3y. Answer: 6xy(2x − 3y) | ✓ 1 | Answer D |
Answer: D
Exam tip: "Completely" means the bracket must have no common factor left. Check the terms inside: 2x and 3y share nothing, so you are done.
Expand and simplify (2x − 5)2
| Marking point | Mark | Notes |
|---|---|---|
| (2x − 5)(2x − 5) = 4x2 − 10x − 10x + 25 = 4x2 − 20x + 25 | ✓ 1 | Answer D |
Answer: D
Common error: (a − b)² is not a² − b². Write the bracket out twice and multiply every term, so you get both middle terms.
A spreadsheet cell calculates T = 2a2 − 5ab.
Find the value of T when a = −3 and b = 4.
| Marking point | Mark | Notes |
|---|---|---|
| 2a2 = 2 × (−3)2 = 2 × 9 = 18 −5ab = −5 × (−3) × 4 = +60 T = 18 + 60 = 78 | ✓ 1 | Answer C |
Answer: C
Common error: Squaring a negative number gives a positive result: (−3)² = 9, not −9. Put negative values in brackets when you substitute so the signs stay visible.
A city bike-share scheme charges a fixed fee of £12 plus £3.50 for each hour a bike is used.
The total cost is £C for h hours.
Write down a formula for C in terms of h.
Work out the total cost for 5 hours.
Mei paid £47. For how many hours did she use the bike?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3.5h or 12 seen as part of an expression for the cost | ✓ 1 | |
| C = 12 + 3.5h | ✓ 1 | |
| Part (b) | ||
| C = 12 + 3.5 × 5 = £29.50 | ✓ 1 | |
| Part (c) | ||
| 47 = 12 + 3.5h or 3.5h = 35 | ✓ 1 | |
| h = 10 | ✓ 1 | |
Common error: A formula must start with the subject and an equals sign: write C = 12 + 3.5h, not just 12 + 3.5h, and not C = 15.5h.
The straight line L1 has equation y = 3x − 4.
The straight line L2 has equation 2x + y = 11.
What are the coordinates of the point where L1 and L2 intersect?
| Marking point | Mark | Notes |
|---|---|---|
| At the intersection both equations hold. Substitute y = 3x − 4 into 2x + y = 11: 2x + 3x − 4 = 11, so 5x = 15 and x = 3. y = 3(3) − 4 = 5. The point is (3, 5). | ✓ 1 | Answer C |
Answer: C
Exam tip: Check your point in BOTH equations: 3(3) − 4 = 5 and 2(3) + 5 = 11. A quick check catches swapped coordinates.
Ana thinks of a number x. When she squares her number and then subtracts 3 times her number, the result is 28.
Which of these gives all the possible values of x?
| Marking point | Mark | Notes |
|---|---|---|
| x2 − 3x = 28 so x2 − 3x − 28 = 0. Two numbers with product −28 and sum −3: −7 and 4. (x − 7)(x + 4) = 0, so x = 7 or x = −4. Check: 49 − 21 = 28 ✓ and 16 + 12 = 28 ✓. | ✓ 1 | Answer C |
Answer: C
Common error: After factorising to (x − 7)(x + 4) = 0, the solutions have the opposite signs to the numbers in the brackets: x = 7 and x = −4. Substitute back to check.
In a number puzzle, three times a number x is subtracted from 7 and the result is greater than 19.
Which inequality gives all the possible values of x?
| Marking point | Mark | Notes |
|---|---|---|
| 7 − 3x > 19 −3x > 12 Divide by −3 and reverse the inequality sign: x < −4. | ✓ 1 | Answer A |
Answer: A
Common error: When you multiply or divide both sides of an inequality by a negative number, the inequality sign must be reversed. Check by substituting a value, e.g. x = −5 gives 7 + 15 = 22 > 19.
A smart thermostat uses a setting x that must satisfy −5 ≤ 2x + 3 < 11.
The solution set is to be shown on a number line. Which description is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Subtract 3 from all three parts: −8 ≤ 2x < 8 Divide all three parts by 2: −4 ≤ x < 4 ≤ includes −4, so a closed (filled) circle at −4; < excludes 4, so an open circle at 4. | ✓ 1 | Answer A |
Answer: A
Exam tip: On a number line, a closed circle means the end value is included (≤ or ≥) and an open circle means it is not (< or >). Do the same operation to all three parts of a double inequality.
Solve each equation.
\(5x - 7 = 18\)
\(3(y + 4) = 33\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for adding 7 to both sides, giving \(5x = 25\) | ✓ 1 | |
| A1 \(x = 5\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(3y + 12 = 33\) or \(y + 4 = 11\) | ✓ 1 | |
| A1 \(y = 7\) | ✓ 1 | |
Exam tip: In part (b) you can divide both sides by 3 first. This avoids expanding the bracket and gives fewer chances to slip.
Solve each equation.
\(7x + 3 = 3x + 23\)
\(\frac{2w - 1}{3} = 5\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for collecting the \(x\) terms on one side and the numbers on the other, e.g. \(4x = 20\) | ✓ 1 | |
| A1 \(x = 5\) | ✓ 1 | |
| Part (b) | ||
| M1 for multiplying both sides by 3, giving \(2w - 1 = 15\) | ✓ 1 | |
| A1 \(w = 8\) | ✓ 1 | |
Common error: In part (b), do not divide only the \(2w\) by 3. Multiply the whole of both sides by 3 to clear the fraction first.
The three angles of a triangle are \(x°\), \((2x + 10)°\) and \((3x - 40)°\).
Show that \(6x - 30 = 180\)
Work out the size of the largest angle of the triangle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(x + 2x + 10 + 3x - 40 = 180\) | ✓ 1 | |
| A1 for simplifying correctly to \(6x - 30 = 180\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(6x = 210\), giving \(x = 35\) | ✓ 1 | |
| A1 for a largest angle of \(80°\) (angles are \(35°\), \(80°\), \(65°\)) | ✓ 1 | |
Exam tip: After finding \(x\), substitute it into all three expressions. Then check that they add to 180° before you choose the largest.
Solve the simultaneous equations
\(3x + 2y = 16\)
\(x - 2y = 0\)
Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for adding the equations to eliminate \(y\), giving \(4x = 16\) | ✓ 1 | |
| A1 \(x = 4\) | ✓ 1 | |
| A1 \(y = 2\) | ✓ 1 | |
Exam tip: The \(y\) terms have opposite signs, so add the equations. Substitute back into the simpler equation to find \(y\).
Solve \(4(2x - 3) = 5x + 9\)
Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for expanding correctly to \(8x - 12\) | ✓ 1 | |
| M1 for rearranging to \(3x = 21\) | ✓ 1 | |
| A1 \(x = 7\) | ✓ 1 | |
Common error: Multiply both terms inside the bracket by 4. Writing \(8x - 3\) instead of \(8x - 12\) loses the accuracy marks.
Solve the equation below by factorising.
\(x^2 + 7x + 12 = 0\)
Factorise \(x^2 + 7x + 12\)
Hence solve \(x^2 + 7x + 12 = 0\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (x + a)(x + b) where a + b = 7 or ab = 12 | ✓ 1 | |
| A1 for (x + 3)(x + 4) | ✓ 1 | |
| Part (b) | ||
| B1 for x = −3 and x = −4 | ✓ 1 | |
Common error: If (x + 3)(x + 4) = 0, the solutions are x = −3 and x = −4, not +3 and +4. Set each bracket equal to zero and solve it.
This question is about linear inequalities.
Solve \(5x + 3 > 2x - 9\)
n is an integer.
List all the values of n that satisfy \(-3 \le 2n + 1 < 9\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 3x > −12, or for −12 in an equivalent form | ✓ 1 | |
| A1 for x > −4 | ✓ 1 | |
| Part (b) | ||
| M1 for subtracting 1 throughout: −4 ≤ 2n < 8 | ✓ 1 | |
| M1 for dividing by 2: −2 ≤ n < 4 | ✓ 1 | |
| A1 for −2, −1, 0, 1, 2, 3 | ✓ 1 | |
Exam tip: Check which end of the inequality is included. Here −2 is included because of ≤, but 4 is not included because of <.
Solve each of these equations.
\(x^2 - 6x = 0\)
\(2x^2 = 72\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for x(x − 6) = 0 | ✓ 1 | |
| A1 for x = 0 and x = 6 | ✓ 1 | |
| Part (b) | ||
| M1 for x2 = 36 | ✓ 1 | |
| A1 for x = 6 and x = −6 | ✓ 1 | |
Common error: Dividing both sides of x² = 6x by x loses the solution x = 0. Factorise instead. Taking a square root also gives a negative solution as well as a positive one.
A formula used for a moving object is
\(v = u + at\)
Make t the subject of the formula.
Find the value of t when v = 25, u = 4 and a = 3.5
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 subtracts u from both sides, e.g. \(v - u = at\) | ✓ 1 | |
| A1 \(t = \frac{v-u}{a}\) | ✓ 1 | |
| Part (b) | ||
| M1 substitutes into a correct rearrangement, e.g. \(\frac{25-4}{3.5}\) | ✓ 1 | |
| A1 6 | ✓ 1 | |
Exam tip: Undo the operations in reverse order: remove the added term first, then divide. Doing the division first leaves you with u/a, which is wrong.
The area, A cm2, of a trapezium with parallel sides a cm and b cm and height h cm is given by
\(A = \frac{(a+b)h}{2}\)
Make h the subject of the formula.
A trapezium has area 52.5 cm2 and parallel sides of length 6.4 cm and 8.6 cm.
Work out the height of the trapezium.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies both sides by 2, e.g. \(2A = (a+b)h\) | ✓ 1 | |
| A1 \(h = \frac{2A}{a+b}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{2 \times 52.5}{6.4 + 8.6}\) or \(\frac{105}{15}\) | ✓ 1 | |
| A1 7 cm | ✓ 1 | |
Common error: Writing h = 2A/a + b loses the accuracy mark. Keep (a + b) together in brackets as the denominator.
8 identical machines can complete a printing job in 15 hours.
The time taken is inversely proportional to the number of machines used.
How long would the same job take using 6 of these machines?
The job must be finished in 5 hours.
How many machines are needed?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 8 × 15 = 120 (machine-hours) | ✓ 1 | |
| A1 20 hours | ✓ 1 | |
| Part (b) | ||
| M1 120 ÷ 5 | ✓ 1 | |
| A1 24 | ✓ 1 | |
Common error: Fewer machines means more time. Using direct proportion here (15 × 6 ÷ 8 = 11.25 hours) gives an answer that should be spotted as unreasonable.
Answer each part.
Simplify 5x + 3y − 2x + y
Expand 4(2x − 5)
Factorise 6x + 15
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1: 3x + 4y | ✓ 1 | |
| Part (b) | ||
| B1: 8x − 20 | ✓ 1 | |
| Part (c) | ||
| B1: 3(2x + 5) | ✓ 1 | |
Exam tip: After factorising, expand your answer in your head to check that you get back the original expression.
Simplify each expression.
m4 × m3
(p5)2
\(\frac{12x^6y^3}{4x^2y}\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1: m7 | ✓ 1 | |
| Part (b) | ||
| B1: p10 | ✓ 1 | |
| Part (c) | ||
| B1 for any two of 3, x4, y2 correct in a single product | ✓ 1 | |
| B1 for fully correct 3x4y2 | ✓ 1 | |
Common error: For (p⁵)², multiply the powers to get p¹⁰; adding them to give p⁷ is wrong.
Answer each part.
Expand and simplify (x + 6)(x − 4)
Factorise x2 − 9x + 20
Hence solve x2 − 9x + 20 = 0
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for at least three of the four terms x2, −4x, +6x, −24 correct | ✓ 1 | |
| A1: x2 + 2x − 24 | ✓ 1 | |
| Part (b) | ||
| M1 for (x ± 4)(x ± 5) or brackets that expand to give two of the three terms correctly | ✓ 1 | |
| A1: (x − 4)(x − 5) | ✓ 1 | |
| Part (c) | ||
| B1: x = 4 and x = 5 (follow through their factors) | ✓ 1 | |
Exam tip: For x² + bx + c, look for two numbers that multiply to c and add to b: here −4 and −5 multiply to 20 and add to −9.
Answer each part.
Factorise fully 8x2y − 12xy2
Expand and simplify 3(2x + 1) − 2(x − 4)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for a correct partial factorisation, e.g. 4x(2xy − 3y2) or xy(8x − 12y) | ✓ 1 | |
| A1: 4xy(2x − 3y) | ✓ 1 | |
| Part (b) | ||
| M1 for 6x + 3 − 2x + 8 (allow one error in sign or value) | ✓ 1 | |
| A1: 4x + 11 | ✓ 1 | |
Common error: −2 × −4 = +8. Writing −8 when expanding −2(x − 4) is the most frequent slip in this type of question.
The velocity, v m/s, of a falling object is given by the formula v = u + at, where u is the starting velocity in m/s, a is the acceleration in m/s2 and t is the time in seconds.
Work out the value of v when u = 4.5, a = 9.8 and t = 3
Make t the subject of the formula v = u + at
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 4.5 + 9.8 × 3 | ✓ 1 | |
| A1: 33.9 | ✓ 1 | |
| Part (b) | ||
| M1 for v − u = at | ✓ 1 | |
| A1: t = \(\frac{v-u}{a}\) | ✓ 1 | |
Common error: Working out (4.5 + 9.8) × 3 = 42.9 ignores the order of operations: the multiplication at is done first.
The cost of notebooks is directly proportional to the number of notebooks bought.
6 notebooks cost £8.70
Work out the cost of 15 notebooks.
Priya spends exactly £30.45 on these notebooks.
How many notebooks does she buy?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 finds the cost of one notebook, 8.70 ÷ 6 (= 1.45), or uses a multiplier such as 15 ÷ 6 (= 2.5) | ✓ 1 | |
| A1 £21.75 | ✓ 1 | |
| Part (b) | ||
| M1 30.45 ÷ 1.45, or 30.45 ÷ 8.70 (= 3.5) then × 6 | ✓ 1 | |
| A1 21 | ✓ 1 | |
Exam tip: Finding the cost of one item first (the unitary method) works for every part of a direct proportion question.
A rectangular rug has a width of x metres and a length of (x + 4) metres.
The area of the rug is 45 m2.
Show that \(x^2 + 4x - 45 = 0\)
Work out the perimeter of the rug.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for x(x + 4) = 45 | ✓ 1 | |
| A1 for x2 + 4x = 45, leading to x2 + 4x − 45 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 for a correct method to solve the quadratic, e.g. (x + 9)(x − 5) = 0, or substitution into the quadratic formula, or (x + 2)2 = 49 | ✓ 1 | |
| A1 for x = 5, with x = −9 rejected because a length cannot be negative | ✓ 1 | |
| A1 for a perimeter of 2(5 + 9) = 28 m | ✓ 1 | |
Exam tip: When a quadratic models a length, state clearly which solution you reject and why. Then use the valid value to answer the question that was actually asked.
y is directly proportional to x.
When x = 12, y = 42
Find a formula for y in terms of x.
Find the value of x when y = 91
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(y = kx\) and substitutes, \(42 = 12k\) | ✓ 1 | |
| A1 \(y = 3.5x\) or equivalent, e.g. \(y = \frac{7}{2}x\) | ✓ 1 | |
| Part (b) | ||
| B1 26 (follow through from their formula) | ✓ 1 | |
Exam tip: Always write the final formula with k replaced by its value, e.g. y = 3.5x, not just k = 3.5.
This question is about expressions.
Expand and simplify 3(2x + 5) − 2(x − 4). (2)
Factorise fully 12a2b − 18ab2. (2)
Expand and simplify (x + 3)(x − 5). (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 6x + 15 − 2x + 8 (at least 3 terms correct) | ✓ 1 | |
| A1 4x + 23 | ✓ 1 | |
| Part (b) | ||
| M1 factor of 6, a or b taken out | ✓ 1 | |
| A1 6ab(2a − 3b) | ✓ 1 | |
| Part (c) | ||
| M1 x2 − 5x + 3x − 15 (3 of 4 terms correct) | ✓ 1 | |
| A1 x2 − 2x − 15 | ✓ 1 | |
Common error: When a bracket has a minus sign in front of it, every term inside changes sign: −2(x − 4) gives −2x + 8.
A taxi company charges a fixed fee of €3.50 plus €1.40 for every kilometre travelled.
A journey costs €24.50. Work out the distance travelled. (3)
Solve 5x − 7 = 2x + 11. (2)
Solve the inequality 3x + 4 < 19. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 24.50 − 3.50 (= 21) | ✓ 1 | |
| M1 21 ÷ 1.40 | ✓ 1 | |
| A1 15 km | ✓ 1 | |
| Part (b) | ||
| M1 3x = 18 (collecting terms correctly) | ✓ 1 | |
| A1 x = 6 | ✓ 1 | |
| Part (c) | ||
| M1 3x < 15 | ✓ 1 | |
| A1 x < 5 | ✓ 1 | |
Exam tip: Solve an inequality as you would an equation, keeping the inequality sign; only reverse it if you multiply or divide by a negative number.
The speed v m/s of a rocket booster is given by v = u + at, where u m/s is the starting speed, a m/s2 is the acceleration and t s is the time.
Work out v when u = 12, a = 3.5 and t = 8. (2)
Rearrange the formula to make t the subject. (2)
Work out t when u = 5, v = 47 and a = 2.8. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 12 + 3.5 × 8 | ✓ 1 | |
| A1 40 | ✓ 1 | |
| Part (b) | ||
| M1 v − u = at | ✓ 1 | |
| A1 t = \(\frac{v - u}{a}\) | ✓ 1 | |
| Part (c) | ||
| M1 (47 − 5) ÷ 2.8 | ✓ 1 | |
| A1 15 s | ✓ 1 | |
Exam tip: Substitute into a rearranged formula only after you have checked each step of the rearranging, or substitute first and solve the resulting equation.
Tickets for a science museum cost €a for an adult and €c for a child. 3 adults and 2 children pay €47 in total. 2 adults and 4 children pay €50 in total.
Write down two equations in a and c. (2)
Solve your equations to find the price of an adult ticket and the price of a child ticket. (3)
Work out the cost of 4 adult tickets and 5 child tickets. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3a + 2c = 47 | ✓ 1 | |
| B1 2a + 4c = 50 | ✓ 1 | |
| Part (b) | ||
| M1 multiply to match coefficients of one letter (e.g. 6a + 4c = 94) | ✓ 1 | |
| M1 subtract or add to eliminate and solve for one letter | ✓ 1 | |
| A1 a = 11 and c = 7 | ✓ 1 | |
| Part (c) | ||
| M1 4 × 11 + 5 × 7 | ✓ 1 | |
| A1 €79 (ECF from (b)) | ✓ 1 | |
Exam tip: After solving simultaneous equations, substitute both values into the equation you did not use to check them.
The extension e cm of a spring is directly proportional to the load F newtons hung on it. A load of 8 N gives an extension of 5.6 cm.
Find a formula for e in terms of F. (2)
Work out the extension for a load of 15 N. (1)
The spring has an extension of 17.5 cm. Work out the load. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 e = kF and 5.6 = 8k | ✓ 1 | |
| A1 e = 0.7F | ✓ 1 | |
| Part (b) | ||
| A1 10.5 cm (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 17.5 = 0.7F or 17.5 ÷ 0.7 | ✓ 1 | |
| A1 25 N | ✓ 1 | |
Common error: Direct proportion means the ratio e : F is constant; do not add the same amount to both quantities.
Solve the equations in this question. Show your working.
Solve 4x − 7 = 2x + 11
A pharmacist has a tablet containing (3x + 5) mg of an active ingredient and a capsule containing (x + 17) mg. Both contain the same amount of the ingredient. Work out x and the amount, in mg, in one tablet.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 collect terms, e.g. 2x = 18 or 4x − 2x = 11 + 7 | ✓ 1 | |
| A1 x = 9 | ✓ 1 | |
| Part (b) | ||
| M1 3x + 5 = x + 17 | ✓ 1 | |
| M1 2x = 12 | ✓ 1 | |
| A1 x = 6 | ✓ 1 | |
| A1 23 mg | ✓ 1 | |
Common error: When you move a term across the equals sign, change its sign; 4x − 2x = 11 + 7, not 11 − 7.
In a video game, 3 swords and 2 shields cost 190 gold coins in total. 2 swords and 3 shields cost 185 gold coins in total. A sword costs s coins and a shield costs h coins.
Write down two equations in s and h.
Solve your equations to find s and h.
Work out the cost of 5 swords and 4 shields.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3s + 2h = 190 | ✓ 1 | |
| B1 2s + 3h = 185 | ✓ 1 | |
| Part (b) | ||
| M1 multiply first by 3 and second by 2 (equal coefficients of h or s), e.g. 9s + 6h = 570 and 4s + 6h = 370 | ✓ 1 | |
| M1 subtract: 5s = 200 | ✓ 1 | |
| A1 s = 40 and h = 35 | ✓ 1 | |
| Part (c) | ||
| M1 5 × 40 + 4 × 35 | ✓ 1 | |
| A1 340 gold coins | ✓ 1 | |
Exam tip: Substitute your answers into the other original equation to check: 2 × 40 + 3 × 35 = 185.
A water pump delivers water at a constant rate. It delivers 480 litres in 12 minutes.
Work out how many litres the pump delivers in 40 minutes.
Work out how long the pump takes to deliver 1800 litres.
Six identical pumps working together fill a storage tank in 8 hours. Work out how long 4 of these pumps take to fill the same tank, working at the same rate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 480 ÷ 12 = 40 litres per minute | ✓ 1 | |
| A1 1600 litres | ✓ 1 | |
| Part (b) | ||
| M1 1800 ÷ 40 | ✓ 1 | |
| A1 45 minutes | ✓ 1 | |
| Part (c) | ||
| M1 6 × 8 = 48 pump-hours, then 48 ÷ 4 | ✓ 1 | |
| A1 12 hours | ✓ 1 | |
Common error: Do not treat the last part as direct proportion: fewer pumps means more time, so the quantities are in inverse proportion.
Questions about equations in a video game.
Solve 7 − 2x = 19
Solve 3(2x − 1) = 4x + 9
A game awards 12 points for each level completed plus a bonus of 40 points. A player has 172 points in total. Form and solve an equation to find the number of levels completed, n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 −2x = 12 | ✓ 1 | |
| A1 x = −6 | ✓ 1 | |
| Part (b) | ||
| M1 expand: 6x − 3 | ✓ 1 | |
| M1 2x = 12 | ✓ 1 | |
| A1 x = 6 | ✓ 1 | |
| Part (c) | ||
| M1 12n + 40 = 172 | ✓ 1 | |
| M1 12n = 132 | ✓ 1 | |
| A1 n = 11 levels | ✓ 1 | |
Common error: In 7 − 2x = 19 the coefficient of x is −2; divide by −2 and do not drop the sign.
Answer the following questions on indices.
Simplify x3 × x5.
Simplify (2y4)3.
Work out the value of 2−3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 x8 | ✓ 1 | |
| Part (b) | ||
| M1 23 or y12 | ✓ 1 | |
| A1 8y12 | ✓ 1 | |
| Part (c) | ||
| B1 \(\frac{1}{8}\) (= 0.125) | ✓ 1 | |
Common error: A common error is to write (2y⁴)³ as 2y¹²; the power applies to the 2 as well.
At a theatre, an adult ticket costs £x and a child ticket costs £(x − 6). A group buys 3 adult tickets and 2 child tickets for a total of £88.
Write an expression, in terms of x, for the total cost in pounds.
Work out the cost of an adult ticket.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3x + 2(x − 6) | ✓ 1 | |
| Part (b) | ||
| M1 3x + 2x − 12 = 88 or 5x − 12 = 88 | ✓ 1 | |
| M1 5x = 100 | ✓ 1 | |
| A1 £20 | ✓ 1 | |
Exam tip: Expand brackets fully before collecting terms: 2(x − 6) = 2x − 12.
A cargo ship changes speed according to the formula v = u + at, where u is the initial speed, v is the final speed, a is the acceleration and t is the time. Speeds are in m/s, acceleration in m/s2 and time in s.
Make t the subject of the formula.
Work out t when u = 4, v = 7 and a = 0.05.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 at = v − u | ✓ 1 | |
| A1 t = \(\frac{v-u}{a}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{7-4}{0.05}\) | ✓ 1 | |
| A1 60 s | ✓ 1 | |
Common error: Subtract u from both sides before dividing by a; dividing only v by a and then subtracting u is a common slip.
Six cranes working at the same rate unload a ship in 15 hours. The number of cranes is inversely proportional to the time taken.
Work out the time taken by 9 cranes.
The ship must be unloaded in 4.5 hours.
Work out the number of cranes needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 6 × 15 = 90 or 15 × 6 ÷ 9 | ✓ 1 | |
| A1 10 hours | ✓ 1 | |
| Part (b) | ||
| M1 90 ÷ 4.5 | ✓ 1 | |
| A1 20 | ✓ 1 | |
Exam tip: With inverse proportion, the product of the two quantities stays constant: cranes × time = 90.
A retailer offers two delivery schemes for parcels. Scheme A charges £4 plus £0.50 per kilogram. Scheme B charges £1 plus £0.80 per kilogram. A parcel of mass w kg costs the same under both schemes.
Write down an equation in w for the two costs being equal and solve it to find w.
Work out the cost of sending this parcel under either scheme.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 4 + 0.5w = 1 + 0.8w | ✓ 1 | |
| M1 collect terms, e.g. 3 = 0.3w | ✓ 1 | |
| A1 w = 10 | ✓ 1 | |
| Part (b) | ||
| A1 £9 (ECF from (a)) | ✓ 1 | |
Exam tip: Write each line of working under the one before it so a slip in one line cannot hide a correct method.
A health analyst simplifies expressions used in a model of hospital admissions.
Factorise fully 15p2q − 10pq2.
Expand and simplify (x + 5)(x − 2).
Solve 3(2y − 5) = 4y + 7.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 partial factorisation, e.g. 5p(3pq − 2q2) or pq(15p − 10q) | ✓ 1 | |
| A1 5pq(3p − 2q) | ✓ 1 | |
| Part (b) | ||
| M1 at least three correct terms from x2 − 2x + 5x − 10 | ✓ 1 | |
| A1 x2 + 3x − 10 | ✓ 1 | |
| Part (c) | ||
| M1 6y − 15 = 4y + 7 | ✓ 1 | |
| M1 2y = 22 | ✓ 1 | |
| A1 y = 11 | ✓ 1 | |
Common error: When factorising fully, check that nothing common is left inside the bracket: 5p(3pq − 2q²) still has a common factor q.
A test site has small and large wind turbines. Every small turbine produces s kW and every large turbine produces l kW. Three small turbines and two large turbines together produce 17 kW. Two small turbines and four large turbines together produce 26 kW.
Set up two equations and solve them simultaneously to find s and l.
Work out the total output of five small turbines and three large turbines.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 3s + 2l = 17 and 2s + 4l = 26 | ✓ 1 | |
| M1 eliminate one variable, e.g. 6s + 4l = 34 then subtract | ✓ 1 | |
| A1 s = 2 | ✓ 1 | |
| A1 l = 5.5 | ✓ 1 | |
| Part (b) | ||
| M1 5 × 2 + 3 × 5.5 | ✓ 1 | |
| A1 26.5 kW (ECF from (a)) | ✓ 1 | |
Exam tip: After solving, substitute both values back into the other equation to check them.
In a stock model for a shop, n is a whole number of boxes satisfying 7 − 2n > −5 and n ≥ 1.
Solve the inequality 7 − 2n > −5.
Write down all the possible values of n.
Solve 4(m − 3) ≤ 2m + 6.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 −2n > −12 | ✓ 1 | |
| A1 n < 6 | ✓ 1 | |
| Part (b) | ||
| M1 uses both n < 6 and n ≥ 1 | ✓ 1 | |
| A1 1, 2, 3, 4, 5 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 4m − 12 ≤ 2m + 6 so 2m ≤ 18 | ✓ 1 | |
| A1 m ≤ 9 | ✓ 1 | |
Common error: Dividing both sides of an inequality by a negative number reverses the inequality sign.
A theatre hires a follow-spot lamp. The hire cost is £45 plus £12 for each hour of use. Let C be the total cost in pounds and h the number of hours of use.
The total cost for one evening is £141. Work out the number of hours the lamp was used.
Rearrange C = 45 + 12h to make h the subject.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for 45 + 12h = 141 or 141 − 45 = 96 | ✓ 1 | |
| h = 8 | ✓ 1 | |
| Part (b) | ||
| M1 for C − 45 = 12h | ✓ 1 | |
| h = (C − 45)/12 or equivalent | ✓ 1 | |
Exam tip: Undo the fixed charge first (subtract), then deal with the multiplier (divide).
A shipping clerk stacks containers in rows on a quayside.
Factorise fully 12n + 18.
The number of containers in one row, n, satisfies n2 − n − 56 = 0. Find the value of n, given that n is positive.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for a common factor of 2, 3 or 6 | ✓ 1 | |
| 6(2n + 3) | ✓ 1 | |
| Part (b) | ||
| M1 for (n − 8)(n + 7) = 0 | ✓ 1 | |
| n = 8 or n = −7 | ✓ 1 | |
| n = 8 (negative value rejected since a number of containers is positive) | ✓ 1 | |
Common error: When a context says the answer must be positive, state which root you reject and why.
The mass M kg of volcanic ash collected on a sensor plate is directly proportional to the depth d mm of the ash layer. When d = 4, M = 36.
Find M when d = 9.
Find d when M = 117.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for M = kd with k = 36 ÷ 4 = 9 | ✓ 1 | |
| M = 81 | ✓ 1 | |
| Part (b) | ||
| M1 for 117 ÷ 9 or 117 × 4 ÷ 36 | ✓ 1 | |
| d = 13 | ✓ 1 | |
Exam tip: Find the constant of proportionality once and reuse it for every later part.
A theatre has 12 chairs on the stage and places 5 chairs in each of x rows in the stalls. The theatre can hold at most 87 chairs in total, so 5x + 12 ≤ 87.
Solve the inequality 5x + 12 ≤ 87.
Write down the greatest possible number of rows in the stalls.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 5x ≤ 75 | ✓ 1 | |
| x ≤ 15 | ✓ 1 | |
| Part (b) | ||
| 15 | ✓ 1 | |
Common error: Do not change the direction of the inequality sign unless you multiply or divide by a negative number.
The Sun is 1.5 × 108 km from Earth. Light travels at 3.0 × 105 km/s.
Work out the time taken for light to reach Earth from the Sun, in seconds.
Write this time in minutes, correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 1.5 × 108 ÷ (3.0 × 105) | ✓ 1 | |
| 500 s | ✓ 1 | |
| Part (b) | ||
| M1 for 500 ÷ 60 | ✓ 1 | |
| 8.33 | ✓ 1 | |
Exam tip: Divide the numbers and subtract the powers of 10 separately: 10⁸ ÷ 10⁵ = 10³.
A sports scientist writes algebraic expressions to model the load on an athlete during training.
Simplify 3x2y × 4xy3.
Expand and simplify (x + 5)(x − 2).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for 12 | ✓ 1 | |
| x3y4 | ✓ 1 | |
| Part (b) | ||
| M1 for x2 − 2x + 5x − 10 | ✓ 1 | |
| x2 + 3x − 10 | ✓ 1 | |
Common error: When multiplying powers of the same letter, add the indices; do not multiply them.
On a regional rail line the fare £F is directly proportional to the distance d km travelled. A journey of 45 km costs £13.50.
Work out the fare for a journey of 120 km.
A passenger pays £21.60. Work out the distance travelled.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for F = 0.30d or 13.50 ÷ 45 | ✓ 1 | |
| £36 | ✓ 1 | |
| Part (b) | ||
| M1 for 21.60 ÷ 0.30 | ✓ 1 | |
| 72 km | ✓ 1 | |
Exam tip: A fare per kilometre (the constant) lets you answer both directions of the question quickly.
A sports scientist models the height h metres of a thrown ball, t seconds after it is thrown, by h = 20t − 5t2.
Show that h = 15 when t2 − 4t + 3 = 0.
Solve t2 − 4t + 3 = 0 and state the time when the ball is at a height of 15 m on its way down.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 20t − 5t2 = 15 rearranged to 5t2 − 20t + 15 = 0 | ✓ 1 | |
| dividing by 5 to reach t2 − 4t + 3 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 for (t − 1)(t − 3) = 0 | ✓ 1 | |
| t = 1 and t = 3 | ✓ 1 | |
| 3 s (the later time, as the ball is falling) | ✓ 1 | |
Common error: A ball passes any height below its maximum twice; the larger time is the one on the way down.
A mountaineering guide hires out tents. Company A charges a fixed fee of £30 plus £12 for each night. Company B charges £18 for each night and no fixed fee.
Write an expression, in terms of n, for the total cost in pounds of hiring a tent from Company A for n nights.
Find the number of nights for which the two companies charge the same amount.
Work out this amount, in pounds.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 30 + 12n (oe) | ✓ 1 | |
| Part (b) | ||
| M1 forming 30 + 12n = 18n | ✓ 1 | |
| A1 n = 5 | ✓ 1 | |
| Part (c) | ||
| B1 £90 | ✓ 1 | |
Exam tip: When two costs are equal, set the two expressions equal to each other and collect the letter terms on one side.
An expedition packs n oxygen cylinders. Each cylinder has a mass of 4 kg and the regulators have a total mass of 3 kg. The total mass of the cylinders and regulators must be less than 23 kg.
Solve the inequality 4n + 3 < 23.
Write down the greatest whole number of cylinders that can be packed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 4n < 20 (or equivalent first step) | ✓ 1 | |
| A1 n < 5 | ✓ 1 | |
| Part (b) | ||
| B1 4 | ✓ 1 | |
Common error: A strict inequality does not include the boundary value, so n = 5 is not allowed here.
A food scientist writes algebraic expressions for the masses, in grams, of ingredients in a sauce.
Factorise fully 12a + 18.
Factorise x2 + 7x + 12.
Solve x2 + 7x + 12 = 0. Write down the smaller solution.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 common factor of 2, 3 or 6 taken out | ✓ 1 | |
| B1 6(2a + 3) | ✓ 1 | |
| Part (b) | ||
| M1 (x + p)(x + q) with pq = 12 or p + q = 7 | ✓ 1 | |
| A1 (x + 3)(x + 4) | ✓ 1 | |
| Part (c) | ||
| B1 −4 (from x = −3 and x = −4) | ✓ 1 | |
Exam tip: Check a factorised answer by expanding the brackets again; the middle term must match.
A café sells 3 pastries at (2a + 5) pence each. It then gives a discount of (a − 4) pence on each of 2 coffees. The total takings, T pence, are the pastry sales minus the two discounts.
Show that T = 4a + 23.
The total takings are 51 pence. Work out the value of a.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expanding 3(2a + 5) = 6a + 15 and 2(a − 4) = 2a − 8 | ✓ 1 | |
| A1 6a + 15 − (2a − 8) = 4a + 23 | ✓ 1 | |
| Part (b) | ||
| M1 4a + 23 = 51 | ✓ 1 | |
| A1 a = 7 | ✓ 1 | |
Common error: When subtracting a bracket, change the sign of every term inside it: −(2a − 8) = −2a + 8.
A volcanologist models the rate of ash fall using algebraic expressions.
Expand and simplify \(3(2x + 5) - 2(x - 4)\).
Factorise fully \(6a^2b + 9ab^2\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expand both brackets, e.g. \(6x + 15 - 2x + 8\) | ✓ 1 | |
| A1 \(4x + 23\) | ✓ 1 | |
| Part (b) | ||
| B1 common factor \(3\), \(a\) or \(b\) taken out | ✓ 1 | |
| A1 \(3ab(2a + 3b)\) | ✓ 1 | |
Common error: When subtracting a bracket, the minus sign changes every term inside it, including the last one.
At a theatre, an adult ticket costs £x and a child ticket costs £4 less than an adult ticket. Three adult tickets and two child tickets cost £47 altogether.
Show that \(5x - 8 = 47\).
Find the cost of one adult ticket.
Find the cost of one child ticket.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 child ticket is \(x - 4\), so \(3x + 2(x - 4) = 47\) | ✓ 1 | |
| A1 expanding to \(3x + 2x - 8 = 47\), so \(5x - 8 = 47\) | ✓ 1 | |
| Part (b) | ||
| M1 \(5x = 55\) | ✓ 1 | |
| A1 £11 | ✓ 1 | |
| Part (c) | ||
| B1 £7 (ECF from (b)) | ✓ 1 | |
Exam tip: Define the unknown clearly first, then write the other quantities in terms of it.
A cargo ship uses fuel at a constant rate per kilometre. On a voyage of 480 km it uses 36 tonnes of fuel.
Calculate the fuel used on a voyage of 700 km.
The fuel costs €640 per tonne. Work out the cost of the fuel for the 700 km voyage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(\frac{36}{480} \times 700\) or \(0.075 \times 700\) | ✓ 1 | |
| A1 52.5 tonnes | ✓ 1 | |
| Part (b) | ||
| M1 \(52.5 \times 640\) | ✓ 1 | |
| A1 €33600 (ECF from (a)) | ✓ 1 | |
Common error: Direct proportion means dividing to find the amount for one unit first; do not add the same number to both quantities.
A shipping agent charges €C to transport a load of w tonnes, where \(C = 250 + 35w\).
Work out the charge for a load of 12 tonnes.
A customer is charged €950. Work out the mass of the load.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(250 + 35 \times 12\) | ✓ 1 | |
| A1 €670 | ✓ 1 | |
| Part (b) | ||
| M1 \(950 - 250 = 700\) and divide by 35 | ✓ 1 | |
| A1 20 tonnes | ✓ 1 | |
Exam tip: Substitute first, then solve; show each step in order so that method marks are not lost.
A sound engineer simplifies some algebraic expressions for a mixing-desk calculator.
Factorise 6x + 15
Factorise fully 12a2b − 8ab2
Expand and simplify 3(2x + 5) − 2(x − 4)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3(2x + 5) | ✓ 1 | |
| Part (b) | ||
| M1 partial factorisation, e.g. 4a(3ab − 2b2) or 2ab(6a − 4b) | ✓ 1 | |
| A1 4ab(3a − 2b) | ✓ 1 | |
| Part (c) | ||
| M1 expands both brackets, e.g. 6x + 15 − 2x + 8 (allow one sign error) | ✓ 1 | |
| A1 4x + 23 | ✓ 1 | |
Common error: When a bracket is multiplied by a negative number, the sign of every term inside changes: −2(x − 4) gives −2x + 8, not −2x − 8.
A small aircraft takes off with 180 litres of fuel and burns fuel at a constant 36 litres per hour. After t hours the fuel remaining is F litres, where F = 180 − 36t.
Work out the value of F when t = 2.5
Rearrange the formula to make t the subject.
Find the value of t when F = 54. You may use your formula from (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 36 × 2.5 (= 90) or 180 − 36 × 2.5 | ✓ 1 | |
| A1 90 | ✓ 1 | |
| Part (b) | ||
| M1 36t = 180 − F | ✓ 1 | |
| A1 t = (180 − F)/36 | ✓ 1 | |
| Part (c) | ||
| M1 (180 − 54)/36 (= 126/36) or 180 − 36t = 54 | ✓ 1 | |
| A1 3.5 (hours) (ECF from (b)) | ✓ 1 | |
Exam tip: Check a rearranged formula by substituting a known pair of values: here t = 2.5 should give F = 90.
On a farm, 3 sacks of seed and 2 sacks of fertiliser cost £73 in total. 1 sack of seed and 2 sacks of fertiliser cost £47 in total. Let s be the cost in pounds of one sack of seed and f the cost in pounds of one sack of fertiliser.
Write down two equations in s and f.
Solve your equations algebraically to find the cost of one sack of seed and the cost of one sack of fertiliser. Show your working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3s + 2f = 73 | ✓ 1 | |
| B1 s + 2f = 47 | ✓ 1 | |
| Part (b) | ||
| M1 eliminates one variable, e.g. subtracts to get 2s = 26 | ✓ 1 | |
| A1 s = 13 | ✓ 1 | |
| M1 substitutes to find the other variable, e.g. 2f = 47 − 13 | ✓ 1 | |
| A1 f = 17 (seed £13, fertiliser £17) | ✓ 1 | |
Exam tip: After solving, substitute both values into the equation you did not use to substitute: 3 × 13 + 2 × 17 = 73.
A music teacher and a choir leader use inequalities when planning.
n is an integer such that −3 < n ≤ 2. Write down all the possible values of n.
A choir buys m copies of a songbook at £6 each. There is also a single delivery charge of £15. The total cost must be no more than £105. By forming and solving an inequality, find the greatest number of songbooks the choir can buy.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 at least four correct values and no more than one incorrect value | ✓ 1 | |
| B1 −2, −1, 0, 1, 2 | ✓ 1 | |
| Part (b) | ||
| B1 6m + 15 ≤ 105 | ✓ 1 | |
| M1 6m ≤ 90 | ✓ 1 | |
| A1 m ≤ 15, so the greatest number is 15 | ✓ 1 | |
Common error: A strict inequality (<) excludes its end value, whereas ≤ includes it. In (a), −3 is not allowed but 2 is.
A stage designer uses indices to model how the brightness of a lamp changes with distance.
Simplify \(3x^2 \times 4x^5\).
Simplify \((2a^3)^4\).
Work out the value of \(2^{-3} \times 16\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for 12 | ✓ 1 | |
| B1 for \(x^7\) | ✓ 1 | |
| Part (b) | ||
| B1 for 16 | ✓ 1 | |
| B1 for \(a^{12}\) | ✓ 1 | |
| Part (c) | ||
| M1 for \(2^{-3}=\frac{1}{8}\) or 0.125 | ✓ 1 | |
| A1 for 2 | ✓ 1 | |
Common error: When raising a bracket to a power, apply the power to the number as well as to the letter: \((2a^3)^4\) is not \(2a^{12}\).
A cargo ship's loading crew solve equations to balance the weight of containers on each side of the deck.
Solve \(5x - 7 = 3x + 9\).
Solve \(3(2y+1) = 4y + 17\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for collecting terms, e.g. \(2x=16\) | ✓ 1 | |
| A1 for \(x=8\) | ✓ 1 | |
| Part (b) | ||
| M1 for expanding the bracket: \(6y+3\) | ✓ 1 | |
| M1 for collecting terms, e.g. \(2y=14\) | ✓ 1 | |
| A1 for \(y=7\) | ✓ 1 | |
Exam tip: Substitute your answer back into both sides of the equation to check that they are equal.
The perimeter, \(P\) metres, of a rectangular theatre stage of length \(l\) metres and width \(w\) metres is given by \(P = 2(l + w)\).
Make \(w\) the subject of the formula.
A stage has perimeter 46 m and length 15.5 m. Work out its width.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for first step, e.g. \(\frac{P}{2}=l+w\) or \(P=2l+2w\) | ✓ 1 | |
| A1 for \(w=\frac{P}{2}-l\) (or equivalent) | ✓ 1 | |
| Part (b) | ||
| M1 for substituting into their formula, e.g. \(23-15.5\) | ✓ 1 | |
| A1 for 7.5 (m) | ✓ 1 | |
Common error: Divide every term on the other side by 2, not just the first: \(\frac{P}{2}-l\), not \(\frac{P-l}{2}\).
In this question you will practise expanding and factorising.
Factorise fully \(12m^2 + 8m\).
Expand and simplify \((x+4)(x-3)\).
Factorise \(x^2 - 5x - 14\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for a correct partial factorisation, e.g. \(2(6m^2+4m)\) or \(m(12m+8)\) | ✓ 1 | |
| B1 for \(4m(3m+2)\) | ✓ 1 | |
| Part (b) | ||
| M1 for four correct terms, e.g. \(x^2-3x+4x-12\) | ✓ 1 | |
| A1 for \(x^2+x-12\) | ✓ 1 | |
| Part (c) | ||
| M1 for two brackets with \(x\) and numbers whose product is \(-14\), or a factor pair of \(-14\) summing to \(-5\) | ✓ 1 | |
| A1 for \((x-7)(x+2)\) | ✓ 1 | |
Exam tip: Expand your factorised answer again to check that it gives back the original expression.
A data centre buys server racks. A small rack costs €s and a large rack costs €l.
3 small racks and 2 large racks cost €2950 altogether.
5 small racks and 1 large rack cost €3050 altogether.
Solve the simultaneous equations \(3s + 2l = 2950\) and \(5s + l = 3050\) to find the cost of one small rack and the cost of one large rack.
Work out the total cost of 4 small racks and 3 large racks.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for a correct method to eliminate one variable, e.g. multiplying the second equation by 2 to get \(10s + 2l = 6100\) | ✓ 1 | |
| A1 for small rack: s = 450 | ✓ 1 | |
| A1 for large rack: l = 800 (ECF if the first value is wrong but the method is correct) | ✓ 1 | |
| Part (b) | ||
| M1 for \(4 \times 450 + 3 \times 800\) (ECF from (a)) | ✓ 1 | |
| A1 for €4200 | ✓ 1 | |
Exam tip: After finding both values, substitute them into the other equation to check that it is satisfied.
A layer of silt covers an archaeological site. The depth, d cm, of the silt after t years is given by the formula
d = 12 + 0.4t
Work out the depth of silt after 35 years.
Rearrange the formula to make t the subject.
The silt is now 50 cm deep. Use your formula from (b) to work out how many years ago the site was covered.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(12 + 0.4 \times 35\) | ✓ 1 | |
| A1 for 26 (cm) | ✓ 1 | |
| Part (b) | ||
| M1 for subtracting 12 from both sides, or dividing every term by 0.4 | ✓ 1 | |
| A1 for \(t = \frac{d - 12}{0.4}\) (accept \(t = 2.5(d - 12)\)) | ✓ 1 | |
| Part (c) | ||
| M1 for \(\frac{50 - 12}{0.4}\) (ECF from (b)) | ✓ 1 | |
| A1 for 95 (years) | ✓ 1 | |
Common error: When dividing both sides by 0.4, divide the whole of d − 12, not just d.
An archaeologist lays 3 identical square sample tiles in a straight line along a strip. Each tile has side length (2x − 5) cm. The strip is (4x + 9) cm long and the three tiles cover it exactly.
Form an equation in x and solve it.
Work out the side length of one tile.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(3(2x - 5) = 4x + 9\) | ✓ 1 | |
| M1 for expanding: \(6x - 15 = 4x + 9\) | ✓ 1 | |
| M1 for collecting terms: \(2x = 24\) | ✓ 1 | |
| A1 for \(x = 12\) | ✓ 1 | |
| Part (b) | ||
| B1 for 19 cm (ECF from (a)) | ✓ 1 | |
Exam tip: Substitute your value of x back into both sides of the equation to confirm they agree.
Answer all parts.
Expand and simplify \(4(2x + 3) - 3(x - 5)\).
Factorise fully \(12a^2b - 18ab^2\).
Factorise \(x^2 - 7x + 12\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(8x + 12 - 3x + 15\) (at least three terms correct) | ✓ 1 | |
| A1 for \(5x + 27\) | ✓ 1 | |
| Part (b) | ||
| B1 for a correct partial factorisation, e.g. \(6a(2ab - 3b^2)\) | ✓ 1 | |
| B1 for \(6ab(2a - 3b)\) | ✓ 1 | |
| Part (c) | ||
| M1 for two numbers with product 12 and sum −7 | ✓ 1 | |
| A1 for \((x - 3)(x - 4)\) | ✓ 1 | |
Common error: Multiply your brackets back out to check that you recover the original expression.
Two vials of a liquid medicine are filled so that they contain the same volume. The first vial holds (4x + 3) ml and the second holds (2x + 19) ml.
Write down an equation in x and solve it.
Work out the volume of liquid in one vial.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 4x + 3 = 2x + 19 | ✓ 1 | |
| M1 collects terms, e.g. 2x = 16 | ✓ 1 | |
| A1 x = 8 | ✓ 1 | |
| Part (b) | ||
| A1 35 ml (ECF from (a): 4x + 3) | ✓ 1 | |
Exam tip: After finding x, substitute into both expressions as a check that they give the same value.
In an online game, 3 skins and 2 booster packs cost £23. 1 skin and 2 booster packs cost £13. Every skin costs the same and every booster pack costs the same.
Let x pounds be the cost of a skin and y pounds the cost of a booster pack.
Write two equations and solve them to find the cost of one skin and the cost of one booster pack.
Work out the cost of 2 skins and 5 booster packs.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3x + 2y = 23 and x + 2y = 13 | ✓ 1 | |
| M1 eliminates a variable, e.g. subtracts to get 2x = 10 | ✓ 1 | |
| A1 x = 5 (skin £5) | ✓ 1 | |
| A1 y = 4 (booster pack £4) | ✓ 1 | |
| Part (b) | ||
| M1 2 × 5 + 5 × 4 | ✓ 1 | |
| A1 £30 (ECF from (a)) | ✓ 1 | |
Common error: After eliminating, students often find only one unknown. Always substitute back to find the second and check in both original equations.
A water company hires a pump. The total cost, C euros, of hiring the pump for t days is given by C = 40 + 12t.
Expand and simplify 3(2x + 5) − 2(x − 4).
Rearrange the formula to make t the subject.
Work out the number of days when the total cost is €148.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 6x + 15 − 2x + 8 (at least three terms correct) | ✓ 1 | |
| A1 4x + 23 | ✓ 1 | |
| Part (b) | ||
| M1 subtracts 40 from both sides, C − 40 = 12t | ✓ 1 | |
| A1 t = (C − 40)/12 | ✓ 1 | |
| Part (c) | ||
| A1 9 (days) | ✓ 1 | |
Common error: When expanding −2(x − 4), the sign of the second term is often wrong: −2 × −4 = +8.
In a game, a level unlocks for a player whose score s satisfies 5s + 3 > 2s + 18. A bonus room opens for a different whole-number setting n that satisfies −3 < 2n + 1 ≤ 9.
Solve 5s + 3 > 2s + 18.
Write down all the whole-number values of n that satisfy −3 < 2n + 1 ≤ 9.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 3s > 15 (collects terms correctly) | ✓ 1 | |
| A1 s > 5 | ✓ 1 | |
| Part (b) | ||
| M1 subtracts 1 from each part, −4 < 2n ≤ 8, or −2 < n ≤ 4 | ✓ 1 | |
| A1 correct range −2 < n ≤ 4 | ✓ 1 | |
| A1 −1, 0, 1, 2, 3, 4 | ✓ 1 | |
Common error: When a double inequality has a strict sign on one side and ≤ on the other, check each end value separately: −2 is excluded but 4 is included.
Coffee beans are roasted in a drum roaster.
Solve 4(t − 3) = 2t + 10.
The time T minutes to roast n kg of beans is given by T = 3n + 4. A batch takes 31 minutes. Find n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expands the bracket: 4t − 12 = 2t + 10 | ✓ 1 | |
| A1 t = 11 | ✓ 1 | |
| Part (b) | ||
| M1 3n + 4 = 31, so 3n = 27 | ✓ 1 | |
| A1 n = 9 (kg) | ✓ 1 | |
Exam tip: After solving, substitute your answer back into the original equation to check that both sides are equal.
During repairs to a footbridge, 3 cable clamps and 2 packs of bolts cost £19.50. One cable clamp and 4 packs of bolts cost £16.50. Let c pounds be the cost of a clamp and b pounds the cost of a pack of bolts.
Write down two equations in c and b.
Solve your equations to find the cost of a clamp and the cost of a pack of bolts.
Work out the cost of 5 clamps and 5 packs of bolts.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3c + 2b = 19.50 | ✓ 1 | |
| B1 c + 4b = 16.50 | ✓ 1 | |
| Part (b) | ||
| M1 a correct method to eliminate one variable (for example 2 × first equation and subtracting, or substitution) | ✓ 1 | |
| A1 b = 3.00 (£3.00 per pack of bolts) | ✓ 1 | |
| A1 c = 4.50 (£4.50 per clamp) | ✓ 1 | |
| Part (c) | ||
| A1 5 × (4.50 + 3.00) = £37.50 (ECF from (b)) | ✓ 1 | |
Exam tip: Check your solution by substituting both values into the equation you did not use for elimination.
The time t seconds a radio signal takes to travel from a space probe to Earth is directly proportional to the distance d km between them. When d = 6.0 × 108, t = 2000.
Find t when d = 9.0 × 108.
Find d when t = 2500.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = kd with k = 2000 ÷ (6.0 × 108), or 9.0 ÷ 6.0 × 2000 | ✓ 1 | |
| A1 3000 (seconds) | ✓ 1 | |
| Part (b) | ||
| M1 2500 ÷ 2000 × 6.0 × 108, or 2500 = kd solved for d | ✓ 1 | |
| A1 7.5 × 108 (km) | ✓ 1 | |
Common error: With direct proportion, the quantity that increases must be multiplied by the same factor as the other; do not add the same amount to both.
Indices and standard form are used in astronomy.
Simplify (p4)3 ÷ p5.
The Sun is 1.5 × 108 km from Earth. Light travels at 3.0 × 105 km/s. Work out the time, in seconds, for light to travel from the Sun to Earth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p12 ÷ p5 | ✓ 1 | |
| A1 p7 | ✓ 1 | |
| Part (b) | ||
| M1 1.5 × 108 ÷ 3.0 × 105 | ✓ 1 | |
| A1 500 (s) | ✓ 1 | |
Common error: When dividing powers of the same base, subtract the indices; when raising a power to a power, multiply them.
A runner trains on three days. On Monday she runs x km, on Tuesday (2x − 3) km and on Wednesday (x + 5) km. She runs 38 km in total.
Show that 4x + 2 = 38.
Solve 4x + 2 = 38.
How far does she run on Tuesday?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x + (2x − 3) + (x + 5) = 38 | ✓ 1 | |
| A1 collects terms to give 4x + 2 = 38 | ✓ 1 | |
| Part (b) | ||
| M1 4x = 36 | ✓ 1 | |
| A1 x = 9 | ✓ 1 | |
| Part (c) | ||
| A1 2 × 9 − 3 = 15 km (ECF from (b)) | ✓ 1 | |
Exam tip: A 'show that' question needs every step written out: the result is given, so the marks are for the working.
A train accelerates uniformly. Its final speed v m/s is given by v = u + at, where u m/s is its initial speed, a m/s2 its acceleration and t s the time.
Work out v when u = 8, a = 0.5 and t = 24.
Rearrange the formula to make t the subject.
Use your formula to find t when v = 32, u = 8 and a = 0.5.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 8 + 0.5 × 24 | ✓ 1 | |
| A1 20 (m/s) | ✓ 1 | |
| Part (b) | ||
| M1 v − u = at | ✓ 1 | |
| A1 t = (v − u)/a | ✓ 1 | |
| Part (c) | ||
| A1 (32 − 8)/0.5 = 48 s (ECF from (b)) | ✓ 1 | |
Common error: When rearranging, do the same operation to every term on both sides; subtract u from both sides before dividing by a.
At a rail station, 3 adult tickets and 2 child tickets cost £34.50. Two adult tickets and 1 child ticket cost £21.00. An adult ticket costs £a and a child ticket costs £c.
Write down two equations in a and c.
Solve the equations to find the cost of an adult ticket and of a child ticket.
Work out the cost of 2 adult tickets and 2 child tickets.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3a + 2c = 34.50 | ✓ 1 | |
| B1 2a + c = 21.00 | ✓ 1 | |
| Part (b) | ||
| M1 a correct method to eliminate a variable (for example 2 × second equation, then subtract) | ✓ 1 | |
| A1 a = 7.50 | ✓ 1 | |
| A1 c = 6.00 | ✓ 1 | |
| Part (c) | ||
| A1 2 × (7.50 + 6.00) = £27.00 (ECF from (b)) | ✓ 1 | |
Exam tip: Check your answers in the equation you did not use to eliminate: 3 × 7.50 + 2 × 6.00 = 34.50.
A sports scientist uses algebra to model an athlete's training load.
Simplify \(3x^2 \times 4x^5\).
Simplify \(15y^8 \div 5y^3\).
Expand and simplify \(3(2p+1) - 2(p-4)\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for 12 | ✓ 1 | |
| B1 for x7, giving 12x7 | ✓ 1 | |
| Part (b) | ||
| B1 for 3 | ✓ 1 | |
| B1 for y5, giving 3y5 | ✓ 1 | |
| Part (c) | ||
| M1 for 6p + 3 and −2p + 8 (at least three of the four terms correct) | ✓ 1 | |
| A1 for 4p + 11 | ✓ 1 | |
Exam tip: Add the indices when multiplying powers of the same letter and subtract them when dividing. Watch the sign when a negative number multiplies a bracket.
A rail operator charges a booking fee of £3.00 plus £0.45 for each stop on a journey. Hamza pays £10.65 for a journey with n stops.
Write down an equation in n.
Solve your equation to find n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for 3 + 0.45n = 10.65 | ✓ 1 | |
| Part (b) | ||
| M1 for subtracting 3 from both sides: 0.45n = 7.65 | ✓ 1 | |
| M1 for dividing by 0.45 | ✓ 1 | |
| A1 for n = 17 | ✓ 1 | |
Common error: Subtract the fixed booking fee before dividing by the cost per stop, not after.
In a sports centre café, 3 protein bars and 2 drinks cost £9.10. Also, 2 protein bars and 3 drinks cost £10.15.
Let a protein bar cost £b and a drink cost £d.
Solve the simultaneous equations to find b and d. You must show your working.
Work out the cost of 4 protein bars and 5 drinks. (ECF from (a))
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for making the coefficients of b (or d) equal, e.g. 6b + 4d = 18.20 and 6b + 9d = 30.45 | ✓ 1 | |
| M1 for subtracting (or adding) to eliminate one variable: 5d = 12.25 | ✓ 1 | |
| A1 for d = 2.45 | ✓ 1 | |
| A1 for b = 1.40 | ✓ 1 | |
| Part (b) | ||
| M1 for 4 × 1.40 + 5 × 2.45 | ✓ 1 | |
| A1 for £17.85 | ✓ 1 | |
Exam tip: Check your answers by substituting both values back into the equation you did not use to find them.
A rail engineer writes expressions for the lengths of track sections, in metres.
Factorise fully \(8a + 12\).
Factorise \(x^2 + 7x + 12\).
Expand and simplify \((x+5)(x-2)\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for a correct common factor of 2 or 4 | ✓ 1 | |
| B1 for 4(2a + 3) | ✓ 1 | |
| Part (b) | ||
| M1 for two numbers with sum 7 and product 12, i.e. 3 and 4 | ✓ 1 | |
| A1 for (x + 3)(x + 4) | ✓ 1 | |
| Part (c) | ||
| M1 for at least three of x², −2x, 5x, −10 correct | ✓ 1 | |
| A1 for x² + 3x − 10 | ✓ 1 | |
Common error: After factorising, expand your answer mentally to check that it returns the original expression.
A wind farm engineer solves some equations.
Solve \(3x + 7 = 25\).
Each turbine on a wind farm produces 2.5 MW. A battery adds a further 4 MW. Together they supply 29 MW. Form and solve an equation to find the number of turbines, \(n\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 subtract 7 from both sides or divide all terms by 3 | ✓ 1 | |
| A1 \(x = 6\) | ✓ 1 | |
| Part (b) | ||
| M1 \(2.5n + 4 = 29\) | ✓ 1 | |
| M1 \(2.5n = 25\) or \(n = (29 - 4) \div 2.5\) | ✓ 1 | |
| A1 \(n = 10\) | ✓ 1 | |
Exam tip: Write the equation first, then undo the operations in reverse order.
At a swimming pool, 2 adult tickets and 3 child tickets cost £19.50 in total. One adult ticket and 2 child tickets cost £11.00 in total.
Work out the cost of one adult ticket and the cost of one child ticket.
Work out the cost of 1 adult ticket and 5 child tickets. ECF from (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 correct equations \(2a + 3c = 19.5\) and \(a + 2c = 11\) and a method to eliminate one variable | ✓ 1 | |
| A1 child ticket costs £2.50 (\(c = 2.5\)) | ✓ 1 | |
| M1 substitute to find the other variable | ✓ 1 | |
| A1 adult ticket costs £6.00 (\(a = 6\)) | ✓ 1 | |
| Part (b) | ||
| B1 £18.50 (ECF from (a): \(a + 5c\)) | ✓ 1 | |
Common error: After finding the first unknown, substitute it back into an original equation and check it fits the other one too.
A library reading room is a rectangle with length \(l\) metres and width \(w\) metres. Its perimeter is \(P\) metres, where \(P = 2(l + w)\).
Expand and simplify \(3(2x + 5) - 2(x - 4)\).
Factorise fully \(12a + 18\).
Rearrange \(P = 2(l + w)\) to make \(w\) the subject.
The perimeter is 38 m and the length is 11.5 m. Work out the width.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(6x + 15 - 2x + 8\) (at most one sign error) | ✓ 1 | |
| A1 \(4x + 23\) | ✓ 1 | |
| Part (b) | ||
| M1 common factor 2 or 3 or 6 seen | ✓ 1 | |
| A1 \(6(2a + 3)\) | ✓ 1 | |
| Part (c) | ||
| M1 divide by 2 or expand brackets first | ✓ 1 | |
| A1 \(w = \frac{P}{2} - l\) or \(w = \frac{P - 2l}{2}\) | ✓ 1 | |
| Part (d) | ||
| B1 7.5 m (ECF from (c)) | ✓ 1 | |
Exam tip: When rearranging, do the same operation to every term on both sides.
The time \(t\) minutes a swimmer takes to complete \(n\) lengths of a pool is directly proportional to \(n\). A swimmer takes 9 minutes to complete 12 lengths.
Work out how long the swimmer takes to complete 20 lengths.
The swimmer swims at the same steady rate for 42 minutes. Work out the number of lengths completed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(t = 0.75n\) or \(9 \div 12\) or \(\frac{20}{12} \times 9\) | ✓ 1 | |
| A1 15 minutes | ✓ 1 | |
| Part (b) | ||
| M1 \(42 \div 0.75\) or \(\frac{42}{9} \times 12\) | ✓ 1 | |
| A1 56 lengths | ✓ 1 | |
Exam tip: For direct proportion, find the constant by dividing, then use it for every other value.
A space probe team solves some equations.
Solve \(7x - 5 = 2x + 20\).
A probe carries 85 kg of fuel and uses 2.5 kg each day. After \(d\) days, 30 kg of fuel is left. Form and solve an equation to find \(d\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 collect terms: \(5x = 25\) | ✓ 1 | |
| A1 \(x = 5\) | ✓ 1 | |
| Part (b) | ||
| M1 \(85 - 2.5d = 30\) | ✓ 1 | |
| M1 \(2.5d = 55\) | ✓ 1 | |
| A1 \(d = 22\) | ✓ 1 | |
Exam tip: Collect the letter terms on one side and the numbers on the other, then check by substituting back.
A coffee roaster mixes \(x\) kg of Arabica beans with \(y\) kg of Robusta beans to make a 40 kg blend. Arabica beans cost £9 per kg and Robusta beans cost £6 per kg. The blend has a total bean cost of £330.
Work out the values of \(x\) and \(y\).
Work out the percentage of the blend that is Arabica. ECF from (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(x + y = 40\) and \(9x + 6y = 330\) | ✓ 1 | |
| M1 eliminate one variable, e.g. \(9x + 6(40 - x) = 330\) | ✓ 1 | |
| A1 \(x = 30\) | ✓ 1 | |
| A1 \(y = 10\) | ✓ 1 | |
| Part (b) | ||
| B1 75% (\(\frac{30}{40} \times 100\)) | ✓ 1 | |
Common error: Always write one equation for the quantities and a second for the costs; mixing them up gives wrong answers.
The deck of a footbridge is a rectangle. Its width is \((x + 3)\) metres and its length is \((x + 4)\) metres.
Show that the area of the deck is \((x^2 + 7x + 12)\) m2.
Factorise fully \(5x^2 - 10x\).
Factorise \(x^2 + 9x + 20\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \((x + 3)(x + 4)\) expanded with at least three correct terms | ✓ 1 | |
| A1 \(x^2 + 7x + 12\) | ✓ 1 | |
| Part (b) | ||
| M1 \(5x\) or \(x\) taken out as a factor | ✓ 1 | |
| A1 \(5x(x - 2)\) | ✓ 1 | |
| Part (c) | ||
| M1 two numbers with product 20 and sum 9, or one correct bracket | ✓ 1 | |
| A1 \((x + 4)(x + 5)\) | ✓ 1 | |
Exam tip: Check a factorisation by expanding the brackets again.
A radio signal from a space probe travels at a constant speed. It travels 144 million km in 8 minutes. The distance travelled is directly proportional to the time taken.
Work out the distance travelled by the signal in 15 minutes.
A second probe is 459 million km away. Work out how long its signal takes to reach Earth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(144 \div 8 = 18\) million km per minute | ✓ 1 | |
| A1 270 million km | ✓ 1 | |
| Part (b) | ||
| M1 \(459 \div 18\) | ✓ 1 | |
| A1 25.5 minutes | ✓ 1 | |
Exam tip: Find the amount per minute first; then multiply or divide by it.
A robot stores its data in memory cells. Use indices to answer these questions.
Simplify \(a^5 \times a^3\).
Work out the value of \(5^{-2}\). Give your answer as a fraction.
The robot has \(2^n\) memory cells and this is equal to 64. Find the value of \(n\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| \(a^8\) | ✓ 1 | |
| Part (b) | ||
| \(\frac{1}{25}\) | ✓ 1 | |
| Part (c) | ||
| M1 writes 64 as a power of 2, e.g. \(2^6\) or \(2\times2\times2\times2\times2\times2\) | ✓ 1 | |
| A1 \(n = 6\) | ✓ 1 | |
Common error: When multiplying powers of the same base, add the indices; do not multiply them.
A fishery buys fish feed. 15 kg of feed costs £42. The cost is directly proportional to the mass of feed bought.
Work out the cost of 40 kg of feed.
Feed for 24 fish lasts 10 days. The same total amount of feed is shared equally among the fish, so the number of days is inversely proportional to the number of fish. Work out how many days the same feed lasts for 40 fish.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(42 \div 15\) (= 2.80) or \(42 \times \frac{40}{15}\) | ✓ 1 | |
| A1 £112 | ✓ 1 | |
| Part (b) | ||
| M1 \(24 \times 10 \div 40\) (or 240 seen) | ✓ 1 | |
| A1 6 days | ✓ 1 | |
Exam tip: Decide first whether the quantities rise together (direct) or one falls as the other rises (inverse) before you calculate.
A tailor is cutting ribbon for a dress.
Solve \(4x + 9 = 33\).
The tailor cuts a second ribbon into two pieces of equal length. The pieces have lengths \((2x + 5)\) cm and \((x + 13)\) cm. Find the value of \(x\).
Work out the length of one of the two pieces in part (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(4x = 24\) or \(x + \frac{9}{4} = \frac{33}{4}\) | ✓ 1 | |
| A1 \(x = 6\) | ✓ 1 | |
| Part (b) | ||
| M1 \(2x + 5 = x + 13\) | ✓ 1 | |
| A1 \(x = 8\) | ✓ 1 | |
| Part (c) | ||
| A1 21 cm (ECF from (b)) | ✓ 1 | |
Exam tip: Check your answer by substituting it back into both sides of the equation.
A designer writes algebraic expressions for the areas of cloth pieces.
Expand and simplify \((x+3)(x-5)\).
Factorise \(x^2 - 7x + 12\).
Factorise fully \(6a^2b + 9ab^2\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 at least three terms correct from \(x^2 - 5x + 3x - 15\) | ✓ 1 | |
| A1 \(x^2 - 2x - 15\) | ✓ 1 | |
| Part (b) | ||
| M1 \((x \pm 3)(x \pm 4)\) | ✓ 1 | |
| A1 \((x-3)(x-4)\) | ✓ 1 | |
| Part (c) | ||
| B1 partial factorisation, e.g. \(3a(2ab+3b^2)\) or \(ab(6a+9b)\) | ✓ 1 | |
| A1 \(3ab(2a+3b)\) | ✓ 1 | |
Common error: 'Factorise fully' means no common factor can remain inside the brackets.
Answer these algebra questions.
Expand and simplify \(4(3x-2)+3(x+5)\).
Factorise completely \(6y^2+9y\).
Simplify \((2p^3q)^3\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(12x-8+3x+15\) (at least three terms correct) | ✓ 1 | |
| A1 \(15x+7\) | ✓ 1 | |
| Part (b) | ||
| M1 a correct common factor, e.g. \(3(2y^2+3y)\) or \(y(6y+9)\) | ✓ 1 | |
| A1 \(3y(2y+3)\) | ✓ 1 | |
| Part (c) | ||
| M1 cubes both the 2 and the powers, e.g. \(8\) or \(p^9\) correct | ✓ 1 | |
| A1 \(8p^9q^3\) | ✓ 1 | |
Common error: When a bracket is raised to a power, the number in front is raised to the power too: \((2p^3)^3\) is \(8p^9\), not \(2p^9\).
A farmer is working out some costs.
Solve \(5x-7=2x+11\).
The farmer buys 3 sacks of feed, each costing £\(y\), and pays a delivery charge of £12. The total cost is £57. Form an equation and solve it to find the cost of one sack.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(3x=18\) (collects terms correctly) | ✓ 1 | |
| A1 \(x=6\) | ✓ 1 | |
| Part (b) | ||
| B1 \(3y+12=57\) | ✓ 1 | |
| M1 \(3y=45\) (or \(y=\frac{57-12}{3}\)) | ✓ 1 | |
| A1 £15 | ✓ 1 | |
Exam tip: After solving, put the answer back into the original equation to check it works.
The fuel, \(F\) litres, needed for a regional flight lasting \(t\) hours is given by \(F=2400t+1500\).
Work out \(F\) when \(t=3.5\).
Rearrange the formula to make \(t\) the subject.
An aircraft is loaded with 14 700 litres. Work out the greatest flight time this allows.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(2400\times3.5+1500\) | ✓ 1 | |
| A1 9900 | ✓ 1 | |
| Part (b) | ||
| M1 \(F-1500=2400t\) | ✓ 1 | |
| A1 \(t=\frac{F-1500}{2400}\) | ✓ 1 | |
| Part (c) | ||
| M1 \(\frac{14700-1500}{2400}\) | ✓ 1 | |
| A1 5.5 hours | ✓ 1 | |
Exam tip: Do the same operation to both sides, and undo the addition before the multiplication when rearranging.
On a farm, the quantity of fertiliser used is directly proportional to the area treated. 1.8 tonnes treat 12 hectares. Separately, the time taken to pick a crop is inversely proportional to the number of pickers.
Work out how much fertiliser is needed to treat 30 hectares.
5 pickers take 12 hours to pick the crop. Work out the time for 8 pickers working at the same rate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(\frac{1.8}{12}\times30\) | ✓ 1 | |
| A1 4.5 tonnes | ✓ 1 | |
| Part (b) | ||
| M1 \(5\times12=60\) picker-hours (or \(\frac{5\times12}{8}\)) | ✓ 1 | |
| A1 7.5 hours | ✓ 1 | |
Common error: With inverse proportion, more workers means less time, so the answer must be smaller than 12 hours.
Two retailers sell garden netting online. Retailer A charges a delivery fee of £4 plus £2.50 for each kilogram. Retailer B charges a delivery fee of £9 plus £1.50 for each kilogram. An order has a weight of w kg.
Write down an expression, in terms of w, for the delivery charge, in pounds, from retailer A.
Find the weight of an order for which both retailers charge the same delivery fee.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 4 + 2.5w | ✓ 1 | |
| Part (b) | ||
| M1 forms the equation 4 + 2.5w = 9 + 1.5w | ✓ 1 | |
| M1 collects w terms on one side and numbers on the other, e.g. 1.0w = 5 | ✓ 1 | |
| A1 w = 5 kg | ✓ 1 | |
Exam tip: Write each line of working below the one before it so that a slip in one step does not hide your method.
Work with the following algebra.
Expand and simplify 3(2x + 5) − 2(x − 4).
Solve 5(2x − 3) = 3x + 20.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expands at least one bracket correctly, e.g. 6x + 15 or −2x + 8 | ✓ 1 | |
| A1 4x + 23 | ✓ 1 | |
| Part (b) | ||
| M1 expands the bracket: 10x − 15 = 3x + 20 | ✓ 1 | |
| M1 collects terms: 7x = 35 | ✓ 1 | |
| A1 x = 5 | ✓ 1 | |
Common error: When a bracket has a negative sign in front, the sign of every term inside changes: −2(x − 4) gives −2x + 8, not −2x − 8.
A service truck on a wind farm track speeds up steadily from u m/s to v m/s while travelling a distance of s metres in t seconds. The quantities are connected by the formula s = ½(u + v)t.
Rearrange the formula to make v the subject.
Find v when s = 90, t = 12 and u = 4.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies both sides by 2 (or divides by t) correctly, e.g. 2s = (u + v)t | ✓ 1 | |
| A1 v = 2s/t − u | ✓ 1 | |
| Part (b) | ||
| M1 substitutes: v = 2 × 90 ÷ 12 − 4 (ECF from (a)) | ✓ 1 | |
| A1 v = 11 m/s | ✓ 1 | |
Exam tip: Substitute into the rearranged formula, then check the answer by putting all four values back into the original formula.
The power output, P watts, of a solar panel array is directly proportional to its area, A m2. An array of area 4.5 m2 produces 900 W in bright sunlight.
Find a formula for P in terms of A.
Work out the area of an array that produces 2300 W in the same sunlight.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = kA and 900 = k × 4.5 | ✓ 1 | |
| A1 P = 200A | ✓ 1 | |
| Part (b) | ||
| M1 2300 = 200A (ECF from (a)) | ✓ 1 | |
| A1 11.5 m2 | ✓ 1 | |
Exam tip: For direct proportion, write the relationship as an equation with a constant (P = kA) before substituting any numbers.
A marine biologist is setting up a research tank for fish.
Solve 3(2x − 5) = 4x + 9
The base of the tank is a rectangle. Its length is (2n + 3) metres and its width is (n − 1) metres. The perimeter of the base is 46 metres.
Work out the value of n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expand the bracket: 6x − 15 = 4x + 9 | ✓ 1 | |
| M1 collect terms: 2x = 24 | ✓ 1 | |
| A1 x = 12 | ✓ 1 | |
| Part (b) | ||
| M1 form an equation: 2(2n + 3 + n − 1) = 46 | ✓ 1 | |
| M1 simplify to 6n + 4 = 46 | ✓ 1 | |
| A1 n = 7 | ✓ 1 | |
Common error: Write the full perimeter as an equation before simplifying; people often add only one length and one width.
Marine biologists model the growth of algae in a bay.
Simplify 3a4 × 2a5
Simplify (x3)4 ÷ x5
The number of algae cells in a sample is 250 × 2n after n hours.
Work out the number of cells after 6 hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 coefficient 6 | ✓ 1 | |
| B1 a9 | ✓ 1 | |
| Part (b) | ||
| M1 x12 seen | ✓ 1 | |
| A1 x7 | ✓ 1 | |
| Part (c) | ||
| M1 250 × 26 | ✓ 1 | |
| A1 16000 | ✓ 1 | |
Exam tip: When multiplying powers of the same letter you add the indices; when raising a power to a power you multiply them.
An architect is designing a rectangular window pane.
Expand and simplify 3(x + 4) + 2(x − 5)
Factorise fully 12a + 18
The area of the window pane is (x2 + 7x + 12) cm2.
Factorise x2 + 7x + 12 to find expressions for the length and the width of the pane.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expand both brackets: 3x + 12 + 2x − 10 | ✓ 1 | |
| A1 5x + 2 | ✓ 1 | |
| Part (b) | ||
| M1 a correct factor, e.g. 2(6a + 9) or 3(4a + 6) | ✓ 1 | |
| A1 6(2a + 3) | ✓ 1 | |
| Part (c) | ||
| M1 attempt at (x + a)(x + b) with ab = 12 or a + b = 7 | ✓ 1 | |
| A1 (x + 3)(x + 4), so the sides are (x + 3) cm and (x + 4) cm | ✓ 1 | |
Exam tip: Check a factorisation by expanding it again; the middle term must match.
A logistics firm charges €C to deliver a pallet a distance of d kilometres, where C = 40 + 2.5d.
Work out C when d = 36.
Rearrange the formula to make d the subject.
A delivery costs €215. Work out the distance, in km.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 40 + 2.5 × 36 | ✓ 1 | |
| A1 130 | ✓ 1 | |
| Part (b) | ||
| M1 C − 40 = 2.5d | ✓ 1 | |
| A1 d = (C − 40) ÷ 2.5 | ✓ 1 | |
| Part (c) | ||
| M1 (215 − 40) ÷ 2.5 (ECF from (b)) | ✓ 1 | |
| A1 70 | ✓ 1 | |
Exam tip: Undo the addition before the multiplication: subtract 40 from both sides, then divide.
A sports scientist writes a training plan for a runner using algebra.
Factorise fully \(12p + 18\).
Solve \(4x + 9 = 33\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 a common factor of 2, 3 or 6 taken out | ✓ 1 | |
| B1 \(6(2p + 3)\) | ✓ 1 | |
| Part (b) | ||
| M1 subtract 9 from both sides, \(4x = 24\) | ✓ 1 | |
| A1 \(x = 6\) | ✓ 1 | |
Common error: 'Fully' means the common factor must be the highest one; 2(6p + 9) is only partly factorised.
The fare, £C, for a rail journey of d kilometres is given by the formula
C = 4.5 + 0.35d
Work out the fare for a journey of 18 km.
A passenger pays a fare of £19.20. Work out the distance of the journey.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(4.5 + 0.35 \times 18\) | ✓ 1 | |
| A1 £10.80 | ✓ 1 | |
| Part (b) | ||
| M1 \(19.2 = 4.5 + 0.35d\) | ✓ 1 | |
| M1 \(0.35d = 14.7\) or \(d = (19.2 - 4.5) \div 0.35\) | ✓ 1 | |
| A1 42 km | ✓ 1 | |
Exam tip: In a formula question, substitute first and then solve; do not try to rearrange and substitute in one step.
A coach models the distance covered in a drill using algebra.
Expand and simplify \(3(2x + 5) - 2(x - 4)\).
Hence, or otherwise, solve \(3(2x + 5) - 2(x - 4) = 51\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(6x + 15 - 2x + 8\) (at least three terms correct) | ✓ 1 | |
| A1 \(4x + 23\) | ✓ 1 | |
| Part (b) | ||
| M1 \(4x + 23 = 51\) so \(4x = 28\) | ✓ 1 | |
| A1 \(x = 7\) | ✓ 1 | |
Common error: Subtracting a bracket changes the sign of every term inside it: \(-2(x - 4)\) gives \(-2x + 8\), not \(-2x - 8\).
Answer the following questions about indices.
Simplify \(x^3 \times x^5\).
Simplify \((2y^4)^3\).
Write \(4^{-2}\) as a fraction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(x^8\) | ✓ 1 | |
| Part (b) | ||
| B1 \(8\) (cubing the 2) | ✓ 1 | |
| B1 \(8y^{12}\) | ✓ 1 | |
| Part (c) | ||
| B1 \(\frac{1}{16}\) | ✓ 1 | |
Common error: \((2y^4)^3\) is \(2^3y^{12}\); the power applies to the number as well as to the letter.
A beekeeper buys 5 hive frames and 1 smoker. The smoker costs £14. The total cost is £61.50. Each frame costs £f.
Write down an equation in f and solve it to find the cost of one frame.
The beekeeper instead buys 8 frames and the smoker. Work out the total cost.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 5f + 14 = 61.50 | ✓ 1 | |
| M1 5f = 47.50 | ✓ 1 | |
| A1 f = 9.50 (£9.50) | ✓ 1 | |
| Part (b) | ||
| M1 8 × 9.50 + 14 (ECF from (a)) | ✓ 1 | |
| A1 £90 (90.00) | ✓ 1 | |
Common error: Subtracting the smoker cost after dividing, instead of before, gives the wrong frame price.
Algebraic expressions.
Expand and simplify 3(2x + 5) − 2(x − 4).
Factorise fully 15y2 − 10y.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 6x + 15 and −2x + 8 (both brackets attempted) | ✓ 1 | |
| A1 4x + 23 | ✓ 1 | |
| Part (b) | ||
| M1 any correct partial factorisation, e.g. 5(3y2 − 2y) or y(15y − 10) | ✓ 1 | |
| A1 5y(3y − 2) | ✓ 1 | |
Common error: Minus a bracket changes the sign of every term inside it, including the second one.
A Formula 1 team uses the model F = 1.8n + 3 to estimate the fuel, F litres, needed for a practice run of n laps.
Make n the subject of the formula.
Use your formula to find the number of laps that can be completed with 75 litres of fuel.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 F − 3 = 1.8n | ✓ 1 | |
| A1 n = (F − 3) / 1.8 | ✓ 1 | |
| Part (b) | ||
| M1 (75 − 3) / 1.8 (ECF from (a)) | ✓ 1 | |
| A1 40 laps | ✓ 1 | |
Exam tip: Undo the addition first, then the multiplication, working in the reverse order to the formula.
Inequalities.
Solve 5x − 8 > 12.
x is also an integer with x ≤ 8. List all integer values of x that satisfy both inequalities.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 5x > 20 | ✓ 1 | |
| A1 x > 4 | ✓ 1 | |
| Part (b) | ||
| M1 4 < x ≤ 8 identified (ECF from (a)) | ✓ 1 | |
| A1 5, 6, 7, 8 | ✓ 1 | |
Common error: Because 4 is not included in x > 4, the list must start at 5.
A beekeeper keeps two hives. Hive A has (5x + 7) frames and hive B has (3x + 19) frames. The two hives contain the same number of frames.
Factorise fully 12b − 18
Solve 5x + 7 = 3x + 19
Work out the number of frames in hive A.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 a common factor of 2, 3 or 6 taken out correctly, e.g. 2(6b − 9) | ✓ 1 | |
| B1 6(2b − 3) | ✓ 1 | |
| Part (b) | ||
| M1 collect terms correctly, e.g. 2x = 12 | ✓ 1 | |
| A1 x = 6 | ✓ 1 | |
| Part (c) | ||
| B1 37 (ECF from (b): 5 × their x + 7) | ✓ 1 | |
Exam tip: Collect the x terms on one side and the numbers on the other, then check your answer by substituting it into both expressions.
A Formula 1 team's simulation software uses expressions written with indices.
Simplify p4 × p7
Simplify (3t2)3
Work out the value of 5−2. Give your answer as a fraction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 p11 | ✓ 1 | |
| Part (b) | ||
| M1 27 or t6 seen, or 33t6 | ✓ 1 | |
| A1 27t6 | ✓ 1 | |
| Part (c) | ||
| B1 1/25 | ✓ 1 | |
Common error: When raising a bracket to a power, the number in front is raised to the power too, so (3t²)³ is not 3t⁶.
A pit crew's engineer sets up equations to find unknown numbers of laps, x.
Solve 3(x + 4) = 27
Solve (2x − 1)/3 = 5
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x + 4 = 9 or 3x + 12 = 27 | ✓ 1 | |
| A1 x = 5 | ✓ 1 | |
| Part (b) | ||
| M1 2x − 1 = 15 | ✓ 1 | |
| A1 x = 8 | ✓ 1 | |
Exam tip: Multiply both sides to clear a fraction first, then solve the two-step equation that remains.
A hospital architect writes expressions for room dimensions.
Expand and simplify (x + 3)(x − 5)
Factorise x2 − 9x + 20
Factorise fully 3a2b − 6ab2
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 at least three terms correct from x2 − 5x + 3x − 15 | ✓ 1 | |
| A1 x2 − 2x − 15 | ✓ 1 | |
| Part (b) | ||
| M1 (x ± 4)(x ± 5) | ✓ 1 | |
| A1 (x − 4)(x − 5) | ✓ 1 | |
| Part (c) | ||
| M1 a correct common factor, e.g. 3ab, 3a or 3b taken out | ✓ 1 | |
| A1 3ab(a − 2b) | ✓ 1 | |
Common error: Factorise fully means taking out every common factor, numbers and letters; check by expanding your answer.
Cycling, and the algebra behind it.
Solve 5x − 8 = 27.
Expand and simplify 3(2y + 1) − 2(y − 4).
A cycle hire shop charges a fixed fee of £4 plus £2.50 for each hour. Katya pays £19. Work out the number of hours she hired the cycle for.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 5x = 35 (adds 8 to both sides) | ✓ 1 | |
| A1 x = 7 | ✓ 1 | |
| Part (b) | ||
| M1 6y + 3 − 2y + 8 (both brackets expanded) | ✓ 1 | |
| A1 4y + 11 | ✓ 1 | |
| Part (c) | ||
| M1 19 − 4 = 15 | ✓ 1 | |
| M1 15 ÷ 2.5 | ✓ 1 | |
| A1 6 hours | ✓ 1 | |
Common error: Forgetting that −2 multiplies both terms in (y − 4), which gives +8 and not −8.
Use the laws of indices in these archaeology questions.
Each of the 23 trenches on a site contains 24 pottery fragments. Write the total number of fragments as a single power of 2.
Simplify 3x2y × 4x5y3.
A glass bead has diameter 2−3 cm. Work out this diameter as a decimal.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 27 | ✓ 1 | |
| Part (b) | ||
| M1 two of 12, x7, y4 correct | ✓ 1 | |
| A1 12x7y4 | ✓ 1 | |
| Part (c) | ||
| B1 0.125 (accept 1/8) | ✓ 1 | |
Common error: Treating 2⁻³ as −8 or −6; a negative power means a reciprocal.
The power used by a data-centre hall is directly proportional to the number of servers running. 15 servers use 9 kW.
Work out the power used by 24 servers.
The hall uses 21.6 kW. Work out the number of servers running.
The full facility runs 250 servers. Work out the power used.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 9 ÷ 15 = 0.6 kW per server, or 24 ÷ 15 × 9 | ✓ 1 | |
| A1 14.4 kW | ✓ 1 | |
| Part (b) | ||
| M1 21.6 ÷ 0.6 | ✓ 1 | |
| A1 36 | ✓ 1 | |
| Part (c) | ||
| B1 250 × 0.6 = 150 kW | ✓ 1 | |
Exam tip: Find the amount for one unit first; every direct proportion question then becomes a multiplication or division.
A cyclist's speed v m/s after t seconds is given by v = u + at, where u m/s is the starting speed and a m/s2 is the acceleration.
Work out v when u = 2, a = 0.6 and t = 5.
Rearrange the formula to make a the subject.
Work out a when v = 9.4, u = 3.4 and t = 8.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 5 | ✓ 1 | |
| Part (b) | ||
| M1 v − u = at | ✓ 1 | |
| A1 a = (v − u) ÷ t | ✓ 1 | |
| Part (c) | ||
| B1 0.75 (ECF from (b)) | ✓ 1 | |
Exam tip: Do the same operation to both sides, and subtract u before dividing by t.
A marine research boat is loading equipment for a survey.
Solve 3x + 5 < 20.
n is an integer with −2 < n ≤ 3. Write down all the possible values of n.
The number of crates, x, loaded must satisfy 4(x − 3) ≥ 2x + 6. Find the smallest integer value of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 3x < 15 | ✓ 1 | |
| A1 x < 5 | ✓ 1 | |
| Part (b) | ||
| B1 at least four of −1, 0, 1, 2, 3 and none outside the range | ✓ 1 | |
| B1 −1, 0, 1, 2, 3 | ✓ 1 | |
| Part (c) | ||
| M1 4x − 12 ≥ 2x + 6 | ✓ 1 | |
| M1 2x ≥ 18 | ✓ 1 | |
| A1 9 | ✓ 1 | |
Common error: When you multiply or divide an inequality by a negative number, the inequality sign reverses. Check your answer by testing a value.
An architect is designing rectangular glass panels for a lobby.
Factorise x2 − 7x + 12.
A panel has area (x2 + 5x − 24) m2 and width (x − 3) m. Find an expression for the length of the panel.
Work out the perimeter of the panel when x = 4.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 two numbers with product 12, or (x ± 3)(x ± 4) | ✓ 1 | |
| A1 (x − 3)(x − 4) | ✓ 1 | |
| Part (b) | ||
| M1 factorises x2 + 5x − 24 as (x + 8)(x − 3), or divides by (x − 3) | ✓ 1 | |
| A1 (x + 8) m | ✓ 1 | |
| Part (c) | ||
| M1 width 1 m and length 12 m, using 2 × (length + width) | ✓ 1 | |
| A1 26 (m) | ✓ 1 | |
Exam tip: Area = length × width, so dividing the area expression by the width gives the length. Factorising makes the division easy.
A logistics firm uses small vans and large vans. A small van carries s tonnes and a large van carries l tonnes. The capacities satisfy
2s + l = 11
s + 3l = 18
Solve the simultaneous equations to find s and l.
Work out the total load that 4 small vans and 3 large vans can carry.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 a correct method to eliminate or substitute, for example 2s + 6l = 36 then subtract | ✓ 1 | |
| A1 one correct value (s = 3 or l = 5) | ✓ 1 | |
| A1 s = 3 and l = 5 | ✓ 1 | |
| Part (b) | ||
| M1 4 × 3 + 3 × 5, using their values (ECF from (a)) | ✓ 1 | |
| A1 27 (tonnes) | ✓ 1 | |
Exam tip: After solving, substitute both values into the equation you did not use for the last step. A correct pair satisfies both equations.
A marine biologist is analysing a dive plan.
Solve 4x − 7 = 21.
Make a the subject of T = 2a + 5.
The depth, d metres, of a diver after n minutes of a descent is given by d = 12 + 3n. Find the value of n when d = 45.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 4x = 28 | ✓ 1 | |
| A1 7 | ✓ 1 | |
| Part (b) | ||
| M1 T − 5 = 2a | ✓ 1 | |
| A1 a = (T − 5) ÷ 2 | ✓ 1 | |
| Part (c) | ||
| M1 45 = 12 + 3n, or 3n = 33 | ✓ 1 | |
| A1 11 (minutes) | ✓ 1 | |
Common error: Undo operations in the reverse order. In T = 2a + 5, subtract 5 first and then divide by 2.
A shop sells scarves for £s each and gloves for £(s − 5) a pair. Maya buys 4 scarves and 3 pairs of gloves.
Show that the total cost, in pounds, is 7s − 15.
Maya pays £90 in total. Work out the value of s.
Work out the cost of one pair of gloves.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 4s + 3(s − 5) seen | ✓ 1 | |
| A1 4s + 3s − 15 = 7s − 15 shown with the bracket expanded | ✓ 1 | |
| Part (b) | ||
| M1 7s − 15 = 90 and adding 15 (or dividing by 7) | ✓ 1 | |
| A1 s = 15 | ✓ 1 | |
| Part (c) | ||
| B1 £10 (15 − 5) (ECF from (b)) | ✓ 1 | |
Exam tip: In a 'show that' question, write every step; the final line must be the expression you were given.
In a simple model of an infection, the average number of new cases caused by one infected person is R = cpd, where c is the number of contacts per day, p is the probability that a contact becomes infected, and d is the number of days the person is infectious.
Work out R when c = 12, p = 0.05 and d = 4.
Rearrange the formula to make d the subject.
Work out d when R = 1.5, c = 10 and p = 0.05.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 12 × 0.05 × 4 | ✓ 1 | |
| A1 2.4 | ✓ 1 | |
| Part (b) | ||
| M1 divide both sides by cp (or by c and by p) | ✓ 1 | |
| A1 d = R ÷ (cp) | ✓ 1 | |
| Part (c) | ||
| M1 1.5 ÷ (10 × 0.05) (or use of the answer to (b)) | ✓ 1 | |
| A1 3 | ✓ 1 | |
Common error: When dividing by a product such as cp, divide by the whole product; dividing by c only leaves p unresolved.
The energy, E kWh, produced by a solar panel array in one day is directly proportional to its area, A m2. An array of area 12 m2 produces 2.4 kWh in one day.
Find a formula for E in terms of A.
Work out the energy produced in one day by an array of area 35 m2.
An array produces 9.5 kWh in one day. Work out its area.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 E = kA and 2.4 = k × 12 (or 2.4 ÷ 12) | ✓ 1 | |
| A1 E = 0.2A | ✓ 1 | |
| Part (b) | ||
| M1 0.2 × 35 (ECF from (a)) | ✓ 1 | |
| A1 7 kWh | ✓ 1 | |
| Part (c) | ||
| M1 9.5 ÷ 0.2 (ECF from (a)) | ✓ 1 | |
| A1 47.5 m2 | ✓ 1 | |
Exam tip: Find the constant of proportionality first; every later part then becomes a single substitution.
Solve the inequalities below.
3x + 4 ≥ 19
2(y − 3) < 14
How many integers n satisfy both 3n + 4 ≥ 19 and 2(n − 3) < 14?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 3x ≥ 15 (subtract 4 from both sides) | ✓ 1 | |
| A1 x ≥ 5 | ✓ 1 | |
| Part (b) | ||
| M1 y − 3 < 7 (or 2y − 6 < 14) | ✓ 1 | |
| A1 y < 10 | ✓ 1 | |
| Part (c) | ||
| M1 integers 5, 6, 7, 8, 9 identified (ECF from (a) and (b)) | ✓ 1 | |
| A1 5 | ✓ 1 | |
Common error: The symbol ≥ includes the end value but < does not: 5 is allowed here, 10 is not.
A wind-farm technician is working out servicing costs. The cost, in £, of servicing the first turbine is 3(2x − 5) + 4(x + 3), where x is the number of hours spent on site.
Expand and simplify 3(2x − 5) + 4(x + 3).
A second turbine costs 5(2x − 3) pounds and a third costs 3x + 20 pounds to service. Given that these two costs are equal, solve 5(2x − 3) = 3x + 20.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expands at least one bracket correctly, e.g. 6x − 15 or 4x + 12 | ✓ 1 | |
| A1 10x − 3 | ✓ 1 | |
| Part (b) | ||
| M1 expands bracket: 10x − 15 = 3x + 20 | ✓ 1 | |
| M1 collects terms: 7x = 35 (or equivalent) | ✓ 1 | |
| A1 x = 5 | ✓ 1 | |
Common error: When expanding a bracket, multiply every term inside by the number outside; −3 × 5 is −15, not +15.
A garden centre sells solar lights and ground stakes. 3 solar lights and 2 stakes cost £19. 2 solar lights and 5 stakes cost £20. Let l be the cost of one light and s the cost of one stake, both in pounds.
Work out the cost of one solar light and the cost of one stake.
Work out the cost of 7 solar lights and 4 stakes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 forms both equations: 3l + 2s = 19 and 2l + 5s = 20 | ✓ 1 | |
| M1 correct method to make one pair of coefficients equal, e.g. 6l + 4s = 38 and 6l + 15s = 60 | ✓ 1 | |
| A1 one variable correct: s = 2 (stake costs £2) | ✓ 1 | |
| A1 other variable correct: l = 5 (light costs £5); accept any valid method such as substitution | ✓ 1 | |
| Part (b) | ||
| M1 7 × 5 + 4 × 2 (ECF from (a)) | ✓ 1 | |
| A1 £43 (ECF from (a)) | ✓ 1 | |
Exam tip: After solving, substitute your values back into the equation you did not use to find the second variable, to check both equations are satisfied.
The energy E kWh generated by a solar array is directly proportional to the number of hours h of full sunshine. In 8 hours of full sunshine the array generates 36 kWh.
Work out the energy generated in 11 hours of full sunshine.
Work out the number of hours of full sunshine needed to generate 90 kWh.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 36 ÷ 8 (= 4.5) or 36 × 11 ÷ 8 or E = 4.5h | ✓ 1 | |
| A1 49.5 kWh | ✓ 1 | |
| Part (b) | ||
| M1 90 ÷ (36 ÷ 8) or 90 × 8 ÷ 36 | ✓ 1 | |
| A1 20 hours | ✓ 1 | |
Exam tip: For direct proportion, find the amount for one unit first (here kWh per hour); the same rate then works in both directions.
A warehouse robot charges its battery. The battery level, B %, after t minutes of charging is given by B = 14 + 2.5t.
Work out the battery level after 18 minutes of charging.
The robot is unplugged when its battery level reaches 96%.
Work out how many minutes of charging this takes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes t = 18, 14 + 2.5 × 18 | ✓ 1 | |
| A1 B = 59 | ✓ 1 | |
| Part (b) | ||
| M1 96 = 14 + 2.5t | ✓ 1 | |
| M1 2.5t = 82, or t = (96 − 14) ÷ 2.5 | ✓ 1 | |
| A1 t = 32.8 | ✓ 1 | |
Exam tip: Write the equation you are solving before you rearrange it, so a slip later still earns the method mark.
A robot accelerates in a straight line. The distance it travels, s metres, is given by
s = ut + ½at2
where u is the initial speed in m/s, a is the acceleration in m/s2 and t is the time in seconds.
Work out the value of s when u = 3, t = 4 and a = 2.5.
The final speed, v m/s, is given by v = u + at.
Make a the subject of this formula.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 3 × 4 + ½ × 2.5 × 42 | ✓ 1 | |
| A1 s = 32 | ✓ 1 | |
| Part (b) | ||
| M1 v − u = at | ✓ 1 | |
| A1 a = (v − u) ÷ t | ✓ 1 | |
Exam tip: Evaluate powers before multiplying: 4² = 16 first, then × 2.5 × ½.
A fashion designer uses algebra to plan the cutting of fabric pieces.
Expand and simplify 3(2x + 5) − 2(x − 4).
Factorise fully 6a2b − 9ab2.
Factorise x2 + x − 20.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 6x + 15 − 2x + 8 with at least 3 terms correct | ✓ 1 | |
| A1 4x + 23 | ✓ 1 | |
| Part (b) | ||
| M1 a correct common factor, e.g. 3a(2ab − 3b2) | ✓ 1 | |
| A1 3ab(2a − 3b) | ✓ 1 | |
| Part (c) | ||
| M1 (x + p)(x + q) with pq = −20 or p + q = 1 | ✓ 1 | |
| A1 (x + 5)(x − 4) | ✓ 1 | |
Common error: When expanding −2(x − 4), the sign of the second term changes: −2 × −4 = +8.
A marine research centre has two tanks, A and B, that hold the same volume of water. Tank A holds (5x + 8) litres and tank B holds (2x + 29) litres.
Work out the value of x.
Work out the volume of water in tank A.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 5x + 8 = 2x + 29 | ✓ 1 | |
| M1 collects terms, 3x = 21 | ✓ 1 | |
| A1 x = 7 | ✓ 1 | |
| Part (b) | ||
| A1 43 litres (ECF from (a)) | ✓ 1 | |
Exam tip: After finding x, substitute it into the other tank's expression as well to check both give the same volume.
A rectangular courtyard in an architect's plan has length (x − 4) metres and width (x − 7) metres, where x > 7.
Factorise fully 6x2y + 9xy2
Factorise x2 − 11x + 28
The area of the courtyard is 18 m2. Work out the value of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 takes out a correct common factor (3, x, y or a product of these) | ✓ 1 | |
| A1 3xy(2x + 3y) | ✓ 1 | |
| Part (b) | ||
| M1 (x ± a)(x ± b) with ab = 28 or a + b = 11 | ✓ 1 | |
| A1 (x − 4)(x − 7) | ✓ 1 | |
| Part (c) | ||
| M1 (x − 4)(x − 7) = 18, giving x2 − 11x + 28 = 18 | ✓ 1 | |
| M1 solves x2 − 11x + 10 = 0, giving x = 1 or x = 10 | ✓ 1 | |
| A1 x = 10 with x = 1 rejected because x > 7 | ✓ 1 | |
Common error: Giving both roots of the quadratic. Always check each root against the condition in the question.
Use the laws of indices to answer this question.
Write 53 × 54 as a single power of 5.
Simplify 12x5y2 ÷ 3x2y.
Simplify (3a2)3.
Work out the value of 2−3. Give your answer as a fraction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 57 | ✓ 1 | |
| Part (b) | ||
| M1 any two of 4, x3, y correct | ✓ 1 | |
| A1 4x3y | ✓ 1 | |
| Part (c) | ||
| M1 33 or a6 seen | ✓ 1 | |
| A1 27a6 | ✓ 1 | |
| Part (d) | ||
| B1 \(\frac{1}{8}\) | ✓ 1 | |
Common error: Writing (3a²)³ as 3a⁶: the power applies to the 3 as well as to a².
In this question, x is a number; in part (b) it must also be an integer.
Solve 3 < 2x + 5 ≤ 17 (treat x as any number for this part).
Work out the sum of all the integer values of x that satisfy the inequality in (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 subtracts 5 from all three parts | ✓ 1 | |
| M1 divides all three parts by 2 | ✓ 1 | |
| A1 −1 < x ≤ 6 | ✓ 1 | |
| Part (b) | ||
| M1 lists integers 0, 1, 2, 3, 4, 5, 6 (ECF from (a)) | ✓ 1 | |
| A1 21 | ✓ 1 | |
Common error: Including −1 in the list. A strict inequality (<) does not include its end value.
A rectangular vegetable garden is (x + 5) metres long and (x − 2) metres wide, where x > 2.
Show that the area of the garden, in m2, is x2 + 3x − 10
The owner increases both the length and the width by 1 metre.
Show that the area increases by (2x + 4) m2.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (x + 5)(x − 2) = x2 − 2x + 5x − 10, at least 3 of the 4 terms correct | ✓ 1 | |
| = x2 + 3x − 10 with all four terms correct before simplifying | ✓ 1 | |
| Part (b) | ||
| New area = (x + 6)(x − 1) | ✓ 1 | |
| = x2 + 5x − 6 | ✓ 1 | |
| (x2 + 5x − 6) − (x2 + 3x − 10) = 2x + 4 | ✓ 1 | |
Common error: When subtracting a whole expression, put it in brackets: −(x² + 3x − 10) gives +10, not −10.
Factorising can make some calculations much quicker.
Factorise x2 − 11x + 28
Factorise y2 − 144
Use the identity a2 − b2 = (a − b)(a + b) to work out 622 − 382 without a calculator.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Two brackets with numbers multiplying to 28 or adding to −11, e.g. (x + 4)(x + 7) | ✓ 1 | |
| (x − 4)(x − 7) | ✓ 1 | |
| Part (b) | ||
| (y − 12)(y + 12) | ✓ 1 | |
| Part (c) | ||
| (62 − 38)(62 + 38) = 24 × 100 | ✓ 1 | |
| = 2400 | ✓ 1 | |
Exam tip: Check a factorisation by expanding it again: the brackets must give back every term, including the signs.
Nikos thinks of a number x. He finds that
(x + 4)/3 − (x − 2)/5 = 2
What is the value of x?
| Marking point | Mark | Notes |
|---|---|---|
| Multiply every term by 15: 5(x + 4) − 3(x − 2) = 30 Expand: 5x + 20 − 3x + 6 = 30 Simplify: 2x + 26 = 30, so 2x = 4 x = 2 | ✓ 1 | Answer C |
Answer: C
Common error: A minus sign in front of a fraction applies to the whole numerator: −3(x − 2) = −3x + 6. Also multiply the right-hand side by the common denominator.
The four interior angles of a quadrilateral, in degrees, are
2x + 10, 3x − 20, x + 40 and 4x − 10
What is the value of x?
| Marking point | Mark | Notes |
|---|---|---|
| Angles in a quadrilateral add up to 360°. (2x + 10) + (3x − 20) + (x + 40) + (4x − 10) = 360 10x + 20 = 360 10x = 340, so x = 34 (Check: the angles are 78°, 82°, 74° and 126°, which add up to 360°.) | ✓ 1 | Answer C |
Answer: C
Exam tip: Collect the number terms carefully, keeping each sign with its number: +10 − 20 + 40 − 10 = +20. Substituting your answer back to check the angle sum takes only a few seconds.
A weather station converts temperatures using the formula
F = 9C/5 + 32
where C is the temperature in degrees Celsius and F is the temperature in degrees Fahrenheit.
Work out the value of F when C = −15
Make C the subject of the formula.
At one temperature the readings in degrees Celsius and degrees Fahrenheit are the same number. Form an equation and solve it to find this temperature.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 9 × (−15)/5 = −27 | ✓ 1 | |
| F = −27 + 32 = 5 | ✓ 1 | |
| Part (b) | ||
| F − 32 = 9C/5 or 5F = 9C + 160 | ✓ 1 | |
| C = 5(F − 32)/9 | ✓ 1 | |
| Part (c) | ||
| C = 9C/5 + 32 or 5C = 9C + 160 | ✓ 1 | |
| −4C = 160, so the temperature is −40 | ✓ 1 | |
Exam tip: When you rearrange, undo the operations in reverse order: subtract 32 first, then multiply by 5, then divide by 9. Test your rearranged formula with the pair you found in part (a).
Ana is x years old. Her father is three times as old as Ana.
In 12 years' time, her father will be twice as old as Ana.
Write down an expression, in terms of x, for her father's age in 12 years' time.
Form an equation in x and solve it to find Ana's age now.
Work out the total of their ages now.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3x + 12 | ✓ 1 | |
| Part (b) | ||
| 3x + 12 = 2(x + 12) | ✓ 1 | |
| 3x + 12 = 2x + 24 | ✓ 1 | |
| x = 12, so Ana is 12 years old | ✓ 1 | |
| Part (c) | ||
| 12 + 36 = 48 years | ✓ 1 | |
Common error: Both people get 12 years older. Ana's future age is x + 12, so 'twice as old' means 2(x + 12), not 2x + 12.
A gardener uses the same total length of edging around two flower beds.
One bed is a rectangle with length (2x + 1) m and width (x + 2) m.
The other bed is an equilateral triangle with sides of length (x + 8) m.
The two beds have the same perimeter.
Show that the perimeter of the rectangle is (6x + 6) m.
Form an equation and solve it to find the value of x.
Work out the area of the rectangular bed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2(2x + 1) + 2(x + 2) = 4x + 2 + 2x + 4 = 6x + 6 | ✓ 1 | |
| Part (b) | ||
| Triangle perimeter 3(x + 8) or 3x + 24 | ✓ 1 | |
| 6x + 6 = 3x + 24 | ✓ 1 | |
| x = 6 | ✓ 1 | |
| Part (c) | ||
| 13 × 8 = 104 m2 | ✓ 1 | |
Exam tip: In a 'show that' part, write every step of the expansion and simplification; the final line must match the given expression exactly.
Two mobile phone plans charge as follows.
Plan A: £15 per month plus 8p per minute of calls.
Plan B: £24 per month plus 5p per minute of calls.
In one month, m minutes of calls are made.
Write down an expression, in pounds, for the monthly cost of Plan A.
Form an equation and solve it to find the number of minutes for which the two plans cost the same.
Lena makes 420 minutes of calls in a month. Which plan is cheaper for her, and by how much?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 15 + 0.08m | ✓ 1 | |
| Part (b) | ||
| 15 + 0.08m = 24 + 0.05m (or 1500 + 8m = 2400 + 5m) | ✓ 1 | |
| 0.03m = 9 or 3m = 900 | ✓ 1 | |
| m = 300 minutes | ✓ 1 | |
| Part (c) | ||
| Plan A £48.60 and Plan B £45.00 | ✓ 1 | |
| Plan B is cheaper by £3.60 | ✓ 1 | |
Common error: Keep the units consistent: either work entirely in pence (1500 + 8m) or entirely in pounds (15 + 0.08m). Mixing £15 with 8m pence gives a wrong equation.
At a cinema, 3 adult tickets and 2 child tickets cost £35.50 in total.
2 adult tickets and 5 child tickets cost £42.00 in total.
What is the total cost of 1 adult ticket and 1 child ticket?
| Marking point | Mark | Notes |
|---|---|---|
| Let an adult ticket cost £a and a child ticket £c. 3a + 2c = 35.5 and 2a + 5c = 42. Multiply the first by 2 and the second by 3: 6a + 4c = 71 and 6a + 15c = 126. Subtract: 11c = 55, so c = 5. Then 3a = 35.5 − 10 = 25.5, so a = 8.5. Total: 8.50 + 5.00 = £13.50. | ✓ 1 | Answer C |
Answer: C
Exam tip: Re-read what the question actually asks for. Here you need a + c, not just one of the prices.
A mobile phone plan charges a fixed monthly fee of €f plus €r for each gigabyte (GB) of data used.
In March, Sofia used 6 GB and paid €27.
In April, she used 10 GB and paid €35.
Write down two equations in f and r.
Solve your equations to find the value of f and the value of r.
In May, Sofia uses 15 GB. Calculate how much she pays.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| f + 6r = 27 | ✓ 1 | |
| f + 10r = 35 | ✓ 1 | |
| Part (b) | ||
| Subtracts: 4r = 8 | ✓ 1 | |
| r = 2, f = 15 | ✓ 1 | |
| Part (c) | ||
| 15 + 15 × 2 = €45 | ✓ 1 | |
Common error: Dividing the bill by the number of GB (27 ÷ 6) ignores the fixed fee. Two unknowns need two equations.
In a basketball match, a team scored only 2-point baskets and 3-point baskets.
The team scored 31 baskets and 72 points in total.
Let x be the number of 2-point baskets and y the number of 3-point baskets.
Write down two equations in x and y.
Solve the equations to find how many baskets of each type the team scored.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| x + y = 31 | ✓ 1 | |
| 2x + 3y = 72 | ✓ 1 | |
| Part (b) | ||
| Correct method to eliminate one variable, e.g. 2x + 2y = 62 subtracted from 2x + 3y = 72 | ✓ 1 | |
| y = 10 | ✓ 1 | |
| x = 21 | ✓ 1 | |
Exam tip: Check your answer in the context: 21 + 10 = 31 baskets and 2(21) + 3(10) = 72 points.
A rectangular photo print has width x cm. Its length is 5 cm more than its width.
The area of the print is 84 cm2.
Show that x2 + 5x − 84 = 0
Solve x2 + 5x − 84 = 0 and hence find the perimeter of the print. Give a reason for rejecting any value of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Area = x(x + 5) = 84 | ✓ 1 | |
| x2 + 5x = 84 so x2 + 5x − 84 = 0 shown with no errors | ✓ 1 | |
| Part (b) | ||
| (x + 12)(x − 7) = 0 | ✓ 1 | |
| x = 7 (x = −12 rejected because a width cannot be negative) | ✓ 1 | |
| Perimeter = 2(7 + 12) = 38 cm | ✓ 1 | |
Exam tip: When a quadratic comes from a length, always say why you reject the negative root, then answer the question actually asked (here the perimeter, not just x).
Two consecutive positive odd numbers have a product of 323.
Let the smaller number be n.
Write down an equation in n and rearrange it into the form n2 + bn + c = 0
Solve your equation to find the two odd numbers.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| n(n + 2) = 323 giving n2 + 2n − 323 = 0 | ✓ 1 | |
| Part (b) | ||
| (n + 19)(n − 17) = 0 or correct substitution into the quadratic formula | ✓ 1 | |
| n = 17 (n = −19 rejected as the numbers are positive) | ✓ 1 | |
| The numbers are 17 and 19 | ✓ 1 | |
Common error: Consecutive odd numbers differ by 2, not 1, so the second number is n + 2, not n + 1.
n is an integer such that
−7 < 3n − 1 ≤ 14
How many possible values of n are there?
| Marking point | Mark | Notes |
|---|---|---|
| Add 1: −6 < 3n ≤ 15 Divide by 3: −2 < n ≤ 5 Integers: −1, 0, 1, 2, 3, 4, 5, so there are 7 values. | ✓ 1 | Answer C |
Answer: C
Common error: Check each end separately: < means the end value is excluded and ≤ means it is included. Remember that 0 and negative integers count too.
Two mobile phone plans charge per month as follows.
Plan A: £15 plus 2p for each text sent.
Plan B: £6 plus 5p for each text sent.
Let t be the number of texts sent in a month.
Write down an inequality, in terms of t, that is true when Plan A costs less than Plan B. Work in pounds.
Solve your inequality.
Find the least number of texts in a month for which Plan A is cheaper.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 15 + 0.02t < 6 + 0.05t (or equivalent in pence: 1500 + 2t < 600 + 5t) | ✓ 1 | |
| Part (b) | ||
| 9 < 0.03t (or 0.03t > 9) | ✓ 1 | |
| t > 300 | ✓ 1 | |
| Part (c) | ||
| 301 | ✓ 1 | |
Common error: t > 300 does not include 300: at exactly 300 texts both plans cost £21. The least whole number satisfying a strict inequality is the next integer.
The sides of a triangle, in centimetres, are 2x + 1, x + 4 and 3x − 2.
The perimeter of the triangle is less than 41 cm.
Show that x < 19/3
Given also that x is an integer and that 3x − 2 is the longest side, find all the possible values of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (2x + 1) + (x + 4) + (3x − 2) < 41 giving 6x + 3 < 41 | ✓ 1 | |
| 6x < 38 so x < 38/6 = 19/3 | ✓ 1 | |
| Part (b) | ||
| 3x − 2 > 2x + 1 gives x > 3 | ✓ 1 | |
| 3x − 2 > x + 4 gives x > 3, so 3 < x < 19/3 | ✓ 1 | |
| 4, 5, 6 | ✓ 1 | |
Exam tip: In a "show that" part, every step must be written down, including the collected expression for the perimeter, so that the given result clearly follows.
At a museum, 3 adult tickets and 2 child tickets cost £36.50 in total.
2 adult tickets and 5 child tickets cost £44.50 in total.
An adult ticket costs £\(a\) and a child ticket costs £\(c\).
Write down two equations in \(a\) and \(c\).
Solve your equations to find the cost of an adult ticket and the cost of a child ticket.
Work out the total cost of 4 adult tickets and 3 child tickets.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(3a + 2c = 36.5\) | ✓ 1 | |
| B1 \(2a + 5c = 44.5\) | ✓ 1 | |
| Part (b) | ||
| M1 for scaling to equal coefficients and eliminating, e.g. \(6a + 4c = 73\) and \(6a + 15c = 133.5\), giving \(11c = 60.5\) | ✓ 1 | |
| A1 \(c = 5.50\) | ✓ 1 | |
| A1 \(a = 8.50\) | ✓ 1 | |
| Part (c) | ||
| B1 £50.50 (ft from their values of \(a\) and \(c\)) | ✓ 1 | |
Exam tip: Check your answers in the equation you did not use to find the second value. Here, \(2(8.50) + 5(5.50) = 44.50\) confirms both values.
\(P = \frac{3(k - 2m)}{5}\)
Make \(m\) the subject of the formula.
Find the value of \(m\) when \(k = 12\) and \(P = 4.2\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for multiplying both sides by 5: \(5P = 3k - 6m\) | ✓ 1 | |
| M1 for isolating the \(m\) term: \(6m = 3k - 5P\) | ✓ 1 | |
| A1 \(m = \frac{3k - 5P}{6}\) or equivalent | ✓ 1 | |
| Part (b) | ||
| M1 for correct substitution, e.g. \(\frac{3 \times 12 - 5 \times 4.2}{6}\) | ✓ 1 | |
| A1 \(m = 2.5\) | ✓ 1 | |
Exam tip: Expand the bracket after clearing the fraction. This puts the \(m\) term on its own and makes the rearrangement much easier.
Line L has equation \(y = 3x - 5\)
Line M has equation \(2y + x = 18\)
Use algebra to find the coordinates of the point where L and M intersect.
Show that the point \((-2, 10)\) lies on line M.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for substituting, e.g. \(2(3x - 5) + x = 18\) | ✓ 1 | |
| M1 for \(7x = 28\) | ✓ 1 | |
| A1 \((4, 7)\) | ✓ 1 | |
| Part (b) | ||
| B1 for \(2 \times 10 + (-2) = 18\), so the point satisfies the equation of M | ✓ 1 | |
Exam tip: When one equation already gives \(y\) in terms of \(x\), substitution is quicker than elimination.
Priya is \(x\) years old.
Her mother is three times as old as Priya.
In 12 years' time, her mother will be twice as old as Priya.
Work out the present age of Priya's mother.
Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for the ages in 12 years: \(x + 12\) and \(3x + 12\) | ✓ 1 | |
| M1 for forming the equation \(3x + 12 = 2(x + 12)\) | ✓ 1 | |
| M1 for solving to get \(x = 12\) | ✓ 1 | |
| A1 the mother is 36 years old | ✓ 1 | |
Common error: Add 12 to both people's ages. Writing \(3x + 12 = 2x + 12\) forgets that the 2 multiplies Priya's future age, \(x + 12\).
The kinetic energy, E joules, of an object of mass m kg moving with speed v m/s is given by
\(E = \frac{1}{2}mv^2\)
Make v the subject of the formula.
Find the speed of an object of mass 12 kg with kinetic energy 540 J.
Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies by 2, \(2E = mv^2\) | ✓ 1 | |
| M1 divides by m, \(v^2 = \frac{2E}{m}\) | ✓ 1 | |
| A1 \(v = \sqrt{\frac{2E}{m}}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{\frac{2 \times 540}{12}}\) or \(\sqrt{90}\) | ✓ 1 | |
| A1 9.49 m/s | ✓ 1 | |
Common error: Forgetting the final square root leaves v² = 90, giving a 'speed' of 90 m/s. The square root must cover the whole fraction.
Answer each part.
Show that (n + 3)2 − (n − 3)2 ≡ 12n
Hence work out the value of 532 − 472
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (n + 3)2 = n2 + 6n + 9 | ✓ 1 | |
| M1 for (n − 3)2 = n2 − 6n + 9 | ✓ 1 | |
| A1 for correct subtraction with brackets shown, leading to 12n | ✓ 1 | |
| Part (b) | ||
| M1 for identifying n = 50 | ✓ 1 | |
| A1: 600 | ✓ 1 | |
Common error: (n + 3)² is not n² + 9. Write it as (n + 3)(n + 3) and expand all four terms.
A ball is kicked straight up into the air.
Its height, h metres, after t seconds is given by the formula
\(h = 20t - 5t^2\)
Work out the height of the ball when t = 1.5
Find the two values of t for which the height of the ball is 15 m.
For how many seconds is the ball higher than 15 m?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 20 × 1.5 − 5 × 1.52 or 30 − 11.25 | ✓ 1 | |
| A1 for 18.75 m | ✓ 1 | |
| Part (b) | ||
| M1 for 20t − 5t2 = 15 rearranged to a correct three-term quadratic = 0, e.g. t2 − 4t + 3 = 0 or 5t2 − 20t + 15 = 0 | ✓ 1 | |
| M1 for a correct method to solve their quadratic, e.g. (t − 1)(t − 3) = 0, or correct substitution into the quadratic formula | ✓ 1 | |
| A1 for t = 1 and t = 3 | ✓ 1 | |
| Part (c) | ||
| B1 for 2 seconds, or ft their larger value of t minus their smaller value of t from part (b) | ✓ 1 | |
Exam tip: Divide every term by 5 before factorising. This gives the simpler equation t² − 4t + 3 = 0, which is much easier to factorise.
Solve \(\frac{x}{4} + \frac{x - 1}{3} = 5\)
Give your answer as a fraction.
Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for multiplying every term by 12 (or another common multiple, e.g. 24) to clear the fractions, giving \(3x + 4(x - 1) = 60\) or equivalent | ✓ 1 | |
| M1 for expanding and collecting correctly, e.g. \(7x - 4 = 60\), so \(7x = 64\) | ✓ 1 | |
| A1 \(x = \frac{64}{7}\) (accept \(9\frac{1}{7}\)) | ✓ 1 | |
Common error: When you multiply \(\frac{x-1}{3}\) by 12, the result is \(4(x - 1) = 4x - 4\), not \(4x - 1\). Keep the numerator in a bracket.
Answer each part.
Make w the subject of P = 2l + 2w
Make x the subject of y = \(\frac{3x-2}{5}\)
Use your answer to the previous part to find x when y = 8
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for P − 2l = 2w or \(\frac{P}{2}\) = l + w | ✓ 1 | |
| A1: w = \(\frac{P-2l}{2}\) oe, e.g. w = \(\frac{P}{2}\) − l | ✓ 1 | |
| Part (b) | ||
| M1 for 5y = 3x − 2 or for \(\frac{3x}{5}\) = y + \(\frac{2}{5}\) | ✓ 1 | |
| A1: x = \(\frac{5y+2}{3}\) oe | ✓ 1 | |
| Part (c) | ||
| B1: 14 | ✓ 1 | |
Exam tip: Undo the operations in reverse order: clear the fraction first by multiplying by 5, then deal with the −2, then divide by 3.
Solve the simultaneous equations
\(5x + 3y = 7.5\)
\(2x - 4y = 16\)
Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for a correct first step towards eliminating one variable. For example, scale to get equal coefficients (\(20x + 12y = 30\) and \(6x - 12y = 48\)). Or rearrange one equation and substitute it into the other (e.g. \(x = 8 + 2y\), so \(5(8 + 2y) + 3y = 7.5\)). Allow one arithmetic slip. | ✓ 1 | |
| M1 for a correct method to reach an equation in one variable and solve it, e.g. \(26x = 78\) or \(13y = -32.5\) | ✓ 1 | |
| A1 \(x = 3\) | ✓ 1 | |
| A1 \(y = -2.5\) (accept \(-\frac{5}{2}\)) | ✓ 1 | |
Common error: After making the coefficients of \(y\) equal, look at their signs. Here they are \(+12y\) and \(-12y\), so you must add the equations, not subtract. Check your answers by substituting into the other equation: \(2(3) - 4(-2.5) = 16\).
A rectangular field is 5 m longer than it is wide. The width of the field is w metres and the area of the field is 126 m2.
Show that w2 + 5w − 126 = 0. (2)
Solve w2 + 5w − 126 = 0 and state the width of the field. (3)
Work out the perimeter of the field. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 w(w + 5) = 126 | ✓ 1 | |
| A1 expands and rearranges to w2 + 5w − 126 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (w + 14)(w − 9) | ✓ 1 | |
| A1 w = −14 or w = 9 | ✓ 1 | |
| B1 width = 9 m, with −14 rejected as a width cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 2 × (9 + 14) | ✓ 1 | |
| A1 46 m (ECF from (b)) | ✓ 1 | |
Common error: A quadratic can give two solutions, but in a context one may not make sense; always say which one you reject and why.
The kinetic energy E joules of a moving object of mass m kg and speed v m/s is given by E = \(\frac{1}{2}mv^2\).
Rearrange the formula to make v the subject. (2)
Work out the speed when m = 0.45 and E = 90. (2)
The mass stays the same but the kinetic energy falls by 36%. Work out the percentage decrease in the speed. (3)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v2 = \(\frac{2E}{m}\) | ✓ 1 | |
| A1 v = \(\sqrt{\frac{2E}{m}}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{\frac{2 \times 90}{0.45}}\) | ✓ 1 | |
| A1 20 m/s | ✓ 1 | |
| Part (c) | ||
| M1 new energy = 0.64E | ✓ 1 | |
| M1 new speed = √0.64 × v = 0.8v | ✓ 1 | |
| A1 20% | ✓ 1 | |
Common error: Speed is proportional to the square root of energy, so a 36% fall in energy is not a 36% fall in speed.
A pack of 2 short cables and 3 long cables costs €12. A pack of 5 short cables and 2 long cables costs €13.50. A short cable costs €s and a long cable costs €l.
Write down two equations in s and l. (2)
Solve your equations to find the cost of one short cable and one long cable. (4)
A bundle of 4 short cables and 5 long cables is sold for €20. Work out how much cheaper this is than buying the cables separately. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 2s + 3l = 12 | ✓ 1 | |
| B1 5s + 2l = 13.5 | ✓ 1 | |
| Part (b) | ||
| M1 multiply to make coefficients of one letter equal (e.g. 4s + 6l = 24 and 15s + 6l = 40.5) | ✓ 1 | |
| M1 subtract to eliminate and solve for one letter | ✓ 1 | |
| A1 s = 1.5 | ✓ 1 | |
| A1 l = 3 | ✓ 1 | |
| Part (c) | ||
| M1 4 × 1.5 + 5 × 3 (= 21) | ✓ 1 | |
| A1 €1 cheaper | ✓ 1 | |
Exam tip: Label your equations and write down each multiplication you do, so a slip in one coefficient is easy to find.
The floor of a rectangular arena in a game has width (x − 5) metres and length x metres.
Factorise x2 − 5x − 24
Hence solve x2 − 5x − 24 = 0
The area of the arena floor is 24 m2. Work out the perimeter of the floor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x + a)(x + b) with ab = −24 or a + b = −5 | ✓ 1 | |
| A1 (x − 8)(x + 3) | ✓ 1 | |
| Part (b) | ||
| B1 x = 8 and x = −3 | ✓ 1 | |
| Part (c) | ||
| M1 x(x − 5) = 24 leading to x2 − 5x − 24 = 0 | ✓ 1 | |
| A1 x = 8 since a length cannot be negative (width = 3) | ✓ 1 | |
| M1 2(8 + 3) | ✓ 1 | |
| A1 22 m (ECF from the value of x chosen) | ✓ 1 | |
Common error: Reject any root that gives a negative length; state why you reject it.
A pharmacist is preparing a dose of medicine.
Solve 5x + 3 ≥ 28
n is an integer such that −4 < 2n − 6 ≤ 8. List all the possible values of n.
Each tablet contains 250 mg of a drug. A patient must not take more than 1750 mg of the drug in one day. Write down an inequality for the number of tablets, n, and hence find the greatest number of tablets the patient can take in one day.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 5x ≥ 25 | ✓ 1 | |
| A1 x ≥ 5 | ✓ 1 | |
| Part (b) | ||
| M1 solve one part, e.g. n > 1 or n ≤ 7 | ✓ 1 | |
| A1 1 < n ≤ 7 | ✓ 1 | |
| A1 2, 3, 4, 5, 6, 7 | ✓ 1 | |
| Part (c) | ||
| M1 250n ≤ 1750 | ✓ 1 | |
| A1 7 tablets | ✓ 1 | |
Common error: In a double inequality, solve both parts and check whether each end point is included.
A pharmacy mixes an 8% solution (solution A) and a 20% solution (solution B) of the same drug to make 600 ml of a 12% solution. The mixture uses a ml of solution A and b ml of solution B.
Write down two equations connecting a and b.
Solve the equations to find the volume of solution B.
Write down the volume of solution A used.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 a + b = 600 | ✓ 1 | |
| B1 0.08a + 0.20b = 0.12 × 600 (= 72) | ✓ 1 | |
| Part (b) | ||
| M1 eliminate or substitute, e.g. 0.08(600 − b) + 0.20b = 72 | ✓ 1 | |
| M1 0.12b = 24 | ✓ 1 | |
| A1 b = 200 ml | ✓ 1 | |
| Part (c) | ||
| B1 400 ml | ✓ 1 | |
Exam tip: Percentages of different volumes cannot be added; convert each to an amount of drug in ml first.
A box of tablets contains two types. Each type A tablet contains 25 mg of a drug and each type B tablet contains 40 mg. The box contains n type B tablets and 6 more type A tablets than type B tablets. The total amount of the drug in the box is 670 mg.
Show that 65n + 150 = 670
Work out the number of type A tablets in the box.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 number of type A tablets is n + 6 | ✓ 1 | |
| M1 25(n + 6) + 40n = 670 | ✓ 1 | |
| A1 25n + 150 + 40n = 670, so 65n + 150 = 670 | ✓ 1 | |
| Part (b) | ||
| M1 65n = 520 so n = 8 | ✓ 1 | |
| A1 14 type A tablets | ✓ 1 | |
Exam tip: Define your letter clearly (here, n = number of type B tablets) before forming the equation.
On a cargo ship, a small crate has mass x tonnes and a large crate has mass y tonnes. 3 small crates and 2 large crates have a total mass of 13.5 tonnes. 2 small crates and 5 large crates have a total mass of 25.5 tonnes.
Write down two equations in x and y.
Solve the simultaneous equations to find the mass of a small crate and the mass of a large crate.
Work out the total mass of 4 small crates and 3 large crates.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3x + 2y = 13.5 | ✓ 1 | |
| B1 2x + 5y = 25.5 | ✓ 1 | |
| Part (b) | ||
| M1 multiplying to make the coefficients of one letter equal, for example 6x + 4y = 27 and 6x + 15y = 76.5 | ✓ 1 | |
| M1 subtracting to eliminate | ✓ 1 | |
| A1 y = 4.5 | ✓ 1 | |
| A1 x = 1.5 | ✓ 1 | |
| Part (c) | ||
| M1 4 × 1.5 + 3 × 4.5 (ECF from (b)) | ✓ 1 | |
| A1 19.5 tonnes | ✓ 1 | |
Exam tip: Check your solution by substituting both values into the equation you did not use for the elimination.
Answer the following questions on algebraic manipulation.
Expand and simplify 3(2x − 5) − 2(x − 4).
Factorise fully 12x2y − 18xy2.
Expand and simplify (x + 3)(x − 5).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 6x − 15 − 2x + 8 (at least three terms correct) | ✓ 1 | |
| A1 4x − 7 | ✓ 1 | |
| Part (b) | ||
| M1 factor of 6, x or y taken out | ✓ 1 | |
| A1 6xy(2x − 3y) | ✓ 1 | |
| Part (c) | ||
| M1 x2 − 5x + 3x − 15 | ✓ 1 | |
| A1 x2 − 2x − 15 | ✓ 1 | |
Common error: When subtracting a bracket, the sign changes on every term: −2(x − 4) gives −2x + 8, not −2x − 8.
Answer the following questions on inequalities.
Solve 5 − 2x ≥ 13.
x is an integer such that −3 ≤ 2x + 1 < 9.
List all possible values of x.
A theatre lighting bar has mass 24 kg. Each light has mass 8 kg. The total mass on the bar must not be more than 120 kg.
Work out the greatest number of lights that can be fitted.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 −2x ≥ 8 | ✓ 1 | |
| A1 x ≤ −4 | ✓ 1 | |
| Part (b) | ||
| M1 −4 ≤ 2x < 8 | ✓ 1 | |
| A1 −2 ≤ x < 4 | ✓ 1 | |
| A1 −2, −1, 0, 1, 2, 3 | ✓ 1 | |
| Part (c) | ||
| M1 24 + 8n ≤ 120 | ✓ 1 | |
| M1 8n ≤ 96 | ✓ 1 | |
| A1 12 | ✓ 1 | |
Common error: When both sides of an inequality are divided by a negative number, the inequality sign must be reversed.
A solar panel array is a rectangle. Its width is w m and its length is 5 m more than its width. The area of the array is 84 m2.
Show that w2 + 5w − 84 = 0.
Solve w2 + 5w − 84 = 0 and write down the width of the array.
Work out the perimeter of the array.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 w(w + 5) = 84 | ✓ 1 | |
| A1 expand to give w2 + 5w − 84 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (w + 12)(w − 7) | ✓ 1 | |
| A1 w = 7 or w = −12 | ✓ 1 | |
| A1 width 7 m, with −12 rejected because a length cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 2(7 + 12) | ✓ 1 | |
| A1 38 m (ECF from (b)) | ✓ 1 | |
Common error: After solving a quadratic from a real context, reject any root that makes no physical sense, such as a negative length.
The cost, €C, of cable for connecting a wind farm is directly proportional to the length, L metres, of cable bought. 45 m of cable costs €63.
Find a formula for C in terms of L.
Work out the cost of 120 m of cable.
A buyer has €245 to spend. Work out the greatest length of cable that can be bought.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 C = kL | ✓ 1 | |
| M1 k = 63 ÷ 45 | ✓ 1 | |
| A1 C = 1.4L | ✓ 1 | |
| Part (b) | ||
| A1 €168 | ✓ 1 | |
| Part (c) | ||
| M1 245 ÷ 1.4 | ✓ 1 | |
| A1 175 m | ✓ 1 | |
Exam tip: Find the constant of proportionality first, then every other value follows from one multiplication or division.
A clinic models the total number of admissions, S, over n days using the formula S = n(a + l)/2, where a is the number of admissions on day 1 and l is the number on day n.
Make a the subject of the formula.
Over 12 days there were 330 admissions in total and 40 admissions on day 12. Work out the number of admissions on day 1.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 2S = n(a + l) | ✓ 1 | |
| M1 2S ÷ n = a + l | ✓ 1 | |
| A1 a = 2S/n − l | ✓ 1 | |
| Part (b) | ||
| M1 2 × 330 ÷ 12 − 40 | ✓ 1 | |
| A1 15 (ECF from (a)) | ✓ 1 | |
Common error: When rearranging, deal with the fraction first, then the bracket; subtracting l too early is a frequent slip.
Vaccine doses delivered each day in two regions are modelled as follows: region A delivers (2x + 1)/3 thousand doses and region B delivers (x − 2)/4 thousand doses. Region A delivers 2 thousand more doses than region B.
Form an equation in x and solve it.
Work out the number of doses delivered by region A.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (2x + 1)/3 − (x − 2)/4 = 2 and multiply by 12: 4(2x + 1) − 3(x − 2) = 24 | ✓ 1 | |
| M1 8x + 4 − 3x + 6 = 24 | ✓ 1 | |
| A1 x = 2.8 | ✓ 1 | |
| Part (b) | ||
| A1 2.2 thousand, which is 2200 doses (ECF from (a)) | ✓ 1 | |
Exam tip: Multiply by the lowest common multiple of the denominators (12 here) to clear all the fractions in one step.
At a theatre, 3 adult tickets and 2 child tickets cost £61 in total. 1 adult ticket and 4 child tickets cost £47 in total. Let an adult ticket cost £a and a child ticket cost £c.
Write down two equations in a and c.
Solve the equations to find a and c. You must show your working.
Work out the total cost of 2 adult tickets and 2 child tickets.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3a + 2c = 61 | ✓ 1 | |
| a + 4c = 47 | ✓ 1 | |
| Part (b) | ||
| M1 for a correct method to match coefficients, e.g. 3a + 12c = 141 | ✓ 1 | |
| M1 for subtracting to get 10c = 80 | ✓ 1 | |
| c = 8 and a = 15 | ✓ 1 | |
| Part (c) | ||
| M1 for 2 × 15 + 2 × 8 (ECF from (b)) | ✓ 1 | |
| £46 | ✓ 1 | |
Exam tip: Check your pair of values in both original equations before moving on.
A ferry travels a distance d km against a current of c km/h. Its speed through the water is v km/h and the time taken is t hours, where t = d/(v − c).
Rearrange the formula to make v the subject.
Work out v when d = 84, t = 3.5 and c = 2.5.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for t(v − c) = d | ✓ 1 | |
| M1 for v − c = d/t | ✓ 1 | |
| v = d/t + c | ✓ 1 | |
| Part (b) | ||
| M1 for 84 ÷ 3.5 + 2.5 | ✓ 1 | |
| v = 26.5 km/h | ✓ 1 | |
Common error: Multiply both sides by the whole bracket (v − c) first; do not cancel part of it.
Two seismometer calibration readings, in microns, are given by the expressions 4(2x − 3) and 5x + 9. The two readings are equal.
Work out the value of x.
Work out the value of the readings.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 8x − 12 = 5x + 9 | ✓ 1 | |
| M1 for 3x = 21 | ✓ 1 | |
| x = 7 | ✓ 1 | |
| Part (b) | ||
| 44 (ECF from (a)) | ✓ 1 | |
Common error: Multiply every term inside the bracket, including the negative: 4(2x − 3) = 8x − 12.
A train accelerates uniformly from 12 m/s and travels 400 m in 20 s. The distance s metres, the initial speed u m/s, the time t seconds and the acceleration a m/s2 are linked by s = ut + ½at2.
Rearrange the formula to make a the subject.
Work out the acceleration of the train.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for s − ut = ½at2 | ✓ 1 | |
| M1 for 2(s − ut) = at2 | ✓ 1 | |
| a = 2(s − ut)/t2 | ✓ 1 | |
| Part (b) | ||
| M1 for 2(400 − 12 × 20)/202 | ✓ 1 | |
| 0.8 m/s2 | ✓ 1 | |
Common error: Subtract ut from both sides before dealing with the ½ and the square; do not divide only part of the right-hand side.
A sports scientist models the training load in week n as L = n2 − 9n + 20.
Factorise n2 − 9n + 20.
Hence find the larger value of n for which L = 0.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (n ± 4)(n ± 5) | ✓ 1 | |
| (n − 4)(n − 5) | ✓ 1 | |
| Part (b) | ||
| n − 4 = 0 or n − 5 = 0 | ✓ 1 | |
| n = 5 | ✓ 1 | |
Exam tip: Check a factorisation by expanding: the two numbers must multiply to +20 and add to −9.
The mass of the Earth is 6.0 × 1024 kg. The mass of Jupiter is 1.9 × 1027 kg.
How many times greater is the mass of Jupiter than the mass of the Earth? Give your answer correct to 3 significant figures.
A planet has a mass equal to 20 times the mass of the Earth. Write its mass in standard form.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 1.9 × 1027 ÷ (6.0 × 1024) | ✓ 1 | |
| 317 | ✓ 1 | |
| Part (b) | ||
| M1 for 20 × 6.0 × 1024 | ✓ 1 | |
| 1.2 × 1026 kg | ✓ 1 | |
Common error: A number such as 12 × 10²⁵ is not in standard form; the first factor must be between 1 and 10.
Rail company A charges £(4n + 15) to hire a carriage for a group of n people. Rail company B charges £(7n − 12) for the same group.
Find the value of n for which both companies charge the same amount.
Work out the amount charged by each company for this group.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 4n + 15 = 7n − 12 | ✓ 1 | |
| M1 for 27 = 3n | ✓ 1 | |
| n = 9 | ✓ 1 | |
| Part (b) | ||
| £51 (ECF from (a)) | ✓ 1 | |
Exam tip: Check by substituting your value into both expressions; they must give the same total.
A telecoms company sells data boosts and roaming passes. Each data boost has the same price and each roaming pass has the same price. Amira buys 3 data boosts and 2 roaming passes for €17. Ben buys 5 data boosts and 1 roaming pass for €19.
Work out the cost of one data boost and the cost of one roaming pass.
Chen buys 4 data boosts and 3 roaming passes. Work out how much Chen pays. (ECF from (a))
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 forming 3x + 2y = 17 and 5x + y = 19 and an attempt to eliminate one variable | ✓ 1 | |
| A1 data boost costs €3 | ✓ 1 | |
| A1 roaming pass costs €4 | ✓ 1 | |
| Part (b) | ||
| B1 €24 | ✓ 1 | |
Common error: After eliminating one variable, substitute back into an original equation and check the answer in the other one.
A food scientist uses the model \(T = 100 + \frac{3m}{w}\) for the boiling temperature, T °C, of a sugar syrup made by dissolving m grams of sugar in w grams of water.
Rearrange the formula to make m the subject.
A syrup is made with 250 g of water and boils at 104.8 °C. Use the model to work out the mass of sugar dissolved.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(T - 100 = \frac{3m}{w}\) (or \(w(T-100) = 3m\)) | ✓ 1 | |
| A1 \(m = \frac{w(T-100)}{3}\) | ✓ 1 | |
| Part (b) | ||
| M1 substituting w = 250 and T = 104.8 | ✓ 1 | |
| A1 400 (g) | ✓ 1 | |
Exam tip: Undo the operations in reverse order: first remove the 100, then multiply by w, then divide by 3.
A telecoms engineer is planning work on mobile masts.
A video call uses data at a constant rate. A call lasting 45 minutes uses 360 MB of data. Work out the data used by a call lasting 2 hours 10 minutes.
6 technicians take 15 hours to fit the antennas on a mast. All technicians work at the same rate. Work out how long 9 technicians would take.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 360 ÷ 45 = 8 MB per minute, or 2 hours 10 minutes written as 130 minutes | ✓ 1 | |
| A1 1040 MB | ✓ 1 | |
| Part (b) | ||
| M1 6 × 15 = 90 hours of work in total (or equivalent) | ✓ 1 | |
| A1 10 hours | ✓ 1 | |
Common error: More workers means less time: this is inverse proportion, so do not scale the time up.
Two climbing groups, A and B, set off together towards a mountain hut. The time taken, in hours, by group A is \(\frac{2x+1}{3}\) and the time taken by group B is \(\frac{x+5}{2}\). Both groups arrive at the hut at the same time.
Solve \(\frac{2x+1}{3} = \frac{x+5}{2}\).
Work out the time taken by each group, in hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplying both sides by 6 (or cross-multiplying) | ✓ 1 | |
| M1 2(2x + 1) = 3(x + 5), i.e. 4x + 2 = 3x + 15 | ✓ 1 | |
| A1 x = 13 | ✓ 1 | |
| Part (b) | ||
| B1 9 (hours) | ✓ 1 | |
Exam tip: To clear fractions, multiply both sides by a common multiple of the denominators, here 6.
At a theatre, stalls tickets cost £s each and balcony tickets cost £b each. The cost of 3 stalls tickets and 2 balcony tickets is £94. The cost of 2 stalls tickets and 5 balcony tickets is £114.
Work out the cost of one stalls ticket and the cost of one balcony ticket. You must show your working.
Work out the cost of 4 stalls tickets and 4 balcony tickets.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 write \(3s + 2b = 94\) and \(2s + 5b = 114\) and multiply to match coefficients, e.g. \(6s + 4b = 188\) and \(6s + 15b = 342\) | ✓ 1 | |
| M1 eliminate one variable, e.g. \(11b = 154\) | ✓ 1 | |
| A1 \(b = 14\) | ✓ 1 | |
| A1 \(s = 22\) | ✓ 1 | |
| Part (b) | ||
| M1 \(4 \times 22 + 4 \times 14\) | ✓ 1 | |
| A1 £144 (ECF from (a)) | ✓ 1 | |
Exam tip: Substitute your answers back into the other equation to check them.
A viewing platform for a volcano is a rectangle. Its length is \((x + 4)\) metres and its width is \((x - 1)\) metres. The area of the platform is 50 m2.
Show that \(x^2 + 3x - 54 = 0\).
Solve \(x^2 + 3x - 54 = 0\).
Work out the perimeter of the platform.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \((x + 4)(x - 1) = 50\) | ✓ 1 | |
| A1 expand to \(x^2 + 3x - 4 = 50\), so \(x^2 + 3x - 54 = 0\) | ✓ 1 | |
| Part (b) | ||
| M1 \((x + 9)(x - 6) = 0\) | ✓ 1 | |
| A1 \(x = 6\) | ✓ 1 | |
| A1 \(x = -9\) | ✓ 1 | |
| Part (c) | ||
| B1 \(x = 6\) only, since \(x = -9\) gives negative lengths | ✓ 1 | |
| A1 30 m (length 10 m, width 5 m) | ✓ 1 | |
Common error: Always check each solution of a quadratic against the context; a length cannot be negative.
A shipping company limits the number of containers, n, loaded in one stack.
Solve the inequality \(5x - 7 > 2x + 11\).
n is an integer and \(-3 \le 2n + 1 < 9\). Write down all the possible values of n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(3x > 18\) | ✓ 1 | |
| A1 \(x > 6\) | ✓ 1 | |
| Part (b) | ||
| M1 \(-4 \le 2n < 8\) | ✓ 1 | |
| M1 \(-2 \le n < 4\) | ✓ 1 | |
| A1 \(-2, -1, 0, 1, 2, 3\) | ✓ 1 | |
Common error: If you multiply or divide an inequality by a negative number, the inequality sign must be reversed; also check whether end values are included.
The stage of a theatre is a triangle. Its sides have lengths \(x\) metres, \((x + 3)\) metres and \((2x - 1)\) metres. The perimeter of the stage is 38 m.
Work out the length of the longest side of the stage.
A lighting technician sets a dimmer value y by solving \(\frac{2y - 3}{5} = \frac{y + 4}{3}\). Find y.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(x + (x + 3) + (2x - 1) = 38\) | ✓ 1 | |
| M1 \(4x + 2 = 38\) so \(x = 9\) | ✓ 1 | |
| A1 17 m (sides 9 m, 12 m, 17 m) | ✓ 1 | |
| Part (b) | ||
| M1 multiply both sides by 15, \(3(2y - 3) = 5(y + 4)\) | ✓ 1 | |
| M1 \(6y - 9 = 5y + 20\) | ✓ 1 | |
| A1 \(y = 29\) | ✓ 1 | |
Common error: When clearing fractions, multiply every term on both sides, and use brackets around whole numerators.
The fuel used by an aircraft is directly proportional to the distance it flies. A flight of 640 km uses 96 litres of fuel.
Work out the fuel used on a flight of 850 km.
The aircraft has 60 litres of fuel left. Work out the greatest distance it can fly.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 96 ÷ 640 (= 0.15 litres per km) or 96 × 850/640 | ✓ 1 | |
| A1 127.5 (litres) | ✓ 1 | |
| Part (b) | ||
| M1 60 ÷ 0.15 or 640 × 60/96 | ✓ 1 | |
| A1 400 (km) | ✓ 1 | |
Exam tip: For direct proportion, find the amount for one unit first (here litres per km); it makes both parts a single multiplication or division.
The number of days needed to harvest a field is inversely proportional to the number of harvesters used. All harvesters work at the same rate. 6 harvesters take 15 days to harvest the field.
How many days would 9 harvesters take?
The farmer wants the field harvested in 5 days. How many harvesters are needed?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 6 × 15 (= 90) ÷ 9 | ✓ 1 | |
| A1 10 (days) | ✓ 1 | |
| Part (b) | ||
| M1 90 ÷ 5 | ✓ 1 | |
| A1 18 (harvesters) | ✓ 1 | |
Common error: Inverse proportion means more workers take fewer days, so the answer to (a) must be less than 15. The product (workers × days) stays constant at 90.
At a concert hall, 4 adult tickets and 3 child tickets cost €69 in total. 3 adult tickets and 5 child tickets cost €71 in total.
Find the cost of one adult ticket and the cost of one child ticket. Show your working.
A family buys 2 adult tickets and 4 child tickets. Work out the total cost.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 forms equations 4a + 3c = 69 and 3a + 5c = 71 and makes one pair of coefficients equal in size (e.g. 20a + 15c = 345 and 9a + 15c = 213) | ✓ 1 | |
| M1 eliminates one variable correctly (e.g. 11a = 132) | ✓ 1 | |
| A1 adult ticket €12 | ✓ 1 | |
| A1 child ticket €7 | ✓ 1 | |
| Part (b) | ||
| M1 2 × 12 + 4 × 7 (ECF from (a)) | ✓ 1 | |
| A1 52 (€) (ECF from (a)) | ✓ 1 | |
Common error: When subtracting two equations, subtract every term, including the right-hand sides, and take care with negative signs.
A theatre sells adult tickets at €\(a\) each and child tickets at €\(c\) each. A booking of 2 adult tickets and 3 child tickets costs €45. A booking of 3 adult tickets and 1 child ticket costs €43.
Write down two equations in \(a\) and \(c\).
Solve your equations to find the price of an adult ticket and the price of a child ticket.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for \(2a+3c=45\) | ✓ 1 | |
| B1 for \(3a+c=43\) | ✓ 1 | |
| Part (b) | ||
| M1 for a method to eliminate one variable, e.g. \(9a+3c=129\) then subtract | ✓ 1 | |
| A1 for \(a=12\) | ✓ 1 | |
| A1 for \(c=7\) | ✓ 1 | |
Exam tip: Use the second equation, or your other original equation, to check both prices.
The rectangular floor of a ship's cargo hold has width \(x\) metres and length \((x+4)\) metres. The area of the floor is 45 m2.
Show that \(x^2 + 4x - 45 = 0\).
Solve \(x^2 + 4x - 45 = 0\) by factorising, and state the width of the hold.
Work out the perimeter of the floor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(x(x+4)=45\) | ✓ 1 | |
| A1 for expanding to \(x^2+4x=45\) and rearranging to \(x^2+4x-45=0\) | ✓ 1 | |
| Part (b) | ||
| M1 for \((x+9)(x-5)\) | ✓ 1 | |
| A1 for \(x=5\) and \(x=-9\) | ✓ 1 | |
| B1 for width 5 m, with \(-9\) rejected because a length cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 for length \(=9\) and \(2(5+9)\) | ✓ 1 | |
| A1 for 28 (m) | ✓ 1 | |
Common error: Always say which solution of a quadratic is rejected and why when the unknown is a length.
At a port, 5 identical cranes take 12 hours to unload a ship. Each crane works at the same constant rate.
How long would 8 cranes take to unload the ship?
How many cranes are needed to unload the ship in 4 hours?
State one assumption made in these calculations.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(5\times12=60\) crane-hours and \(60\div8\) | ✓ 1 | |
| A1 for 7.5 hours | ✓ 1 | |
| Part (b) | ||
| M1 for \(60\div4\) | ✓ 1 | |
| A1 for 15 (cranes) | ✓ 1 | |
| Part (c) | ||
| B1 for e.g. all cranes work at the same rate for the whole time / no delays or breaks | ✓ 1 | |
Exam tip: For inverse proportion, multiply the two quantities to find the constant, then divide by the new value.
A volcano observatory raises an alert level when a quantity \(x\) satisfies the inequalities below.
Solve \(4x - 7 < 2x + 9\).
\(n\) is an integer such that \(-3 \le 2n + 1 < 9\). List all the possible values of \(n\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for collecting terms, e.g. \(2x<16\) | ✓ 1 | |
| A1 for \(x<8\) | ✓ 1 | |
| Part (b) | ||
| M1 for subtracting 1 from every part: \(-4\le 2n<8\) | ✓ 1 | |
| A1 for \(-2\le n<4\) | ✓ 1 | |
| A1 for \(-2, -1, 0, 1, 2, 3\) | ✓ 1 | |
Common error: With a double inequality, do each step to all three parts, and check whether each end is included (≤) or excluded (<).
The time, t hours, taken to process a data job is inversely proportional to the number, n, of identical servers used. Assume the servers share the work equally.
8 servers take 45 hours to process the job.
Find an expression for t in terms of n.
Work out the time taken by 15 servers.
The job must be finished in 9 hours. Work out the least number of servers needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(t = \frac{k}{n}\) with substitution, or \(k = 8 \times 45\) | ✓ 1 | |
| A1 for \(t = \frac{360}{n}\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(\frac{360}{15}\) (ECF from (a)) | ✓ 1 | |
| A1 for 24 hours | ✓ 1 | |
| Part (c) | ||
| M1 for \(9 = \frac{360}{n}\) or \(n = \frac{360}{9}\) (ECF from (a)) | ✓ 1 | |
| A1 for 40 (servers) | ✓ 1 | |
Exam tip: For inverse proportion, the product of the two quantities stays constant, which gives a quick check.
The number of racks, n, installed in a hall must satisfy the inequality \(5 \le 3n - 4 < 20\), where n is a whole number.
Solve the inequality \(5 \le 3n - 4 < 20\).
Write down all the possible values of n.
Solve the inequality \(5(x - 2) > 3x + 4\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for adding 4 to all three parts: \(9 \le 3n < 24\) | ✓ 1 | |
| A1 for \(3 \le n < 8\) | ✓ 1 | |
| Part (b) | ||
| B1 for 3, 4, 5, 6, 7 | ✓ 1 | |
| Part (c) | ||
| M1 for \(5x - 10 > 3x + 4\) and collecting terms to get \(2x > 14\) | ✓ 1 | |
| A1 for \(x > 7\) | ✓ 1 | |
Common error: When listing integers, check carefully which ends of the range are included and which are excluded.
On an excavation, one team buys 4 trowels and 3 brushes for £34.50. A second team buys 5 trowels and 2 brushes for £37.00. A trowel costs £t and a brush costs £b.
Form two equations and solve them to find the cost of one trowel and one brush.
Work out the cost of 7 trowels and 6 brushes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for \(4t + 3b = 34.5\) and \(5t + 2b = 37\) | ✓ 1 | |
| M1 for a correct method to eliminate one variable, e.g. \(8t + 6b = 69\) and \(15t + 6b = 111\) | ✓ 1 | |
| A1 for trowel: t = 6.00 | ✓ 1 | |
| A1 for brush: b = 3.50 | ✓ 1 | |
| Part (b) | ||
| M1 for \(7 \times 6 + 6 \times 3.5\) (ECF from (a)) | ✓ 1 | |
| A1 for £63.00 | ✓ 1 | |
Exam tip: Write money answers to 2 decimal places, and check by substituting into the equation you did not use.
A child's dose of a medicine is directly proportional to the child's mass. A child of mass 20 kg is given a 15 mg dose.
A hospital pump system delivers a fixed volume of saline. 3 identical pumps working together take 14 hours to deliver it.
Work out the dose for a child of mass 34 kg.
The number of hours is inversely proportional to the number of pumps. Work out the time taken by 7 identical pumps to deliver the same volume.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 15 ÷ 20 × 34 (or 0.75 mg per kg) | ✓ 1 | |
| A1 25.5 mg | ✓ 1 | |
| Part (b) | ||
| M1 3 × 14 = 42 (pump-hours) or 42 ÷ 7 | ✓ 1 | |
| A1 6 hours | ✓ 1 | |
Exam tip: For inverse proportion, the product of the two quantities stays constant: 3 × 14 = 7 × t.
A pharmacist uses the formula D = Am/70 to estimate a child's dose, where D mg is the child's dose, A mg is the adult dose and m kg is the child's mass.
Work out D when A = 420 and m = 35.
Rearrange the formula to make m the subject.
A child is given a dose of 150 mg when the adult dose is 525 mg. Work out the child's mass.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 420 × 35 ÷ 70 | ✓ 1 | |
| A1 210 | ✓ 1 | |
| Part (b) | ||
| M1 multiplies both sides by 70, 70D = Am | ✓ 1 | |
| A1 m = 70D/A | ✓ 1 | |
| Part (c) | ||
| M1 70 × 150 ÷ 525 | ✓ 1 | |
| A1 20 kg | ✓ 1 | |
Exam tip: Rearranging first then substituting in (c) is quicker than substituting into the original formula and solving.
Water flows through two pipes. The flow rate in the first pipe is (2x − 3)/5 litres per minute and the flow rate in the second pipe is (x + 4)/3 litres per minute. The two flow rates are equal.
Solve (2x − 3)/5 = (x + 4)/3.
Work out the flow rate in the first pipe.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies both sides by 15, 3(2x − 3) = 5(x + 4) | ✓ 1 | |
| M1 6x − 9 = 5x + 20 | ✓ 1 | |
| A1 x = 29 | ✓ 1 | |
| Part (b) | ||
| A1 11 litres per minute (ECF from (a)) | ✓ 1 | |
Exam tip: Multiplying both sides by the lowest common multiple of the denominators (15) removes both fractions in one step.
A small pump delivers x thousand litres per hour and a large pump delivers y thousand litres per hour. 5 small pumps and 2 large pumps together deliver 22 thousand litres per hour. 3 small pumps deliver 2 thousand litres per hour more than 2 large pumps.
Write two equations and solve them to find x and y.
Work out how much 4 small pumps and 3 large pumps deliver together in one hour.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 5x + 2y = 22 and 3x − 2y = 2 | ✓ 1 | |
| M1 adds the equations to eliminate y, 8x = 24 | ✓ 1 | |
| A1 x = 3 | ✓ 1 | |
| A1 y = 3.5 | ✓ 1 | |
| Part (b) | ||
| M1 4 × 3 + 3 × 3.5 | ✓ 1 | |
| A1 22.5 thousand litres (ECF from (a)) | ✓ 1 | |
Exam tip: When the y terms have equal size and opposite signs, add the equations; when they have the same sign, subtract.
A rectangular footbridge deck has width w metres and length (w + 9) metres. The area of the deck is 136 m2.
Show that w2 + 9w − 136 = 0.
Solve w2 + 9w − 136 = 0 by factorising, and state the width of the deck.
Work out the length of the deck.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 w(w + 9) = 136 | ✓ 1 | |
| A1 expands to w2 + 9w = 136 and rearranges to the given equation | ✓ 1 | |
| Part (b) | ||
| M1 (w + 17)(w − 8) = 0 | ✓ 1 | |
| A1 w = 8 or w = −17 | ✓ 1 | |
| A1 width = 8 m, as a width cannot be negative | ✓ 1 | |
| Part (c) | ||
| A1 8 + 9 = 17 m (ECF from (b)) | ✓ 1 | |
Exam tip: When a quadratic arises from a length or area, reject any negative solution and say why.
A space probe moves in a straight line with constant acceleration. The formula v2 = u2 + 2as links its initial speed u m/s, final speed v m/s, acceleration a m/s2 and distance s m.
Work out v when u = 40, a = 0.5 and s = 3000. Give your answer correct to 3 significant figures.
Rearrange the formula to make a the subject.
Work out a when v = 100, u = 60 and s = 4000.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v2 = 402 + 2 × 0.5 × 3000 = 4600 | ✓ 1 | |
| A1 v = 67.8 (m/s) | ✓ 1 | |
| Part (b) | ||
| M1 v2 − u2 = 2as | ✓ 1 | |
| A1 a = (v2 − u2) ÷ (2s) | ✓ 1 | |
| Part (c) | ||
| M1 (1002 − 602) ÷ (2 × 4000) = 6400 ÷ 8000 | ✓ 1 | |
| A1 0.8 (m/s2) | ✓ 1 | |
Common error: Square the speeds first and then subtract: v² − u² is not the same as (v − u)².
Mission planners use inequalities to decide how long a probe's battery must charge for x hours.
Solve 7 − 3x < −5.
Solve 2x + 1 ≤ 17.
x is a whole number that satisfies both inequalities. How many possible values of x are there?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 −3x < −12 | ✓ 1 | |
| A1 x > 4 (the inequality sign is reversed on dividing by −3) | ✓ 1 | |
| Part (b) | ||
| M1 2x ≤ 16 | ✓ 1 | |
| A1 x ≤ 8 | ✓ 1 | |
| Part (c) | ||
| A1 5, 6, 7, 8, so 4 values | ✓ 1 | |
Common error: When you multiply or divide an inequality by a negative number, the inequality sign must be reversed.
A space probe carries cameras, each of mass (x + 3) kg, and sensors, each of mass (2x − 1) kg.
The probe carries 4 cameras and 3 sensors. Show that their total mass is (10x + 9) kg.
The total mass of these instruments is 79 kg. Work out x.
Work out the mass of one sensor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 4(x + 3) + 3(2x − 1) | ✓ 1 | |
| A1 4x + 12 + 6x − 3 = 10x + 9 | ✓ 1 | |
| Part (b) | ||
| M1 10x + 9 = 79 | ✓ 1 | |
| A1 x = 7 | ✓ 1 | |
| Part (c) | ||
| A1 2 × 7 − 1 = 13 kg | ✓ 1 | |
Common error: When expanding 3(2x − 1), multiply every term in the bracket: the result is 6x − 3, not 6x − 1.
In a fitness drill, the total number of repetitions R after n sets is given by R = n2 + 4n.
Show that R = 45 gives n2 + 4n − 45 = 0.
Solve n2 + 4n − 45 = 0.
How many sets give a total of 45 repetitions?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 sets n2 + 4n = 45 and rearranges to the given equation | ✓ 1 | |
| Part (b) | ||
| M1 (n + a)(n + b) with ab = −45 | ✓ 1 | |
| A1 (n + 9)(n − 5) = 0 | ✓ 1 | |
| A1 n = −9 or n = 5 | ✓ 1 | |
| Part (c) | ||
| A1 5 sets, as the number of sets cannot be negative | ✓ 1 | |
Common error: Always say which solution is rejected and why; writing both answers without a decision loses the final mark.
A fitness app uses inequalities to judge training plans.
Solve −2 ≤ 3x + 4 < 16.
A plan is suitable when the number of weekly sessions n satisfies 20 − 3n > 2n − 5. Solve this inequality.
What is the largest whole number of sessions that makes a plan suitable?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 subtracts 4 from all three parts: −6 ≤ 3x < 12 | ✓ 1 | |
| M1 divides all three parts by 3 | ✓ 1 | |
| A1 −2 ≤ x < 4 | ✓ 1 | |
| Part (b) | ||
| M1 25 > 5n (or equivalent rearrangement) | ✓ 1 | |
| A1 n < 5 | ✓ 1 | |
| Part (c) | ||
| A1 4 | ✓ 1 | |
Common error: With a strict inequality such as n < 5, the value 5 itself is not allowed, so the largest whole number is 4.
Algebra is used to design solar panel arrays at an observatory.
Factorise 6y2 + 9y.
Expand and simplify (3x − 2)(x + 4).
A rectangular array has length (3x − 2) m and width (x + 4) m. Work out its area when x = 3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 takes out a common factor, for example 3 or y | ✓ 1 | |
| A1 3y(2y + 3) | ✓ 1 | |
| Part (b) | ||
| M1 at least three of 3x2, 12x, −2x, −8 correct | ✓ 1 | |
| A1 3x2 + 10x − 8 | ✓ 1 | |
| Part (c) | ||
| A1 (3 × 3 − 2)(3 + 4) = 7 × 7 = 49 m2 | ✓ 1 | |
Common error: A factorised expression must be completely factorised: 3(2y² + 3y) is only partly factorised.
Two satellite tracks are modelled by the straight lines 3x + 2y = 16 and 5x − 4y = 12, where distances are in thousands of km. The lines cross at point P.
Find the coordinates of P.
Work out the gradient of the line from the origin O to P.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies the first equation by 2 to match the y coefficients: 6x + 4y = 32 | ✓ 1 | |
| M1 adds the equations: 11x = 44 | ✓ 1 | |
| A1 x = 4 | ✓ 1 | |
| A1 y = 2 | ✓ 1 | |
| Part (b) | ||
| A1 2/4 = 0.5 (ECF from (a)) | ✓ 1 | |
Exam tip: If the signs of the matched coefficients are opposite, add the equations; if they are the same, subtract.
A sports scientist models changes in an athlete's resting heart rate, in beats per minute, using inequalities.
Solve \(5x - 7 \leq 3x + 9\).
n is an integer such that \(-3 < 2n + 1 \leq 9\). List all the possible values of n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for collecting terms: 2x ≤ 16 | ✓ 1 | |
| A1 for x ≤ 8 | ✓ 1 | |
| Part (b) | ||
| M1 for subtracting 1 throughout: −4 < 2n ≤ 8 | ✓ 1 | |
| A1 for −2 < n ≤ 4 | ✓ 1 | |
| A1 for −1, 0, 1, 2, 3, 4 | ✓ 1 | |
Common error: With a strict inequality such as −2 < n, the end value −2 is not allowed, but with ≤ the end value 4 is.
The energy E kJ used by a cyclist is directly proportional to the distance d km cycled. When d = 3, E = 210.
Find a formula for E in terms of d.
Work out E when d = 12.5.
Work out d when E = 1050.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for E = kd and 210 = k × 3 | ✓ 1 | |
| A1 for E = 70d | ✓ 1 | |
| Part (b) | ||
| M1 for 70 × 12.5 | ✓ 1 | |
| A1 for 875 | ✓ 1 | |
| Part (c) | ||
| M1 for 1050 ÷ 70 | ✓ 1 | |
| A1 for 15 | ✓ 1 | |
Exam tip: For direct proportion, find the constant k first, then use the full formula for every later part.
On one night, Priya records (5n − 3) objects and Joe records 2(n + 6) objects, where n is the number of hours spent observing. They record the same number of objects.
Find the value of n.
How many objects does each person record?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 5n − 3 = 2(n + 6) | ✓ 1 | |
| M1 for expanding and collecting terms: 3n = 15 | ✓ 1 | |
| A1 for n = 5 | ✓ 1 | |
| Part (b) | ||
| B1 for 22 | ✓ 1 | |
Exam tip: Substitute your value of n into both expressions: if they do not match, there is an error.
A coach uses the formula \(T = 0.6(220 - a) + 40\) to find an athlete's target heart rate T (beats per minute) at age a years.
Find T when a = 30.
Rearrange the formula to make a the subject.
An athlete has a target heart rate of 148 beats per minute. Find the athlete's age.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 0.6 × 190 + 40 | ✓ 1 | |
| A1 for 154 | ✓ 1 | |
| Part (b) | ||
| M1 for T − 40 = 0.6(220 − a) | ✓ 1 | |
| M1 for (T − 40) ÷ 0.6 = 220 − a | ✓ 1 | |
| A1 for a = 220 − (T − 40)/0.6, or an equivalent form | ✓ 1 | |
| Part (c) | ||
| M1 for 220 − (148 − 40)/0.6 | ✓ 1 | |
| A1 for 40 years | ✓ 1 | |
Exam tip: When rearranging, undo the operations in reverse order: first remove the +40, then divide by 0.6.
A library lends out books under the following rules.
A member may borrow \(n\) books, where \(4n + 2 \leq 30\). Solve this inequality.
Write down all the integer values of \(x\) that satisfy \(-3 < 2x - 1 \leq 7\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(4n \leq 28\) | ✓ 1 | |
| A1 \(n \leq 7\) | ✓ 1 | |
| Part (b) | ||
| M1 add 1 and divide by 2 on all three parts | ✓ 1 | |
| A1 \(-1 < x \leq 4\) | ✓ 1 | |
| A1 0, 1, 2, 3, 4 | ✓ 1 | |
Common error: Check each end of a range: the integer at a strict inequality (\(<\)) is not included, but one at (\(\leq\)) is.
Two sizes of turbine are installed on a wind farm. A farm has 6 small turbines and 2 large turbines, producing a total of 15.4 MW. A second farm has 4 small turbines and 5 large turbines, producing a total of 22 MW. All small turbines produce the same power as each other, and all large turbines produce the same power as each other.
Work out the power produced by one small turbine and by one large turbine.
A third farm has 10 small turbines and 10 large turbines. Work out its total power. ECF from (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 equations \(6s + 2l = 15.4\) and \(4s + 5l = 22\) | ✓ 1 | |
| M1 multiply to match coefficients and eliminate | ✓ 1 | |
| A1 small turbine 1.5 MW | ✓ 1 | |
| A1 large turbine 3.2 MW | ✓ 1 | |
| Part (b) | ||
| B1 47 MW (ECF from (a)) | ✓ 1 | |
Exam tip: Number your equations and show the subtraction or addition step clearly so method marks are secure.
Six engineers can service all the turbines on a wind farm in 15 days. The time taken, \(T\) days, is inversely proportional to the number of engineers, \(n\). All engineers work at the same rate.
Work out the time taken if 9 engineers do the work.
The work must be finished in 4.5 days. Work out the number of engineers needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(6 \times 15 = 90\) or \(T = \frac{90}{n}\) | ✓ 1 | |
| A1 10 days | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{90}{4.5}\) | ✓ 1 | |
| A1 20 engineers | ✓ 1 | |
Common error: In inverse proportion the product of the two quantities stays constant, so do not divide the values the wrong way round.
The energy \(E\) kWh produced by a wind turbine is given by \(E = 0.4pt\), where \(p\) kW is the rated power of the turbine and \(t\) is the time in hours.
Rearrange the formula to make \(t\) the subject.
A turbine has rated power 2500 kW and produces 36 000 kWh. Work out the value of \(t\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 divide both sides by \(0.4p\) (or by 0.4 and by \(p\)) | ✓ 1 | |
| A1 \(t = \frac{E}{0.4p}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{36000}{0.4 \times 2500}\) | ✓ 1 | |
| A1 36 hours | ✓ 1 | |
Exam tip: Rearrange first and substitute afterwards; it is easier to check and spot errors.
A coffee roasting company tests recipes using inequalities.
Solve \(3(2x - 1) > 4x + 9\).
Solve \(10 - 3x \leq 22\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expand: \(6x - 3 > 4x + 9\) | ✓ 1 | |
| M1 \(2x > 12\) | ✓ 1 | |
| A1 \(x > 6\) | ✓ 1 | |
| Part (b) | ||
| M1 divide by \(-3\) or rearrange to \(-12 \leq 3x\) | ✓ 1 | |
| A1 \(x \geq -4\) | ✓ 1 | |
Common error: Dividing or multiplying an inequality by a negative number reverses the inequality sign.
A bridge is built from short steel beams and long steel beams. All short beams have the same mass as each other, and all long beams have the same mass as each other. 5 short beams and 2 long beams have a total mass of 3.4 tonnes. 3 short beams and 4 long beams have a total mass of 4.0 tonnes.
Work out the mass of one short beam and the mass of one long beam.
Work out the total mass of 12 short beams and 8 long beams. ECF from (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 equations \(5s + 2l = 3.4\) and \(3s + 4l = 4.0\) | ✓ 1 | |
| M1 multiply the first by 2 and subtract to eliminate \(l\): \(7s = 2.8\) | ✓ 1 | |
| A1 short beam 0.4 tonnes | ✓ 1 | |
| A1 long beam 0.7 tonnes | ✓ 1 | |
| Part (b) | ||
| B1 10.4 tonnes | ✓ 1 | |
Exam tip: Substitute both answers into the other original equation to check they satisfy it.
A space probe travels a fixed distance to a planet. The time taken, \(t\) days, is inversely proportional to the speed, \(v\) km/s. At a speed of 12 km/s the journey takes 30 days.
Work out the time taken at a speed of 18 km/s.
The journey must take 45 days. Work out the speed needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(vt = 12 \times 30 = 360\) or \(t = \frac{360}{v}\) | ✓ 1 | |
| A1 20 days | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{360}{45}\) | ✓ 1 | |
| A1 8 km/s | ✓ 1 | |
Common error: A faster speed means a shorter time. If your time increases when speed increases, you have used direct proportion by mistake.
A probe reaches a speed \(v\) m/s after travelling a distance \(s\) m with constant acceleration \(a\) m/s2 from an initial speed \(u\) m/s. These are linked by \(v^2 = u^2 + 2as\).
Rearrange the formula to make \(a\) the subject.
Work out \(a\) when \(u = 4\), \(v = 10\) and \(s = 21\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 subtract \(u^2\) from both sides: \(v^2 - u^2 = 2as\) | ✓ 1 | |
| A1 \(a = \frac{v^2 - u^2}{2s}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{10^2 - 4^2}{2 \times 21}\) | ✓ 1 | |
| A1 \(a = 2\) | ✓ 1 | |
Exam tip: Subtract \(u^2\) from \(v^2\) before dividing; do not subtract \(u\) and \(v\) first.
A fishery sells fish by the crate. 3 crates of cod and 2 crates of haddock cost €198. 2 crates of cod and 5 crates of haddock cost €297.
Work out the cost of one crate of cod and the cost of one crate of haddock. You must show your working.
Work out the cost of 4 crates of cod and 3 crates of haddock.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 forms \(3c + 2h = 198\) and \(2c + 5h = 297\) and makes one pair of coefficients equal | ✓ 1 | |
| M1 correct addition or subtraction to eliminate one variable | ✓ 1 | |
| A1 one correct value | ✓ 1 | |
| A1 cod €36 and haddock €45 | ✓ 1 | |
| Part (b) | ||
| M1 \(4 \times 36 + 3 \times 45\) (or their values) | ✓ 1 | |
| A1 €279 (ECF from (a)) | ✓ 1 | |
Exam tip: Substitute your answers back into the other original equation to check them.
A robot's controller uses the number \(x\) as a setting.
Solve the inequality \(3x - 5 < 16\).
\(n\) is an integer such that \(-7 < 3n + 2 \leqslant 11\). List all the possible values of \(n\).
Work out the sum of all the values of \(n\) in part (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(3x < 21\) | ✓ 1 | |
| A1 \(x < 7\) | ✓ 1 | |
| Part (b) | ||
| M1 subtracts 2 from all three parts, giving \(-9 < 3n \leqslant 9\) | ✓ 1 | |
| A1 \(-3 < n \leqslant 3\) | ✓ 1 | |
| A1 \(-2, -1, 0, 1, 2, 3\) | ✓ 1 | |
| Part (c) | ||
| A1 3 (ECF from (b)) | ✓ 1 | |
Common error: On a number line or in a list, check whether each end of the range is included: \(<\) excludes the end value, \(\leqslant\) includes it.
A robot moves in a straight line. Its final speed \(v\) m/s is given by \(v^2 = u^2 + 2as\), where \(u\) m/s is its initial speed, \(a\) m/s2 is its acceleration and \(s\) m is the distance travelled.
Rearrange the formula to make \(a\) the subject.
Work out the value of \(a\) when \(u = 1.5\), \(v = 4.5\) and \(s = 6.0\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(2as = v^2 - u^2\) | ✓ 1 | |
| A1 \(a = \dfrac{v^2 - u^2}{2s}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\dfrac{4.5^2 - 1.5^2}{2 \times 6.0}\) (or \(\dfrac{18}{12}\)) | ✓ 1 | |
| A1 1.5 m/s2 | ✓ 1 | |
Exam tip: Rearrange step by step, doing the same to both sides, before you substitute any numbers.
A rectangular fish pond is \(x\) metres wide. Its length is 4 metres more than its width. The area of the pond is 96 m2.
Show that \(x^2 + 4x - 96 = 0\).
Solve \(x^2 + 4x - 96 = 0\) and write down the width of the pond.
Work out the perimeter of the pond.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(x(x+4) = 96\) | ✓ 1 | |
| A1 expands and rearranges to \(x^2 + 4x - 96 = 0\) with no errors | ✓ 1 | |
| Part (b) | ||
| M1 \((x+12)(x-8)\) or correct substitution into the quadratic formula | ✓ 1 | |
| A1 \(x = 8\) and \(x = -12\) | ✓ 1 | |
| A1 width 8 m, with \(-12\) rejected because a length cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 length = 12 and \(2(8 + 12)\) (or their values) | ✓ 1 | |
| A1 40 m (ECF from (b)) | ✓ 1 | |
Common error: After solving a quadratic from a real situation, reject any root that makes a length negative.
Tickets for a school music concert cost £\(a\) for an adult and £\(c\) for a child. Three adult tickets and two child tickets cost £61. Two adult tickets and five child tickets cost £70.
Write down two equations in \(a\) and \(c\).
Solve the equations to find the cost of a child ticket.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(3a+2c=61\) | ✓ 1 | |
| B1 \(2a+5c=70\) | ✓ 1 | |
| Part (b) | ||
| M1 multiplies to match coefficients, e.g. \(6a+4c=122\) and \(6a+15c=210\) | ✓ 1 | |
| A1 \(11c=88\) | ✓ 1 | |
| A1 £8 | ✓ 1 | |
Exam tip: Check your values in the equation you did not use to eliminate: \(2\times15+5\times8=70\).
A rectangular field has width \(x\) metres. Its length is \(5\) m more than its width. The area of the field is 336 m2.
Show that \(x^2+5x-336=0\).
Solve \(x^2+5x-336=0\) by factorising and state the width of the field.
Work out the perimeter of the field.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(x(x+5)=336\) | ✓ 1 | |
| A1 expands and rearranges to \(x^2+5x-336=0\) | ✓ 1 | |
| Part (b) | ||
| M1 \((x+21)(x-16)=0\) | ✓ 1 | |
| A1 \(x=16\) or \(x=-21\) | ✓ 1 | |
| B1 width 16 m, with \(-21\) rejected as a length cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 \(2(16+21)\) (ECF from (b)) | ✓ 1 | |
| A1 74 m | ✓ 1 | |
Common error: Always reject a negative root when the unknown is a length.
An airline sets a baggage rule using inequalities.
Solve \(4x-7>2x+9\).
List all the integers \(n\) that satisfy \(-3\le 2n+1<9\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(2x>16\) | ✓ 1 | |
| A1 \(x>8\) | ✓ 1 | |
| Part (b) | ||
| M1 \(-4\le2n<8\) | ✓ 1 | |
| A1 \(-2\le n<4\) | ✓ 1 | |
| A1 \(-2, -1, 0, 1, 2, 3\) | ✓ 1 | |
Exam tip: For a double inequality, do the same operation to all three parts, and check whether each end is included or excluded before listing integers.
A farmer keeps cows, chickens and sheep. There are \(x\) cows, \(3x\) chickens and \((x+8)\) sheep. Cows and sheep have 4 legs each and chickens have 2 legs each. The animals have 144 legs in total.
Show that \(14x+32=144\).
Work out the number of cows.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 legs: \(4x\) (cows), \(6x\) (chickens) | ✓ 1 | |
| M1 \(4(x+8)\) for the sheep | ✓ 1 | |
| A1 \(4x+6x+4x+32=144\), so \(14x+32=144\) | ✓ 1 | |
| Part (b) | ||
| M1 \(14x=112\) | ✓ 1 | |
| A1 8 | ✓ 1 | |
Exam tip: Count legs, not animals: multiply each number of animals by its legs per animal before adding.
At a vaccination clinic each adult dose contains a ml of vaccine and each child dose contains c ml of vaccine. On Monday the clinic gave 4 adult doses and 3 child doses and used 2.9 ml of vaccine in total. On Tuesday it gave 2 adult doses and 5 child doses and used 2.5 ml in total.
Write down two equations in a and c to show this information.
Solve your equations simultaneously to find a and c.
Work out the total volume of vaccine, in ml, needed for 6 adult doses and 6 child doses. Use the exact values from (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 4a + 3c = 2.9 | ✓ 1 | |
| B1 2a + 5c = 2.5 | ✓ 1 | |
| Part (b) | ||
| M1 multiplies one or both equations to match coefficients of one letter, e.g. 8a + 6c = 5.8 and 8a + 20c = 10 | ✓ 1 | |
| M1 adds or subtracts correctly to eliminate one letter, e.g. 14c = 4.2 | ✓ 1 | |
| A1 c = 0.3 | ✓ 1 | |
| A1 a = 0.5, found by correct substitution | ✓ 1 | |
| Part (c) | ||
| M1 6 × 0.5 + 6 × 0.3 (ECF from (b)) | ✓ 1 | |
| A1 4.8 ml | ✓ 1 | |
Exam tip: After solving, substitute both values into the equation you did not use for the final step to check them.
Algebra and indices.
Simplify fully (2a3b)4 ÷ (8a5b2).
Solve 5x = 1/125.
Solve 2x + 1 = 32.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expands the bracket: 16a12b4 (at least two terms correct) | ✓ 1 | |
| M1 divides coefficients and subtracts powers of a and b | ✓ 1 | |
| A1 2a7b2 | ✓ 1 | |
| Part (b) | ||
| M1 writes 125 as 53 or 1/125 as 5−3 | ✓ 1 | |
| A1 x = −3 | ✓ 1 | |
| Part (c) | ||
| M1 writes 32 as 25, so x + 1 = 5 | ✓ 1 | |
| A1 x = 4 | ✓ 1 | |
Common error: When raising a bracket to a power, the coefficient must be raised to that power as well: (2a³)⁴ has coefficient 16, not 2.
A solar farm is built on a rectangular field. The length of the field is 30 m more than its width. The width of the field is w metres and its area is 4000 m2.
Show that w2 + 30w − 4000 = 0.
Solve w2 + 30w − 4000 = 0 and state the width of the field.
The field is to be surrounded by a fence. Work out the total length of the fence.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 w(w + 30) = 4000 | ✓ 1 | |
| A1 expands and rearranges to w2 + 30w − 4000 = 0 with no errors | ✓ 1 | |
| Part (b) | ||
| M1 factorises as (w + 80)(w − 50) or substitutes correctly into the quadratic formula | ✓ 1 | |
| A1 w = 50 and w = −80 | ✓ 1 | |
| A1 width = 50 m, with −80 rejected because a length cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 2 × (50 + 80) (ECF from (b)) | ✓ 1 | |
| A1 260 m | ✓ 1 | |
Exam tip: When a quadratic comes from a real situation, check each solution against the context and reject any that are impossible.
Inequalities.
Solve 5x − 8 > 2x + 13.
n is an integer such that −3 ≤ 2n + 1 < 9. List all the possible values of n.
A bookshop can spend at most £500 on an order of x boxes of paperbacks. Each box costs £12 and delivery of the whole order costs £20. Work out the greatest whole number of boxes the shop can order.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 3x > 21 | ✓ 1 | |
| A1 x > 7 | ✓ 1 | |
| Part (b) | ||
| M1 subtracts 1 from each part: −4 ≤ 2n < 8 | ✓ 1 | |
| M1 divides each part by 2: −2 ≤ n < 4 | ✓ 1 | |
| A1 −2, −1, 0, 1, 2, 3 | ✓ 1 | |
| Part (c) | ||
| M1 12x + 20 ≤ 500, so x ≤ 40 | ✓ 1 | |
| A1 40 boxes | ✓ 1 | |
Exam tip: On a number line or in a list, check whether each end of the range is included: ≤ includes the end value, < does not.
A logistics company ships two sizes of crate. Let a small crate have mass s kg and a large crate have mass l kg.
Two small crates and three large crates have a total mass of 39 kg.
Four small crates and one large crate have a total mass of 33 kg.
Write down two equations in s and l.
Solve the equations to find the mass of one small crate and the mass of one large crate.
A lorry carries 15 small crates and 12 large crates. Work out their total mass.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 2s + 3l = 39 | ✓ 1 | |
| B1 4s + l = 33 | ✓ 1 | |
| Part (b) | ||
| M1 a correct method to eliminate one variable, e.g. 4s + 6l = 78 then subtract | ✓ 1 | |
| A1 l = 9 | ✓ 1 | |
| A1 s = 6 | ✓ 1 | |
| Part (c) | ||
| M1 15 × 6 + 12 × 9 (ECF from (b)) | ✓ 1 | |
| A1 198 kg | ✓ 1 | |
Exam tip: Check solutions by substituting both values into the equation you did not use for elimination.
The cost, €C, of shipping a container is directly proportional to its mass, m tonnes. A container of mass 12 tonnes costs €540 to ship.
Find a formula for C in terms of m.
Work out the cost of shipping a container of mass 18.5 tonnes.
A container costs €1170 to ship. Work out its mass.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 C = km and 540 = k × 12 | ✓ 1 | |
| A1 C = 45m | ✓ 1 | |
| Part (b) | ||
| M1 45 × 18.5 (ECF from (a)) | ✓ 1 | |
| A1 €832.50 | ✓ 1 | |
| Part (c) | ||
| M1 1170 ÷ 45 | ✓ 1 | |
| A1 26 tonnes | ✓ 1 | |
Exam tip: For direct proportion, divide cost by mass to find the constant once, then reuse it.
The time, t hours, taken to paint a large building is inversely proportional to the number of painters, p. All painters work at the same rate. Six painters take 15 hours.
Find a formula for t in terms of p.
Work out how long 8 painters would take. Give your answer in hours and minutes.
The building must be finished in less than 5 hours. Work out the smallest number of painters needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = k/p or k = 6 × 15 | ✓ 1 | |
| A1 t = 90/p | ✓ 1 | |
| Part (b) | ||
| M1 90 ÷ 8 = 11.25 (ECF from (a)) | ✓ 1 | |
| A1 11 hours 15 minutes | ✓ 1 | |
| Part (c) | ||
| M1 90/p < 5, so p > 18 | ✓ 1 | |
| A1 19 painters | ✓ 1 | |
Exam tip: The number of painters must be a whole number, and exactly 18 painters would take exactly 5 hours, which is not less than 5.
Solve these inequalities.
5x − 8 > 2x + 13
n is an integer such that −3 ≤ 2n + 1 < 9. List all the possible values of n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 3x > 21 | ✓ 1 | |
| A1 x > 7 | ✓ 1 | |
| Part (b) | ||
| M1 subtract 1 throughout: −4 ≤ 2n < 8 | ✓ 1 | |
| A1 −2 ≤ n < 4 | ✓ 1 | |
| A1 −2, −1, 0, 1, 2, 3 | ✓ 1 | |
Exam tip: For a double inequality, do the same operation to all three parts, and check whether each end is included.
On a rail excursion, an adult ticket costs £a and a child ticket costs £c.
Three adults and two children pay a total of £31.50.
Two adults and five children pay a total of £37.50.
Work out the cost of one adult ticket and the cost of one child ticket. You must show your working.
Work out the total cost of four adult tickets and three child tickets.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 correct equations \(3a + 2c = 31.5\) and \(2a + 5c = 37.5\), and a method to make coefficients equal | ✓ 1 | |
| A1 one correct value after eliminating, e.g. \(11a = 82.5\) | ✓ 1 | |
| M1 substitute to find the other unknown | ✓ 1 | |
| A1 \(a = 7.50\) and \(c = 4.50\) | ✓ 1 | |
| Part (b) | ||
| M1 \(4 \times 7.5 + 3 \times 4.5\) (ECF from (a)) | ✓ 1 | |
| A1 £43.50 | ✓ 1 | |
Exam tip: After solving a pair of simultaneous equations, check your answers in the other equation: \(2(7.5) + 5(4.5) = 37.5\).
Answer the following about inequalities.
Solve the inequality \(5x - 8 < 3x + 6\).
n is an integer such that \(-3 \le 2n + 1 < 9\). Write down all the possible values of n.
Work out the sum of the values of n in your answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(2x < 14\) | ✓ 1 | |
| A1 \(x < 7\) | ✓ 1 | |
| Part (b) | ||
| M1 \(-4 \le 2n < 8\) | ✓ 1 | |
| A1 \(-2 \le n < 4\) | ✓ 1 | |
| A1 \(-2, -1, 0, 1, 2, 3\) | ✓ 1 | |
| Part (c) | ||
| B1 3 (ECF from (b)) | ✓ 1 | |
Common error: In a double inequality, whatever you do to the middle you must do to both ends; an end value with ≤ is included in the list, one with < is not.
The amount of fuel, F litres, used by a rail engine is directly proportional to the distance, d km, that it travels.
The engine uses 85 litres of fuel to travel 340 km.
Work out the amount of fuel needed to travel 500 km.
The engine has 60 litres of fuel. Work out the greatest distance it can travel.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(F = kd\) with \(85 = 340k\) or a ratio method | ✓ 1 | |
| A1 \(k = 0.25\) | ✓ 1 | |
| A1 125 litres | ✓ 1 | |
| Part (b) | ||
| M1 \(60 = 0.25d\) or \(60 \div 0.25\) | ✓ 1 | |
| A1 240 km | ✓ 1 | |
Exam tip: For direct proportion write \(F = kd\), find k first, and then use it for every later part.
The time, t hours, needed to repaint a set of rail carriages is inversely proportional to the number of workers, n.
6 workers take 16 hours.
Work out the time that 8 workers would take.
The job must be finished in 4 hours. Work out the number of workers needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(t = k/n\) or \(tn\) constant | ✓ 1 | |
| M1 \(k = 96\) | ✓ 1 | |
| A1 12 hours | ✓ 1 | |
| Part (b) | ||
| M1 \(96 \div 4\) | ✓ 1 | |
| A1 24 | ✓ 1 | |
Common error: With inverse proportion the product \(n \times t\) stays constant; more workers means less time, so check that your answer to (a) is smaller than 16.
A beekeeper sells honey in small jars and large jars.
3 small jars and 2 large jars cost £13.80.
5 small jars and 1 large jar cost £15.30.
Work out the cost of one small jar and the cost of one large jar.
Work out the total cost of 4 small jars and 4 large jars.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3s + 2l = 13.80 and 5s + l = 15.30 | ✓ 1 | |
| M1 multiply to equalise one coefficient, e.g. 10s + 2l = 30.60 | ✓ 1 | |
| M1 subtract (or substitute) to eliminate one variable, e.g. 7s = 16.80 | ✓ 1 | |
| A1 small jar £2.40 | ✓ 1 | |
| A1 large jar £3.30 | ✓ 1 | |
| Part (b) | ||
| M1 4 × 2.40 + 4 × 3.30 (ECF from (a)) | ✓ 1 | |
| A1 £22.80 | ✓ 1 | |
Exam tip: Substitute your answers back into both original equations to check them.
In a hospital pharmacy, the dose D mg of a drug is directly proportional to the patient’s mass m kg.
When m = 60, D = 240.
Find the dose for a patient of mass 85 kg.
A patient is given a dose of 290 mg. Find the patient’s mass.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 D = 4m (or 240 ÷ 60 × 85) | ✓ 1 | |
| A1 340 mg | ✓ 1 | |
| Part (b) | ||
| M1 290 ÷ 4 | ✓ 1 | |
| A1 72.5 kg | ✓ 1 | |
Common error: With direct proportion, divide to find the constant first; multiplying 240 by 85 is not valid.
During a Formula 1 weekend, the time T hours needed to analyse the car’s data is inversely proportional to the number n of engineers working on it.
6 engineers need 10 hours.
Find the time needed by 8 engineers.
The team wants the analysis finished in 4 hours. How many engineers are needed?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 T = k/n with 10 = k/6 | ✓ 1 | |
| A1 k = 60 | ✓ 1 | |
| A1 7.5 hours | ✓ 1 | |
| Part (b) | ||
| M1 4 = 60/n (ECF from (a)) | ✓ 1 | |
| A1 15 engineers | ✓ 1 | |
Exam tip: For inverse proportion, T × n stays constant, which gives a quick check.
A Formula 1 car accelerates uniformly. Its final speed v m/s, initial speed u m/s, acceleration a m/s2 and time t s are related by v = u + at.
Make a the subject of the formula.
A car goes from 12 m/s to 60 m/s in 4 seconds. Work out its acceleration.
The car keeps this acceleration from 12 m/s. How long does it take to reach 84 m/s?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v − u = at | ✓ 1 | |
| A1 a = (v − u)/t | ✓ 1 | |
| Part (b) | ||
| M1 (60 − 12) / 4 | ✓ 1 | |
| A1 12 m/s2 | ✓ 1 | |
| Part (c) | ||
| M1 84 = 12 + 12t or (84 − 12)/12 (ECF from (b)) | ✓ 1 | |
| A1 6 s | ✓ 1 | |
Exam tip: State the units for each part: m/s for speed, m/s² for acceleration and s for time.
A hospital pharmacy buys boxes of gloves and boxes of masks. A box of gloves costs £x and a box of masks costs £y.
3 boxes of gloves and 2 boxes of masks cost £27.50 in total.
2 boxes of gloves and 5 boxes of masks cost £33.00 in total.
Solve these simultaneous equations to find the cost of one box of gloves and the cost of one box of masks.
You must show clear algebraic working.
Work out the cost of 4 boxes of gloves and 3 boxes of masks.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 forming 3x + 2y = 27.5 and 2x + 5y = 33 (or equivalent in pence) | ✓ 1 | |
| M1 a correct method to equalise coefficients and eliminate one variable, e.g. 11x = 71.5 | ✓ 1 | |
| A1 gloves £6.50 | ✓ 1 | |
| A1 masks £4.00 | ✓ 1 | |
| Part (b) | ||
| M1 4 × their 6.50 + 3 × their 4.00 | ✓ 1 | |
| A1 £38.00 (ECF from (a)) | ✓ 1 | |
Exam tip: After finding one unknown, substitute it into an original equation and check the other equation as well.
A hospital ward needs staff cover. Let n be the number of extra nurses called in.
Solve 5n − 8 > 12
Solve −3 ≤ 2y + 1 < 9
Write down all the integer values of y that satisfy −3 ≤ 2y + 1 < 9
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 5n > 20 | ✓ 1 | |
| A1 n > 4 | ✓ 1 | |
| Part (b) | ||
| M1 −4 ≤ 2y < 8 | ✓ 1 | |
| A1 −2 ≤ y < 4 | ✓ 1 | |
| Part (c) | ||
| B1 at least four correct integers from the list and none outside the range | ✓ 1 | |
| B1 −2, −1, 0, 1, 2, 3 (ECF from (b)) | ✓ 1 | |
Common error: When both sides of an inequality are divided by a negative number, the inequality sign reverses; check the end values of an integer list against the inequality.
The mass, M kg, of honey collected in a season is directly proportional to the number of hives, n. A beekeeper with 8 hives collects 120 kg of honey.
Work out the mass of honey collected from 15 hives.
The beekeeper wants to collect 390 kg of honey. Work out how many hives are needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 120 ÷ 8 = 15 kg per hive, or M = 15n | ✓ 1 | |
| A1 225 kg | ✓ 1 | |
| Part (b) | ||
| M1 390 ÷ 15 or 390 ÷ (120 ÷ 8) | ✓ 1 | |
| A1 26 hives | ✓ 1 | |
Exam tip: For direct proportion find the value for one unit first, or write M = kn and find k from the given pair.
The final speed v m/s of a Formula 1 car that accelerates uniformly from speed u m/s over a distance s m with acceleration a m/s2 is given by
v2 = u2 + 2as
Make s the subject of the formula.
A car has u = 12, v = 30 and a = 4.5. Work out the value of s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 first step correct, e.g. v2 − u2 = 2as | ✓ 1 | |
| A1 s = (v2 − u2)/(2a) | ✓ 1 | |
| Part (b) | ||
| M1 substitution into their formula, e.g. (900 − 144)/9 | ✓ 1 | |
| A1 84 m | ✓ 1 | |
Exam tip: Undo operations in reverse order: subtract u² before dividing by 2a, and keep the whole numerator over the whole denominator.
In archaeology, a small amphora has mass s kg and a large amphora has mass l kg. Three small and two large amphorae have a total mass of 19.5 kg. Five small and one large amphora have a total mass of 18.5 kg.
Solve the simultaneous equations 3s + 2l = 19.5 and 5s + l = 18.5.
Work out the total mass of four small and three large amphorae.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies to match coefficients, e.g. 10s + 2l = 37 | ✓ 1 | |
| M1 subtracts to obtain 7s = 17.5 | ✓ 1 | |
| A1 s = 2.5 | ✓ 1 | |
| A1 l = 6 | ✓ 1 | |
| Part (b) | ||
| M1 4 × 2.5 + 3 × 6 (ECF from (a)) | ✓ 1 | |
| A1 28 kg | ✓ 1 | |
Exam tip: Substitute your answers back into both original equations to check them.
A data centre uses n cooling units. The power limit requires 4n + 6 ≤ 38 and the cooling requirement needs 2n − 3 > 5.
Solve 4n + 6 ≤ 38.
Solve 2n − 3 > 5.
Write down all the possible whole-number values of n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 4n ≤ 32 | ✓ 1 | |
| A1 n ≤ 8 | ✓ 1 | |
| Part (b) | ||
| M1 2n > 8 | ✓ 1 | |
| A1 n > 4 | ✓ 1 | |
| Part (c) | ||
| B1 5, 6, 7, 8 (ECF from (a) and (b)) | ✓ 1 | |
Exam tip: The inequality sign stays the same unless you multiply or divide by a negative number.
The time t hours to copy a data set is inversely proportional to the number n of servers used. With 8 servers, copying takes 15 hours.
Show that t = 120 ÷ n.
Work out the time taken with 12 servers.
The copy must be completed in 4 hours. Work out the number of servers needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = k ÷ n and 15 = k ÷ 8 | ✓ 1 | |
| A1 k = 120, so t = 120 ÷ n | ✓ 1 | |
| Part (b) | ||
| B1 10 hours | ✓ 1 | |
| Part (c) | ||
| M1 4 = 120 ÷ n | ✓ 1 | |
| A1 30 | ✓ 1 | |
Common error: Treating inverse proportion as direct: more servers means less time, so the product nt stays constant.
A marine biologist filters sea water to collect plankton. The mass of plankton collected is directly proportional to the volume of water filtered. 40 g of plankton is collected from 250 litres of water.
Work out the mass of plankton collected from 700 litres of water.
A logistics team loads a research ship. The time, t hours, to load the ship is inversely proportional to the number of workers, n. 6 workers take 15 hours. Work out the time taken by 9 workers.
Work out the smallest number of workers needed to load the ship in 7.5 hours or less.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 40 ÷ 250 × 700, or a constant of 0.16 g per litre | ✓ 1 | |
| A1 112 (g) | ✓ 1 | |
| Part (b) | ||
| M1 nt = 90, or 6 × 15 ÷ 9 | ✓ 1 | |
| A1 10 (hours) | ✓ 1 | |
| Part (c) | ||
| M1 n = 90 ÷ t or 90 ÷ 7.5 | ✓ 1 | |
| A1 12 | ✓ 1 | |
| B1 states that fewer than 12 workers would take longer than 7.5 hours, or checks that 12 workers take exactly 7.5 hours | ✓ 1 | |
Common error: With inverse proportion the product of the two quantities stays constant. Do not divide the time by the ratio of the workers the wrong way round.
A courier firm charges £s for each standard parcel and £e for each express parcel. 3 standard parcels and 2 express parcels cost £21.50 in total. 5 standard parcels and 3 express parcels cost £34.00 in total.
Work out the cost of one standard parcel and the cost of one express parcel.
Work out the cost of 8 standard parcels and 4 express parcels.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 forms 3s + 2e = 21.5 and 5s + 3e = 34 and scales to equal coefficients, for example 9s + 6e = 64.5 and 10s + 6e = 68 | ✓ 1 | |
| M1 subtracts or substitutes correctly | ✓ 1 | |
| A1 s = 3.50 | ✓ 1 | |
| A1 e = 5.50 | ✓ 1 | |
| Part (b) | ||
| M1 8 × 3.5 + 4 × 5.5, using their values (ECF from (a)) | ✓ 1 | |
| A1 £50.00 | ✓ 1 | |
Exam tip: Write the equations one above the other and label them. Multiply so that one letter has matching coefficients, then eliminate it.
A marine biologist records data from a tank of fish and a dive survey.
Solve 5(2x − 3) = 3x + 20.
Solve (x + 5)/3 − (x − 1)/6 = 3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 10x − 15 = 3x + 20 | ✓ 1 | |
| M1 7x = 35 | ✓ 1 | |
| A1 5 | ✓ 1 | |
| Part (b) | ||
| M1 multiplies every term by 6, 2(x + 5) − (x − 1) = 18 | ✓ 1 | |
| M1 expands brackets correctly, 2x + 10 − x + 1 = 18 | ✓ 1 | |
| A1 7 | ✓ 1 | |
Common error: When subtracting a fraction with a bracket in the numerator, the minus sign changes every term in that bracket.
A courier firm charges a fixed fee plus a rate per kilometre. A delivery of 2 km costs £7 and a delivery of 6 km costs £19.
Work out the cost per kilometre and the fixed fee.
A second firm charges a fixed fee of £10 plus £1.20 per kilometre. Find the distance at which both firms charge the same amount.
Work out the cost of a delivery of this distance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (19 − 7) ÷ (6 − 2) | ✓ 1 | |
| A1 £3 per kilometre | ✓ 1 | |
| A1 £1 fixed fee | ✓ 1 | |
| Part (b) | ||
| M1 3x + 1 = 1.2x + 10, using their values from (a) | ✓ 1 | |
| M1 1.8x = 9 | ✓ 1 | |
| A1 5 (km) | ✓ 1 | |
| Part (c) | ||
| B1 £16 | ✓ 1 | |
Exam tip: For a linear cost, the difference in cost divided by the difference in distance gives the rate per kilometre.
A bakery sells loaves at £l each and cakes at £c each. Three loaves and two cakes cost £12.80. Two loaves and five cakes cost £24.30.
Write down two equations in l and c.
Solve the equations to find the cost of one loaf and the cost of one cake.
Work out the cost of 4 loaves and 3 cakes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3l + 2c = 12.80 | ✓ 1 | |
| B1 2l + 5c = 24.30 | ✓ 1 | |
| Part (b) | ||
| M1 multiply to make one set of coefficients equal (e.g. 15l + 10c = 64 and 4l + 10c = 48.60) | ✓ 1 | |
| M1 subtract (or add) to eliminate one variable (11l = 15.40) | ✓ 1 | |
| A1 loaf £1.40 | ✓ 1 | |
| A1 cake £4.30 | ✓ 1 | |
| Part (c) | ||
| M1 4 × 1.40 + 3 × 4.30 (ECF from (b)) | ✓ 1 | |
| A1 £18.50 | ✓ 1 | |
Exam tip: After finding the second unknown, substitute both values into the equation you did not use to check them.
A laboratory tests infection samples using identical machines. The time, T hours, to test all the samples is inversely proportional to the number of machines, m, in use. With 4 machines the testing takes 15 hours.
Find a formula for T in terms of m.
Work out the time needed with 6 machines.
The testing must be finished in 2.5 hours. Work out the number of machines needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 T = k ÷ m and 15 = k ÷ 4 | ✓ 1 | |
| A1 T = 60 ÷ m | ✓ 1 | |
| Part (b) | ||
| M1 60 ÷ 6 (ECF from (a)) | ✓ 1 | |
| A1 10 hours | ✓ 1 | |
| Part (c) | ||
| M1 2.5 = 60 ÷ m (ECF from (a)) | ✓ 1 | |
| A1 24 | ✓ 1 | |
Exam tip: For inverse proportion, the product T × m stays constant, which gives a quick check on each answer.
An energy supplier offers two tariffs. Tariff A charges a fixed £12 plus £0.18 for each kWh used. Tariff B charges a fixed £5 plus £0.25 for each kWh used. A household uses k kWh.
Write down an expression, in terms of k, for the cost in pounds under each tariff.
Find the value of k for which the two tariffs cost the same.
The household uses 140 kWh. Which tariff is cheaper, and by how much?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 Tariff A: 12 + 0.18k | ✓ 1 | |
| B1 Tariff B: 5 + 0.25k | ✓ 1 | |
| Part (b) | ||
| M1 12 + 0.18k = 5 + 0.25k | ✓ 1 | |
| M1 7 = 0.07k (collecting terms) | ✓ 1 | |
| A1 k = 100 | ✓ 1 | |
| Part (c) | ||
| M1 evaluate one cost (e.g. 12 + 0.18 × 140) | ✓ 1 | |
| A1 £37.20 and £40.00 | ✓ 1 | |
| A1 Tariff A cheaper by £2.80 | ✓ 1 | |
Exam tip: Where two quantities are being compared, state clearly which one is larger or cheaper in your final sentence.
A rectangular plot for solar panels has width x metres and length (x + 5) metres. The area of the plot is 84 m².
Show that x2 + 5x − 84 = 0.
Solve x2 + 5x − 84 = 0 by factorising.
Work out the perimeter of the plot.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 5) = 84 | ✓ 1 | |
| A1 expands to x2 + 5x = 84 and rearranges to the given equation | ✓ 1 | |
| Part (b) | ||
| M1 attempt to factorise with brackets (x ± a)(x ± b) where ab = 84 | ✓ 1 | |
| A1 (x + 12)(x − 7) = 0 | ✓ 1 | |
| A1 x = 7 and x = −12 | ✓ 1 | |
| Part (c) | ||
| B1 rejects x = −12 (a length cannot be negative), so width 7 m and length 12 m | ✓ 1 | |
| A1 38 m | ✓ 1 | |
Exam tip: After solving a quadratic from a real situation, check each solution against the context and reject any that make a length negative.
The kinetic energy E joules of a trolley of mass m kg moving at speed v m/s is given by E = ½mv2.
Rearrange the formula to make v the subject.
A trolley has mass 12 kg and kinetic energy 4500 J. Work out its speed. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 2E = mv2 (or v2 = 2E/m) | ✓ 1 | |
| A1 v = √(2E/m) | ✓ 1 | |
| Part (b) | ||
| M1 √(2 × 4500 ÷ 12) = √750 | ✓ 1 | |
| A1 27.4 m/s | ✓ 1 | |
Exam tip: Undo the operations in reverse order: first multiply by 2 and divide by m, then take the square root of the whole expression.
At a water treatment plant the time t hours taken to fill a storage tank is inversely proportional to the number of pumps n in use. When 5 pumps are used the tank takes 12 hours to fill.
Find a formula for t in terms of n.
Work out the time taken when 8 pumps are used. Give your answer in hours and minutes.
The tank must be filled in less than 5 hours. Work out the smallest number of pumps needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = k/n with k = 5 × 12 = 60 | ✓ 1 | |
| A1 t = 60/n | ✓ 1 | |
| Part (b) | ||
| B1 7 hours 30 minutes (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 60/n < 5, so n > 12 (or tests n = 12 giving exactly 5 hours) | ✓ 1 | |
| A1 13 pumps | ✓ 1 | |
Common error: Inverse proportion means the product n × t stays constant, so more pumps means less time; do not divide when you should multiply.
A fashion shop sells scarves and hats. All scarves cost the same and all hats cost the same.
3 scarves and 2 hats cost £67.
2 scarves and 5 hats cost £96.
Work out the cost of one scarf and the cost of one hat.
You must show all your working.
Work out the cost of 4 scarves and 3 hats.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 3s + 2h = 67 and 2s + 5h = 96 | ✓ 1 | |
| M1 multiplies to make one pair of coefficients equal, e.g. 6s + 4h = 134 and 6s + 15h = 288, and eliminates one variable | ✓ 1 | |
| A1 one price correct (h = 14 or s = 13) | ✓ 1 | |
| A1 scarf £13 and hat £14 | ✓ 1 | |
| Part (b) | ||
| M1 4 × 13 + 3 × 14 | ✓ 1 | |
| A1 £94 (ECF from (a)) | ✓ 1 | |
Common error: After finding the first unknown, substitute into one of the original equations and check the answer in the other.
A fashion designer buys fabric. The cost of the fabric is directly proportional to its length.
3.5 m of fabric costs £28.
Work out the cost of 8 m of fabric.
Work out the length of fabric that can be bought for £100.
The time, t hours, taken to sew a batch of coats is inversely proportional to the number of workers, n.
4 workers take 9 hours.
Work out the time taken by 6 workers.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 28 ÷ 3.5 = 8 per metre, or 28 × 8 ÷ 3.5 | ✓ 1 | |
| A1 £64 | ✓ 1 | |
| Part (b) | ||
| M1 100 ÷ 8 | ✓ 1 | |
| A1 12.5 m | ✓ 1 | |
| Part (c) | ||
| M1 t = 36 ÷ n, or 4 × 9 ÷ 6 | ✓ 1 | |
| A1 6 hours | ✓ 1 | |
Exam tip: For inverse proportion, find the constant by multiplying the two linked values, here workers × hours.
This question is about indices.
Write 3−2 as a fraction.
Simplify fully (3a2b)2 × 2ab3.
5n ÷ 52 = 125
Work out the value of n.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 1/9 | ✓ 1 | |
| Part (b) | ||
| M1 (3a2b)2 = 9a4b2 | ✓ 1 | |
| M1 multiplies coefficients 9 × 2 and adds powers of a or b | ✓ 1 | |
| A1 18a5b5 | ✓ 1 | |
| Part (c) | ||
| M1 5n−2 = 53, or 5n = 3125 | ✓ 1 | |
| A1 n = 5 | ✓ 1 | |
Common error: A negative power means a reciprocal, not a negative number: 3⁻² is 1/9, not −9.
A courier van carries small boxes and large boxes. Every small box has the same mass and every large box has the same mass.
3 small boxes and 2 large boxes have a total mass of 46 kg.
5 small boxes and 4 large boxes have a total mass of 84 kg.
Work out the mass of one small box and the mass of one large box.
Work out the total mass of 7 small boxes and 3 large boxes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 forms two equations, 3s + 2l = 46 and 5s + 4l = 84, and equalises coefficients of one letter | ✓ 1 | |
| M1 eliminates a letter correctly | ✓ 1 | |
| A1 small box 8 kg | ✓ 1 | |
| A1 large box 11 kg | ✓ 1 | |
| Part (b) | ||
| A1 89 kg (ECF from (a)) | ✓ 1 | |
Exam tip: Substitute both values back into the other original equation to check them before moving on.
The time d days taken to unload a cargo ship is inversely proportional to the number n of crane operators working. When 6 operators work, the ship takes 15 days to unload.
Find a formula for d in terms of n.
A port manager wants the ship unloaded in 4.5 days. Work out the number of crane operators needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 d = k/n with 15 = k/6, or dn = constant | ✓ 1 | |
| A1 d = 90/n | ✓ 1 | |
| Part (b) | ||
| M1 4.5 = 90/n (ECF from (a)) | ✓ 1 | |
| A1 20 | ✓ 1 | |
Exam tip: For inverse proportion, the product of the two quantities stays constant, so check that 6 × 15 = 20 × 4.5.
A window in a building design is a trapezium. Its area A is given by the formula
A = ½(a + b)h
where a and b are the lengths of the parallel sides and h is the height of the window.
Rearrange the formula to make h the subject.
A window has area 2.7 m2, with parallel sides of length 1.5 m and 0.9 m. Work out its height.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies both sides by 2, or divides by ½(a + b) | ✓ 1 | |
| A1 h = 2A/(a + b) | ✓ 1 | |
| Part (b) | ||
| M1 h = 2 × 2.7/(1.5 + 0.9) | ✓ 1 | |
| A1 2.25 m | ✓ 1 | |
Exam tip: Keep the bracket (a + b) together when dividing; divide by the whole bracket, not just by a.
Solve \(\frac{3x - 2}{4} - \frac{2x + 5}{6} = \frac{x - 1}{3}\)
Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for multiplying every term by 12 (or another common multiple), e.g. \(3(3x - 2) - 2(2x + 5) = 4(x - 1)\) | ✓ 1 | |
| M1 for expanding correctly: \(9x - 6 - 4x - 10 = 4x - 4\) | ✓ 1 | |
| M1 for simplifying to \(5x - 16 = 4x - 4\) | ✓ 1 | |
| A1 \(x = 12\) | ✓ 1 | |
Common error: The minus sign in front of the second fraction multiplies the whole numerator. So \(-2(2x + 5) = -4x - 10\), not \(-4x + 10\).
A straight line has equation \(y = mx + c\)
The line passes through the points \((2, 11)\) and \((-1, 2)\).
Form and solve a pair of simultaneous equations to find the values of \(m\) and \(c\).
The line meets the line with equation \(2x + y = 30\) at the point P.
Find the coordinates of P.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(11 = 2m + c\) and \(2 = -m + c\) | ✓ 1 | |
| A1 \(m = 3\) | ✓ 1 | |
| A1 \(c = 5\) | ✓ 1 | |
| Part (b) | ||
| M1 for substituting, e.g. \(2x + 3x + 5 = 30\) (ft their \(m\) and \(c\)) | ✓ 1 | |
| M1 for \(5x = 25\), so \(x = 5\) | ✓ 1 | |
| A1 \((5, 20)\) | ✓ 1 | |
Exam tip: Subtract the two equations in part (a) to eliminate \(c\) straight away: \(9 = 3m\).
A right-angled triangle has two shorter sides of length x cm and (x + 7) cm.
Its hypotenuse has length (2x + 1) cm.
Show that \(x^2 - 5x - 24 = 0\)
Work out the area of the triangle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for using Pythagoras: x2 + (x + 7)2 = (2x + 1)2 | ✓ 1 | |
| M1 for correct expansion: 2x2 + 14x + 49 = 4x2 + 4x + 1 | ✓ 1 | |
| A1 for rearranging to 2x2 − 10x − 48 = 0 and dividing by 2 to give x2 − 5x − 24 = 0, with no errors | ✓ 1 | |
| Part (b) | ||
| M1 for solving x2 − 5x − 24 = 0 by any valid method, e.g. (x − 8)(x + 3) = 0, the quadratic formula or completing the square | ✓ 1 | |
| A1 for x = 8, with x = −3 rejected because a length cannot be negative | ✓ 1 | |
| A1 for shorter sides 8 cm and 15 cm, giving area ½ × 8 × 15 = 60 cm2 | ✓ 1 | |
Common error: (2x + 1)² is not 4x² + 1. Write it as (2x + 1)(2x + 1) and expand fully to get 4x² + 4x + 1.
A theatre sells 240 tickets for a Saturday show. Stalls tickets cost £45 and balcony tickets cost £32. The total income from ticket sales is £8980.
Let s be the number of stalls tickets and b the number of balcony tickets. Write down two equations in s and b.
Work out the number of stalls tickets sold.
For the next show the stalls price rises by 10% and the balcony price falls to £30. The numbers of tickets sold are the same as for Saturday.
Work out the total income for the next show.
Work out the percentage change in the total income compared with Saturday.
Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 s + b = 240 | ✓ 1 | |
| B1 45s + 32b = 8980 | ✓ 1 | |
| Part (b) | ||
| M1 45s + 32(240 − s) = 8980 or equal coefficients, for example 32s + 32b = 7680 | ✓ 1 | |
| M1 13s = 1300 (or 13s = 8980 − 7680) | ✓ 1 | |
| A1 s = 100 | ✓ 1 | |
| A1 100 stalls tickets and 140 balcony tickets | ✓ 1 | |
| Part (c) | ||
| M1 45 × 1.1 = 49.50 and 100 × 49.50 + 140 × 30 (ECF from (b)) | ✓ 1 | |
| A1 £9150 | ✓ 1 | |
| Part (d) | ||
| M1 (9150 − 8980) ÷ 8980 × 100 | ✓ 1 | |
| A1 1.89% increase (1.893...) | ✓ 1 | |
Exam tip: Define your letters clearly, then check the final numbers against both original equations.
A rectangular enclosure for a battery storage unit is to be fenced using 70 m of fencing for the whole perimeter. The enclosure has width x metres and the area enclosed must be 300 m².
Show that x2 − 35x + 300 = 0.
Solve x2 − 35x + 300 = 0.
Write down the length of the longer side of the enclosure.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 length = 35 − x (from half the perimeter) | ✓ 1 | |
| M1 x(35 − x) = 300 | ✓ 1 | |
| A1 expands and rearranges to x2 − 35x + 300 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (x − 15)(x − 20) = 0 (or correct use of the formula) | ✓ 1 | |
| A1 x = 15 and x = 20 | ✓ 1 | |
| Part (c) | ||
| B1 20 m | ✓ 1 | |
Exam tip: The perimeter of a rectangle gives the sum of two adjacent sides: halve it first, then write the second side as (half-perimeter − x).
This question is about quadratic inequalities.
Solve \(x^2 > 49\)
n is an integer.
Find all the values of n such that \(n^2 < 10\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for the critical values 7 and −7 | ✓ 1 | |
| A1 for x < −7 or x > 7 | ✓ 1 | |
| Part (b) | ||
| M1 for −√10 < n < √10, or for listing values that include 3 or −3 | ✓ 1 | |
| A1 for −3, −2, −1, 0, 1, 2, 3 | ✓ 1 | |
Common error: Writing x > ±7 is wrong. The inequality x² > 49 has two separate regions: x 7.
Find the value of 27−2/3
| Marking point | Mark | Notes |
|---|---|---|
| 271/3 = 3, so 272/3 = 32 = 9. The negative index means the reciprocal: 27−2/3 = 1/9 | ✓ 1 | Answer B |
Answer: B
Exam tip: Deal with a fractional negative index in three steps: the denominator is a root, the numerator is a power, and the minus sign means take the reciprocal.
Answer each part without a calculator. Show your working clearly.
Find the value of 163/4
Simplify fully (4x6)−1/2
Find the value of n such that 9n = 27
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 161/4 = 2 | ✓ 1 | |
| 23 = 8 | ✓ 1 | |
| Part (b) | ||
| (4x6)1/2 = 2x3, or one part of the answer correct, e.g. 2−1 or x−3 | ✓ 1 | |
| 1/(2x3) or ½x−3 | ✓ 1 | |
| Part (c) | ||
| Writes both sides as powers of 3: 32n = 33 | ✓ 1 | |
| 2n = 3, so n = 3/2 | ✓ 1 | |
Exam tip: When the unknown is in the index, rewrite both sides as powers of the same base, then equate the indices.
In a ball sport, the energy E joules carried by a ball of mass m kg moving at speed v metres per second is given by
E = mv2/2
Make v the subject of the formula. Give your answer for a positive value of v.
A ball of mass 0.45 kg carries 360 joules of energy. Find its speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v2 = 2E/m | ✓ 1 | |
| v = √(2E/m) | ✓ 1 | |
| Part (b) | ||
| v = √(2 × 360 / 0.45) or √1600 | ✓ 1 | |
| v = 40 m/s | ✓ 1 | |
Common error: Isolate v² completely before taking the square root, and take the root of the whole right-hand side: √(2E/m), not √(2E)/m.
On a dry test track, the braking distance, d metres, of a car is directly proportional to the square of its speed, v km/h.
When v = 30, d = 18.
What is the braking distance when v = 50?
| Marking point | Mark | Notes |
|---|---|---|
| d = kv2. 18 = k × 302 gives k = 18 ÷ 900 = 0.02. When v = 50: d = 0.02 × 502 = 0.02 × 2500 = 50.0 m. | ✓ 1 | Answer C |
Answer: C
Common error: Using simple direct proportion (scaling 18 by 50/30) ignores the square. Always write the equation with the constant k, find k, then substitute.
In a puzzle game, the number of bonus points, P, awarded for a level is inversely proportional to the square root of the time taken, T seconds.
When T = 16, P = 6.
Find the value of T when P = 8.
| Marking point | Mark | Notes |
|---|---|---|
| P = k ÷ √T. 6 = k ÷ √16 = k ÷ 4, so k = 24. When P = 8: 8 = 24 ÷ √T, so √T = 3 and T = 9. | ✓ 1 | Answer B |
Answer: B
Exam tip: When the unknown is under a square root, find √T first and then square it as a separate, clearly written step.
A factory makes solid ball bearings. The mass, m grams, of a ball bearing is directly proportional to the cube of its radius, r cm.
A ball bearing of radius 3 cm has a mass of 270 g.
Find a formula for m in terms of r.
Calculate the mass of a ball bearing of radius 4.5 cm. Give your answer to 3 significant figures.
Another ball bearing has a mass of 2160 g. Find its radius.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| m = kr3 and 270 = k × 33 | ✓ 1 | |
| m = 10r3 | ✓ 1 | |
| Part (b) | ||
| 10 × 4.53 = 911.25, so 911 g | ✓ 1 | |
| Part (c) | ||
| r3 = 2160 ÷ 10 = 216 | ✓ 1 | |
| r = ∛216 = 6 cm | ✓ 1 | |
Exam tip: Write the general equation with k first and show the substitution. That line earns credit even if a later calculation slips.
The force of attraction, F newtons, between a planet and a space probe is inversely proportional to the square of the distance, d thousand kilometres, between the probe and the centre of the planet.
When d = 5, F = 720.
Find a formula for F in terms of d.
Calculate F when d = 12.
Find the distance d at which the force is 20 N.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = k ÷ d2 and 720 = k ÷ 52 | ✓ 1 | |
| F = 18 000 ÷ d2 | ✓ 1 | |
| Part (b) | ||
| 18 000 ÷ 144 = 125 N | ✓ 1 | |
| Part (c) | ||
| d2 = 18 000 ÷ 20 = 900 | ✓ 1 | |
| d = 30 (thousand km) | ✓ 1 | |
Common error: Multiplying by d instead of d² when finding k: for inverse square, k = F × d² = 720 × 25.
Solve 3x2 + 5x − 4 = 0, giving your solutions correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| a = 3, b = 5, c = −4. x = (−5 ± √(52 − 4 × 3 × (−4))) ÷ (2 × 3) = (−5 ± √73) ÷ 6. √73 = 8.544…, so x = 3.544… ÷ 6 = 0.5907… or x = −13.544… ÷ 6 = −2.257… x = 0.591 or x = −2.26 (3 s.f.) | ✓ 1 | Answer C |
Answer: C
Exam tip: Write down a, b and c with their signs before substituting, and put brackets round negative values: −4 × 3 × (−4) = +48.
Leo solves the equation x2 + 8x + 3 = 0 by completing the square.
Which of these is the correct exact solution?
| Marking point | Mark | Notes |
|---|---|---|
| x2 + 8x = (x + 4)2 − 16. So (x + 4)2 − 16 + 3 = 0, giving (x + 4)2 = 13. x + 4 = ±√13, so x = −4 ± √13. | ✓ 1 | Answer C |
Answer: C
Common error: When completing the square, (x + 4)² = x² + 8x + 16, so you must subtract 16 to keep the expression the same.
A market stall sells handmade candles. When each candle is priced at x euros, the stall's daily profit, P euros, is modelled by
P = −x2 + 10x − 18
Write x2 − 10x + 18 in the form (x − a)2 − b, where a and b are integers.
Hence find the two prices at which the stall makes zero profit. Give your answers in the form p ± √q and also correct to 3 significant figures.
Write down the price that gives the greatest daily profit and the value of this greatest profit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (x − 5)2 − 25 + 18 | ✓ 1 | |
| (x − 5)2 − 7 | ✓ 1 | |
| Part (b) | ||
| (x − 5)2 = 7 so x − 5 = ±√7 | ✓ 1 | |
| x = 5 ± √7, i.e. 2.35 euros and 7.65 euros | ✓ 1 | |
| Part (c) | ||
| P = −[(x − 5)2 − 7] = 7 − (x − 5)2, so price 5 euros | ✓ 1 | |
| Greatest profit 7 euros | ✓ 1 | |
Exam tip: "Hence" means use your completed-square form: (x − a)² = b gives x = a ± √b in one step. The same form gives the turning point: y = (x − a)² − b has a minimum at (a, −b), so P = b − (x − a)² has a maximum at (a, b).
Solve the inequality x2 − 3x − 28 ≤ 0
| Marking point | Mark | Notes |
|---|---|---|
| Factorise: (x − 7)(x + 4) ≤ 0, so the critical values are −4 and 7. The graph of y = x2 − 3x − 28 is a U-shaped curve, so it is on or below the x-axis between the roots: −4 ≤ x ≤ 7. | ✓ 1 | Answer B |
Answer: B
Exam tip: After finding the critical values, sketch the parabola (or test a value such as x = 0) to decide whether the solution lies between the roots or outside them.
In a number game, a player chooses an integer n and works out (2n + 3)2 − (2n − 1)2.
Show that (2n + 3)2 − (2n − 1)2 simplifies to 16n + 8.
Hence prove that the result is always a multiple of 8 but never a multiple of 16.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4n2 + 12n + 9 − (4n2 − 4n + 1) with at least one bracket expanded correctly | ✓ 1 | |
| = 16n + 8 | ✓ 1 | |
| Part (b) | ||
| 16n + 8 = 8(2n + 1), so it is a multiple of 8 | ✓ 1 | |
| 2n + 1 is always odd, so 8(2n + 1) is not divisible by 16 | ✓ 1 | |
Exam tip: A proof needs a general argument: factorise to show the factor you need and state the conclusion in words. Testing a few numbers is not a proof.
This question is about the expression \(2x^2 - 7x - 15\)
Factorise \(2x^2 - 7x - 15\)
Hence solve \(2x^2 - 7x - 15 \ge 0\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (2x + a)(x + b) where ab = −15 and a + 2b = −7, or for a correct partial factorisation | ✓ 1 | |
| A1 for (2x + 3)(x − 5) | ✓ 1 | |
| Part (b) | ||
| M1 for the critical values −3/2 and 5 | ✓ 1 | |
| A1 for x ≤ −1.5 or x ≥ 5 | ✓ 1 | |
Exam tip: Sketch the U-shaped curve mentally. For '≥ 0' you need the parts of the curve on or above the x-axis. These are the two outer regions, not the region between the roots.
\(y = x^2 - 8x + 3\)
Write \(x^2 - 8x + 3\) in the form \((x - a)^2 - b\), where a and b are integers.
Write down the coordinates of the turning point of the curve \(y = x^2 - 8x + 3\)
Hence solve \(x^2 - 8x + 3 = 0\)
Give your solutions in the form \(p \pm \sqrt{q}\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (x − 4)2 seen, or for −16 + 3 | ✓ 1 | |
| A1 for (x − 4)2 − 13 | ✓ 1 | |
| Part (b) | ||
| B1 for (4, −13) | ✓ 1 | |
| Part (c) | ||
| M1 for (x − 4)2 = 13, leading to x − 4 = ±√13 | ✓ 1 | |
| A1 for x = 4 ± √13 | ✓ 1 | |
Common error: The turning point of (x − 4)² − 13 is at x = +4, not x = −4. The x-coordinate is the value that makes the bracket equal to zero.
Solve the equation below.
\(\frac{3}{x - 1} + \frac{2}{x + 2} = 1\)
Give your solutions correct to 3 significant figures.
Solve the equation. Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for multiplying through by (x − 1)(x + 2): 3(x + 2) + 2(x − 1) = (x − 1)(x + 2) | ✓ 1 | |
| M1 for simplifying to x2 − 4x − 6 = 0 | ✓ 1 | |
| M1 for correct substitution into the quadratic formula, or for (x − 2)2 = 10 | ✓ 1 | |
| A1 for x = 5.16 and x = −1.16 | ✓ 1 | |
Common error: Every term must be multiplied by the common denominator, including the 1 on the right-hand side. That side becomes (x − 1)(x + 2), not 1.
The distance, d metres, that a stone falls is directly proportional to the square of the time, t seconds, for which it has been falling.
When t = 2, d = 19.6
Find a formula for d in terms of t.
Find the distance fallen when t = 3.5
Find the time taken for the stone to fall 100 metres.
Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(d = kt^2\) | ✓ 1 | |
| M1 \(19.6 = k \times 2^2\) | ✓ 1 | |
| A1 \(d = 4.9t^2\) | ✓ 1 | |
| Part (b) | ||
| B1 60.0 m (accept 60.025) | ✓ 1 | |
| Part (c) | ||
| M1 \(t^2 = \frac{100}{4.9}\) | ✓ 1 | |
| A1 4.52 s | ✓ 1 | |
Common error: Using d = kt instead of d = kt² gives k = 9.8 and every later answer is wrong. Read 'square of' carefully.
The intensity of light, I lux, from a lamp is inversely proportional to the square of the distance, d metres, from the lamp.
When d = 5, I = 36
Find a formula for I in terms of d.
Find the distance from the lamp at which the intensity is 4 lux.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(I = \frac{k}{d^2}\) | ✓ 1 | |
| M1 \(36 = \frac{k}{25}\), so \(k = 900\) | ✓ 1 | |
| A1 \(I = \frac{900}{d^2}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(d^2 = \frac{900}{4} = 225\) | ✓ 1 | |
| A1 15 m | ✓ 1 | |
Exam tip: Check the answer makes sense: the intensity has fallen to one ninth, so the distance should have trebled (5 m to 15 m).
The time, T seconds, for one swing of a pendulum is directly proportional to the square root of its length, L metres.
When L = 0.81, T = 1.8
Find a formula for T in terms of L.
Find the length of a pendulum for which T = 3
The length of a pendulum is multiplied by 4.
Describe the effect on T.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(T = k\sqrt{L}\) | ✓ 1 | |
| M1 \(1.8 = k\sqrt{0.81} = 0.9k\) | ✓ 1 | |
| A1 \(T = 2\sqrt{L}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{L} = 1.5\) | ✓ 1 | |
| A1 2.25 m | ✓ 1 | |
| Part (c) | ||
| B1 T is doubled (multiplied by 2) | ✓ 1 | |
Exam tip: For part (c), use the proportion directly: multiplying L by 4 multiplies √L by √4 = 2. No new constant is needed.
For a lens, the focal length f cm, the object distance u cm and the image distance v cm are related by
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)
Make v the subject of the formula.
Give your answer as a single fraction.
Find v when f = 12 and u = 30
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\) | ✓ 1 | |
| M1 combines into a single fraction, \(\frac{1}{v} = \frac{u - f}{fu}\) | ✓ 1 | |
| A1 \(v = \frac{fu}{u - f}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{12 \times 30}{30 - 12}\) | ✓ 1 | |
| A1 20 cm | ✓ 1 | |
Common error: The reciprocal of 1/f − 1/u is NOT f − u. Combine into one fraction first, then turn it upside down.
Answer each part.
Factorise 2x2 + 7x − 15
Hence solve 2x2 + 7x − 15 = 0
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (2x ± 3)(x ± 5) or brackets that expand to give two of the three terms correctly | ✓ 1 | |
| A1: (2x − 3)(x + 5) | ✓ 1 | |
| Part (b) | ||
| B1: x = 1.5 and x = −5 (follow through their factors) | ✓ 1 | |
Exam tip: Split the middle term: find two numbers that multiply to 2 × −15 = −30 and add to 7 (10 and −3), then factorise in pairs.
Expand and simplify (x + 3)(2x − 1)(x − 4)
Show all your working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for correct expansion of any two brackets, e.g. 2x2 + 5x − 3 | ✓ 1 | |
| M1 for multiplying their quadratic by the third bracket with at least 4 of the 6 terms correct, e.g. 2x3 − 8x2 + 5x2 − 20x − 3x + 12 | ✓ 1 | |
| A1: 2x3 − 3x2 − 23x + 12 | ✓ 1 | |
Exam tip: Expand two brackets first and simplify before multiplying by the third; write all six terms out before collecting like terms.
Simplify fully \(\frac{x^2-16}{x^2+x-12}\)
Show clear algebraic working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for numerator (x − 4)(x + 4) | ✓ 1 | |
| M1 for denominator (x + 4)(x − 3) | ✓ 1 | |
| A1: \(\frac{x-4}{x-3}\) | ✓ 1 | |
Common error: Only whole factors can be cancelled. Crossing out the x² terms from top and bottom is not valid.
Answer each part.
Make a the subject of 3a + b = ac − 2
Find the value of a when b = 5.4 and c = 7.5
Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for collecting terms in a on one side, e.g. 3a − ac = −2 − b | ✓ 1 | |
| M1 for factorising, e.g. a(3 − c) = −2 − b | ✓ 1 | |
| A1: a = \(\frac{b+2}{c-3}\) oe | ✓ 1 | |
| Part (b) | ||
| M1 for \(\frac{5.4+2}{7.5-3}\) | ✓ 1 | |
| A1: 1.64 | ✓ 1 | |
Exam tip: When the new subject appears twice, gather every term containing it on one side, then take it out as a common factor.
\(a = \frac{3b + 2}{b - 4}\)
Make b the subject of the formula.
Find the value of b when a = 5
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies both sides by (b − 4): \(a(b-4) = 3b + 2\) | ✓ 1 | |
| M1 expands and collects the b terms on one side: \(ab - 3b = 4a + 2\), or equivalent such as \(3b - ab = -4a - 2\) | ✓ 1 | |
| M1 factorises out b: \(b(a - 3) = 4a + 2\), or equivalent such as \(b(3 - a) = -4a - 2\) | ✓ 1 | |
| A1 \(b = \frac{4a+2}{a-3}\), or any equivalent form such as \(\frac{-4a-2}{3-a}\) | ✓ 1 | |
| Part (b) | ||
| B1 11, from substituting into the rearranged formula or solving \(5 = \frac{3b+2}{b-4}\) directly | ✓ 1 | |
Exam tip: When the subject appears twice, gather every term containing it on one side and factorise it out. Check by substituting back: (3×11 + 2)/(11 − 4) = 35/7 = 5.
The line with equation \(y = 2x - 5\) meets the curve with equation \(y = x^2 - 3x + 1\) at the points A and B.
Find the coordinates of A and B.
Find the exact length of AB.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for equating: x2 − 3x + 1 = 2x − 5 | ✓ 1 | |
| M1 for rearranging to x2 − 5x + 6 = 0 and using a valid method to solve it, e.g. factorising to (x − 2)(x − 3) = 0, using the quadratic formula or completing the square | ✓ 1 | |
| A1 for x = 2 and x = 3 | ✓ 1 | |
| A1 for (2, −1) and (3, 1), with y-values found from either equation | ✓ 1 | |
| Part (b) | ||
| M1 for √((3 − 2)2 + (1 − (−1))2) or equivalent | ✓ 1 | |
| A1 for √5 | ✓ 1 | |
Exam tip: Substitute the x-values into the linear equation to find the y-values. It is quicker than using the quadratic and less likely to cause errors.
A number x satisfies \(x(3x - 5) = 4\)
Show that \(3x^2 - 5x - 4 = 0\)
Solve \(3x^2 - 5x - 4 = 0\)
Give your solutions correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for expanding to 3x2 − 5x = 4 and rearranging to 3x2 − 5x − 4 = 0, with no errors seen | ✓ 1 | |
| Part (b) | ||
| M1 for correct substitution into the quadratic formula, \(\frac{5 \pm \sqrt{(-5)^2 - 4 \times 3 \times (-4)}}{2 \times 3}\), or a correct first step in completing the square, e.g. \(3\left[\left(x - \frac{5}{6}\right)^2 - \frac{25}{36}\right] - 4 = 0\) | ✓ 1 | |
| M1 for reaching \(\frac{5 \pm \sqrt{73}}{6}\) or equivalent, e.g. \(\frac{5}{6} \pm \sqrt{\frac{73}{36}}\) | ✓ 1 | |
| A1 for both x = 2.26 and x = −0.591 | ✓ 1 | |
Common error: With c = −4, the term −4ac is −4 × 3 × (−4) = +48. This gives 25 + 48 = 73, not 25 − 48. Put negative values in brackets when you substitute.
This question is about a quadratic expression.
Factorise 2x2 − 5x − 12. (2)
Solve the inequality 2x2 − 5x − 12 ≤ 0. (3)
Write down the number of integers that satisfy the inequality in (b). (1)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (2x ± 3)(x ± 4) | ✓ 1 | |
| A1 (2x + 3)(x − 4) | ✓ 1 | |
| Part (b) | ||
| M1 critical values found from the factors | ✓ 1 | |
| A1 −\(\frac{3}{2}\) and 4 | ✓ 1 | |
| B1 −\(\frac{3}{2}\) ≤ x ≤ 4 | ✓ 1 | |
| Part (c) | ||
| B1 6 (−1, 0, 1, 2, 3, 4) | ✓ 1 | |
Exam tip: Sketching the shape of the quadratic with its roots shows at once whether the solution lies between the roots or outside them.
For a thin lens, the object distance u cm, the image distance v cm and the focal length f cm are connected by \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\).
Rearrange the formula to make v the subject. (3)
Work out v when u = 30 and f = 12. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\) | ✓ 1 | |
| M1 \(\frac{1}{v} = \frac{u - f}{uf}\) | ✓ 1 | |
| A1 v = \(\frac{uf}{u - f}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{30 \times 12}{30 - 12}\) | ✓ 1 | |
| A1 20 cm | ✓ 1 | |
Common error: Do not take reciprocals term by term: \(\frac{1}{v}\) = \(\frac{1}{f}\) − \(\frac{1}{u}\) does not mean v = f − u.
The illuminance E lux of a lamp is inversely proportional to the square of the distance d metres from the lamp. When d = 2, E = 75.
Find a formula for E in terms of d. (2)
Work out the illuminance at a distance of 5 m. (2)
The distance from the lamp is multiplied by 3. State the factor by which the illuminance changes. (1)
Work out the distance at which the illuminance is 3 lux. (2)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 E = \(\frac{k}{d^2}\) and 75 = \(\frac{k}{4}\) | ✓ 1 | |
| A1 E = \(\frac{300}{d^2}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{300}{5^2}\) | ✓ 1 | |
| A1 12 lux | ✓ 1 | |
| Part (c) | ||
| B1 multiplied by \(\frac{1}{9}\) | ✓ 1 | |
| Part (d) | ||
| M1 3 = \(\frac{300}{d^2}\) so d2 = 100 | ✓ 1 | |
| A1 10 m | ✓ 1 | |
Common error: Inverse square means the quantity depends on 1 ÷ d²; trebling the distance divides by 9, not by 3.
The concentration, c mg per litre, of a drug in a patient's blood t hours after a dose is given by c = 18t − 2t2 for 0 ≤ t ≤ 9.
Show that c = 30 when t2 − 9t + 15 = 0
Solve t2 − 9t + 15 = 0. Give your answers correct to 2 decimal places.
For how many hours is the concentration at least 30 mg per litre? Use unrounded values from (b) and give your answer correct to 2 decimal places.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 30 = 18t − 2t2 | ✓ 1 | |
| A1 rearrange to 2t2 − 18t + 30 = 0 and divide by 2 to reach t2 − 9t + 15 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the quadratic formula | ✓ 1 | |
| A1 t = (9 ± √21) ÷ 2 | ✓ 1 | |
| A1 t = 2.21 and t = 6.79 | ✓ 1 | |
| Part (c) | ||
| M1 difference of the two roots (the concentration is above 30 between them) | ✓ 1 | |
| A1 √21 = 4.58 hours | ✓ 1 | |
Exam tip: Write the quadratic formula before substituting, and keep √21 unrounded until the final answer.
In a game, the path of a flying character is the curve with equation y = x2 − 2x − 3. A laser beam travels along the straight line y = 2x + 2. Distances are in metres.
Show that the x coordinates of the points where the path meets the beam satisfy x2 − 4x − 5 = 0
Find the coordinates of the points where the path meets the beam.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x2 − 2x − 3 = 2x + 2 | ✓ 1 | |
| A1 rearrange to x2 − 4x − 5 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (x − 5)(x + 1) = 0 or correct use of formula | ✓ 1 | |
| A1 x = 5 and x = −1 | ✓ 1 | |
| M1 substitute into the equation of the line | ✓ 1 | |
| A1 (5, 12) and (−1, 0) | ✓ 1 | |
Common error: After finding the x values, substitute into the linear equation to get y; do not leave the answer as x values only.
The flow rate, Q litres per minute, through a water pipe is directly proportional to the square of the pipe diameter, d cm. When d = 3, Q = 45.
Show that Q = 5d2
Work out Q when d = 4.2
Work out d when Q = 180
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 Q = kd2 | ✓ 1 | |
| M1 45 = k × 9 | ✓ 1 | |
| A1 k = 5 so Q = 5d2 | ✓ 1 | |
| Part (b) | ||
| M1 5 × 4.22 | ✓ 1 | |
| A1 88.2 litres per minute | ✓ 1 | |
| Part (c) | ||
| M1 180 = 5d2 leading to d2 = 36 | ✓ 1 | |
| A1 d = 6 cm | ✓ 1 | |
Common error: Squaring is part of the relationship: doubling the diameter multiplies the flow by 4, not 2.
During an eruption, the height h metres above the vent of a lava bomb is modelled by h = 30t − 5t2, where t seconds is the time after ejection and t ≥ 0.
Find the two times at which the lava bomb is 40 m above the vent.
Find the time at which the lava bomb lands at the height of the vent.
Work out the greatest height reached above the vent.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 5t2 − 30t + 40 = 0 or t2 − 6t + 8 = 0 | ✓ 1 | |
| M1 (t − 2)(t − 4) = 0 or correct use of the formula | ✓ 1 | |
| A1 t = 2 s | ✓ 1 | |
| A1 t = 4 s | ✓ 1 | |
| Part (b) | ||
| M1 30t − 5t2 = 0 or 5t(6 − t) = 0 | ✓ 1 | |
| A1 6 s | ✓ 1 | |
| Part (c) | ||
| M1 greatest height at t = 3 (midway between 0 and 6) | ✓ 1 | |
| A1 45 m | ✓ 1 | |
Exam tip: A quadratic with a negative t² term is symmetrical, so its maximum lies halfway between its two zeros.
The rectangular floor of a cargo hold has width x metres and length (x + 4) metres. The area of the floor is 50 m2.
Show that x2 + 4x − 50 = 0.
Solve x2 + 4x − 50 = 0 to find the width of the floor.
Give your answer in metres correct to 3 significant figures. You must show your working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 4) = 50 | ✓ 1 | |
| A1 x2 + 4x = 50, so x2 + 4x − 50 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the quadratic formula, with a = 1, b = 4, c = −50 | ✓ 1 | |
| A1 \(\frac{-4\pm\sqrt{216}}{2}\) | ✓ 1 | |
| A1 5.35 m (5.3485...) | ✓ 1 | |
| B1 negative root −9.35 rejected because a width must be positive | ✓ 1 | |
Common error: After using the quadratic formula, check which root makes sense in the context; a length cannot be negative.
In a model of falling volcanic debris, the speed v m/s is directly proportional to the square root of the fall height h metres. When h = 36, v = 12.
Find the speed when h = 100.
Find the height when v = 15.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v = k√h | ✓ 1 | |
| A1 k = 2 | ✓ 1 | |
| A1 20 m/s | ✓ 1 | |
| Part (b) | ||
| M1 15 = 2√h or √h = 7.5 | ✓ 1 | |
| A1 56.25 m | ✓ 1 | |
Common error: The proportion is with √h, not with h; doubling h does not double v.
Answer the following questions on algebraic fractions.
Simplify \(\frac{x^2-9}{x^2+5x+6}\).
Write \(\frac{2}{x-1}+\frac{3}{x+2}\) as a single fraction in its simplest form.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 factorising numerator (x − 3)(x + 3) | ✓ 1 | |
| M1 factorising denominator (x + 2)(x + 3) | ✓ 1 | |
| A1 \(\frac{x-3}{x+2}\) | ✓ 1 | |
| Part (b) | ||
| M1 common denominator (x − 1)(x + 2) | ✓ 1 | |
| M1 numerator 2(x + 2) + 3(x − 1) | ✓ 1 | |
| A1 \(\frac{5x+1}{(x-1)(x+2)}\) | ✓ 1 | |
Common error: Only factors can be cancelled, not terms: \(\frac{x-3}{x+2}\) cannot be simplified further by cancelling the x.
A wind farm has a safety zone: the circle x2 + y2 = 25, where x and y are distances in km east and north of a control hub at the origin. A straight service road has equation y = x + 1.
Solve the equations x2 + y2 = 25 and y = x + 1 simultaneously to find the coordinates of the points where the road crosses the boundary of the zone.
Work out the length of the part of the road inside the zone, giving your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitute: x2 + (x + 1)2 = 25 | ✓ 1 | |
| M1 expand and rearrange to 2x2 + 2x − 24 = 0 | ✓ 1 | |
| M1 (x + 4)(x − 3) = 0 | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
| Part (b) | ||
| M1 √((3 − (−4))2 + (4 − (−3))2) | ✓ 1 | |
| A1 9.90 km (accept 9.9) (ECF from (a)) | ✓ 1 | |
Exam tip: Substituting the linear equation into the circle always gives a quadratic; set it equal to zero before solving.
A technician on a platform 3 m above the ground throws a bolt upwards. The height, h metres, of the bolt above the ground t seconds after it is thrown is h = 3 + 7t − 2t2 until it lands.
Show that the bolt reaches the ground when 2t2 − 7t − 3 = 0.
Use the quadratic formula to find how long the bolt takes to reach the ground. Give your answer to 2 decimal places.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 sets h = 0 and rearranges 3 + 7t − 2t2 = 0 to 2t2 − 7t − 3 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 t = (7 ± √(49 + 24)) ÷ 4 | ✓ 1 | |
| A1 t = 3.886… or −0.386… | ✓ 1 | |
| A1 3.89 s, with the negative root rejected | ✓ 1 | |
Common error: In the quadratic formula, −b is +7 here and the term under the root is b² − 4ac = 49 + 24, not 49 − 24.
The daily cost, £C, of running a shop's chiller at setting x is C = x2 − 12x + 50.
Write x2 − 12x + 50 in the form (x − a)2 + b.
Hence write down the minimum daily cost and the setting at which it occurs.
Find the two settings at which the daily cost is £18.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x − 6)2 | ✓ 1 | |
| A1 (x − 6)2 + 14 | ✓ 1 | |
| Part (b) | ||
| B1 minimum cost £14 | ✓ 1 | |
| B1 at setting x = 6 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 (x − 6)2 + 14 = 18 so (x − 6)2 = 4 | ✓ 1 | |
| M1 x − 6 = ±2 | ✓ 1 | |
| A1 x = 4 and x = 8 | ✓ 1 | |
Exam tip: The vertex form (x − a)² + b gives the minimum value b at x = a immediately.
A volcanic bomb is thrown upwards from a crater rim. Its height h metres above the ground, t seconds after launch, is modelled by h = 5 + 24t − 5t2.
Work out the height of the bomb when t = 2.
Work out the time at which the bomb hits the ground.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 5 + 24 × 2 − 5 × 22 | ✓ 1 | |
| 33 m | ✓ 1 | |
| Part (b) | ||
| M1 for setting h = 0, i.e. 5t2 − 24t − 5 = 0 | ✓ 1 | |
| M1 for (5t + 1)(t − 5) = 0 or a correct substitution into the quadratic formula | ✓ 1 | |
| t = 5 or t = −0.2 | ✓ 1 | |
| t = 5 s (negative time rejected) | ✓ 1 | |
Common error: Setting the expression equal to zero for the ground is the key step; rearrange so the t² term is positive before factorising.
The time t hours taken to unload a cargo ship is inversely proportional to the number of cranes n in use. With 4 cranes the ship is unloaded in 9 hours.
Work out the time taken with 6 cranes.
The ship must be unloaded in 2.25 hours. Work out the number of cranes needed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for t = k/n with k = 36 | ✓ 1 | |
| t = 6 hours | ✓ 1 | |
| Part (b) | ||
| M1 for 36 ÷ 2.25 | ✓ 1 | |
| 16 | ✓ 1 | |
Exam tip: For inverse proportion, n × t stays constant, which gives a quick check.
A theatre manager models the weekly profit, in £ thousands, from selling x hundred programmes as P = 14 + 5x − x2, where x ≥ 0.
Show that P = (7 − x)(x + 2).
Find the values of x for which the weekly profit is positive.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for expanding (7 − x)(x + 2) = 7x + 14 − x2 − 2x | ✓ 1 | |
| simplifying to 14 + 5x − x2 | ✓ 1 | |
| Part (b) | ||
| B1 for critical values x = 7 and x = −2 | ✓ 1 | |
| M1 for a sign check showing (7 − x)(x + 2) > 0 between the roots | ✓ 1 | |
| 0 ≤ x < 7 | ✓ 1 | |
Common error: A quadratic with a negative x² term is positive between its roots, not outside them.
In a simulation of a star cluster, the number of dust grains after n steps is N = 5 × 2n.
Work out N after 6 steps.
Find the value of n for which N = 2560.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 5 × 26 | ✓ 1 | |
| 320 | ✓ 1 | |
| Part (b) | ||
| M1 for 2n = 2560 ÷ 5 = 512 | ✓ 1 | |
| M1 for 512 = 29 | ✓ 1 | |
| n = 9 | ✓ 1 | |
Common error: Divide by the 5 first; 5 × 2ⁿ is not 10ⁿ.
The brightness B (in arbitrary units) of a star seen from a distance d light-years is inversely proportional to d2. When d = 3, B = 16.
Show that B = 144/d2.
Work out B when d = 6.
Work out d when B = 2.25.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for B = k/d2 | ✓ 1 | |
| k = 16 × 9 = 144, giving B = 144/d2 | ✓ 1 | |
| Part (b) | ||
| M1 for 144 ÷ 36 | ✓ 1 | |
| B = 4 | ✓ 1 | |
| Part (c) | ||
| M1 for d2 = 144 ÷ 2.25 = 64 | ✓ 1 | |
| d = 8 | ✓ 1 | |
Common error: Remember to take the square root at the end when the proportion involves d².
A rectangular railway platform has length (x + 4) m and width x m. Its area is 700 m2.
Show that x2 + 4x − 700 = 0.
Solve the equation to find the width of the platform. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for x(x + 4) = 700 | ✓ 1 | |
| expanding and rearranging to x2 + 4x − 700 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 for substituting into the quadratic formula correctly | ✓ 1 | |
| x = (−4 ± √2816)/2 | ✓ 1 | |
| x = 24.5 m (negative root rejected) | ✓ 1 | |
Common error: In the quadratic formula, −4ac becomes +2800 here because c is negative.
A mountain base camp is a rectangle. The width is x metres and the length is (x + 6) metres. The area of the camp is 50 m2.
Show that x2 + 6x − 50 = 0.
Solve x2 + 6x − 50 = 0. Give your solutions correct to 3 significant figures.
Work out the perimeter of the camp, in metres, to 3 significant figures. Use the unrounded positive solution of the equation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 6) = 50 | ✓ 1 | |
| A1 expanding and rearranging to x2 + 6x − 50 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the quadratic formula (allow one error) | ✓ 1 | |
| A1 x = −10.7 | ✓ 1 | |
| A1 x = 4.68 | ✓ 1 | |
| Part (c) | ||
| M1 2(x + x + 6) with the positive value of x | ✓ 1 | |
| A1 30.7 (m) | ✓ 1 | |
Exam tip: A length cannot be negative, so reject the negative root and say why.
A cheese maker's profit, in pounds, on a day when x kilograms of cheese are made and sold is modelled by P = 40x − x2 − 300.
Factorise x2 − 40x + 300.
Hence solve P > 0. Give your answer as an inequality.
The cheese maker only sells whole numbers of kilograms. How many different values of x give a positive profit?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x − a)(x − b) with ab = 300 or a + b = 40 | ✓ 1 | |
| A1 (x − 10)(x − 30) | ✓ 1 | |
| Part (b) | ||
| M1 critical values 10 and 30 identified | ✓ 1 | |
| M1 correct method to choose the region (for example a sketch of an upward-opening parabola for x2 − 40x + 300 and the region below the axis) | ✓ 1 | |
| A1 10 < x < 30 | ✓ 1 | |
| Part (c) | ||
| B1 19 (x = 11 to 29) | ✓ 1 | |
Common error: P is the negative of the factorised quadratic, so the profit is positive between the roots, not outside them.
The signal strength S, in microwatts per square metre, received from a mobile mast is inversely proportional to the square of the distance d metres from the mast. At a distance of 200 m the signal strength is 0.72.
Find a formula for S in terms of d.
Work out the signal strength at a distance of 300 m.
A phone needs a signal strength of at least 0.05 to work. Work out the greatest distance from the mast at which it works. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(S = \frac{k}{d^2}\) with d = 200 and S = 0.72 substituted | ✓ 1 | |
| A1 \(S = \frac{28800}{d^2}\) | ✓ 1 | |
| Part (b) | ||
| B1 0.32 | ✓ 1 | |
| Part (c) | ||
| M1 0.05 = 28800 ÷ d2 or d2 = 576000 | ✓ 1 | |
| A1 759 (m) | ✓ 1 | |
Common error: Inverse proportion to the square means S = k/d², not S = k/d.
The time of swing, T seconds, of a theatre lamp hung on a cable of length L metres is given by \(T = 2\pi\sqrt{\frac{L}{g}}\), where g = 9.8 m/s2.
Make L the subject of the formula.
Work out L when T = 3.0. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 divide by \(2\pi\) to get \(\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\) | ✓ 1 | |
| M1 square both sides, \(\frac{T^2}{4\pi^2} = \frac{L}{g}\) | ✓ 1 | |
| A1 \(L = \frac{gT^2}{4\pi^2}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{9.8 \times 3^2}{4\pi^2}\) | ✓ 1 | |
| A1 2.23 m (calculator value 2.234...) | ✓ 1 | |
Common error: When rearranging, square both sides of the equation, not just the term under the root sign.
A prop is thrown upwards on a theatre stage. Its height above the stage floor, h metres, at time t seconds after release is modelled by \(h = 1.5 + 8t - 5t^2\).
Show that the prop lands on the stage floor when \(10t^2 - 16t - 3 = 0\).
Use the quadratic formula to solve \(10t^2 - 16t - 3 = 0\). Give your solution that fits the context correct to 3 significant figures.
Explain why the other solution of the equation is not used.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 set \(h = 0\), so \(1.5 + 8t - 5t^2 = 0\) | ✓ 1 | |
| A1 multiply by \(-2\) to get \(10t^2 - 16t - 3 = 0\) | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the formula, \(t = \frac{16 \pm \sqrt{(-16)^2 - 4(10)(-3)}}{2 \times 10}\) | ✓ 1 | |
| M1 discriminant \(376\) so \(t = \frac{16 \pm 19.39...}{20}\) | ✓ 1 | |
| A1 \(t = 1.77\) (other solution \(-0.17\) rejected) | ✓ 1 | |
| Part (c) | ||
| B1 it is negative, and time after release cannot be negative | ✓ 1 | |
Common error: Keep the full calculator value inside the formula and write brackets round negative numbers such as \((-16)^2\).
In a ship's engineering log, the load ratio of an engine is given by the expression \(\frac{x^2 - 9}{x^2 + 5x + 6}\).
Simplify \(\frac{x^2 - 9}{x^2 + 5x + 6}\) fully.
Hence solve \(\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{2}{3}\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 numerator \((x - 3)(x + 3)\) | ✓ 1 | |
| B1 denominator \((x + 2)(x + 3)\) | ✓ 1 | |
| A1 \(\frac{x - 3}{x + 2}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{x - 3}{x + 2} = \frac{2}{3}\) | ✓ 1 | |
| M1 \(3(x - 3) = 2(x + 2)\), so \(3x - 9 = 2x + 4\) | ✓ 1 | |
| A1 \(x = 13\) | ✓ 1 | |
Common error: Cancel only common factors of the whole numerator and denominator, never separate terms.
A rectangular stage has length (x + 8) metres and width (x − 3) metres, where x > 3.
Show that the area of the stage, in m2, is x2 + 5x − 24
The area of the stage is 60 m2. Find the value of x.
Work out the perimeter of the stage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 expands (x + 8)(x − 3), e.g. x2 − 3x + 8x − 24 | ✓ 1 | |
| A1 collects terms to reach x2 + 5x − 24 with no errors | ✓ 1 | |
| Part (b) | ||
| M1 x2 + 5x − 24 = 60, rearranged to x2 + 5x − 84 = 0 | ✓ 1 | |
| M1 factorises: (x + 12)(x − 7) = 0 (or correct use of formula) | ✓ 1 | |
| A1 x = 7 (and x = −12 rejected because x > 3) | ✓ 1 | |
| Part (c) | ||
| M1 length 15 and width 4, with 2(15 + 4) | ✓ 1 | |
| A1 38 (m) (ECF from (b)) | ✓ 1 | |
Exam tip: When a quadratic gives two solutions, check each against the conditions in the question and reject any that are impossible.
A flight-control program accepts a setting x only when x satisfies both 5 − 2x < 13 and x2 − x − 12 ≤ 0.
Solve 5 − 2x < 13
Solve x2 − x − 12 ≤ 0
Work out how many integer values of x satisfy both inequalities.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 −2x < 8 | ✓ 1 | |
| A1 x > −4 (the inequality sign is reversed on dividing by −2) | ✓ 1 | |
| Part (b) | ||
| M1 factorises: (x − 4)(x + 3) | ✓ 1 | |
| M1 critical values −3 and 4, and identifies that the region between them is required (e.g. by sketch) | ✓ 1 | |
| A1 −3 ≤ x ≤ 4 | ✓ 1 | |
| Part (c) | ||
| B1 8 (the integers −3, −2, −1, 0, 1, 2, 3, 4) (ECF from (a) and (b)) | ✓ 1 | |
Exam tip: Sketching a quadratic curve shows that ≤ 0 means the part of the curve on or below the x-axis, that is, between the roots.
The illumination I lux from a runway light is inversely proportional to the square of the distance d metres from the light. When d = 30, I = 20.
Show that I = 18 000/d2
Work out I when d = 20.
Find the distance at which the illumination is 5 lux.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = k/d2 and 20 = k/302 | ✓ 1 | |
| A1 k = 18 000 shown, so I = 18 000/d2 | ✓ 1 | |
| Part (b) | ||
| M1 18 000/202 | ✓ 1 | |
| A1 45 (lux) | ✓ 1 | |
| Part (c) | ||
| M1 5 = 18 000/d2 | ✓ 1 | |
| M1 d2 = 3600 | ✓ 1 | |
| A1 60 (m) | ✓ 1 | |
Exam tip: 'Inversely proportional to the square' means y = k/x². Find k first, then rearrange to make the unknown the subject.
The frequency f hertz of a note from a guitar string is directly proportional to the square root of the tension T newtons in the string. When T = 81, f = 270.
Show that f = 30√T
Work out the frequency when T = 144.
The string is tuned to 420 Hz. Work out the tension in the string.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 f = k√T and 270 = k × 9 | ✓ 1 | |
| A1 k = 30, giving f = 30√T | ✓ 1 | |
| Part (b) | ||
| M1 30 × √144 (= 30 × 12) | ✓ 1 | |
| A1 360 (Hz) | ✓ 1 | |
| Part (c) | ||
| M1 420 = 30√T | ✓ 1 | |
| M1 √T = 14 | ✓ 1 | |
| A1 196 (N) | ✓ 1 | |
Common error: To undo a square root, square both sides after isolating it: from √T = 14 get T = 14² = 196, not 14 ÷ 2.
The rectangular apron at the front of a theatre stage has width \(x\) metres and length \((2x+5)\) metres. Its area is 9 m2.
Show that \(2x^2 + 5x - 9 = 0\).
Use the quadratic formula to solve \(2x^2 + 5x - 9 = 0\). Give your solutions correct to 3 significant figures.
Work out the perimeter of the apron. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(x(2x+5)=9\) | ✓ 1 | |
| A1 for expanding and rearranging to \(2x^2+5x-9=0\) with no errors | ✓ 1 | |
| Part (b) | ||
| M1 for substituting \(a=2\), \(b=5\), \(c=-9\) into the formula | ✓ 1 | |
| M1 for evaluating the discriminant \(25+72=97\) | ✓ 1 | |
| A1 for \(x=1.21\) | ✓ 1 | |
| A1 for \(x=-3.71\) | ✓ 1 | |
| Part (c) | ||
| M1 for using the positive solution, length \(=2x+5\approx7.42\), and \(2(x+\text{length})\) | ✓ 1 | |
| A1 for 17.3 (m) | ✓ 1 | |
Exam tip: Write the formula and each substitution on separate lines, and put negative numbers in brackets such as \(-(5)\) and \(-4(2)(-9)\).
In this question you will use completing the square.
Write \(x^2 + 8x + 3\) in the form \((x+p)^2 + q\).
Hence solve \(x^2 + 8x + 3 = 0\). Give your solutions in the form \(a \pm \sqrt{b}\).
Write down the minimum value of \(x^2 + 8x + 3\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for \(p=4\) | ✓ 1 | |
| B1 for \(q=-13\) | ✓ 1 | |
| Part (b) | ||
| M1 for \((x+4)^2=13\) | ✓ 1 | |
| M1 for taking the square root of both sides, with \(\pm\) | ✓ 1 | |
| A1 for \(x=-4\pm\sqrt{13}\) | ✓ 1 | |
| Part (c) | ||
| B1 for \(-13\) | ✓ 1 | |
Exam tip: Expand \((x+p)^2+q\) to check that it matches the original expression.
On a ship's radar screen, one unit is 1 km and the lighthouse is at the origin. The ship sails along the straight line \(y = x + 2\). A restricted zone around the lighthouse has boundary \(x^2 + y^2 = 20\).
Find the coordinates of the two points where the ship's path crosses the boundary of the restricted zone.
Calculate the distance, in kilometres, between the two points. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for substituting \(y=x+2\) into \(x^2+y^2=20\) | ✓ 1 | |
| M1 for expanding \((x+2)^2=x^2+4x+4\) | ✓ 1 | |
| A1 for \(x^2+2x-8=0\) (or \(2x^2+4x-16=0\)) | ✓ 1 | |
| M1 for solving, e.g. \((x+4)(x-2)=0\) | ✓ 1 | |
| A1 for \((2, 4)\) and \((-4, -2)\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(\sqrt{(2-(-4))^2+(4-(-2))^2}\) | ✓ 1 | |
| A1 for 8.49 (km) | ✓ 1 | |
Common error: After finding the \(x\) values, substitute into the linear equation, not the quadratic one, to find each \(y\) value.
A rectangular cycle-training area has width x metres and length (x + 5) metres. Its area is 84 m2.
Show that \(x^2 + 5x - 84 = 0\).
Solve \(x^2 + 5x - 84 = 0\) and state the width of the area.
Work out the perimeter of the area.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(x(x + 5) = 84\) | ✓ 1 | |
| A1 for expanding and rearranging to reach \(x^2 + 5x - 84 = 0\) with no errors | ✓ 1 | |
| Part (b) | ||
| M1 for \((x + 12)(x - 7)\) or a correct use of the quadratic formula | ✓ 1 | |
| A1 for width 7 m, with the root \(x = -12\) rejected because a length cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 for \(2(7 + 12)\) (ECF from (b)) | ✓ 1 | |
| A1 for 38 m | ✓ 1 | |
Common error: A quadratic gives two roots; always check each one against the context before giving a final answer.
The air resistance force, F newtons, on a cyclist is directly proportional to the square of her speed, v m/s.
When v = 6.0, F = 18.
Show that \(F = 0.5v^2\).
Work out the force when her speed is 10 m/s.
Work out her speed when the force is 32 N.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(F = kv^2\) and substituting: \(18 = k \times 36\) | ✓ 1 | |
| A1 for \(k = 0.5\), so \(F = 0.5v^2\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(0.5 \times 10^2\) | ✓ 1 | |
| A1 for 50 N | ✓ 1 | |
| Part (c) | ||
| M1 for \(32 = 0.5v^2\) leading to \(v^2 = 64\) | ✓ 1 | |
| A1 for 8 m/s | ✓ 1 | |
Common error: Proportional to the square means the constant multiplies v²; do not square the constant or double the speed.
The time period, T seconds, of a pendulum of length L metres is given by the formula \(T = 2\pi\sqrt{\frac{L}{g}}\), where g m/s2 is the acceleration due to gravity.
Rearrange the formula to make L the subject.
A museum pendulum has a time period of 4.0 s. Use g = 9.8 to work out its length. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for dividing by \(2\pi\): \(\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\) | ✓ 1 | |
| M1 for squaring both sides | ✓ 1 | |
| A1 for \(L = \frac{gT^2}{4\pi^2}\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(\frac{9.8 \times 4.0^2}{4\pi^2}\) (ECF from (a)) | ✓ 1 | |
| A1 for 3.97 m | ✓ 1 | |
Exam tip: Undo operations in reverse order: deal with the multiplier, then the square root, then the division.
A data centre's cooling cost, £C per hour, at fan setting x (where \(0 \le x \le 15\)) is given by \(C = x^2 - 12x + 50\).
Write \(x^2 - 12x + 50\) in the form \((x - a)^2 + b\).
Use your answer to state the minimum cooling cost and the fan setting at which it occurs.
Find the fan setting at which the cost is £63 per hour.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \((x - 6)^2\) | ✓ 1 | |
| A1 for \((x - 6)^2 + 14\) | ✓ 1 | |
| Part (b) | ||
| B1 for minimum cost £14 per hour (ECF from (a)) | ✓ 1 | |
| B1 for fan setting 6 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 for \((x - 6)^2 + 14 = 63\) leading to \((x - 6)^2 = 49\) | ✓ 1 | |
| A1 for \(x = 13\), with \(x = -1\) rejected as outside the allowed range | ✓ 1 | |
Exam tip: The completed-square form shows the turning point directly: the minimum is b, reached when x = a.
The plan of a rectangular reservoir shows its length as (x + 5) metres and its width as (x − 2) metres. The area of the reservoir is 120 m2.
Show that x2 + 3x − 130 = 0.
Solve x2 + 3x − 130 = 0.
Work out the perimeter of the reservoir.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x + 5)(x − 2) = 120 | ✓ 1 | |
| A1 x2 + 3x − 10 = 120 leading to x2 + 3x − 130 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (x + 13)(x − 10) (or correct use of formula) | ✓ 1 | |
| A1 x = −13 | ✓ 1 | |
| A1 x = 10 | ✓ 1 | |
| Part (c) | ||
| M1 chooses x = 10 and finds length 15 and width 8 (rejects x = −13) | ✓ 1 | |
| A1 46 m | ✓ 1 | |
Exam tip: A negative root can give a negative length. State why it is rejected before using the other root.
A rectangular conservation pool has width x metres and length (x + 6) metres. Its area is 50 m2.
Show that x2 + 6x − 50 = 0.
Solve x2 + 6x − 50 = 0. Give your solutions correct to 3 significant figures.
Explain why only one of the solutions in (b) can be the width of the pool.
Work out the perimeter of the pool. Use the unrounded value of x and give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 6) = 50 | ✓ 1 | |
| A1 x2 + 6x = 50 leading to x2 + 6x − 50 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the quadratic formula, (−6 ± √(36 + 200))/2 | ✓ 1 | |
| A1 x = 4.68 | ✓ 1 | |
| A1 x = −10.7 | ✓ 1 | |
| Part (c) | ||
| B1 a width cannot be negative, so x = −10.7 is rejected | ✓ 1 | |
| Part (d) | ||
| M1 2(x + x + 6) = 4x + 12 with x = (−6 + √236)/2 = 4.6811... | ✓ 1 | |
| A1 30.7 m | ✓ 1 | |
Common error: Quadratic formula errors usually come from the sign of −b and from the discriminant: here 6² − 4(1)(−50) = 36 + 200, not 36 − 200.
A coastal sensor at the origin detects any boat on the circle x2 + y2 = 25 (distances in km). A boat sails along the straight line y = x − 1.
Find the coordinates of the two points where the boat's path crosses the circle.
Work out the distance between the two points. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes y = x − 1, x2 + (x − 1)2 = 25 | ✓ 1 | |
| A1 2x2 − 2x − 24 = 0 or x2 − x − 12 = 0 | ✓ 1 | |
| M1 (x − 4)(x + 3) = 0 | ✓ 1 | |
| A1 x = 4 and x = −3 | ✓ 1 | |
| A1 (4, 3) and (−3, −4) | ✓ 1 | |
| Part (b) | ||
| M1 √(72 + 72) (ECF from (a)) | ✓ 1 | |
| A1 9.90 km | ✓ 1 | |
Exam tip: Find both x values, then use the linear equation to find each matching y value, so the coordinates stay paired correctly.
The height of a cable above the level of a walkway, in metres, is modelled by f(x) = x2 − 3x − 10, where x ≥ 0 is the horizontal distance in metres from the left tower. A negative value means the cable is below the walkway.
Factorise x2 − 3x − 10.
Solve the inequality x2 − 3x − 10 < 0.
State the values of x for which the cable is below the walkway.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x + a)(x + b) with ab = −10 or a + b = −3 | ✓ 1 | |
| A1 (x − 5)(x + 2) | ✓ 1 | |
| Part (b) | ||
| B1 critical values x = −2 and x = 5 | ✓ 1 | |
| M1 sketch of an upward parabola, or a sign test, showing the expression is negative between the roots | ✓ 1 | |
| A1 −2 < x < 5 | ✓ 1 | |
| Part (c) | ||
| A1 0 ≤ x < 5 | ✓ 1 | |
Exam tip: For a quadratic inequality with a positive x² term, sketch the curve: the expression is negative between the roots and positive outside them.
A probe follows the straight path y = x + 1 past a planet at the origin O. A danger zone around the planet has boundary the circle x2 + y2 = 25. Distances are in thousands of km.
Show that the x-coordinates of the points where the path meets the boundary satisfy x2 + x − 12 = 0.
Solve the equation in (a) and find the coordinates of both points.
Work out the distance between the two points. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes y = x + 1 into the circle equation | ✓ 1 | |
| M1 expands (x + 1)2 = x2 + 2x + 1 | ✓ 1 | |
| A1 2x2 + 2x − 24 = 0, then divides by 2 to reach the given equation | ✓ 1 | |
| Part (b) | ||
| M1 (x + 4)(x − 3) = 0 or use of the quadratic formula | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| M1 substitutes into y = x + 1 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
| Part (c) | ||
| M1 √(72 + 72) = √98 | ✓ 1 | |
| A1 9.90 thousand km (9.8995…) | ✓ 1 | |
Exam tip: After finding x values, always substitute into the linear equation (not the quadratic) to find y, so that you do not create false pairs.
The time d days for n workers to repaint a bridge is inversely proportional to n. Twelve workers take 15 days.
Express d in terms of n.
How many days would 9 workers take?
How many workers are needed to repaint the bridge in 6 days?
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 d = k/n and 15 = k/12 (so k = 180) | ✓ 1 | |
| A1 d = 180/n | ✓ 1 | |
| Part (b) | ||
| A1 180/9 = 20 days | ✓ 1 | |
| Part (c) | ||
| M1 6 = 180/n | ✓ 1 | |
| A1 30 workers | ✓ 1 | |
Common error: Inverse proportion means d × n is constant. Doing 12 → 9 workers does not take 15 − 3 days.
The brightness B of a star, as seen from a planet, is inversely proportional to the square of the distance d light-years from the planet. When d = 4, B = 50 (in suitable units).
Find a formula for B in terms of d.
Find B when d = 10.
Find d when B = 2.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 B = k/d2 and 50 = k/16 | ✓ 1 | |
| A1 B = 800/d2 | ✓ 1 | |
| Part (b) | ||
| A1 800/100 = 8 | ✓ 1 | |
| Part (c) | ||
| M1 2 = 800/d2, so d2 = 400 | ✓ 1 | |
| A1 d = 20 (light-years) | ✓ 1 | |
Common error: In inverse square proportion, doubling the distance divides the value by 4, not by 2.
A comet's path is modelled by the curve y = x2 − 4x + 3. A tracking line is y = x − 1. Distances are in millions of km.
Show that the x-coordinates of the points where the curve and line meet satisfy x2 − 5x + 4 = 0.
Solve x2 − 5x + 4 = 0.
Find the y-coordinate of the meeting point with the greater x-coordinate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x2 − 4x + 3 = x − 1 | ✓ 1 | |
| A1 rearranges to the given equation | ✓ 1 | |
| Part (b) | ||
| M1 (x − 1)(x − 4) = 0 | ✓ 1 | |
| A1 x = 1 or x = 4 | ✓ 1 | |
| Part (c) | ||
| A1 y = 4 − 1 = 3 | ✓ 1 | |
Exam tip: Substitute into the linear equation to find y; this is quicker and cannot give values that lie on only one of the graphs.
Body mass index is given by B = m/h2, where m is a person's mass in kg and h is their height in m.
Work out B for m = 72 and h = 1.80, correct to 1 decimal place.
Make h the subject of the formula.
A person of mass 80 kg has B = 24. Work out their height, correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 72 ÷ 1.802 = 72 ÷ 3.24 | ✓ 1 | |
| A1 22.2 | ✓ 1 | |
| Part (b) | ||
| M1 h2 = m/B | ✓ 1 | |
| A1 h = √(m/B) | ✓ 1 | |
| Part (c) | ||
| M1 √(80/24) = √3.333… | ✓ 1 | |
| A1 1.83 m | ✓ 1 | |
Common error: Divide by B and multiply by h² first, and take the square root of the whole fraction last; do not square-root only one term.
The number x of telescope bookings in a week satisfies \(x^2 - 5x - 14 < 0\), where x is a positive integer.
Solve the inequality \(x^2 - 5x - 14 < 0\).
List all the possible values of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for factorising: (x − 7)(x + 2) | ✓ 1 | |
| A1 for critical values x = 7 and x = −2 | ✓ 1 | |
| A1 for −2 < x < 7 | ✓ 1 | |
| Part (b) | ||
| B1 for greatest value 6 or smallest value 1 | ✓ 1 | |
| B1 for 1, 2, 3, 4, 5, 6 | ✓ 1 | |
Common error: For a quadratic with a positive x² term and '< 0', the solution lies between the roots. Sketching the U-shaped curve avoids giving the outside region.
A rectangular solar panel on a satellite has length \((3x + 1)\) m and width \((x - 3)\) m, where x > 3. The area of the panel is 12 m2.
Show that \(3x^2 - 8x - 15 = 0\).
Use the quadratic formula to solve \(3x^2 - 8x - 15 = 0\). Give the value of x that fits the situation to 3 significant figures.
Find the width of the panel to 3 significant figures. Use the unrounded value of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for expanding (3x + 1)(x − 3) = 3x² − 9x + x − 3 | ✓ 1 | |
| M1 for setting the expression equal to 12 | ✓ 1 | |
| A1 for rearranging to 3x² − 8x − 15 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 for substituting into the formula with a = 3, b = −8, c = −15 | ✓ 1 | |
| A1 for x = (8 ± √244)/6 | ✓ 1 | |
| A1 for x = 3.94, with the negative root (−1.27) rejected because the width x − 3 would be negative | ✓ 1 | |
| Part (c) | ||
| B1 for 0.937 m | ✓ 1 | |
Exam tip: The quadratic formula needs the equation in the form ax² + bx + c = 0. Reject any root that does not fit the situation.
A student simplifies algebraic fractions.
Factorise \(2x^2 + 5x - 12\).
Simplify \(\frac{2x^2 + 5x - 12}{x^2 - 16}\).
Solve \(\frac{2x - 3}{x - 4} = 3\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (2x ± a)(x ± b) with ab = 12 | ✓ 1 | |
| A1 for (2x − 3)(x + 4) | ✓ 1 | |
| Part (b) | ||
| M1 for factorising the denominator: (x − 4)(x + 4) | ✓ 1 | |
| M1 for cancelling the common factor (x + 4) | ✓ 1 | |
| A1 for (2x − 3)/(x − 4) | ✓ 1 | |
| Part (c) | ||
| M1 for 2x − 3 = 3(x − 4) | ✓ 1 | |
| A1 for x = 9 | ✓ 1 | |
Common error: Only factors can be cancelled in an algebraic fraction, never separate terms. Factorise the top and bottom fully first.
A rectangular swimming pool has width \(w\) metres and length \((w + 14)\) metres. The area of the pool surface is 351 m2.
Show that \(w^2 + 14w - 351 = 0\).
Solve \(w^2 + 14w - 351 = 0\) by factorising and state the width of the pool.
Work out the perimeter of the pool. ECF from (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(w(w + 14) = 351\) and expansion to \(w^2 + 14w = 351\), leading to the given result | ✓ 1 | |
| Part (b) | ||
| M1 \((w + 27)(w - 13)\) | ✓ 1 | |
| A1 \(w = 13\) and \(w = -27\) | ✓ 1 | |
| B1 width is 13 m, rejecting \(-27\) as a width cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 length \(= 13 + 14 = 27\) m | ✓ 1 | |
| A1 80 m | ✓ 1 | |
Common error: A quadratic often gives two solutions; always check which one makes sense in the context.
A library designer models a rectangular display wall. Its width is \(x\) metres and its height is \((3x - 8)\) metres. Its area is 2 m2.
Show that \(3x^2 - 8x - 2 = 0\).
Solve \(3x^2 - 8x - 2 = 0\). Give your answers correct to 3 significant figures.
Write down the width of the display wall in metres, correct to 3 significant figures. Use your answers to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(x(3x - 8) = 2\) leading to \(3x^2 - 8x = 2\) and hence the given result | ✓ 1 | |
| Part (b) | ||
| M1 substitution into the quadratic formula with \(a = 3\), \(b = -8\), \(c = -2\) (or equivalent completing the square) | ✓ 1 | |
| A1 \(x = 2.90\) | ✓ 1 | |
| A1 \(x = -0.230\) | ✓ 1 | |
| Part (c) | ||
| A1 2.90 m (the other solution is negative, and a width cannot be negative) | ✓ 1 | |
Exam tip: With the quadratic formula, put brackets round negative values when you substitute.
The brightness \(B\) of a library reading lamp is inversely proportional to the square of the distance \(d\) metres from the lamp. When \(d = 1.5\), \(B = 80\).
Find a formula for \(B\) in terms of \(d\).
Work out \(B\) when \(d = 2.5\).
Work out the value of \(d\) when \(B = 45\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(B = \frac{k}{d^2}\) | ✓ 1 | |
| A1 \(B = \frac{180}{d^2}\) | ✓ 1 | |
| Part (b) | ||
| A1 28.8 | ✓ 1 | |
| Part (c) | ||
| M1 \(45 = \frac{180}{d^2}\) or \(d^2 = 4\) | ✓ 1 | |
| A1 \(d = 2\) | ✓ 1 | |
Common error: The word 'square' means \(d\) must be squared in the formula; do not just use \(B = k/d\).
A space probe has a rectangular solar array. Its width is \((2x - 1)\) metres and its length is \((x + 3)\) metres. The area of the array is 15 m2.
Show that \(2x^2 + 5x - 18 = 0\).
Solve \(2x^2 + 5x - 18 = 0\) by factorising and state the value of \(x\) for the array.
Work out the perimeter of the array. ECF from (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \((2x - 1)(x + 3) = 15\) with expansion to \(2x^2 + 5x - 3 = 15\), leading to the given result | ✓ 1 | |
| Part (b) | ||
| M1 \((2x + 9)(x - 2) = 0\) | ✓ 1 | |
| A1 \(x = 2\) and \(x = -4.5\) | ✓ 1 | |
| B1 \(x = 2\), since \(x = -4.5\) would give a negative width | ✓ 1 | |
| Part (c) | ||
| A1 16 m (width 3 m, length 5 m) | ✓ 1 | |
Common error: Always reject solutions that give negative lengths, and state clearly why.
This question is about the quadratic expression \(x^2 + 6x - 5\), which models the height of a bridge arch support.
Write \(x^2 + 6x - 5\) in the form \((x + a)^2 + b\).
Hence solve \(x^2 + 6x - 5 = 0\). Give your answers correct to 2 decimal places.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \((x + 3)^2\) seen | ✓ 1 | |
| A1 \((x + 3)^2 - 14\) | ✓ 1 | |
| Part (b) | ||
| M1 \((x + 3)^2 = 14\) | ✓ 1 | |
| A1 \(x = -3 + \sqrt{14}\) or \(x = -3 - \sqrt{14}\) (accept \(-3 \pm 3.7417\)) | ✓ 1 | |
| A1 \(x = 0.74\) and \(x = -6.74\) | ✓ 1 | |
Exam tip: Half the coefficient of \(x\) gives the number inside the bracket when you complete the square.
The equations of a parabola and a straight line, which describe two cables on a bridge model, are \(y = x^2 - 3x\) and \(y = 2x - 6\).
Solve these simultaneous equations.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(x^2 - 3x = 2x - 6\) | ✓ 1 | |
| A1 \(x^2 - 5x + 6 = 0\) | ✓ 1 | |
| M1 factorise \((x - 2)(x - 3) = 0\) | ✓ 1 | |
| A1 \(x = 2\) and \(x = 3\) | ✓ 1 | |
| A1 \((2, -2)\) and \((3, 0)\) | ✓ 1 | |
Common error: Give the answers as pairs of values; each \(x\)-value needs its matching \(y\)-value.
A rectangular panel of silk has width \(x\) metres and length \((2x - 5)\) metres. The area of the panel is 6 m2.
Show that \(2x^2 - 5x - 6 = 0\).
Solve \(2x^2 - 5x - 6 = 0\) using the quadratic formula. Give your answers correct to 3 significant figures.
Work out the length of the panel. Use the unrounded value of \(x\) from part (b) and give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(x(2x-5) = 6\) | ✓ 1 | |
| A1 expands and rearranges to \(2x^2 - 5x - 6 = 0\) with no errors | ✓ 1 | |
| Part (b) | ||
| M1 \(x = \dfrac{5 \pm \sqrt{(-5)^2 - 4(2)(-6)}}{2 \times 2}\) | ✓ 1 | |
| A1 \(x = 3.39\) | ✓ 1 | |
| A1 \(x = -0.886\) | ✓ 1 | |
| Part (c) | ||
| M1 \(2 \times \dfrac{5 + \sqrt{73}}{4} - 5\) (or \(2 \times 3.386... - 5\)) | ✓ 1 | |
| A1 1.77 m | ✓ 1 | |
Exam tip: Only the positive root can be a width, so state clearly which solution you are using.
The power \(P\) watts used by a robot's motor is directly proportional to the square of its speed \(v\) m/s. When \(v = 2\), \(P = 48\).
Find a formula for \(P\) in terms of \(v\).
Work out the power used when \(v = 3.5\).
Work out the speed when the power used is 300 W.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(P = kv^2\) and \(k = 48 \div 4\) | ✓ 1 | |
| A1 \(P = 12v^2\) | ✓ 1 | |
| Part (b) | ||
| B1 147 W | ✓ 1 | |
| Part (c) | ||
| M1 \(300 = 12v^2\) or \(v^2 = 25\) | ✓ 1 | |
| A1 \(v = 5\) m/s | ✓ 1 | |
Common error: Direct proportion to the square means \(P = kv^2\), not \(P = kv\); the constant must be found using the squared value.
A robot moves along the straight line \(y = x + 2\) across a flat floor, where \(x\) and \(y\) are distances in metres from a fixed point. A circular safety boundary is described by \(x^2 + y^2 = 34\).
Find the coordinates of the two points where the robot's path crosses the boundary. You must show your working.
Work out the distance between the two points in part (a). Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes \(y = x+2\) to get \(x^2 + (x+2)^2 = 34\) | ✓ 1 | |
| A1 \(2x^2 + 4x - 30 = 0\) or \(x^2 + 2x - 15 = 0\) | ✓ 1 | |
| M1 \((x+5)(x-3) = 0\) or correct use of the quadratic formula | ✓ 1 | |
| A1 \(x = 3\) and \(x = -5\) | ✓ 1 | |
| A1 \((3, 5)\) and \((-5, -3)\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{8^2 + 8^2}\) (or \(\sqrt{128}\)) (ECF from (a)) | ✓ 1 | |
| A1 11.3 m | ✓ 1 | |
Exam tip: After finding each \(x\) value, substitute into the linear equation (the simpler one) to find \(y\).
An aircraft accelerates along a runway. Its final speed \(v\) m/s, initial speed \(u\) m/s, acceleration \(a\) m/s2 and distance \(s\) m are related by \(v^2=u^2+2as\).
Make \(a\) the subject of the formula.
Work out \(a\) when \(u=20\), \(v=60\) and \(s=800\).
The formula \(P=2(l+w)\) gives the perimeter of a rectangle. Make \(w\) the subject.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(v^2-u^2=2as\) | ✓ 1 | |
| A1 \(a=\frac{v^2-u^2}{2s}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{60^2-20^2}{2\times800}\) (ECF from (a)) | ✓ 1 | |
| A1 2 m/s2 | ✓ 1 | |
| Part (c) | ||
| M1 \(\frac{P}{2}=l+w\) | ✓ 1 | |
| A1 \(w=\frac{P}{2}-l\) | ✓ 1 | |
Common error: When a formula contains \(v^2\), subtract \(u^2\) from both sides first; the square root is not needed when \(a\) is the subject.
A model aircraft is launched from a platform. Its height \(h\) metres above the ground after \(t\) seconds is \(h=2+14t-5t^2\).
Show that \(h=0\) gives \(5t^2-14t-2=0\).
Use the quadratic formula to find the time when the aircraft lands. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 sets \(2+14t-5t^2=0\) and multiplies by \(-1\) to reach \(5t^2-14t-2=0\) | ✓ 1 | |
| Part (b) | ||
| M1 substitutes \(a=5\), \(b=-14\), \(c=-2\) into the formula correctly | ✓ 1 | |
| M1 discriminant \(196+40=236\) | ✓ 1 | |
| A1 \(t=\frac{14+\sqrt{236}}{10}=2.94\) s | ✓ 1 | |
| B1 rejects the other root \(-0.136\) as time cannot be negative | ✓ 1 | |
Exam tip: Write the formula down, then substitute with brackets around negative values, especially \(b\).
Some questions on indices.
Simplify \(\frac{12a^5b^2}{3a^2b^5}\).
Solve \(3^{2x-1}=81\).
Work out the value of \(27^{-\frac{2}{3}}\). Give your answer as a fraction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(4\) and \(a^3\) correct | ✓ 1 | |
| A1 \(4a^3b^{-3}\) or \(\frac{4a^3}{b^3}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(81=3^4\) | ✓ 1 | |
| A1 \(2x-1=4\) so \(x=2.5\) | ✓ 1 | |
| Part (c) | ||
| M1 \(27^{\frac13}=3\) | ✓ 1 | |
| A1 \(3^{-2}=\frac{1}{9}\) (0.111) | ✓ 1 | |
Common error: A negative power means a reciprocal, not a negative number: \(27^{-2/3}\) is positive.
The number of people waiting in a queue outside a clinic, q, is modelled by q = 3t2 + 4t, where t is the time in hours after the clinic opens. The queue reaches 60 people.
Show that 3t2 + 4t − 60 = 0.
Solve 3t2 + 4t − 60 = 0. Give your answers correct to 3 significant figures.
Explain why only one of your solutions gives a suitable value of t.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 sets 3t2 + 4t = 60 and rearranges to 3t2 + 4t − 60 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 substitutes a = 3, b = 4, c = −60 correctly into the quadratic formula | ✓ 1 | |
| M1 evaluates the discriminant: 16 + 720 = 736 | ✓ 1 | |
| A1 t = 3.85 | ✓ 1 | |
| A1 t = −5.19 | ✓ 1 | |
| Part (c) | ||
| B1 t is a time after opening so cannot be negative; the queue reaches 60 people when t = 3.85 hours | ✓ 1 | |
Common error: In the quadratic formula the term −b is +4 × −1 = −4 here, and −4ac becomes +720 because c is negative; keep each substitution in brackets.
The power output, P kW, of a wind turbine is directly proportional to the cube of the wind speed, v m/s. When v = 6, P = 270.
Find a formula for P in terms of v.
Work out the power output when the wind speed is 10 m/s.
Work out the wind speed needed for an output of 640 kW.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = kv3 | ✓ 1 | |
| M1 270 = k × 63 | ✓ 1 | |
| A1 P = 1.25v3 | ✓ 1 | |
| Part (b) | ||
| M1 1.25 × 103 (ECF from (a)) | ✓ 1 | |
| A1 1250 kW | ✓ 1 | |
| Part (c) | ||
| M1 640 = 1.25v3 (ECF from (a)) | ✓ 1 | |
| M1 v3 = 512 and takes the cube root | ✓ 1 | |
| A1 8 m/s | ✓ 1 | |
Common error: 'Proportional to the cube' means P = kv³; doubling the speed multiplies the power by 8, not by 2 or 3.
The line with equation y = 2x + 3 meets the curve with equation y = x2 − 2x − 2 at the points A and B.
Find the coordinates of A and B.
Calculate the length of AB. Give your answer correct to 3 significant figures. Use the exact coordinates from (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 equates: x2 − 2x − 2 = 2x + 3 | ✓ 1 | |
| A1 x2 − 4x − 5 = 0 | ✓ 1 | |
| M1 factorises (x − 5)(x + 1) or uses the quadratic formula correctly | ✓ 1 | |
| A1 x = 5 and x = −1 | ✓ 1 | |
| A1 (5, 13) and (−1, 1) | ✓ 1 | |
| Part (b) | ||
| M1 (5 − (−1))2 + (13 − 1)2 (ECF from (a)) | ✓ 1 | |
| M1 AB = √180 | ✓ 1 | |
| A1 13.4 | ✓ 1 | |
Exam tip: Once you have the x-values, substitute them into the linear equation to find y; it is simpler than using the curve.
A rectangular courtyard has width x metres and length (x + 5) metres. Its area is 84 m2.
Show that x2 + 5x − 84 = 0.
Solve x2 + 5x − 84 = 0 by factorising.
Work out the perimeter of the courtyard.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 5) = 84 | ✓ 1 | |
| A1 expand to x2 + 5x = 84 and rearrange to x2 + 5x − 84 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 attempt at (x + a)(x + b) with ab = −84 | ✓ 1 | |
| A1 (x + 12)(x − 7) | ✓ 1 | |
| A1 x = −12 or x = 7 | ✓ 1 | |
| Part (c) | ||
| B1 reject x = −12 because a length cannot be negative, so x = 7 | ✓ 1 | |
| M1 2 × (7 + 12) | ✓ 1 | |
| A1 38 m | ✓ 1 | |
Exam tip: After solving a quadratic from a real situation, always check whether each solution is possible.
A sea lion leaps from a platform that is 1 m above the water. Its height h metres above the water, t seconds after leaving the platform, is given by
h = 1 + 8t − 5t2
Work out the height of the sea lion when t = 0.5.
The sea lion lands in the water when h = 0. Use the quadratic formula to find the value of t when it lands. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 1 + 8(0.5) − 5(0.5)2 | ✓ 1 | |
| A1 3.75 m | ✓ 1 | |
| Part (b) | ||
| M1 rearrange to 5t2 − 8t − 1 = 0 | ✓ 1 | |
| M1 substitute a = 5, b = −8, c = −1 into the formula, e.g. t = (8 ± √84) ÷ 10 | ✓ 1 | |
| B1 reject the negative root t = −0.117 (negative time) | ✓ 1 | |
| A1 t = 1.72 s | ✓ 1 | |
Exam tip: Write the equation in the form ax² + bx + c = 0 before using the formula, and keep the signs of b and c with them.
An architect models the curve of an arch by y = 9 − x2 for −3 ≤ x ≤ 3, and a straight support beam by y = x + 3. Both x and y are in metres.
Find, by algebra, the coordinates of the two points where the beam meets the arch.
Work out the length of the beam between these two points. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 9 − x2 = x + 3 | ✓ 1 | |
| M1 rearrange to x2 + x − 6 = 0 | ✓ 1 | |
| A1 (x + 3)(x − 2) = 0, so x = −3 and x = 2 | ✓ 1 | |
| M1 substitute into y = x + 3 | ✓ 1 | |
| A1 (−3, 0) and (2, 5) | ✓ 1 | |
| Part (b) | ||
| M1 √(52 + 52) (ECF from (a)) | ✓ 1 | |
| A1 7.07 m | ✓ 1 | |
Exam tip: Always substitute the x-values into the simpler (linear) equation to find the y-values.
A rectangular training pitch has a width of x metres. Its length is 7 metres more than its width. The area of the pitch is 330 m2.
Show that \(x^2 + 7x - 330 = 0\).
Solve \(x^2 + 7x - 330 = 0\) by factorising, and write down the width of the pitch.
Work out the perimeter of the pitch.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(x(x + 7) = 330\) leading to \(x^2 + 7x - 330 = 0\) | ✓ 1 | |
| Part (b) | ||
| M1 \((x + a)(x + b)\) with \(ab = -330\) | ✓ 1 | |
| A1 \((x + 22)(x - 15) = 0\) | ✓ 1 | |
| A1 width 15 m, with \(x = -22\) rejected | ✓ 1 | |
| Part (c) | ||
| M1 \(2(15 + 22)\) (ECF from (b)) | ✓ 1 | |
| A1 74 m | ✓ 1 | |
Exam tip: When a quadratic comes from a length, reject any negative root and say why.
A football is kicked upwards. Its height, h metres, above the ground after t seconds is modelled by
h = 18t − 4.9t2
Show that \(h = 10\) leads to \(4.9t^2 - 18t + 10 = 0\).
Solve \(4.9t^2 - 18t + 10 = 0\). Give your answers to 3 significant figures.
Work out how long the ball is at least 10 m above the ground. Use the unrounded values from (b) and give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(10 = 18t - 4.9t^2\) rearranged to \(4.9t^2 - 18t + 10 = 0\) | ✓ 1 | |
| Part (b) | ||
| M1 correct use of the quadratic formula with \(a = 4.9\), \(b = -18\), \(c = 10\) | ✓ 1 | |
| A1 \(t = 0.682\) | ✓ 1 | |
| A1 \(t = 2.99\) | ✓ 1 | |
| Part (c) | ||
| M1 difference of the two roots, \(\sqrt{128} \div 4.9\) or \(2.9912 - 0.6823\) | ✓ 1 | |
| A1 2.31 s | ✓ 1 | |
Common error: Subtracting the rounded answers 2.99 and 0.682 gives 2.308, which is not accurate enough; keep the unrounded roots until the end.
The kinetic energy, E joules, of a sprinter of mass m kg moving at speed v m/s is given by
E = ½mv2
Make v the subject of the formula.
A sprinter of mass 65 kg has kinetic energy 1560 J. Work out the speed of the sprinter. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(2E = mv^2\) or \(v^2 = 2E/m\) | ✓ 1 | |
| A1 \(v = \sqrt{\frac{2E}{m}}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{2 \times 1560 \div 65}\) (\(= \sqrt{48}\)) | ✓ 1 | |
| A1 6.93 m/s | ✓ 1 | |
Exam tip: Do the same operation to both sides in order, and take the square root of the whole expression, not of each part separately.
A rectangular garden at a hospital has length (x + 5) m and width (x − 2) m.
The area of the garden is 60 m2.
Show that x2 + 3x − 70 = 0.
Solve x2 + 3x − 70 = 0.
Work out the perimeter of the garden.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x + 5)(x − 2) = 60 with expansion attempted, e.g. x2 − 2x + 5x − 10 = 60 | ✓ 1 | |
| A1 x2 + 3x − 70 = 0 shown with no errors | ✓ 1 | |
| Part (b) | ||
| M1 attempt to factorise or use the formula | ✓ 1 | |
| A1 (x + 10)(x − 7) = 0 | ✓ 1 | |
| A1 x = 7 and x = −10 | ✓ 1 | |
| Part (c) | ||
| B1 x = 7 chosen, since x = −10 would give a negative width, so length 12 m and width 5 m | ✓ 1 | |
| A1 34 m | ✓ 1 | |
Exam tip: When a quadratic comes from a real shape, check each root against the context before using it.
Solve 3x2 − 7x + 1 = 0.
Give your answers correct to 2 decimal places.
Solve the equation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitute a = 3, b = −7, c = 1 correctly into the quadratic formula | ✓ 1 | |
| A1 x = (7 ± √37) / 6 | ✓ 1 | |
| A1 x = 2.18 | ✓ 1 | |
| A1 x = 0.15 | ✓ 1 | |
Common error: Write −b as +7 when b = −7, and square −7 as +49.
Quadratic inequalities.
Solve x2 − 2x − 15 < 0.
Write down all the integer values of x that satisfy both x2 − 2x − 15 < 0 and 2x + 1 > 5.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 attempt to factorise, e.g. (x − 5)(x + 3) | ✓ 1 | |
| A1 critical values −3 and 5 | ✓ 1 | |
| A1 −3 < x < 5 | ✓ 1 | |
| Part (b) | ||
| M1 x > 2 found | ✓ 1 | |
| A1 3, 4 (ECF from (a)) | ✓ 1 | |
Exam tip: A “less than zero” quadratic with positive x² gives the values between the roots.
A rectangular bed bay in a hospital ward has length (x + 5) metres and width (x + 2) metres. The area of the bay is 54 m2.
Show that x2 + 7x − 44 = 0
Solve x2 + 7x − 44 = 0 and state the value of x that fits this situation.
Work out the perimeter of the bay.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x + 5)(x + 2) = 54 | ✓ 1 | |
| A1 x2 + 7x + 10 = 54 leading to x2 + 7x − 44 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (x + 11)(x − 4) or a correct use of the quadratic formula | ✓ 1 | |
| A1 x = 4 and x = −11 | ✓ 1 | |
| A1 x = 4 chosen because a length cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 2 × (their 9 + their 6) or 2 × ((4 + 5) + (4 + 2)) | ✓ 1 | |
| A1 30 m (ECF from (b)) | ✓ 1 | |
Common error: After solving a quadratic that models a real length, always reject any root that would make a length negative.
In a hospital X-ray room the radiation intensity, I units, is inversely proportional to the square of the distance, d metres, from the source. When d = 2, I = 45.
Find a formula for I in terms of d.
Work out the intensity when d = 5.
Work out the distance at which the intensity is 20 units.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = k/d2 | ✓ 1 | |
| M1 45 = k/4 | ✓ 1 | |
| A1 I = 180/d2 | ✓ 1 | |
| Part (b) | ||
| B1 7.2 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 20 = 180/d2 leading to d2 = 9 | ✓ 1 | |
| A1 3 m (ECF from (a)) | ✓ 1 | |
Common error: Inverse square means the distance is squared before dividing, so doubling the distance divides the intensity by 4, not by 2.
A beekeeper plots hive positions on a map of the apiary. Units on both axes are 10 metres, with the origin at the apiary entrance. A circular fence has equation x2 + y2 = 25. A straight path has equation y = x + 1.
Find the coordinates of the two points where the path meets the fence. You must show your working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitute y = x + 1 into x2 + y2 = 25 | ✓ 1 | |
| A1 2x2 + 2x − 24 = 0 or x2 + x − 12 = 0 | ✓ 1 | |
| M1 (x − 3)(x + 4) = 0 or a correct use of the quadratic formula | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
Exam tip: Substitute the linear equation into the quadratic one to get a single quadratic, then find both y-values using the linear equation, not the circle.
Quadratic equations in a cycling setting.
Factorise x2 + 5x − 24.
Solve 2x2 − 7x − 15 = 0.
A cycling club marks out a rectangular training area with length (x + 8) m and width (x − 3) m. The area is 60 m2. Show that x2 + 5x − 84 = 0.
Hence work out the perimeter of the training area.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x + a)(x + b) with ab = −24 or a + b = 5 | ✓ 1 | |
| A1 (x + 8)(x − 3) | ✓ 1 | |
| Part (b) | ||
| M1 attempt to factorise or use the quadratic formula | ✓ 1 | |
| A1 (2x + 3)(x − 5) | ✓ 1 | |
| A1 x = 5 and x = −1.5 | ✓ 1 | |
| Part (c) | ||
| M1 (x + 8)(x − 3) = 60 | ✓ 1 | |
| A1 x2 + 5x − 24 = 60, leading to x2 + 5x − 84 = 0 | ✓ 1 | |
| Part (d) | ||
| M1 (x + 12)(x − 7) = 0 or use of the formula | ✓ 1 | |
| A1 x = 7 (reject −12 as a length is positive) | ✓ 1 | |
| M1 2 × [(7 + 8) + (7 − 3)] | ✓ 1 | |
| A1 38 m | ✓ 1 | |
Common error: Keeping both roots when one gives a negative length; always check the answer against the context.
On an archaeological site map (units in metres) a circular exclusion zone is centred at the origin and has equation x2 + y2 = 25. A straight survey line has equation y = x + 1.
Show that x2 + x − 12 = 0 at the points where the line meets the circle.
Work out the coordinates of the two points where the line meets the circle.
Work out the distance between these two points. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes y = x + 1 into x2 + y2 = 25 | ✓ 1 | |
| M1 x2 + x2 + 2x + 1 = 25 | ✓ 1 | |
| A1 2x2 + 2x − 24 = 0, dividing by 2 to give x2 + x − 12 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (x + 4)(x − 3) = 0 or use of the formula | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
| Part (c) | ||
| M1 √[(3 − (−4))2 + (4 − (−3))2] = √98 | ✓ 1 | |
| A1 9.90 m | ✓ 1 | |
Common error: Substituting y = x + 1 and writing (x + 1)² as x² + 1.
The kinetic energy E joules of a cyclist of mass m kg travelling at v m/s is given by E = ½mv2.
Rearrange the formula to make v the subject.
Make x the subject of y = (3x + 2) ÷ (x − 1).
A cyclist of mass 80 kg has kinetic energy 1000 J. Work out the speed of the cyclist.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 v2 = 2E ÷ m | ✓ 1 | |
| A1 v = √(2E ÷ m) | ✓ 1 | |
| Part (b) | ||
| M1 y(x − 1) = 3x + 2 | ✓ 1 | |
| M1 xy − y = 3x + 2, collecting x terms: xy − 3x = y + 2 | ✓ 1 | |
| M1 factorises: x(y − 3) = y + 2 | ✓ 1 | |
| A1 x = (y + 2) ÷ (y − 3) | ✓ 1 | |
| Part (c) | ||
| M1 √(2 × 1000 ÷ 80) | ✓ 1 | |
| A1 5 m/s | ✓ 1 | |
Exam tip: When the new subject appears twice, expand, gather its terms on one side and factorise.
A rectangular courtyard has length (x + 5) m and width (x − 2) m. Its area is 60 m2.
Show that x2 + 3x − 70 = 0.
Solve x2 + 3x − 70 = 0 and hence find the value of x.
Work out the perimeter of the courtyard.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x + 5)(x − 2) = 60 expanded to x2 + 3x − 10 = 60 | ✓ 1 | |
| A1 x2 + 3x − 70 = 0 with the rearrangement shown | ✓ 1 | |
| Part (b) | ||
| M1 (x + 10)(x − 7) = 0 | ✓ 1 | |
| A1 x = −10 or x = 7 | ✓ 1 | |
| A1 x = 7, rejecting −10 because the width cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 length 12 m and width 5 m, using 2 × (length + width) | ✓ 1 | |
| A1 34 (m) | ✓ 1 | |
Common error: A quadratic can give two roots, but in a context one of them may be impossible. State clearly why you reject it.
An architect models the ratio of two beam lengths in a roof truss using algebra.
Simplify fully (3x2 − 12) ÷ (x2 + x − 6).
The ratio of the two lengths is 5 : 2, so the simplified expression in (a) equals 5/2. Find the value of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 factorises numerator as 3(x − 2)(x + 2) | ✓ 1 | |
| M1 factorises denominator as (x + 3)(x − 2) | ✓ 1 | |
| A1 3(x + 2) ÷ (x + 3) | ✓ 1 | |
| Part (b) | ||
| M1 2 × 3(x + 2) = 5(x + 3), using their expression from (a) | ✓ 1 | |
| M1 6x + 12 = 5x + 15 | ✓ 1 | |
| A1 3 | ✓ 1 | |
Exam tip: Always factorise the top and bottom of an algebraic fraction before cancelling. You can only cancel factors, never single terms.
A rectangular array of solar panels has width x metres and length (x + 6) metres. The area of the array is 55 m2.
Show that x2 + 6x − 55 = 0.
Solve x2 + 6x − 55 = 0 by factorising.
Work out the perimeter of the array.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 6) = 55 | ✓ 1 | |
| A1 x2 + 6x = 55 then subtract 55 to reach the given equation | ✓ 1 | |
| Part (b) | ||
| M1 attempt to factorise | ✓ 1 | |
| A1 (x + 11)(x − 5) = 0 | ✓ 1 | |
| A1 x = −11 and x = 5 | ✓ 1 | |
| Part (c) | ||
| B1 reject x = −11 because a width cannot be negative | ✓ 1 | |
| M1 2(5 + 11) (using x = 5) | ✓ 1 | |
| A1 32 m | ✓ 1 | |
Common error: After solving a quadratic from a real situation, check each root: a negative length must be rejected.
The power, P watts, produced by a wind turbine is given by P = ½ρAv3, where ρ kg/m3 is the density of the air, A m2 is the area swept by the blades and v m/s is the wind speed.
Work out P when ρ = 1.2, A = 50 and v = 6.
Rearrange the formula to make v the subject.
A turbine with A = 50 produces 15 360 W in air of density 1.2 kg/m3. Work out the wind speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 ½ × 1.2 × 50 × 63 | ✓ 1 | |
| A1 6480 | ✓ 1 | |
| Part (b) | ||
| M1 multiply both sides by 2 (2P = ρAv3) | ✓ 1 | |
| M1 divide both sides by ρA | ✓ 1 | |
| A1 v = ∛(2P ÷ ρA) | ✓ 1 | |
| Part (c) | ||
| M1 2 × 15360 ÷ (1.2 × 50) = 512 (ECF from (b)) | ✓ 1 | |
| A1 8 m/s | ✓ 1 | |
Common error: Take the cube root of the whole fraction, not just of the numerator.
The power, P kW, produced by a wind turbine is directly proportional to the cube of the wind speed, v m/s. When v = 4, P = 48.
Show that P = 0.75v3.
Work out the power produced when v = 10.
Work out the wind speed when the power is 162 kW.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = kv3 | ✓ 1 | |
| M1 48 = k × 43 = 64k | ✓ 1 | |
| A1 k = 0.75 so P = 0.75v3 | ✓ 1 | |
| Part (b) | ||
| M1 0.75 × 103 | ✓ 1 | |
| A1 750 kW | ✓ 1 | |
| Part (c) | ||
| M1 162 = 0.75v3 | ✓ 1 | |
| M1 v3 = 216 | ✓ 1 | |
| A1 6 m/s | ✓ 1 | |
Exam tip: For y ∝ x³, doubling x multiplies y by 8; use this to sanity-check your answers.
Consider the equation 3x2 − 8x − 2 = 0.
Solve 3x2 − 8x − 2 = 0. Give both solutions correct to 3 significant figures.
Hence, or otherwise, find the larger solution of 3(y − 1)2 − 8(y − 1) − 2 = 0. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 correct substitution into the quadratic formula, e.g. (8 ± √(64 + 24)) / 6 | ✓ 1 | |
| M1 simplifies to (8 ± √88) / 6 | ✓ 1 | |
| A1 x = 2.90 and x = −0.230 | ✓ 1 | |
| Part (b) | ||
| M1 y − 1 = x, so y = x + 1 (using 2.8968... or 2.90) | ✓ 1 | |
| A1 3.90 | ✓ 1 | |
Common error: In the quadratic formula the term −b is the opposite of b, and the −4ac inside the root becomes +24 here because c is negative.
Consider the expression x2 − 3x − 10.
Factorise x2 − 3x − 10.
Solve the inequality x2 − 3x − 10 < 0.
Write down the number of integers that satisfy the inequality in (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x ± 2)(x ± 5) | ✓ 1 | |
| A1 (x − 5)(x + 2) | ✓ 1 | |
| Part (b) | ||
| M1 critical values x = 5 and x = −2 (ECF from (a)) | ✓ 1 | |
| A1 −2 < x < 5 | ✓ 1 | |
| Part (c) | ||
| B1 6 (the integers −1, 0, 1, 2, 3, 4) (ECF from (b)) | ✓ 1 | |
Exam tip: For a quadratic inequality with < 0, the solution lies between the two critical values; a quick sketch of the U-shaped curve helps.
Simplify fully the algebraic fraction (x2 − 9) / (x2 + x − 12), working through the stages below.
Factorise x2 − 9.
Factorise x2 + x − 12.
Hence simplify (x2 − 9) / (x2 + x − 12).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 (x − 3)(x + 3) | ✓ 1 | |
| Part (b) | ||
| M1 (x ± 3)(x ± 4) | ✓ 1 | |
| A1 (x + 4)(x − 3) | ✓ 1 | |
| Part (c) | ||
| M1 cancels the common factor (x − 3) | ✓ 1 | |
| A1 (x + 3)/(x + 4) | ✓ 1 | |
Common error: Only common factors can be cancelled; you cannot cancel the x² terms or the numbers 9 and 12 before factorising.
In a model of an outbreak, the value of x satisfies the equation (x + 3)/4 − (x − 1)/6 = 2.
Solve (x + 3)/4 − (x − 1)/6 = 2.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies every term by 12: 3(x + 3) − 2(x − 1) = 24 | ✓ 1 | |
| M1 expands correctly: 3x + 9 − 2x + 2 = 24 | ✓ 1 | |
| A1 x = 13 | ✓ 1 | |
Common error: When subtracting a fraction such as (x − 1)/6, the minus sign applies to the whole numerator, so −2(x − 1) becomes −2x + 2.
A robotic arm moves a tool along a straight rail. The distance, d metres, of the tool from its start after t seconds is d = 3t2 − 4t, for t ≥ 0.
Show that the tool is 11 m from its start when 3t2 − 4t − 11 = 0.
Solve 3t2 − 4t − 11 = 0.
Give your answers correct to 3 significant figures.
Write down the time, in seconds, at which the tool is 11 m from its start. Give a reason for your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 sets 3t2 − 4t = 11 and rearranges to 3t2 − 4t − 11 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 substitutes a = 3, b = −4, c = −11 into the quadratic formula | ✓ 1 | |
| M1 simplifies to (4 ± √148) ÷ 6 | ✓ 1 | |
| A1 t = 2.69 and t = −1.36 | ✓ 1 | |
| Part (c) | ||
| A1 2.69 s, because time cannot be negative (ECF from (b)) | ✓ 1 | |
Common error: Put brackets round negative values such as b = −4 when substituting into the quadratic formula.
The strength, S, of a radio signal received by a robot is inversely proportional to the square of its distance, d metres, from the transmitter.
S = 40 when d = 3.
Find a formula for S in terms of d.
Work out the value of d when S = 2.5.
The distance of the robot from the transmitter increases from 3 m to 4.5 m.
Work out the percentage decrease in the signal strength.
Give your answer correct to 3 significant figures. Use unrounded values in your working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 S = k ÷ d2 | ✓ 1 | |
| A1 k = 40 × 9 = 360 | ✓ 1 | |
| A1 S = 360 ÷ d2 | ✓ 1 | |
| Part (b) | ||
| M1 2.5 = 360 ÷ d2, so d2 = 144 | ✓ 1 | |
| A1 d = 12 | ✓ 1 | |
| Part (c) | ||
| M1 S = 360 ÷ 4.52 = 17.77… | ✓ 1 | |
| M1 (40 − 17.77…) ÷ 40 × 100 | ✓ 1 | |
| A1 55.6% | ✓ 1 | |
Common error: Inverse square means dividing by d²; do not use k ÷ d or kd².
In a model of a fishery, the catch rate, R, is given by
R = (3n + 5) ÷ (n − 2)
where n is the number of nets in use and n > 2.
Make n the subject of the formula.
Work out the number of nets in use when R = 5.75.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R(n − 2) = 3n + 5 | ✓ 1 | |
| M1 Rn − 2R = 3n + 5 | ✓ 1 | |
| M1 collects terms in n and factorises: n(R − 3) = 2R + 5 | ✓ 1 | |
| A1 n = (2R + 5) ÷ (R − 3) | ✓ 1 | |
| Part (b) | ||
| M1 (2 × 5.75 + 5) ÷ (5.75 − 3) = 16.5 ÷ 2.75 | ✓ 1 | |
| A1 6 nets | ✓ 1 | |
Exam tip: When the subject appears twice, expand, gather all those terms on one side, then factorise.
Show all your working in this question.
Work out the value of \(\left(\frac{27}{8}\right)^{-\frac{2}{3}}\). Give your answer as a fraction.
Solve \(4^{x+1} = 8^{x-1}\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 reciprocal and cube root, \(\frac{2}{3}\) seen | ✓ 1 | |
| A1 \(\frac{4}{9}\) | ✓ 1 | |
| Part (b) | ||
| M1 writes both sides as powers of 2: \(2^{2(x+1)}\) and \(2^{3(x-1)}\) | ✓ 1 | |
| M1 equates indices: \(2x + 2 = 3x - 3\) | ✓ 1 | |
| A1 \(x = 5\) | ✓ 1 | |
Common error: Treating a negative index as a negative answer. A negative index means take the reciprocal.
The mass M kg of a reef fish is directly proportional to the cube of its length L cm. A fish of length 40 cm has mass 1.6 kg.
Find a formula for M in terms of L.
Work out the mass of a fish of length 55 cm. Give your answer in kg correct to 3 significant figures.
A fish has mass 0.5 kg. Work out its length. Give your answer in cm correct to 3 significant figures. Use the exact value of k from part (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 M = kL3 and 1.6 = k × 403 | ✓ 1 | |
| A1 M = 2.5 × 10−5L3 | ✓ 1 | |
| Part (b) | ||
| M1 2.5 × 10−5 × 553 (ECF from (a)) | ✓ 1 | |
| A1 4.16 kg (from 4.159375) | ✓ 1 | |
| Part (c) | ||
| M1 0.5 = 2.5 × 10−5 × L3 | ✓ 1 | |
| M1 L3 = 20 000 and takes the cube root | ✓ 1 | |
| A1 27.1 cm (from 27.144...) | ✓ 1 | |
Common error: Using a direct proportion with L instead of L³. Write the formula with the correct power before substituting.
A rectangular garden in a landscape plan has width w metres. Its length is 3 m more than twice its width. The area of the garden is 40 m2.
Show that 2w2 + 3w − 40 = 0.
Solve 2w2 + 3w − 40 = 0 using the quadratic formula. Give the width of the garden in metres correct to 3 significant figures.
Work out the length of the garden. Use the unrounded width from (b). Give your answer in metres correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 w(2w + 3) = 40 | ✓ 1 | |
| A1 expands and rearranges to 2w2 + 3w − 40 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the formula, \(w = \frac{-3 \pm \sqrt{9 + 320}}{4}\) | ✓ 1 | |
| A1 roots 3.78 and −5.28 | ✓ 1 | |
| A1 w = 3.78 with the negative root rejected | ✓ 1 | |
| Part (c) | ||
| M1 2 × 3.7846... + 3 | ✓ 1 | |
| A1 10.6 m | ✓ 1 | |
Common error: Forgetting to reject the negative root. A width cannot be negative.
Simplify fully (x2 − 9) / (2x2 + 5x − 3)
| Marking point | Mark | Notes |
|---|---|---|
| Numerator (difference of two squares): x2 − 9 = (x − 3)(x + 3). Denominator: 2x2 + 5x − 3 = (2x − 1)(x + 3). Cancel the common factor (x + 3): (x − 3)/(2x − 1) | ✓ 1 | Answer A |
Answer: A
Exam tip: Only whole factors can be cancelled. Factorise the top and the bottom completely first, then look for an identical bracket in both.
A small bakery models its daily running cost, in hundreds of euros, as
C = 2x2 − 12x + 23
where x is the number of hours the large oven is switched on (0 ≤ x ≤ 8).
Write 2x2 − 12x + 23 in the form a(x + b)2 + c, where a, b and c are integers.
Hence write down the minimum daily running cost, in euros, and the number of hours for which it occurs.
Explain why the model never gives a negative cost for any value of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2(x2 − 6x) + 23 | ✓ 1 | |
| 2[(x − 3)2 − 9] + 23 | ✓ 1 | |
| 2(x − 3)2 + 5 | ✓ 1 | |
| Part (b) | ||
| Minimum cost €500 | ✓ 1 | |
| when x = 3 hours | ✓ 1 | |
| Part (c) | ||
| 2(x − 3)2 ≥ 0 for all x, so C ≥ 5 > 0 | ✓ 1 | |
Common error: When the x² coefficient is not 1, take it out of the x terms first and remember to multiply the −9 by that factor (2 × 9 = 18) before combining with 23.
A cyclist rides 3 km on a flat road at (x + 2) km/h, then 2 km up a hill at (x − 1) km/h, where x > 1.
Show that the time on the flat road minus the time on the hill, in hours, is
(x − 7) / ((x + 2)(x − 1))
Simplify fully
[(x2 + 4x + 4) / (x2 − 4)] ÷ [(3x + 6) / (x − 2)]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Times: 3/(x + 2) and 2/(x − 1) | ✓ 1 | |
| [3(x − 1) − 2(x + 2)] / ((x + 2)(x − 1)) | ✓ 1 | |
| (3x − 3 − 2x − 4) / ((x + 2)(x − 1)) = (x − 7) / ((x + 2)(x − 1)) | ✓ 1 | |
| Part (b) | ||
| Factorises: (x + 2)2, (x + 2)(x − 2) and 3(x + 2) | ✓ 1 | |
| Inverts and multiplies: (x + 2)2/((x + 2)(x − 2)) × (x − 2)/(3(x + 2)) | ✓ 1 | |
| 1/3 | ✓ 1 | |
Common error: When subtracting fractions, the minus sign applies to the whole second numerator: −2(x + 2) = −2x − 4, not −2x + 4.
An engineer calibrates a sensor using the formula
y = (3x + 2) / (x − 5)
Which of these is a correct rearrangement to make x the subject?
| Marking point | Mark | Notes |
|---|---|---|
| Multiply both sides by (x − 5): y(x − 5) = 3x + 2 Expand: xy − 5y = 3x + 2 Collect the x terms on one side: xy − 3x = 5y + 2 Factorise: x(y − 3) = 5y + 2 Divide: x = (5y + 2) / (y − 3) | ✓ 1 | Answer A |
Answer: A
Exam tip: When the new subject appears twice, follow four steps: clear the fraction, expand, collect every term containing the subject on one side, then factorise it out.
An observer stands on a cliff above the sea. The distance to the horizon, D km, is directly proportional to the square root of the observer's height above sea level, h metres.
When h = 25, D = 18.
Find a formula for D in terms of h.
Find the distance to the horizon when the observer is 49 m above sea level.
Find the height the observer must be above sea level to see a horizon 30 km away. Give your answer to 3 significant figures.
The observer climbs to a point whose height above sea level is 4 times as great. Describe the effect on the distance to the horizon.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| D = k√h and 18 = k × √25 | ✓ 1 | |
| D = 3.6√h | ✓ 1 | |
| Part (b) | ||
| 3.6 × √49 = 3.6 × 7 = 25.2 km | ✓ 1 | |
| Part (c) | ||
| √h = 30 ÷ 3.6 (= 8.333…) | ✓ 1 | |
| h = 69.4 m (69.44…) | ✓ 1 | |
| Part (d) | ||
| The distance is multiplied by 2 (doubles), since √4 = 2 | ✓ 1 | |
Exam tip: For a 'what happens if' part, apply the proportion rule to the multiplier: h × 4 gives D × √4 = D × 2.
Show your working clearly. Answers without working may not gain full marks.
Solve the simultaneous equations
4x + 3y = 5
6x − 5y = 36
Hence solve the simultaneous equations
8u + 3(v + 1) = 5
12u − 5(v + 1) = 36
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Matches coefficients, e.g. 12x + 9y = 15 and 12x − 10y = 72 | ✓ 1 | |
| Eliminates to get 19y = −57, so y = −3 | ✓ 1 | |
| x = 3.5, y = −3 | ✓ 1 | |
| Part (b) | ||
| Recognises 2u = x = 3.5 and v + 1 = y = −3 | ✓ 1 | |
| u = 1.75, v = −4 | ✓ 1 | |
Exam tip: 'Hence' means use the result you have just found. Spot which expressions play the roles of x and y instead of starting again.
A delivery drone flies 36 km from a depot to a farm at an average speed of v km/h.
It flies the same 36 km back at an average speed of (v + 6) km/h.
The outward journey takes 1 hour longer than the return journey.
Show that v2 + 6v − 216 = 0
Solve the equation to find v, and hence find the total time for the round trip.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Outward time 36/v and return time 36/(v + 6) | ✓ 1 | |
| 36/v − 36/(v + 6) = 1 | ✓ 1 | |
| 36(v + 6) − 36v = v(v + 6), so 216 = v2 + 6v, giving v2 + 6v − 216 = 0 | ✓ 1 | |
| Part (b) | ||
| (v + 18)(v − 12) = 0 | ✓ 1 | |
| v = 12 (−18 rejected because speed must be positive) | ✓ 1 | |
| Total time = 36/12 + 36/18 = 3 + 2 = 5 hours | ✓ 1 | |
Exam tip: Time = distance ÷ speed. Write each time as a fraction first, form the equation in words (longer time − shorter time = 1), then multiply every term by the common denominator.
On a plan of a park, the edge of a circular fountain has equation x2 + y2 = 34.
A straight walkway across the fountain follows the line y = x + 2.
Units on both axes are metres.
Find the coordinates of the two points where the walkway meets the edge of the fountain. Show your working clearly.
Calculate the length of walkway that lies inside the fountain. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| x2 + (x + 2)2 = 34 | ✓ 1 | |
| 2x2 + 4x − 30 = 0, i.e. x2 + 2x − 15 = 0 | ✓ 1 | |
| (x + 5)(x − 3) = 0 so x = 3 or x = −5 | ✓ 1 | |
| (3, 5) and (−5, −3) | ✓ 1 | |
| Part (b) | ||
| √((3 − (−5))2 + (5 − (−3))2) = √128 | ✓ 1 | |
| 11.3 m (11.31…; exact 8√2) | ✓ 1 | |
Common error: Expanding (x + 2)² as x² + 4 loses the middle term 4x. Write (x + 2)(x + 2) and expand fully.
An engineer models the curved underside of a footbridge by y = x2 − 4x + 1 and a straight support beam by y = 2x − 7.
Find the coordinates of the points where the beam meets the curve. Show your working clearly.
A second beam is planned along y = 2x − 10. Show that this beam does not meet the curve.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| x2 − 4x + 1 = 2x − 7 | ✓ 1 | |
| x2 − 6x + 8 = 0 | ✓ 1 | |
| (x − 2)(x − 4) = 0 so x = 2 or x = 4 | ✓ 1 | |
| (2, −3) and (4, 1) | ✓ 1 | |
| Part (b) | ||
| x2 − 4x + 1 = 2x − 10 gives x2 − 6x + 11 = 0 | ✓ 1 | |
| (x − 3)2 + 2 = 0 so (x − 3)2 = −2, which has no real solutions (or b2 − 4ac = 36 − 44 = −8 < 0), so the beam does not meet the curve | ✓ 1 | |
Exam tip: To show a line and a curve do not meet, form the quadratic and show it has no real solutions: a square equal to a negative number, or a negative value under the square root in the formula.
A volleyball is hit upwards. Its height, h metres, above the ground t seconds after it is hit is modelled by
h = 1.5 + 12t − 4.9t2
Use the quadratic formula to find the time at which the ball reaches the ground. Give your answer correct to 3 significant figures.
Find, correct to 3 significant figures, the length of time for which the ball is more than 6 m above the ground.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.9t2 − 12t − 1.5 = 0 | ✓ 1 | |
| t = (12 ± √(144 + 29.4)) ÷ 9.8 = (12 ± √173.4) ÷ 9.8 | ✓ 1 | |
| t = 2.57 s (2.568…); the negative root (−0.119) is rejected | ✓ 1 | |
| Part (b) | ||
| 4.9t2 − 12t + 4.5 = 0 | ✓ 1 | |
| t = (12 ± √55.8) ÷ 9.8, so t = 0.4623… or t = 1.9867… | ✓ 1 | |
| 1.9867… − 0.4623… = 1.52 s (1.524…) | ✓ 1 | |
Exam tip: Keep unrounded values in your calculator for the subtraction in the last step. Rounding each time first (1.99 − 0.462 = 1.528) can make the final answer wrong.
A road-safety model gives the stopping distance d metres of a delivery van travelling at v metres per second as
d = v + v2/20
The van must be able to stop within a distance of less than 40 m.
Show that v2 + 20v − 800 < 0
Solve v2 + 20v − 800 < 0. Show your working clearly.
Hence state the range of speeds of the van for which it can stop within the required distance. Give a reason for your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v + v2/20 < 40, multiply by 20: 20v + v2 < 800 so v2 + 20v − 800 < 0 | ✓ 1 | |
| Part (b) | ||
| (v + 40)(v − 20) | ✓ 1 | |
| critical values −40 and 20 | ✓ 1 | |
| −40 < v < 20 | ✓ 1 | |
| Part (c) | ||
| 0 < v < 20 (or 0 ≤ v < 20) because speed cannot be negative | ✓ 1 | |
Exam tip: Solving a quadratic inequality gives a mathematical answer; in a context question, always check whether part of it is impossible (here a negative speed) and reject that part with a reason.
A gardener plants x rows of tomatoes and y rows of beans. The numbers of rows satisfy
x ≥ 2, y > 1, x + y < 8
where x and y are integers.
The lines x = 2, y = 1 and x + y = 8 form a triangle. Find the coordinates of its three vertices.
Find the number of possible combinations (x, y).
Each row of tomatoes earns £30 and each row of beans earns £60. Find the greatest possible total earnings.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (2, 1) and (7, 1) | ✓ 1 | |
| (2, 6) | ✓ 1 | |
| Part (b) | ||
| y = 2: 4 points, y = 3: 3, y = 4: 2, y = 5: 1, or a full correct list | ✓ 1 | |
| 10 | ✓ 1 | |
| Part (c) | ||
| £360 at (2, 5) | ✓ 1 | |
Common error: Points on a boundary line count only for ≤ or ≥. Here y > 1 and x + y < 8 are strict, so points on those lines, such as (2, 1) and (2, 6), are not allowed.
A tile designer studies patterns built from whole numbers.
Two consecutive even numbers can be written as 2n and 2n + 2, where n is an integer. Show that their product plus 1 is equal to (2n + 1)2.
Hence explain why the product of two consecutive even numbers, plus 1, is always the square of an odd number.
Prove that the sum of the squares of any two consecutive integers is always odd.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2n(2n + 2) + 1 = 4n2 + 4n + 1 | ✓ 1 | |
| (2n + 1)2 = 4n2 + 4n + 1, so they are equal | ✓ 1 | |
| Part (b) | ||
| 2n + 1 is one more than a multiple of 2, so it is odd, and the expression is its square | ✓ 1 | |
| Part (c) | ||
| m2 + (m + 1)2 = 2m2 + 2m + 1 | ✓ 1 | |
| = 2(m2 + m) + 1, which is one more than a multiple of 2, so it is odd | ✓ 1 | |
Common error: Using n and n + 1 for two even numbers does not work: start by writing a general even number as 2n and an odd number as 2n + 1.
This question is about the inequalities \(x^2 - 2x - 15 < 0\) and \(2x - 1 \ge 3\)
Solve \(x^2 - 2x - 15 < 0\)
Find all the integers that satisfy both \(x^2 - 2x - 15 < 0\) and \(2x - 1 \ge 3\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for (x − 5)(x + 3) | ✓ 1 | |
| A1 for the critical values 5 and −3 | ✓ 1 | |
| A1 for −3 < x < 5 | ✓ 1 | |
| Part (b) | ||
| M1 for x ≥ 2 | ✓ 1 | |
| A1 for 2, 3, 4 | ✓ 1 | |
Exam tip: Put both solution sets on a number line. The integers you need lie where the two regions overlap. Remember that 5 is excluded because the inequality is strict.
y is directly proportional to the cube of x.
x is increased by 20%.
Find the percentage increase in y.
Find the percentage decrease in x that would make y halve.
Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(y = kx^3\) with x replaced by 1.2x | ✓ 1 | |
| M1 \(1.2^3 = 1.728\) | ✓ 1 | |
| A1 72.8% | ✓ 1 | |
| Part (b) | ||
| M1 \((\text{multiplier})^3 = 0.5\) | ✓ 1 | |
| M1 multiplier \(= \sqrt[3]{0.5} = 0.7937...\) | ✓ 1 | |
| A1 20.6% | ✓ 1 | |
Common error: A 20% increase in x does not give a 60% increase in y. Multipliers are cubed, not percentages multiplied by 3.
a is directly proportional to the square of b.
b is inversely proportional to c.
When c = 2, b = 6
When b = 6, a = 72
Find a formula for a in terms of c.
Find the positive value of c when a = 8
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(b = \frac{m}{c}\) with \(m = 12\), so \(b = \frac{12}{c}\) | ✓ 1 | |
| M1 \(a = kb^2\) with \(72 = 36k\), so \(k = 2\) | ✓ 1 | |
| M1 substitutes, \(a = 2\left(\frac{12}{c}\right)^2\) | ✓ 1 | |
| A1 \(a = \frac{288}{c^2}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(c^2 = \frac{288}{8} = 36\) | ✓ 1 | |
| A1 6 | ✓ 1 | |
Exam tip: Use a different letter for each constant of proportionality so the two constants are not confused when you combine the formulae.
Answer each part.
Simplify fully \(\frac{2x^2-5x-3}{4x^2-1} \div \frac{x^2-9}{6x-3}\)
Hence solve \(\frac{2x^2-5x-3}{4x^2-1} \div \frac{x^2-9}{6x-3} = \frac{3}{5}\)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 2x2 − 5x − 3 = (2x + 1)(x − 3) | ✓ 1 | |
| M1 for 4x2 − 1 = (2x − 1)(2x + 1) and x2 − 9 = (x − 3)(x + 3) | ✓ 1 | |
| M1 for inverting the second fraction and multiplying, with 6x − 3 = 3(2x − 1) | ✓ 1 | |
| A1: \(\frac{3}{x+3}\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(\frac{3}{x+3} = \frac{3}{5}\) leading to x + 3 = 5 | ✓ 1 | |
| A1: x = 2 | ✓ 1 | |
Exam tip: Factorise every numerator and denominator completely before cancelling; a difference of two squares often hides in expressions like 4x² − 1.
Here is a formula.
\(k = \sqrt{\frac{m+n}{m-n}}\)
Make m the subject of the formula.
Find the value of m when k = 3 and n = 4
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 squares both sides, \(k^2 = \frac{m+n}{m-n}\) | ✓ 1 | |
| M1 multiplies both sides by \((m-n)\) and expands, \(k^2m - k^2n = m + n\) | ✓ 1 | |
| M1 collects the m terms on one side and factorises, \(m(k^2 - 1) = n(k^2 + 1)\) or equivalent, e.g. \(m(1-k^2) = -n(k^2+1)\) | ✓ 1 | |
| A1 \(m = \frac{n(k^2+1)}{k^2-1}\) or any equivalent form, e.g. \(\frac{nk^2+n}{k^2-1}\) or \(\frac{n(1+k^2)}{k^2-1}\) | ✓ 1 | |
| Part (b) | ||
| B1 5 | ✓ 1 | |
Exam tip: Remove the square root first by squaring, then clear the fraction. Check your answer by substituting back: with m = 5 and n = 4, √(9/1) = 3 as required.
The line with equation \(y = x + 1\) meets the circle with equation \(x^2 + y^2 = 25\) at the points A and B.
Use algebra to find the coordinates of A and B.
Find the length of AB. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for substituting to eliminate one variable, e.g. \(x^2 + (x + 1)^2 = 25\) or \((y - 1)^2 + y^2 = 25\) | ✓ 1 | |
| M1 for simplifying to a three-term quadratic, e.g. \(x^2 + x - 12 = 0\) or \(2x^2 + 2x - 24 = 0\) (or \(y^2 - y - 12 = 0\)) | ✓ 1 | |
| M1 for a correct method of solving, e.g. \((x + 4)(x - 3) = 0\) (or \((y - 4)(y + 3) = 0\)) | ✓ 1 | |
| A1 \(x = 3\) and \(x = -4\) (or \(y = 4\) and \(y = -3\)) | ✓ 1 | |
| A1 \((3, 4)\) and \((-4, -3)\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(\sqrt{(3 - (-4))^2 + (4 - (-3))^2}\), i.e. \(\sqrt{98}\) | ✓ 1 | |
| A1 9.90 | ✓ 1 | |
Common error: Expand \((x + 1)^2\) as \(x^2 + 2x + 1\), not \(x^2 + 1\). Missing the middle term gives the wrong quadratic.
(ax + 5)(3x − b) ≡ 12x2 + cx − 15, where a, b and c are integers.
Find the values of a, b and c.
Hence factorise fully 12x2 + cx − 15, using your value of c.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for expanding to 3ax2 − abx + 15x − 5b, or for writing 3a = 12 or −5b = −15 | ✓ 1 | |
| A1 for a = 4 and b = 3 | ✓ 1 | |
| A1 for c = 15 − ab = 15 − 12 = 3 | ✓ 1 | |
| Part (b) | ||
| M1 for (4x + 5)(3x − 3), or 3(4x2 + x − 5), or another correct partial factorisation | ✓ 1 | |
| A1 for 3(4x + 5)(x − 1) | ✓ 1 | |
Exam tip: In an identity, compare coefficients of each power of x separately: the x² and constant terms give a and b directly, then the x term gives c. When asked to factorise fully, check for a common factor in every bracket.
Answer each part.
Write 2x2 − 12x + 23 in the form a(x + b)2 + c, where a, b and c are integers.
Hence write down the minimum value of 2x2 − 12x + 23 and the value of x at which it occurs.
Explain why the equation 2x2 − 12x + 23 = 0 has no real solutions.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 2(x2 − 6x) + 23 | ✓ 1 | |
| M1 for 2(x − 3)2 − 18 + 23 | ✓ 1 | |
| A1: 2(x − 3)2 + 5 | ✓ 1 | |
| Part (b) | ||
| B1: minimum value 5 | ✓ 1 | |
| B1: x = 3 | ✓ 1 | |
| Part (c) | ||
| B1 for 2(x − 3)2 ≥ 0, so the expression is always at least 5 and can never equal 0; OR the discriminant (−12)2 − 4 × 2 × 23 = −40 < 0, so there are no real roots | ✓ 1 | |
Common error: When taking out the factor 2, remember the −9 inside the bracket also gets doubled: 2(x − 3)² means subtracting 18, not 9.
A ball is thrown upwards from a height of 1.5 m above the ground. Its height h metres above the ground after t seconds is modelled by h = 1.5 + 12t − 5t2.
Show that the ball hits the ground when 5t2 − 12t − 1.5 = 0. (2)
Use the quadratic formula to find the time the ball hits the ground. Give your answer to 3 significant figures. (3)
Write 1.5 + 12t − 5t2 in the form p − 5(t − q)2, and hence state the greatest height of the ball. (3)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 h = 0 | ✓ 1 | |
| A1 1.5 + 12t − 5t2 = 0 rearranged to 5t2 − 12t − 1.5 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 substitution into the formula with a = 5, b = −12, c = −1.5 | ✓ 1 | |
| M1 \(\frac{12 \pm \sqrt{174}}{10}\) | ✓ 1 | |
| A1 2.52 s (2.5190...; the other root is rejected as negative) | ✓ 1 | |
| Part (c) | ||
| M1 1.5 − 5(t2 − 2.4t) | ✓ 1 | |
| M1 (t − 1.2)2 − 1.44 used | ✓ 1 | |
| A1 8.7 − 5(t − 1.2)2 so greatest height 8.7 m | ✓ 1 | |
Exam tip: In the quadratic formula, put brackets round negative values of b and c when you substitute.
A skate ramp has a cross-section modelled by the curve y = x2 − 3x. A straight support beam is modelled by the line 2x + y = 6. The units are metres. The beam meets the curve at the points P and Q.
Find the coordinates of P and Q. (5)
Work out the length of the beam PQ. Give your answer to 3 significant figures. (3)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 y = 6 − 2x and x2 − 3x = 6 − 2x | ✓ 1 | |
| M1 x2 − x − 6 = 0 | ✓ 1 | |
| M1 (x − 3)(x + 2) = 0 | ✓ 1 | |
| A1 x = 3 and x = −2 | ✓ 1 | |
| A1 (3, 0) and (−2, 10) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{5^2 + 10^2}\) | ✓ 1 | |
| A1 √125 = 5√5 | ✓ 1 | |
| A1 11.2 m (11.18 m) | ✓ 1 | |
Common error: After finding the x values, substitute into the linear equation to find y; using the quadratic as well can give values that do not lie on the line.
This question is about an algebraic equation.
Show that the equation \(\frac{3}{x + 2} + \frac{2}{x - 1} = 1\) can be written as x2 − 4x − 3 = 0. (4)
Solve x2 − 4x − 3 = 0. Give your solutions correct to 3 significant figures. (3)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 common denominator (x + 2)(x − 1) | ✓ 1 | |
| M1 3(x − 1) + 2(x + 2) = (x + 2)(x − 1) | ✓ 1 | |
| A1 5x + 1 = x2 + x − 2 | ✓ 1 | |
| A1 rearranged to x2 − 4x − 3 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{4 \pm \sqrt{28}}{2}\) | ✓ 1 | |
| A1 4.65 | ✓ 1 | |
| A1 −0.646 (that is 2 − √7 = −0.6457...) | ✓ 1 | |
Exam tip: Multiply every term by the common denominator, including the term on the right-hand side, to clear the fractions in one step.
The volume, V, of water in a straight cylindrical pipe of radius r and length L is given by V = πr2L.
Make L the subject of the formula.
The pipe holds 0.45 m3 of water and has radius 0.12 m. Work out the length of the pipe. Give your answer correct to 3 significant figures.
Make x the subject of y = (3x + 2) ÷ (x − 1)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 divide both sides by πr2 | ✓ 1 | |
| A1 L = V ÷ (πr2) | ✓ 1 | |
| Part (b) | ||
| M1 0.45 ÷ (π × 0.122) | ✓ 1 | |
| A1 9.95 m | ✓ 1 | |
| Part (c) | ||
| M1 multiply both sides by (x − 1) | ✓ 1 | |
| M1 expand: yx − y = 3x + 2 | ✓ 1 | |
| M1 collect x terms and factorise: x(y − 3) = y + 2 | ✓ 1 | |
| A1 x = (y + 2) ÷ (y − 3) | ✓ 1 | |
Exam tip: When the subject appears twice, collect every term containing it on one side and factorise.
The depth of water in a reservoir, h metres, t days after a storm is modelled by h = 12 + 7t − t2 for 0 ≤ t ≤ 7.
Show that h > 18 when t2 − 7t + 6 < 0
Solve t2 − 7t + 6 < 0
Solve x2 − x − 12 ≥ 0
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 12 + 7t − t2 > 18 | ✓ 1 | |
| A1 rearrange: t2 − 7t + 6 < 0 (reversing the inequality sign) | ✓ 1 | |
| Part (b) | ||
| M1 (t − 1)(t − 6) or correct use of formula | ✓ 1 | |
| A1 critical values 1 and 6 | ✓ 1 | |
| A1 1 < t < 6 | ✓ 1 | |
| Part (c) | ||
| M1 (x − 4)(x + 3) | ✓ 1 | |
| A1 critical values −3 and 4 | ✓ 1 | |
| A1 x ≤ −3 or x ≥ 4 | ✓ 1 | |
Common error: For an inequality such as x² − x − 12 ≥ 0 the answer is two separate regions, not one combined interval.
The light intensity I lux on a theatre stage is inversely proportional to the square of the distance d metres from a spotlight. When d = 5, I = 64.
Find a formula for I in terms of d.
Find the distance at which the intensity is 100 lux.
The distance from the spotlight is increased by 25%.
Work out the percentage decrease in the intensity.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = \(\frac{k}{d^2}\) with 64 = \(\frac{k}{25}\) | ✓ 1 | |
| A1 I = \(\frac{1600}{d^2}\) | ✓ 1 | |
| Part (b) | ||
| M1 100 = \(\frac{1600}{d^2}\) or d2 = 16 | ✓ 1 | |
| A1 4 m | ✓ 1 | |
| Part (c) | ||
| M1 new distance = 1.25d | ✓ 1 | |
| M1 new intensity = \(\frac{1600}{(1.25d)^2}\) = 0.64 × I | ✓ 1 | |
| A1 36% | ✓ 1 | |
Common error: A 25% increase in distance does not give a 25% decrease in intensity; the distance is squared, so the multiplier is 1 ÷ 1.25² = 0.64.
A harbour's radar shows a circular no-entry zone of radius √13 km. Taking the radar station as the origin, the boundary of the zone has equation x2 + y2 = 13. A ship sails along the straight line y = 2x − 1. Distances are in kilometres.
Show that the x-coordinates of the points where the ship's course meets the boundary satisfy 5x2 − 4x − 12 = 0.
Solve 5x2 − 4x − 12 = 0.
Work out the distance between the two points where the course meets the boundary.
Give your answer in kilometres correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x2 + (2x − 1)2 = 13 | ✓ 1 | |
| M1 x2 + 4x2 − 4x + 1 = 13 | ✓ 1 | |
| A1 5x2 − 4x − 12 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (5x + 6)(x − 2) = 0 or correct use of the formula | ✓ 1 | |
| A1 x = 2 | ✓ 1 | |
| A1 x = −1.2 | ✓ 1 | |
| Part (c) | ||
| M1 points (2, 3) and (−1.2, −3.4) | ✓ 1 | |
| M1 √(3.22 + 6.42) | ✓ 1 | |
| A1 7.16 km (√51.2 = 7.155...) | ✓ 1 | |
Exam tip: Substitute the linear equation into the circle equation to get a single quadratic; then use the line to find each y-value.
In a model of two teams tracing contacts during an outbreak, a positive number x satisfies 3/(x + 1) + 2/(x − 2) = 1.
Show that the equation can be written as x2 − 6x + 2 = 0.
Solve x2 − 6x + 2 = 0. Give both solutions to 2 decimal places.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiply through by (x + 1)(x − 2) | ✓ 1 | |
| M1 3(x − 2) + 2(x + 1) = (x + 1)(x − 2) | ✓ 1 | |
| M1 5x − 4 = x2 − x − 2 | ✓ 1 | |
| A1 x2 − 6x + 2 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 x = (6 ± √(36 − 8)) ÷ 2 | ✓ 1 | |
| A1 5.65 | ✓ 1 | |
| A1 0.35 | ✓ 1 | |
Common error: Multiply every term by the common denominator, including the term on the right-hand side.
The power, P watts, received by a solar panel from a lamp in a laboratory is inversely proportional to the square of the distance, d metres, between the lamp and the panel. When d = 2.0, P = 45.
Find a formula for P in terms of d.
Work out P when d = 3.0.
Work out the value of d when P = 5.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = k/d2 | ✓ 1 | |
| M1 45 = k/22 so k = 180 | ✓ 1 | |
| A1 P = 180/d2 | ✓ 1 | |
| Part (b) | ||
| A1 20 W | ✓ 1 | |
| Part (c) | ||
| M1 d2 = 180 ÷ 5 = 36 | ✓ 1 | |
| A1 d = 6 m (ECF from (a)) | ✓ 1 | |
Common error: Inverse square means P = k/d², not k/d; remember to square the distance before dividing.
The proportion p of a population protected in a model depends on a vaccination ratio r (with r ≠ 4) according to p = (3r + 2)/(r − 4).
Make r the subject of the formula.
Use your answer to (a) to find r when p = 7.
Show that no value of r gives p = 3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p(r − 4) = 3r + 2 | ✓ 1 | |
| M1 pr − 3r = 4p + 2 | ✓ 1 | |
| M1 r(p − 3) = 4p + 2 | ✓ 1 | |
| A1 r = (4p + 2)/(p − 3) | ✓ 1 | |
| Part (b) | ||
| A1 r = 7.5 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 3(r − 4) = 3r + 2 | ✓ 1 | |
| R1 this gives −12 = 2, which is impossible, so p = 3 never occurs | ✓ 1 | |
Exam tip: Collect every term containing the new subject on one side before factorising it out.
On a harbour map, each unit is 1 km and the origin is at a lighthouse. A restricted zone is the region inside the circle x2 + y2 = 25. A straight shipping lane has equation y = x + 1.
Show that the x-coordinates of the points where the lane meets the boundary of the zone satisfy x2 + x − 12 = 0.
Find the coordinates of the two points where the lane meets the boundary of the zone.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for substituting to get x2 + (x + 1)2 = 25 | ✓ 1 | |
| M1 for expanding to 2x2 + 2x + 1 = 25 | ✓ 1 | |
| dividing by 2 to reach x2 + x − 12 = 0 with no errors | ✓ 1 | |
| Part (b) | ||
| M1 for (x + 4)(x − 3) = 0 or a correct use of the formula | ✓ 1 | |
| x = 3 and x = −4 | ✓ 1 | |
| (3, 4) and (−4, −3) | ✓ 1 | |
Common error: After finding the x-values, substitute into the linear equation to get y; do not leave the answers as x-values only.
A rectangular stage platform has length (x + 6) m and width x m. Its area is 50 m2.
Show that x2 + 6x − 50 = 0.
Solve x2 + 6x − 50 = 0 to find the width of the platform. Give your answer correct to 3 significant figures.
Work out the perimeter of the platform. Use the unrounded value of x and give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for x(x + 6) = 50 | ✓ 1 | |
| expanding and rearranging to x2 + 6x − 50 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 for substituting into the quadratic formula correctly | ✓ 1 | |
| x = (−6 ± √236)/2 | ✓ 1 | |
| x = 4.68 m (negative root rejected) | ✓ 1 | |
| Part (c) | ||
| M1 for perimeter = 2(x + x + 6) or 4x + 12 | ✓ 1 | |
| M1 for 4 × 4.6811… + 12 | ✓ 1 | |
| 30.7 m | ✓ 1 | |
Common error: Keep the full calculator value of the width for the next part; rounding early can change the third figure.
A shipping company models the efficiency index E of a loading team with w workers by the formula E = (4w + 5)/(w − 2), where w > 2.
Rearrange the formula to make w the subject.
Work out the value of w when E = 7. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for E(w − 2) = 4w + 5 | ✓ 1 | |
| M1 for Ew − 2E = 4w + 5 | ✓ 1 | |
| M1 for collecting w terms: Ew − 4w = 2E + 5 and factorising w(E − 4) | ✓ 1 | |
| w = (2E + 5)/(E − 4) | ✓ 1 | |
| Part (b) | ||
| M1 for (2 × 7 + 5)/(7 − 4) | ✓ 1 | |
| 6.33 | ✓ 1 | |
Common error: When the subject appears twice, collect all terms containing it on one side before factorising.
The kinetic energy E joules of a runner is directly proportional to the square of the runner's speed v m/s. When v = 5, E = 900.
Find a formula for E in terms of v.
Work out E when v = 8.
Work out v when E = 1764.
The runner's speed increases by 50%. Work out the percentage increase in E.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for E = kv2 | ✓ 1 | |
| k = 900 ÷ 25 = 36, so E = 36v2 | ✓ 1 | |
| Part (b) | ||
| M1 for 36 × 64 | ✓ 1 | |
| E = 2304 J | ✓ 1 | |
| Part (c) | ||
| M1 for v2 = 1764 ÷ 36 = 49 | ✓ 1 | |
| v = 7 m/s | ✓ 1 | |
| Part (d) | ||
| M1 for 1.52 = 2.25 | ✓ 1 | |
| 125% | ✓ 1 | |
Common error: A 50% increase in speed multiplies the speed by 1.5, so the energy is multiplied by 1.5², not 1.5.
In a model of star formation, a parameter x satisfies the equations given in parts (a) and (b).
Solve 2x+1 × 4x = 128.
Solve 27x−1 = 9x+2.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 4x = 22x | ✓ 1 | |
| M1 for 23x+1 = 27, giving 3x + 1 = 7 | ✓ 1 | |
| x = 2 | ✓ 1 | |
| Part (b) | ||
| M1 for writing 27 = 33 and 9 = 32 | ✓ 1 | |
| M1 for 3(x − 1) = 2(x + 2) | ✓ 1 | |
| x = 7 | ✓ 1 | |
Common error: Write both sides with the same base before comparing indices; multiplying powers of the same base means adding the indices.
A train completes a 240 km journey at a constant speed of v km/h. If its speed were 20 km/h greater, the journey would take 1 hour less.
Show that v2 + 20v − 4800 = 0.
Solve the equation to find the speed of the train.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 for time = 240/v | ✓ 1 | |
| B1 for time at greater speed = 240/(v + 20) | ✓ 1 | |
| M1 for 240/v − 240/(v + 20) = 1 | ✓ 1 | |
| M1 for multiplying through by v(v + 20) to get 240(v + 20) − 240v = v(v + 20) | ✓ 1 | |
| simplifying to v2 + 20v − 4800 = 0 with no errors | ✓ 1 | |
| Part (b) | ||
| M1 for (v + 80)(v − 60) = 0 or a correct use of the formula | ✓ 1 | |
| v = 60 or v = −80 | ✓ 1 | |
| v = 60 km/h (negative speed rejected) | ✓ 1 | |
Common error: Put the two journey times over a common denominator before simplifying; a speed cannot be negative, so reject that root.
On a map grid, 1 unit represents 1 km. A circular no-fly zone has its centre at the origin O and radius 5 km, so its edge has equation x2 + y2 = 25. A cable-car line follows the straight line y = x + 1.
Find the coordinates of the two points where the cable-car line crosses the edge of the no-fly zone.
Work out the distance between these two points, in km, to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substituting y = x + 1 into x2 + y2 = 25 | ✓ 1 | |
| M1 expanding to 2x2 + 2x − 24 = 0 (or x2 + x − 12 = 0) | ✓ 1 | |
| M1 solving the three-term quadratic (factorising or formula) | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{7^2 + 7^2}\) (or \(7\sqrt{2}\)) | ✓ 1 | |
| A1 9.90 (km) | ✓ 1 | |
Exam tip: Substitute the linear equation into the quadratic one, and remember to find the matching y value for every x.
A food scientist models the ratio R of two ingredient masses in a batch by \(R = \frac{x^2-9}{x^2+5x+6}\), where x is a positive number.
Simplify the expression for R fully.
Hence solve \(R = \frac{2}{3}\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 numerator factorised as (x − 3)(x + 3) | ✓ 1 | |
| B1 denominator factorised as (x + 2)(x + 3) | ✓ 1 | |
| A1 \(\frac{x-3}{x+2}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{x-3}{x+2} = \frac{2}{3}\) (using the simplified form) | ✓ 1 | |
| M1 cross-multiplying to 3(x − 3) = 2(x + 2) | ✓ 1 | |
| A1 x = 13 | ✓ 1 | |
Exam tip: Factorise both the top and the bottom before cancelling; only whole factors can be cancelled, never single terms.
A mountain hut's heating cost index is modelled by \(y = \frac{2x+3}{x-4}\), where x > 4.
Make x the subject of the formula.
Work out the value of x when y = 5. Give your answer to 3 significant figures. (ECF from (a))
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 y(x − 4) = 2x + 3 | ✓ 1 | |
| M1 expanding to xy − 4y = 2x + 3 | ✓ 1 | |
| M1 collecting x terms on one side and factorising, x(y − 2) = 4y + 3 | ✓ 1 | |
| A1 \(x = \frac{4y+3}{y-2}\) | ✓ 1 | |
| Part (b) | ||
| M1 substituting y = 5 to get 23 ÷ 3 | ✓ 1 | |
| A1 7.67 | ✓ 1 | |
Common error: When x appears in two places, collect every x term on one side and factorise before dividing.
On a map of a volcano, coordinates are in kilometres and the vent is at the origin. The boundary of a circular exclusion zone has equation \(x^2 + y^2 = 25\). A straight road has equation \(y = x + 1\).
Find the coordinates of the two points where the road crosses the boundary of the exclusion zone.
Work out the length of the road inside the exclusion zone. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitute \(y = x + 1\) into \(x^2 + y^2 = 25\) | ✓ 1 | |
| M1 expand to \(2x^2 + 2x - 24 = 0\) or \(x^2 + x - 12 = 0\) | ✓ 1 | |
| M1 \((x + 4)(x - 3) = 0\) | ✓ 1 | |
| A1 \(x = 3\) and \(x = -4\) | ✓ 1 | |
| A1 \((3, 4)\) and \((-4, -3)\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{(3 - (-4))^2 + (4 - (-3))^2}\) | ✓ 1 | |
| M1 \(\sqrt{98}\) | ✓ 1 | |
| A1 9.90 km | ✓ 1 | |
Exam tip: After finding the x-values, substitute into the linear equation (not the quadratic) to find the matching y-values.
A stage spotlight shines on a screen. The light intensity, I lux, is inversely proportional to the square of the distance, d metres, from the spotlight. When d = 3, I = 80.
Find a formula for I in terms of d.
Work out the intensity when d = 5.
Work out the distance at which the intensity is 45 lux.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 \(I = \frac{k}{d^2}\) | ✓ 1 | |
| M1 \(80 = \frac{k}{3^2}\) | ✓ 1 | |
| A1 \(I = \frac{720}{d^2}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\frac{720}{5^2}\) | ✓ 1 | |
| A1 28.8 lux | ✓ 1 | |
| Part (c) | ||
| M1 \(45 = \frac{720}{d^2}\), so \(d^2 = 16\) | ✓ 1 | |
| A1 \(d = 4\) m | ✓ 1 | |
Common error: Inverse square proportion uses \(\frac{k}{d^2}\); students often use \(\frac{k}{d}\) or forget to take the square root at the end.
A volcanologist models the net heat gain, H units, of a lava sample by \(H = 2x^2 - 5x - 12\) where x is a parameter set by the experiment.
Solve the inequality \(2x^2 - 5x - 12 \ge 0\).
x is an integer and \(H < 0\). Work out how many possible values x can take.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 factorise, \((2x + 3)(x - 4)\) | ✓ 1 | |
| A1 critical values \(x = -\frac{3}{2}\) and \(x = 4\) | ✓ 1 | |
| M1 use of a sketch of the upward parabola or sign test to choose outside the roots | ✓ 1 | |
| A1 \(x \le -1.5\) or \(x \ge 4\) | ✓ 1 | |
| Part (b) | ||
| M1 \(-1.5 < x < 4\) (complement of (a)) | ✓ 1 | |
| A1 integers \(-1, 0, 1, 2, 3\), so 5 values | ✓ 1 | |
Common error: A quadratic inequality with a positive x² term and '≥ 0' gives two separate regions, not one range.
A farmer's rectangular field has a perimeter of 140 m. Its diagonal is 50 m. The length of the field is l metres and the width is w metres, where l > w.
Show that l + w = 70 and l2 + w2 = 2500
Solve these equations simultaneously to find l and w.
Work out the area of the field.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 2l + 2w = 140, so l + w = 70 | ✓ 1 | |
| B1 Pythagoras: l2 + w2 = 502 = 2500 | ✓ 1 | |
| Part (b) | ||
| M1 substitutes w = 70 − l into l2 + w2 = 2500 | ✓ 1 | |
| M1 expands and rearranges to l2 − 70l + 1200 = 0 | ✓ 1 | |
| M1 factorises (l − 30)(l − 40) = 0 or uses the formula | ✓ 1 | |
| A1 l = 40 | ✓ 1 | |
| A1 w = 30 | ✓ 1 | |
| Part (c) | ||
| A1 1200 (m2) (ECF from (b)) | ✓ 1 | |
Common error: When substituting w = 70 − l, the bracket must be squared as a whole: (70 − l)² = 4900 − 140l + l², not 4900 + l².
Let f = (2x2 + 7x − 15)/(4x2 − 9)
Factorise 2x2 + 7x − 15
Show that f simplifies to (x + 5)/(2x + 3)
Find the value of x for which f = 1.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (2x ± 3)(x ± 5) or equivalent attempt | ✓ 1 | |
| A1 (2x − 3)(x + 5) | ✓ 1 | |
| Part (b) | ||
| B1 4x2 − 9 = (2x − 3)(2x + 3) | ✓ 1 | |
| M1 cancels the common factor (2x − 3) (ECF from (a)) | ✓ 1 | |
| A1 (x + 5)/(2x + 3) | ✓ 1 | |
| Part (c) | ||
| M1 x + 5 = 2x + 3 | ✓ 1 | |
| A1 x = 2 | ✓ 1 | |
Exam tip: Only factors can be cancelled in an algebraic fraction, never separate terms. Factorise the top and bottom fully first.
In a sound-engineering formula, G = (3p + 2)/(p − 1), where p ≠ 1.
Rearrange the formula to make p the subject.
Work out the value of p when G = 7. Give your answer as a decimal.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 G(p − 1) = 3p + 2 | ✓ 1 | |
| M1 expands: Gp − G = 3p + 2 | ✓ 1 | |
| M1 collects the p terms on one side and factorises: p(G − 3) = G + 2 | ✓ 1 | |
| A1 p = (G + 2)/(G − 3) | ✓ 1 | |
| Part (b) | ||
| M1 (7 + 2)/(7 − 3) (= 9/4) (ECF from (a)) | ✓ 1 | |
| A1 2.25 | ✓ 1 | |
Common error: When the new subject appears twice, expand first, gather all terms containing it on one side, then factorise it out.
A lava bomb is approximately spherical. Its mass, \(M\) kilograms, is directly proportional to the cube of its radius, \(r\) centimetres. A lava bomb with radius 12 cm has mass 7.2 kg.
Find a formula for \(M\) in terms of \(r\).
Work out the mass of a lava bomb with radius 18 cm.
A lava bomb has mass 1.8 kg. Work out its radius. Use the exact value of \(k\) and give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(M=kr^3\) and \(7.2=k\times12^3\) | ✓ 1 | |
| A1 for \(M=\frac{r^3}{240}\) (or \(k=0.00417\)) | ✓ 1 | |
| Part (b) | ||
| M1 for \(18^3\div240\) | ✓ 1 | |
| A1 for 24.3 (kg) | ✓ 1 | |
| Part (c) | ||
| M1 for \(1.8=\frac{r^3}{240}\) | ✓ 1 | |
| M1 for \(r^3=432\) and cube root | ✓ 1 | |
| A1 for 7.56 (cm) | ✓ 1 | |
Common error: If \(M\) is proportional to \(r^3\), doubling the radius multiplies the mass by 8, not by 2.
A shipping company models its route cost index \(y\) using the formula \(y = \frac{3x+2}{x-4}\), where \(x > 4\).
Rearrange the formula to make \(x\) the subject.
Use your answer to (a) to find the value of \(x\) when \(y = 5\).
Simplify \(\frac{x^2-9}{x^2+x-6}\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for multiplying both sides by \((x-4)\): \(y(x-4)=3x+2\) | ✓ 1 | |
| M1 for expanding: \(xy-4y=3x+2\) | ✓ 1 | |
| M1 for collecting the \(x\) terms and factorising: \(x(y-3)=4y+2\) | ✓ 1 | |
| A1 for \(x=\frac{4y+2}{y-3}\) | ✓ 1 | |
| Part (b) | ||
| B1 for \(x=11\) | ✓ 1 | |
| Part (c) | ||
| M1 for factorising the numerator: \((x-3)(x+3)\) | ✓ 1 | |
| M1 for factorising the denominator: \((x+3)(x-2)\) | ✓ 1 | |
| A1 for \(\frac{x-3}{x-2}\) | ✓ 1 | |
Exam tip: When the new subject appears twice, gather all its terms on one side so that you can factorise it out.
A ship travels 60 km upstream against a river current of 4 km/h and then returns the same 60 km downstream with the current. The ship's speed in still water is \(x\) km/h, where \(x > 4\). The total travelling time is 8 hours.
Show that \(x^2 - 15x - 16 = 0\).
Solve \(x^2 - 15x - 16 = 0\) and state the speed of the ship in still water.
Work out the time taken for the upstream journey.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for time upstream \(\frac{60}{x-4}\) and downstream \(\frac{60}{x+4}\), forming \(\frac{60}{x-4}+\frac{60}{x+4}=8\) | ✓ 1 | |
| A1 for combining over a common denominator: \(\frac{120x}{x^2-16}=8\) | ✓ 1 | |
| M1 for multiplying out: \(120x=8(x^2-16)\) | ✓ 1 | |
| A1 for \(8x^2-120x-128=0\) and dividing by 8 to reach \(x^2-15x-16=0\) | ✓ 1 | |
| Part (b) | ||
| M1 for \((x-16)(x+1)\) | ✓ 1 | |
| A1 for \(x=16\) and \(x=-1\) | ✓ 1 | |
| B1 for 16 km/h, with \(-1\) rejected because a speed cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 for \(60\div(16-4)\) | ✓ 1 | |
| A1 for 5 (hours) | ✓ 1 | |
Common error: The upstream speed is the still-water speed minus the current; the downstream speed is the still-water speed plus the current.
A cyclist rides 60 km at an average speed of x km/h. She then rides a further 60 km at an average speed of (x − 5) km/h. The total time for the two stages is 7 hours.
Show that \(7x^2 - 155x + 300 = 0\).
Solve \(7x^2 - 155x + 300 = 0\) and state the value of x that fits the situation.
Work out her average speed in the second stage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for \(\frac{60}{x} + \frac{60}{x - 5} = 7\) | ✓ 1 | |
| M1 for multiplying through by \(x(x - 5)\): \(60(x - 5) + 60x = 7x(x - 5)\) | ✓ 1 | |
| A1 for expanding to \(120x - 300 = 7x^2 - 35x\) and rearranging to the required result | ✓ 1 | |
| Part (b) | ||
| M1 for \((7x - 15)(x - 20)\) or a correct use of the quadratic formula | ✓ 1 | |
| A1 for \(x = 20\) and \(x = \frac{15}{7}\) | ✓ 1 | |
| B1 for rejecting \(\frac{15}{7}\) because then \(x - 5\) would be negative; so \(x = 20\) | ✓ 1 | |
| Part (c) | ||
| B1 for 15 km/h | ✓ 1 | |
Exam tip: In a time-from-speed problem, write each stage as distance ÷ speed before combining the fractions.
A security camera covers a circular area. Taking the camera as the origin, with units in metres, the edge of the area is the circle \(x^2 + y^2 = 25\). A straight fence follows the line \(y = x + 1\). The fence crosses the edge of the covered area at two points.
Show that \(x^2 + x - 12 = 0\) at the points where the fence crosses the edge.
Solve \(x^2 + x - 12 = 0\).
Find the coordinates of the two points.
Work out the distance between the two points. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for substituting \(y = x + 1\) to get \(x^2 + (x + 1)^2 = 25\) | ✓ 1 | |
| M1 for expanding: \(2x^2 + 2x + 1 = 25\) | ✓ 1 | |
| A1 for dividing by 2 and rearranging to \(x^2 + x - 12 = 0\) | ✓ 1 | |
| Part (b) | ||
| M1 for \((x + 4)(x - 3)\) or a correct use of the quadratic formula | ✓ 1 | |
| A1 for \(x = 3\) and \(x = -4\) | ✓ 1 | |
| Part (c) | ||
| M1 for substituting both x values into \(y = x + 1\) | ✓ 1 | |
| A1 for (3, 4) and (−4, −3) | ✓ 1 | |
| Part (d) | ||
| M1 for \(\sqrt{(3 - (-4))^2 + (4 - (-3))^2}\) (ECF from (c)) | ✓ 1 | |
| A1 for 9.90 m | ✓ 1 | |
Common error: After substituting a line into a circle, remember to find the y value for every x value; each solution gives one point.
Answer both parts.
Simplify fully \(\frac{x^2 - 9}{x^2 + x - 12}\).
Hence solve \(\frac{x^2 - 9}{x^2 + x - 12} = \frac{2}{3}\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for factorising the numerator: \((x - 3)(x + 3)\) | ✓ 1 | |
| M1 for factorising the denominator: \((x + 4)(x - 3)\) | ✓ 1 | |
| A1 for \(\frac{x + 3}{x + 4}\) | ✓ 1 | |
| Part (b) | ||
| M1 for \(3(x + 3) = 2(x + 4)\) (ECF from (a)) | ✓ 1 | |
| A1 for \(x = -1\) | ✓ 1 | |
Common error: Cancel only common factors, never common terms: you cannot cancel the 3s in \(x + 3\) with those in \(x - 3\).
A game designer models the net gain in a player's points with the expression 2x2 − x − 15, where x is a whole number. A bonus is not triggered when 2x2 − x − 15 ≤ 0.
Factorise 2x2 − x − 15.
Solve 2x2 − x − 15 ≤ 0.
Work out how many whole-number values of x satisfy the inequality in (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (2x + a)(x + b) with ab = −15 or equivalent attempt | ✓ 1 | |
| A1 (2x + 5)(x − 3) | ✓ 1 | |
| Part (b) | ||
| M1 critical values x = −2.5 and x = 3 (ECF from (a)) | ✓ 1 | |
| M1 sketch or sign analysis showing the expression is negative between the roots | ✓ 1 | |
| A1 −2.5 ≤ x ≤ 3 | ✓ 1 | |
| Part (c) | ||
| B1 6 (the values −2, −1, 0, 1, 2, 3) | ✓ 1 | |
Common error: For a quadratic inequality with a positive x² coefficient, ≤ 0 gives the interval between the roots, not the two outer regions.
In a game, the efficiency rating E of a character after t minutes is given by E = (3t + 2)/(t − 4), for t > 4.
Rearrange the formula to make t the subject.
Work out the time when E = 7.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 E(t − 4) = 3t + 2 | ✓ 1 | |
| M1 expands and collects the t terms, Et − 3t = 4E + 2 | ✓ 1 | |
| M1 factorises, t(E − 3) = 4E + 2 | ✓ 1 | |
| A1 t = (4E + 2)/(E − 3) | ✓ 1 | |
| Part (b) | ||
| A1 7.5 minutes (ECF from (a)) | ✓ 1 | |
Common error: When the subject appears in two places, collect all terms containing it on one side and factorise before dividing.
The signal strength S of a gaming router is inversely proportional to the square of the distance d metres from the router. When d = 4, S = 36.
Find a formula for S in terms of d.
Work out S when d = 6.
Work out the distance at which S = 9.
The distance from the router is increased by 25%. Work out the percentage decrease in signal strength.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 S = k/d2 with 36 = k/16 | ✓ 1 | |
| A1 S = 576/d2 | ✓ 1 | |
| Part (b) | ||
| A1 16 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 9 = 576/d2 so d2 = 64 | ✓ 1 | |
| A1 8 m | ✓ 1 | |
| Part (d) | ||
| M1 new S = k/(1.25d)2 = 0.64 × original (or equivalent numerical check) | ✓ 1 | |
| A1 36% | ✓ 1 | |
Exam tip: For a percentage change in an inverse square relationship, use a multiplier: distance × 1.25 gives strength × 1/1.25² = 0.64.
The maximum load W kN that a beam supports is directly proportional to the square of its thickness t m and inversely proportional to its length L m. A beam with t = 0.20 and L = 4.0 supports 15 kN.
Show that W = 1500t2/L.
Find the maximum load for a beam with t = 0.30 and L = 6.0.
The thickness of a beam is increased by 20% and its length is unchanged. Work out the percentage increase in the maximum load.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 W = kt2/L | ✓ 1 | |
| M1 15 = k × 0.202/4.0 = 0.01k | ✓ 1 | |
| A1 k = 1500, giving the stated formula | ✓ 1 | |
| Part (b) | ||
| M1 1500 × 0.302/6.0 | ✓ 1 | |
| A1 22.5 kN | ✓ 1 | |
| Part (c) | ||
| M1 factor 1.22 for the thickness squared | ✓ 1 | |
| A1 1.44 | ✓ 1 | |
| A1 44% increase | ✓ 1 | |
Common error: A 20% increase in a quantity that is squared does not give a 40% increase; the multiplier 1.2 must itself be squared.
A mission analyst models a ratio R = (2x2 + 5x − 12)/(x2 − 16) for x > 4.
Factorise 2x2 + 5x − 12.
Show that R simplifies to (2x − 3)/(x − 4).
Find the value of x for which R = 3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (2x + a)(x + b) with ab = −12 | ✓ 1 | |
| A1 (2x − 3)(x + 4) | ✓ 1 | |
| Part (b) | ||
| M1 x2 − 16 = (x − 4)(x + 4) | ✓ 1 | |
| A1 cancels the common factor (x + 4) to reach the given result | ✓ 1 | |
| Part (c) | ||
| M1 (2x − 3) = 3(x − 4) | ✓ 1 | |
| M1 2x − 3 = 3x − 12 and collects terms | ✓ 1 | |
| A1 x = 9 | ✓ 1 | |
Exam tip: Factorise numerator and denominator before cancelling; you can only cancel whole factors, never individual terms.
A bridge engineer uses the ratio R = (3x + 2)/(x − 4), where x > 4 is a distance in metres.
Make x the subject of the formula.
Find x when R = 5.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies both sides by (x − 4): R(x − 4) = 3x + 2 | ✓ 1 | |
| M1 expands: Rx − 4R = 3x + 2 | ✓ 1 | |
| M1 collects x terms on one side and factorises: x(R − 3) = 4R + 2 | ✓ 1 | |
| A1 x = (4R + 2)/(R − 3) | ✓ 1 | |
| Part (b) | ||
| A1 (4 × 5 + 2)/(5 − 3) = 11 (m) | ✓ 1 | |
Common error: When the subject appears in two places, collect all the terms containing it on one side and factorise before dividing.
A rectangular storage plot beside a railway track has width x metres and length (x + 6) metres. Its area is 18 m2.
Show that x2 + 6x − 18 = 0.
Solve x2 + 6x − 18 = 0. Give both solutions correct to 2 decimal places.
Work out the length of the plot, giving your answer to 2 decimal places.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 6) = 18 and rearranges to the given equation | ✓ 1 | |
| Part (b) | ||
| M1 substitutes a = 1, b = 6, c = −18 correctly into the quadratic formula | ✓ 1 | |
| A1 x = (−6 ± √108)/2 | ✓ 1 | |
| A1 x = 2.20 and x = −8.20 | ✓ 1 | |
| Part (c) | ||
| B1 rejects −8.20 as a width cannot be negative | ✓ 1 | |
| A1 length = 2.196… + 6 = 8.20 m | ✓ 1 | |
Exam tip: Keep the full calculator value in the formula until the end and round only the final answers.
A train completes a 120 km journey at an average speed of v km/h. If its average speed were 10 km/h greater, the journey would take 0.4 hours less.
Show that v2 + 10v − 3000 = 0.
Solve the equation to find the original average speed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 times 120/v and 120/(v + 10) hours | ✓ 1 | |
| M1 120/v − 120/(v + 10) = 0.4 | ✓ 1 | |
| M1 multiplies by v(v + 10): 120(v + 10) − 120v = 0.4v(v + 10) | ✓ 1 | |
| A1 1200 = 0.4v2 + 4v, leading to the given equation | ✓ 1 | |
| Part (b) | ||
| M1 (v + 60)(v − 50) = 0 or use of the quadratic formula | ✓ 1 | |
| A1 v = 50 or v = −60 | ✓ 1 | |
| A1 50 km/h, as a speed cannot be negative | ✓ 1 | |
Common error: Set up the time equation as 'time at lower speed − time at higher speed = time saved', and then multiply every term by the common denominator.
In a planetary system, the square of the orbital period T days of a planet is directly proportional to the cube of its orbital distance r, measured in units of 107 km. Planet A has r = 4.0 and T = 20.
Find the constant k in T2 = kr3.
Planet B has r = 9.0. Find its orbital period.
Planet C has an orbital period of 100 days. Find its orbital distance, correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 T2 = kr3 | ✓ 1 | |
| M1 202 = k × 43, so 400 = 64k | ✓ 1 | |
| A1 k = 6.25 | ✓ 1 | |
| Part (b) | ||
| M1 T2 = 6.25 × 93 = 4556.25 | ✓ 1 | |
| A1 T = 67.5 days | ✓ 1 | |
| Part (c) | ||
| M1 r3 = 1002 ÷ 6.25 = 1600 | ✓ 1 | |
| A1 r = 11.7 (× 107 km) | ✓ 1 | |
Common error: The proportion involves a square and a cube: remember to take a square root for T and a cube root for r.
A railway track is shown on a map by the straight line \(y = x + 1\). The station is at the origin O. A restricted zone around the station has boundary the circle \(x^2 + y^2 = 25\). Distances are in km.
Find the coordinates of the two points where the track crosses the boundary of the restricted zone.
Work out the distance between these two points. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for substituting y = x + 1: x² + (x + 1)² = 25 | ✓ 1 | |
| M1 for expanding and rearranging: x² + x − 12 = 0 | ✓ 1 | |
| M1 for factorising or using the formula: (x + 4)(x − 3) = 0 | ✓ 1 | |
| A1 for x = 3 and x = −4 | ✓ 1 | |
| A1 for (3, 4) and (−4, −3) | ✓ 1 | |
| Part (b) | ||
| M1 for √((3 − (−4))² + (4 − (−3))²) | ✓ 1 | |
| A1 for 9.90 km (exact value 7√2) | ✓ 1 | |
Exam tip: To find where a line meets a curve, substitute the line's equation into the curve's equation, then solve. Find both coordinates of each point.
The brightness B of a star, measured by a telescope in arbitrary units, is inversely proportional to the square of its distance d light-years from the telescope. When d = 3, B = 40.
Find a formula for B in terms of d.
Find B when d = 12.
Find d when B = 0.9.
The distance of the star from the telescope is increased by 50%. Find the percentage decrease in B. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for B = k/d² | ✓ 1 | |
| M1 for 40 = k/9, so k = 360 | ✓ 1 | |
| A1 for B = 360/d² | ✓ 1 | |
| Part (b) | ||
| M1 for 360 ÷ 12² | ✓ 1 | |
| A1 for 2.5 | ✓ 1 | |
| Part (c) | ||
| M1 for 0.9 = 360/d² | ✓ 1 | |
| M1 for d² = 400 | ✓ 1 | |
| A1 for d = 20 (the negative root is rejected as a distance cannot be negative) | ✓ 1 | |
| Part (d) | ||
| M1 for new distance 1.5d, so new B = 360/(1.5d)² | ✓ 1 | |
| M1 for new B = (1/2.25) × old B ≈ 0.444 × old B | ✓ 1 | |
| A1 for a decrease of 55.6% | ✓ 1 | |
Common error: With inverse square proportion, multiplying the distance by 1.5 divides the brightness by 1.5², not by 1.5.
A freight train travels 150 km at an average speed of x km/h and then 100 km at an average speed of \((x + 20)\) km/h. The total time for the journey is 3.75 hours.
Show that \(3x^2 - 140x - 2400 = 0\).
Solve \(3x^2 - 140x - 2400 = 0\) and state the value of x that fits the situation.
Find the time taken for the 100 km section. Give your answer in minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 for 150/x + 100/(x + 20) = 3.75 | ✓ 1 | |
| M1 for multiplying through by x(x + 20): 150(x + 20) + 100x = 3.75x(x + 20) | ✓ 1 | |
| M1 for expanding: 250x + 3000 = 3.75x² + 75x | ✓ 1 | |
| A1 for rearranging to 3.75x² − 175x − 3000 = 0 and dividing by 1.25 to give 3x² − 140x − 2400 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 for factorising: (3x + 40)(x − 60) | ✓ 1 | |
| A1 for x = 60 and x = −40/3 | ✓ 1 | |
| B1 for x = 60, since a speed must be positive | ✓ 1 | |
| Part (c) | ||
| M1 for 100 ÷ (60 + 20) | ✓ 1 | |
| A1 for 75 minutes | ✓ 1 | |
Exam tip: In a 'show that' question every step must appear; do not skip the line where you multiply through by the denominators.
A circle has equation \(x^2 + y^2 = 25\). A straight line has equation \(y = x + 1\). They meet at two points, \(P\) and \(Q\).
Find the coordinates of \(P\) and \(Q\).
Work out the distance \(PQ\). Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitute \(y = x + 1\) into \(x^2 + y^2 = 25\) | ✓ 1 | |
| A1 \(2x^2 + 2x - 24 = 0\) or \(x^2 + x - 12 = 0\) | ✓ 1 | |
| M1 solve the quadratic by factorising, \((x + 4)(x - 3) = 0\), or by formula | ✓ 1 | |
| A1 \(x = 3\) and \(x = -4\) | ✓ 1 | |
| A1 \((3, 4)\) and \((-4, -3)\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{(3 - (-4))^2 + (4 - (-3))^2}\) or \(\sqrt{98}\) | ✓ 1 | |
| A1 9.90 | ✓ 1 | |
Common error: After finding the \(x\)-values, substitute into the linear equation (not the circle) to get \(y\), or you may create extra false solutions.
This question is about inequalities in a model of a library's weekly budget.
Solve \(2x^2 - x - 15 < 0\).
Solve \(3x - 2 > 1\).
Find the integer values of \(x\) that satisfy both inequalities. ECF from (a) and (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 factorise \((2x + 5)(x - 3)\) or use the quadratic formula | ✓ 1 | |
| A1 critical values \(x = -\frac{5}{2}\) and \(x = 3\) | ✓ 1 | |
| M1 quadratic opening upwards, so negative between the roots | ✓ 1 | |
| A1 \(-\frac{5}{2} < x < 3\) | ✓ 1 | |
| Part (b) | ||
| A1 \(x > 1\) | ✓ 1 | |
| Part (c) | ||
| M1 identifies the overlap \(1 < x < 3\) | ✓ 1 | |
| A1 \(x = 2\) | ✓ 1 | |
Common error: A quadratic inequality is solved between or outside the roots, not by treating it like a linear inequality.
The cost, £\(y\), per swimmer of hiring a coach for a swimming club trip is given by \(y = \frac{3x + 2}{x - 4}\), where \(x\) is the number of swimmers and \(x > 4\).
Make \(x\) the subject of the formula.
The cost is £5 per swimmer. Work out the number of swimmers.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiply by \((x - 4)\) and expand: \(yx - 4y = 3x + 2\) | ✓ 1 | |
| M1 collect the \(x\) terms on one side: \(yx - 3x = 4y + 2\) | ✓ 1 | |
| M1 factorise: \(x(y - 3) = 4y + 2\) | ✓ 1 | |
| A1 \(x = \frac{4y + 2}{y - 3}\) | ✓ 1 | |
| Part (b) | ||
| A1 11 (ECF from (a)) | ✓ 1 | |
Exam tip: When the subject appears twice, collect those terms on one side and factorise.
This question is about the inequalities \(5x - 3 \geq 2x + 9\) and \(x^2 - 2x - 24 < 0\) in a coffee roasting model.
Solve \(5x - 3 \geq 2x + 9\).
Solve \(x^2 - 2x - 24 < 0\).
Find the sum of the integer values of \(x\) that satisfy both inequalities. ECF from (a) and (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(3x \geq 12\) | ✓ 1 | |
| A1 \(x \geq 4\) | ✓ 1 | |
| Part (b) | ||
| M1 \((x - 6)(x + 4)\) or the quadratic formula | ✓ 1 | |
| A1 critical values \(x = 6\) and \(x = -4\) | ✓ 1 | |
| A1 \(-4 < x < 6\) | ✓ 1 | |
| Part (c) | ||
| M1 overlap \(4 \leq x < 6\) identified, integers 4 and 5 | ✓ 1 | |
| A1 9 | ✓ 1 | |
Common error: Check the ends: 4 is included because of \(\geq\) but 6 is not included because of \(<\).
Algebraic work on a space probe's signal model.
Simplify fully \(\frac{x^2 - 9}{2x^2 + 5x - 3}\).
Solve \(\frac{3}{x + 1} + \frac{2}{x - 2} = 1\). Give the larger solution correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 numerator factorised \((x - 3)(x + 3)\) | ✓ 1 | |
| M1 denominator factorised \((2x - 1)(x + 3)\) | ✓ 1 | |
| A1 \(\frac{x - 3}{2x - 1}\) | ✓ 1 | |
| Part (b) | ||
| M1 multiply through by \((x + 1)(x - 2)\) or use a common denominator | ✓ 1 | |
| A1 \(3(x - 2) + 2(x + 1) = (x + 1)(x - 2)\) or \(5x - 4 = x^2 - x - 2\) | ✓ 1 | |
| A1 \(x^2 - 6x + 2 = 0\) | ✓ 1 | |
| M1 quadratic formula with \(a = 1\), \(b = -6\), \(c = 2\) | ✓ 1 | |
| A1 \(x = 5.65\) (the other solution is \(x = 0.35\)) | ✓ 1 | |
Exam tip: Factorise before cancelling; you can only cancel whole brackets, never single terms.
A model of a satellite orbit says that \(T^2\) is directly proportional to \(r^3\), where \(T\) hours is the time for one orbit and \(r\) is the orbital radius in units of 104 km. In the model, \(T = 4\) when \(r = 2\).
Show that \(T^2 = 2r^3\).
Work out \(T\) when \(r = 5\). Give your answer correct to 3 significant figures.
Work out \(r\) when \(T = 9\). Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(T^2 = kr^3\) with substitution \(16 = 8k\), giving \(k = 2\) | ✓ 1 | |
| Part (b) | ||
| M1 \(T^2 = 2 \times 125 = 250\) | ✓ 1 | |
| A1 15.8 | ✓ 1 | |
| Part (c) | ||
| M1 \(81 = 2r^3\) so \(r^3 = 40.5\) | ✓ 1 | |
| M1 cube root of 40.5 | ✓ 1 | |
| A1 3.43 | ✓ 1 | |
Common error: Take the square or cube root only at the end, after rearranging for the squared or cubed quantity.
Answer both parts of this question on algebraic fractions.
Write \(\dfrac{3}{x+2} + \dfrac{2}{x-1}\) as a single fraction in its simplest form.
Hence, or otherwise, solve \(\dfrac{3}{x+2} + \dfrac{2}{x-1} = 1\). Give your solutions correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 common denominator \((x+2)(x-1)\) | ✓ 1 | |
| M1 numerator \(3(x-1) + 2(x+2)\) | ✓ 1 | |
| A1 \(\dfrac{5x+1}{(x+2)(x-1)}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(5x + 1 = (x+2)(x-1)\) | ✓ 1 | |
| A1 \(x^2 - 4x - 3 = 0\) | ✓ 1 | |
| M1 correct use of the quadratic formula or completing the square, e.g. \(x = 2 \pm \sqrt{7}\) | ✓ 1 | |
| A1 \(x = 4.65\) and \(x = -0.646\) | ✓ 1 | |
Common error: Do not add the denominators; write both fractions over a common denominator first.
The gear ratio \(y\) of a robot's joint is given by \(y = \dfrac{3x+2}{x-4}\), where \(x > 4\).
Make \(x\) the subject of the formula. You must show each step of your working.
Use your formula to find the value of \(x\) when \(y = 5\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(y(x-4) = 3x + 2\) | ✓ 1 | |
| M1 expands and collects the \(x\) terms on one side, e.g. \(yx - 3x = 4y + 2\) | ✓ 1 | |
| M1 factorises, \(x(y-3) = 4y + 2\) | ✓ 1 | |
| A1 \(x = \dfrac{4y+2}{y-3}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\dfrac{4(5)+2}{5-3}\) (ECF from (a)) | ✓ 1 | |
| A1 11 | ✓ 1 | |
Exam tip: When the subject appears in a fraction and in the numerator, multiply out first, then gather every term in the subject on one side so you can factorise.
A fishing company's profit \(P\) (in thousands of euros) from catching \(x\) hundred tonnes of fish in a season is modelled by \(P = 12x - x^2 - 20\).
Solve the inequality \(12x - x^2 - 20 > 0\) to find the values of \(x\) for which the company makes a profit.
Write \(P = 12x - x^2 - 20\) in the form \(a - (x-b)^2\).
Use your answer to part (b) to state the greatest profit and the number of tonnes of fish caught to achieve it.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 rearranges to \(x^2 - 12x + 20 < 0\) and factorises or uses the formula, e.g. \((x-2)(x-10)\) | ✓ 1 | |
| A1 critical values 2 and 10 | ✓ 1 | |
| A1 \(2 < x < 10\) | ✓ 1 | |
| Part (b) | ||
| M1 attempt to complete the square, e.g. \(-[(x-6)^2 - 16]\) | ✓ 1 | |
| A1 \(P = 16 - (x-6)^2\) | ✓ 1 | |
| Part (c) | ||
| B1 greatest profit €16 000 | ✓ 1 | |
| B1 600 tonnes | ✓ 1 | |
Exam tip: In the form \(a-(x-b)^2\) the greatest value is \(a\), and it occurs when \(x = b\).
An irrigation pipe runs along the line \(y=2x-1\). A stream follows the curve \(y=x^2-4x+4\). Coordinates are in metres.
Find the coordinates of the two points where the pipe meets the stream.
Work out the distance between the two points. Give your answer in the form \(a\sqrt{b}\) metres.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 sets \(x^2-4x+4=2x-1\) | ✓ 1 | |
| A1 \(x^2-6x+5=0\) | ✓ 1 | |
| M1 \((x-1)(x-5)=0\) | ✓ 1 | |
| A1 \((1, 1)\) and \((5, 9)\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{4^2+8^2}=\sqrt{80}\) | ✓ 1 | |
| A1 \(4\sqrt{5}\) m (8.94 m) | ✓ 1 | |
Common error: After finding the \(x\) values, substitute into the linear equation to get \(y\): the quadratic gives extra, wrong values.
Work with algebraic fractions.
Simplify fully \(\frac{x^2-9}{x^2+x-12}\).
Show that \(\frac{3}{x+1}+\frac{2}{x-2}=2\) can be written as \(2x^2-7x=0\).
Hence find the positive solution of \(\frac{3}{x+1}+\frac{2}{x-2}=2\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 factorises numerator \((x-3)(x+3)\) | ✓ 1 | |
| M1 factorises denominator \((x+4)(x-3)\) | ✓ 1 | |
| A1 \(\frac{x+3}{x+4}\) | ✓ 1 | |
| Part (b) | ||
| M1 combines over the common denominator: \(\frac{3(x-2)+2(x+1)}{(x+1)(x-2)}\) (numerator \(5x-4\)) | ✓ 1 | |
| M1 multiplies out: \(5x-4=2(x+1)(x-2)=2x^2-2x-4\) | ✓ 1 | |
| A1 rearranges to \(2x^2-7x=0\) | ✓ 1 | |
| Part (c) | ||
| M1 \(x(2x-7)=0\) | ✓ 1 | |
| A1 \(x=3.5\) | ✓ 1 | |
Exam tip: In a "show that" question every line of working must appear; do not jump from the fractions to the final result.
The drag force \(F\) newtons on an aircraft is directly proportional to the square of its speed \(v\) m/s. When \(v=60\), \(F=7200\).
Find a formula for \(F\) in terms of \(v\).
Work out the speed when \(F=20\,000\).
The speed increases by 20%. Work out the percentage increase in \(F\).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(F=kv^2\) and \(7200=k\times60^2\) | ✓ 1 | |
| A1 \(F=2v^2\) | ✓ 1 | |
| Part (b) | ||
| M1 \(v^2=10\,000\) | ✓ 1 | |
| A1 100 m/s | ✓ 1 | |
| Part (c) | ||
| M1 new speed \(1.2v\) | ✓ 1 | |
| M1 new force \(2(1.2v)^2=1.44\times2v^2\) (a factor of 1.44) | ✓ 1 | |
| A1 44% | ✓ 1 | |
Common error: A 20% increase in speed does not give a 20% increase in force: the factor 1.2 must be squared first.
Solve the equation 3/(x + 1) + 2/(x − 2) = 1.
Show that the equation can be written as x2 − 6x + 2 = 0.
Solve x2 − 6x + 2 = 0. Give your answers correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies every term by (x + 1)(x − 2), or writes the fractions over a common denominator | ✓ 1 | |
| M1 3(x − 2) + 2(x + 1) = (x + 1)(x − 2) | ✓ 1 | |
| A1 5x − 4 = x2 − x − 2 | ✓ 1 | |
| A1 rearranges to x2 − 6x + 2 = 0 with no errors | ✓ 1 | |
| Part (b) | ||
| M1 substitutes a = 1, b = −6, c = 2 correctly into the quadratic formula | ✓ 1 | |
| A1 x = 5.65 | ✓ 1 | |
| A1 x = 0.354 | ✓ 1 | |
Common error: Multiply every term, including the right-hand side, by the common denominator; leaving the 1 unmultiplied is a frequent source of lost marks.
Algebraic fractions.
Factorise 2x2 + x − 15.
Simplify fully (2x2 + x − 15)/(x2 − 9).
Solve (2x2 + x − 15)/(x2 − 9) = 3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (2x ± 5)(x ± 3) | ✓ 1 | |
| A1 (2x − 5)(x + 3) | ✓ 1 | |
| Part (b) | ||
| B1 x2 − 9 = (x − 3)(x + 3) | ✓ 1 | |
| A1 (2x − 5)/(x − 3) | ✓ 1 | |
| Part (c) | ||
| M1 2x − 5 = 3(x − 3) (ECF from (b)) | ✓ 1 | |
| A1 x = 4 | ✓ 1 | |
Exam tip: Factorise both the top and the bottom of an algebraic fraction before cancelling; only common factors can be cancelled.
A café sells a fruit smoothie at a price of £x each, where 0 < x < 10. The daily profit, £P, is modelled by P = −4x2 + 40x − 64.
Show that P > 0 when x2 − 10x + 16 < 0.
Find the range of prices for which the café makes a profit.
Find the price that gives the greatest daily profit and work out that profit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 divides by −4 or factors out −4: P = −4(x2 − 10x + 16) | ✓ 1 | |
| A1 reverses the inequality sign to give x2 − 10x + 16 < 0 | ✓ 1 | |
| Part (b) | ||
| M1 solves x2 − 10x + 16 = 0 to get critical values 2 and 8 | ✓ 1 | |
| M1 uses a U-shaped graph or a test value to show the expression is negative between the roots | ✓ 1 | |
| A1 2 < x < 8 | ✓ 1 | |
| Part (c) | ||
| M1 uses the midpoint of the roots, or completes the square, or −b/2a | ✓ 1 | |
| A1 price = £5 | ✓ 1 | |
| A1 greatest profit = £36 | ✓ 1 | |
Common error: Dividing or multiplying an inequality by a negative number reverses the inequality sign; forgetting this gives the wrong range.
The engine power P kW needed to drive a research boat at speed v knots is directly proportional to v3. At 8 knots the power needed is 96 kW.
Find a formula for P in terms of v.
Work out the power needed at 12 knots.
The engine can deliver at most 500 kW. Work out the greatest speed of the boat, correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = kv3 | ✓ 1 | |
| M1 96 = k × 83, so k = 96/512 | ✓ 1 | |
| A1 P = 0.1875v3 (or 3v3/16) | ✓ 1 | |
| Part (b) | ||
| M1 0.1875 × 123 (ECF from (a)) | ✓ 1 | |
| A1 324 kW | ✓ 1 | |
| Part (c) | ||
| M1 500 = 0.1875v3 | ✓ 1 | |
| M1 v3 = 2666.67... and cube root | ✓ 1 | |
| A1 13.9 knots | ✓ 1 | |
Exam tip: Use the unrounded value of k, or the fraction 3/16, to avoid rounding errors in the final answer.
A rectangular roof panel has width x metres and length (x + 3) metres. The area of the panel must be less than 40 m2.
Show that x2 + 3x − 40 < 0.
Solve x2 + 3x − 40 < 0.
The width of each panel is a whole number of metres. List all the possible widths.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 area = x(x + 3) | ✓ 1 | |
| A1 x(x + 3) < 40 rearranged to x2 + 3x − 40 < 0 | ✓ 1 | |
| Part (b) | ||
| M1 attempt to factorise or solve, e.g. (x + 8)(x − 5) | ✓ 1 | |
| A1 critical values −8 and 5 | ✓ 1 | |
| M1 use of the shape of the graph (positive x2, so negative between the roots) | ✓ 1 | |
| A1 −8 < x < 5 | ✓ 1 | |
| Part (c) | ||
| B1 widths must be positive, so x > 0 used | ✓ 1 | |
| A1 1, 2, 3, 4 | ✓ 1 | |
Exam tip: For a quadratic inequality, find the critical values first, then decide which region satisfies it, and use the context to cut out impossible values.
A formula used in a model of tides is y = (3x + 2) ÷ (x − 4).
Make x the subject of the formula.
Simplify fully (x2 − 9) ÷ (2x2 + 5x − 3).
Use your answer to part (a) to find the value of x when y = 5.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiply both sides: y(x − 4) = 3x + 2 | ✓ 1 | |
| M1 expand: xy − 4y = 3x + 2 | ✓ 1 | |
| M1 collect x terms: xy − 3x = 2 + 4y | ✓ 1 | |
| A1 x = (4y + 2) ÷ (y − 3) | ✓ 1 | |
| Part (b) | ||
| M1 numerator (x − 3)(x + 3) | ✓ 1 | |
| M1 denominator (2x − 1)(x + 3) | ✓ 1 | |
| A1 (x − 3) ÷ (2x − 1) | ✓ 1 | |
| Part (c) | ||
| M1 (4 × 5 + 2) ÷ (5 − 3) (ECF from (a)) | ✓ 1 | |
| A1 11 | ✓ 1 | |
Exam tip: When the subject appears twice, gather all the terms containing it on one side, then factorise.
This question is about algebraic fractions.
Write \(\frac{3}{x + 2} + \frac{2}{x - 1}\) as a single fraction in its simplest form.
Hence find the positive solution of \(\frac{3}{x + 2} + \frac{2}{x - 1} = 1\). Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 common denominator \((x + 2)(x - 1)\) | ✓ 1 | |
| M1 \(3(x - 1) + 2(x + 2)\) | ✓ 1 | |
| A1 \(\frac{5x + 1}{(x + 2)(x - 1)}\) | ✓ 1 | |
| Part (b) | ||
| M1 \(5x + 1 = (x + 2)(x - 1)\) (ECF from (a)) | ✓ 1 | |
| A1 \(x^2 - 4x - 3 = 0\) | ✓ 1 | |
| M1 correct use of the quadratic formula or completing the square | ✓ 1 | |
| A1 \(x = 4.65\) (the other solution \(-0.646\) is rejected) | ✓ 1 | |
Common error: Do not add the denominators; fractions must be written over a common denominator before the numerators are added.
On a map grid with units of 1 km, a straight railway line has equation y = x + 1. A river follows the curve y = x2 − 2x − 3. The railway line crosses the river at two points, P and Q.
Work out the coordinates of P and Q.
Work out the distance PQ. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 \(x + 1 = x^2 - 2x - 3\) | ✓ 1 | |
| A1 \(x^2 - 3x - 4 = 0\) | ✓ 1 | |
| M1 factorise or use the formula, \((x - 4)(x + 1) = 0\) | ✓ 1 | |
| A1 \(x = 4\) and \(x = -1\) | ✓ 1 | |
| A1 \((4, 5)\) and \((-1, 0)\) | ✓ 1 | |
| Part (b) | ||
| M1 \(\sqrt{5^2 + 5^2}\) (ECF from (a)) | ✓ 1 | |
| A1 7.07 km | ✓ 1 | |
Exam tip: To find where a line meets a curve, set the two expressions for y equal and solve; remember to find the y-value for each x-value.
Solve the simultaneous equations
y = x + 1
x2 + y2 = 25
Find all pairs of values (x, y) that satisfy both equations.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitute to get x2 + (x + 1)2 = 25 | ✓ 1 | |
| A1 x2 + x − 12 = 0 (or 2x2 + 2x − 24 = 0) | ✓ 1 | |
| M1 attempt to solve the quadratic | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
Common error: (x + 1)² is not x² + 1; the middle term 2x must be included.
The downforce D newtons on a Formula 1 car is directly proportional to the square of its speed v m/s.
At 50 m/s the downforce is 6000 N.
Calculate the downforce at 80 m/s.
Calculate the speed at which the downforce is 9000 N. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 D = kv2 with 6000 = k × 502 | ✓ 1 | |
| A1 k = 2.4 | ✓ 1 | |
| A1 15 360 N | ✓ 1 | |
| Part (b) | ||
| M1 9000 = 2.4v2 (ECF from (a)) | ✓ 1 | |
| M1 v = √3750 | ✓ 1 | |
| A1 61.2 m/s | ✓ 1 | |
Common error: Squaring the speed is needed in both directions: take a square root when finding v.
(a) and (b) concern the expression (2x2 − 8) / (x2 + x − 6).
Simplify (2x2 − 8) / (x2 + x − 6) fully.
Hence solve (2x2 − 8) / (x2 + x − 6) = 3/2.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 numerator factorised, 2(x − 2)(x + 2) | ✓ 1 | |
| M1 denominator factorised, (x + 3)(x − 2) | ✓ 1 | |
| A1 2(x + 2)/(x + 3) | ✓ 1 | |
| Part (b) | ||
| M1 2(x + 2)/(x + 3) = 3/2 (ECF from (a)) | ✓ 1 | |
| M1 cross-multiply: 4x + 8 = 3x + 9 | ✓ 1 | |
| A1 x = 1 | ✓ 1 | |
Exam tip: Factorise the numerator and denominator before cancelling; you can only cancel common factors, not common terms.
A Formula 1 pit garage has a rectangular floor. The width of the floor is x metres and the length is (x + 4) metres. The floor area is 30 m2.
Show that x2 + 4x − 30 = 0
Solve x2 + 4x − 30 = 0. Give both solutions correct to 3 significant figures, then state which solution gives the width of the garage.
Edging strip is fitted all the way round the floor and costs £6.50 per metre. Work out the cost of the strip. Use the unrounded value of x in your working and give your answer to the nearest penny.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 x(x + 4) = 30 | ✓ 1 | |
| A1 x2 + 4x = 30 leading to x2 + 4x − 30 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the quadratic formula (or completing the square: (x + 2)2 = 34) | ✓ 1 | |
| A1 x = 3.83 and x = −7.83 | ✓ 1 | |
| B1 width = 3.83 m because a width cannot be negative | ✓ 1 | |
| Part (c) | ||
| M1 perimeter = 2(x + x + 4) = 4x + 8 with x = −2 + √34 | ✓ 1 | |
| M1 perimeter × 6.50 (perimeter = 23.32...) | ✓ 1 | |
| A1 £151.60 | ✓ 1 | |
Exam tip: Keep the exact or unrounded value on your calculator between parts so that rounding does not change the final penny.
In a hospital model, the ratio R of occupied beds to free beds after d days (d > 1) is given by
R = (3d + 2)/(d − 1)
Make d the subject of the formula.
Work out the number of days after which R = 7.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R(d − 1) = 3d + 2 | ✓ 1 | |
| M1 expand: Rd − R = 3d + 2 | ✓ 1 | |
| M1 collect the d terms on one side and factorise: d(R − 3) = R + 2 | ✓ 1 | |
| A1 d = (R + 2)/(R − 3) | ✓ 1 | |
| Part (b) | ||
| M1 substitution of 7 into their rearranged formula, e.g. 9/4 | ✓ 1 | |
| A1 2.25 days (ECF from (a)) | ✓ 1 | |
Common error: When the subject appears in more than one place, collect all the terms containing it on one side and factorise before dividing.
A Formula 1 team models tyre grip with algebraic fractions.
Simplify fully (x2 − 9)/(x2 + x − 6)
Solve 3/(x − 1) + 2/(x + 2) = 1
Give your solutions correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 numerator factorised as (x − 3)(x + 3) | ✓ 1 | |
| M1 denominator factorised as (x + 3)(x − 2) | ✓ 1 | |
| A1 (x − 3)/(x − 2) | ✓ 1 | |
| Part (b) | ||
| M1 common denominator, e.g. 3(x + 2) + 2(x − 1) = (x − 1)(x + 2) | ✓ 1 | |
| A1 x2 − 4x − 6 = 0 | ✓ 1 | |
| M1 correct use of the quadratic formula or completing the square | ✓ 1 | |
| A1 x = 5.16 and x = −1.16 | ✓ 1 | |
Exam tip: Only cancel factors, never terms: first factorise the top and bottom completely, then cancel the matching brackets.
The air resistance force F newtons on a cyclist is directly proportional to the square of the speed v m/s. When v = 6, F = 18.
Find a formula for F in terms of v.
Work out F when v = 10.
Work out v when F = 32.
The speed of the cyclist increases by 20%. Work out the percentage increase in F.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 F = kv2 | ✓ 1 | |
| M1 18 = k × 36 | ✓ 1 | |
| A1 F = 0.5v2 | ✓ 1 | |
| Part (b) | ||
| B1 50 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 32 = 0.5v2, so v2 = 64 | ✓ 1 | |
| A1 8 m/s | ✓ 1 | |
| Part (d) | ||
| M1 uses F ∝ v2 so the multiplier is 1.22 | ✓ 1 | |
| A1 1.44 | ✓ 1 | |
| A1 44% | ✓ 1 | |
Common error: Applying the 20% increase directly to F. The force scales with the square of the speed.
A data-centre modeller uses a profit function P = x2 − 2x − 15 (in £ thousand), where x is a scaled load setting that can take any real value.
Solve the inequality x2 − 2x − 15 > 0.
Simplify fully (x2 − x − 12) ÷ (x2 − 9).
Solve 3 ÷ (x + 2) + 2 ÷ (x − 1) = 1. Give your answers correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 (x − 5)(x + 3) = 0 or critical values −3 and 5 | ✓ 1 | |
| M1 correct region selected for > 0 (outside the roots) | ✓ 1 | |
| A1 x < −3 or x > 5 | ✓ 1 | |
| Part (b) | ||
| M1 numerator factorised: (x − 4)(x + 3) | ✓ 1 | |
| M1 denominator factorised: (x − 3)(x + 3) | ✓ 1 | |
| A1 (x − 4) ÷ (x − 3) | ✓ 1 | |
| Part (c) | ||
| M1 multiplies through by (x + 2)(x − 1) or uses a common denominator | ✓ 1 | |
| A1 3(x − 1) + 2(x + 2) = (x + 2)(x − 1) | ✓ 1 | |
| A1 x2 − 4x − 3 = 0 | ✓ 1 | |
| A1 x = 2 + √7 = 4.65 and x = 2 − √7 = −0.646 | ✓ 1 | |
Exam tip: For a quadratic inequality, find the roots first and then decide which region satisfies the sign.
Indices and an archaeological model.
Work out the value of (27/8)−2/3.
Solve 2x+1 × 4x−1 = 32.
An archaeologist models the difference in the number of sherds between two layers as 5x+2 − 5x. Show that this equals 24 × 5x.
The difference is 600 sherds. Hence work out the value of x.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 reciprocal taken: (8/27)2/3 | ✓ 1 | |
| M1 cube root 2/3 then squared: (2/3)2 | ✓ 1 | |
| A1 4/9 | ✓ 1 | |
| Part (b) | ||
| M1 4x−1 written as 22x−2 | ✓ 1 | |
| M1 23x−1 = 25 | ✓ 1 | |
| A1 x = 2 | ✓ 1 | |
| Part (c) | ||
| M1 5x+2 = 25 × 5x | ✓ 1 | |
| A1 25 × 5x − 5x = 24 × 5x | ✓ 1 | |
| Part (d) | ||
| M1 24 × 5x = 600, so 5x = 25 | ✓ 1 | |
| A1 x = 2 | ✓ 1 | |
Common error: Writing 5x⁺² as 5x + 5²; adding to an index means multiplying the powers.
In an architect's model of a roof truss, the ratio r of two lengths is given by r = (3x + 2)/(x − 4), where x > 4.
Make x the subject of the formula.
Find the value of x when r = 5.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 r(x − 4) = 3x + 2 | ✓ 1 | |
| M1 rx − 4r = 3x + 2 | ✓ 1 | |
| M1 collects the x terms on one side, x(r − 3) = 4r + 2 | ✓ 1 | |
| A1 x = (4r + 2)/(r − 3) | ✓ 1 | |
| Part (b) | ||
| M1 substitutes r = 5 into their formula from (a) | ✓ 1 | |
| A1 11 | ✓ 1 | |
Common error: When the subject appears in more than one place, expand, collect every term containing it on one side and then factorise.
A circular window has its centre at the origin and radius 5 m. Coordinates are in metres. A straight beam lies along the line with equation y = x + 1. The equation of the window frame is x2 + y2 = 25.
Find the coordinates of the two points where the beam meets the frame.
Work out the length of the beam between these two points. Give your answer correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes y = x + 1 into x2 + y2 = 25 | ✓ 1 | |
| M1 2x2 + 2x − 24 = 0, or x2 + x − 12 = 0 | ✓ 1 | |
| M1 (x + 4)(x − 3) = 0 | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
| Part (b) | ||
| M1 √(72 + 72), using their points from (a) | ✓ 1 | |
| A1 9.90 (m) | ✓ 1 | |
Exam tip: For a line and a circle, substitute the linear equation into the circle equation, then solve the quadratic and find both y values.
A wind-farm monitoring station is at the origin O of a coordinate grid, with units in kilometres. Its signal covers every point whose distance from O is at most 5 km, so the edge of the covered region has equation x2 + y2 = 25. A straight service road follows the line y = x + 1.
Show that the points where the road meets the edge of the covered region satisfy x2 + x − 12 = 0.
Solve x2 + x − 12 = 0.
Work out the coordinates of the two points where the road meets the edge of the covered region.
Work out the exact distance between the two points, in the form a√2 km.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitute y = x + 1 into x2 + y2 = 25 | ✓ 1 | |
| M1 x2 + x2 + 2x + 1 = 25 (bracket expanded) | ✓ 1 | |
| A1 2x2 + 2x − 24 = 0 then divide by 2 to reach the given equation | ✓ 1 | |
| Part (b) | ||
| M1 (x + 4)(x − 3) = 0 (or correct use of the quadratic formula) | ✓ 1 | |
| A1 x = −4 and x = 3 | ✓ 1 | |
| Part (c) | ||
| B1 (3, 4) | ✓ 1 | |
| B1 (−4, −3) | ✓ 1 | |
| Part (d) | ||
| M1 differences in coordinates of 7 and 7 (ECF from (c)) | ✓ 1 | |
| M1 √(72 + 72) | ✓ 1 | |
| A1 7√2 km | ✓ 1 | |
Common error: Substitute into the circle equation, not the other way round, and remember (x + 1)² has a middle term 2x.
In this question, x ≠ 3 and x ≠ −3.
Factorise 2x2 + 7x + 3.
Simplify (2x2 + 7x + 3) ÷ (x2 − 9).
Solve (2x + 1) ÷ (x − 3) = 5. Give your answer as a fraction.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 attempt to factorise | ✓ 1 | |
| A1 (2x + 1)(x + 3) | ✓ 1 | |
| Part (b) | ||
| M1 x2 − 9 = (x − 3)(x + 3) | ✓ 1 | |
| M1 cancel the common factor (x + 3) | ✓ 1 | |
| A1 (2x + 1) ÷ (x − 3) | ✓ 1 | |
| Part (c) | ||
| M1 2x + 1 = 5(x − 3) (ECF from (b)) | ✓ 1 | |
| M1 collect terms: 16 = 3x | ✓ 1 | |
| A1 x = 16/3 | ✓ 1 | |
Common error: Cancel only common factors, never common terms: (2x + 1) ÷ (x − 3) cannot be simplified further.
A shop's profit, £P, from selling x items in one day is modelled by P = 40x − x2 − 300, for 0 ≤ x ≤ 40.
Solve the inequality P > 0.
Work out how many whole numbers of items give a positive profit.
Find the greatest possible profit in one day.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 rearrange to x2 − 40x + 300 < 0 | ✓ 1 | |
| M1 attempt to factorise or use the quadratic formula: (x − 10)(x − 30) | ✓ 1 | |
| A1 critical values 10 and 30 | ✓ 1 | |
| A1 10 < x < 30 | ✓ 1 | |
| Part (b) | ||
| M1 integers from 11 to 29 inclusive (ECF from (a)) | ✓ 1 | |
| A1 19 | ✓ 1 | |
| Part (c) | ||
| M1 attempt to complete the square or use symmetry to find x = 20 | ✓ 1 | |
| A1 P = 100 − (x − 20)2 or P(20) evaluated | ✓ 1 | |
| A1 £100 | ✓ 1 | |
Common error: When dividing or rearranging an inequality, a negative coefficient of x² reverses the inequality sign.
The circle x2 + y2 = 25 and the straight line y = x + 1 intersect at two points, P and Q.
Solve the equations simultaneously to find the coordinates of P and Q.
Work out the distance PQ. Give your answer in the form a√2.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes y = x + 1 into the circle equation | ✓ 1 | |
| M1 x2 + (x + 1)2 = 25 expanded to 2x2 + 2x − 24 = 0 (or x2 + x − 12 = 0) | ✓ 1 | |
| M1 attempt to solve: (x + 4)(x − 3) = 0 | ✓ 1 | |
| A1 x = 3 and x = −4 | ✓ 1 | |
| A1 (3, 4) and (−4, −3) | ✓ 1 | |
| Part (b) | ||
| M1 √(72 + 72) (ECF from (a)) | ✓ 1 | |
| A1 7√2 | ✓ 1 | |
Common error: Always substitute back into the linear equation to find y; pairing the wrong x and y values gives points that are not on the line.
This question is about algebraic fractions.
Simplify fully (x2 − 9) ÷ (x2 + 5x + 6).
Write 3 ÷ (x + 1) + 2 ÷ (x − 2) as a single fraction in its simplest form.
Hence solve (5x − 4) ÷ ((x + 1)(x − 2)) = 1.
Give your answers correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 numerator factorised as (x − 3)(x + 3) | ✓ 1 | |
| M1 denominator factorised as (x + 2)(x + 3) | ✓ 1 | |
| A1 (x − 3) ÷ (x + 2) | ✓ 1 | |
| Part (b) | ||
| M1 common denominator (x + 1)(x − 2) | ✓ 1 | |
| M1 numerator 3(x − 2) + 2(x + 1) | ✓ 1 | |
| A1 (5x − 4) ÷ ((x + 1)(x − 2)) | ✓ 1 | |
| Part (c) | ||
| M1 5x − 4 = (x + 1)(x − 2) | ✓ 1 | |
| M1 rearranges to x2 − 6x + 2 = 0 | ✓ 1 | |
| M1 correct use of the quadratic formula, (6 ± √28) ÷ 2 | ✓ 1 | |
| A1 x = 5.65 and x = 0.354 | ✓ 1 | |
Exam tip: Factorise before cancelling; you can only cancel whole brackets, never single terms.
A robot moves along the straight path with equation y = 2x − 1. A sensor creates a circular restricted zone with equation x2 + y2 = 13. All distances are in metres.
Show that the x coordinates of the points where the path meets the circle satisfy 5x2 − 4x − 12 = 0.
Solve the equations simultaneously to find the coordinates of the two points where the path meets the circle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes y = 2x − 1 into x2 + y2 = 13 | ✓ 1 | |
| M1 expands (2x − 1)2 = 4x2 − 4x + 1 | ✓ 1 | |
| A1 5x2 − 4x + 1 = 13 leading to 5x2 − 4x − 12 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 (5x + 6)(x − 2) = 0, or correct use of the formula | ✓ 1 | |
| A1 x = 2 and x = −1.2 | ✓ 1 | |
| M1 substitutes into y = 2x − 1 | ✓ 1 | |
| A1 (2, 3) and (−1.2, −3.4) | ✓ 1 | |
Common error: Substitute back into the linear equation, not the circle, to find each y value; the circle gives extra incorrect pairs.
A fishery's daily profit, in hundreds of pounds, when it uses x nets is P = 12 + 5x − 2x2.
Show that the fishery makes a profit when 2x2 − 5x − 12 < 0.
Solve 2x2 − 5x − 12 < 0.
The fishery uses a whole number of nets, with at least 1 net.
Write down all the numbers of nets for which it makes a profit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P > 0, so 12 + 5x − 2x2 > 0 | ✓ 1 | |
| A1 rearranges, reversing the inequality sign, to 2x2 − 5x − 12 < 0 | ✓ 1 | |
| Part (b) | ||
| M1 (2x + 3)(x − 4), or correct use of the formula | ✓ 1 | |
| A1 critical values x = −1.5 and x = 4 | ✓ 1 | |
| A1 −1.5 < x < 4 | ✓ 1 | |
| Part (c) | ||
| M1 uses the interval from (b) | ✓ 1 | |
| A1 1, 2, 3 (ECF from (b)) | ✓ 1 | |
Exam tip: Sketch a quick "u-shaped" curve in your working: for a positive x² term, the values below zero lie between the roots.
On a site plan, 1 unit represents 1 metre. The edge of a circular fountain base has equation x2 + y2 = 34. A straight path has equation y = x + 2. The path meets the edge of the fountain base at two points, A and B.
Work out the coordinates of A and B.
Work out the distance AB. Give your answer in metres correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 substitutes: x2 + (x + 2)2 = 34 | ✓ 1 | |
| A1 2x2 + 4x − 30 = 0, or x2 + 2x − 15 = 0 | ✓ 1 | |
| M1 solves the quadratic | ✓ 1 | |
| A1 x = 3 and x = −5 | ✓ 1 | |
| A1 (3, 5) and (−5, −3) | ✓ 1 | |
| Part (b) | ||
| M1 √(82 + 82) | ✓ 1 | |
| A1 11.3 m (from 11.3137...) | ✓ 1 | |
Exam tip: When one equation is linear, substitute it into the other to leave a single quadratic, then find the second coordinate using the linear equation.
In a logistics model, y = (2x + 1)/(x − 3), where x > 3.
Make x the subject of the formula.
Use your answer to (a) to work out the value of x when y = 7.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies out the denominator: y(x − 3) = 2x + 1 | ✓ 1 | |
| M1 expands: yx − 3y = 2x + 1 | ✓ 1 | |
| M1 collects x terms and factorises: x(y − 2) = 3y + 1 | ✓ 1 | |
| A1 x = (3y + 1)/(y − 2) | ✓ 1 | |
| Part (b) | ||
| A1 4.4 (ECF from (a)) | ✓ 1 | |
Common error: Stopping at yx − 3y = 2x + 1. The subject must appear once only, so collect the x terms and factorise.
x is a number with x ≠ −1 and x ≠ 2, and
3/(x + 1) + 2/(x − 2) = 1
Show that x2 − 6x + 2 = 0.
Solve x2 − 6x + 2 = 0. Give your solutions correct to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 multiplies through by (x + 1)(x − 2), or uses a common denominator | ✓ 1 | |
| M1 3(x − 2) + 2(x + 1) = (x + 1)(x − 2), giving 5x − 4 = x2 − x − 2 | ✓ 1 | |
| A1 rearranges to x2 − 6x + 2 = 0 | ✓ 1 | |
| Part (b) | ||
| M1 correct substitution into the quadratic formula, \(x = \frac{6 \pm \sqrt{28}}{2}\) | ✓ 1 | |
| A1 5.65 | ✓ 1 | |
| A1 0.354 | ✓ 1 | |
Exam tip: Multiply every term, including the 1 on the right-hand side, by the common denominator.
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