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International GCSE Maths · Pearson Edexcel 4MA1 · Unit 2

Equations, formulae and identities

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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1IGM-2-N1-012.1 Indices Paper FEasy1 mark

Simplify fully (2x3y)4

Show mark scheme
Marking pointMarkNotes
Raise every factor to the power 4: 24 = 16, (x3)4 = x3×4 = x12, y4.
Answer: 16x12y4
✓ 1Answer 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.

2IGM-2-N1-032.2 Algebraic manipulation Paper FEasy1 mark

Factorise completely 12x2y − 18xy2

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Marking pointMarkNotes
The highest common factor of 12x2y and 18xy2 is 6xy.
12x2y ÷ 6xy = 2x, 18xy2 ÷ 6xy = 3y.
Answer: 6xy(2x − 3y)
✓ 1Answer 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.

3IGM-2-N1-042.2 Algebraic manipulation Paper FEasy1 mark

Expand and simplify (2x − 5)2

Show mark scheme
Marking pointMarkNotes
(2x − 5)(2x − 5) = 4x2 − 10x − 10x + 25 = 4x2 − 20x + 25✓ 1Answer 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.

4IGM-2-N2-012.3 Expressions and formulae Paper FEasy1 mark

A spreadsheet cell calculates T = 2a2 − 5ab.
Find the value of T when a = −3 and b = 4.

Show mark scheme
Marking pointMarkNotes
2a2 = 2 × (−3)2 = 2 × 9 = 18
−5ab = −5 × (−3) × 4 = +60
T = 18 + 60 = 78
✓ 1Answer 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.

5IGM-2-N2-052.3 Expressions and formulae Paper FEasy5 marks

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.

(a)

Write down a formula for C in terms of h.

(2)
(b)

Work out the total cost for 5 hours.

(1)
(c)

Mei paid £47. For how many hours did she use the bike?

(2)
Show mark scheme
Marking pointMarkNotes
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.

6IGM-2-N3-042.6 Simultaneous linear equations Paper FEasy1 mark

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?

Show mark scheme
Marking pointMarkNotes
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).
✓ 1Answer 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.

7IGM-2-N4-012.7 Quadratic equations Paper FEasy1 mark

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?

Show mark scheme
Marking pointMarkNotes
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 ✓.
✓ 1Answer 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.

8IGM-2-N5-012.8 Inequalities Paper FEasy1 mark

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?

Show mark scheme
Marking pointMarkNotes
7 − 3x > 19
−3x > 12
Divide by −3 and reverse the inequality sign: x < −4.
✓ 1Answer 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.

9IGM-2-N5-022.8 Inequalities Paper FEasy1 mark

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?

Show mark scheme
Marking pointMarkNotes
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.
✓ 1Answer 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.

10IGME-059-01-01Linear and simultaneous equations Paper FEasy4 marks

Solve each equation.

(a)

\(5x - 7 = 18\)

(2)
(b)

\(3(y + 4) = 33\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

11IGME-059-01-02Linear and simultaneous equations Paper FEasy4 marks

Solve each equation.

(a)

\(7x + 3 = 3x + 23\)

(2)
(b)

\(\frac{2w - 1}{3} = 5\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

12IGME-059-01-03Linear and simultaneous equations Paper FEasy4 marks

The three angles of a triangle are \(x°\), \((2x + 10)°\) and \((3x - 40)°\).

(a)

Show that \(6x - 30 = 180\)

(2)
(b)

Work out the size of the largest angle of the triangle.

(2)
Show mark scheme
Marking pointMarkNotes
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.

13IGME-059-01-04Linear and simultaneous equations Paper FEasy3 marks

Solve the simultaneous equations
\(3x + 2y = 16\)
\(x - 2y = 0\)

(a)

Show clear algebraic working.

(3)
Show mark scheme
Marking pointMarkNotes
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 \(y\).

14IGME-059-01-05Linear and simultaneous equations Paper FEasy3 marks

Solve \(4(2x - 3) = 5x + 9\)

(a)

Show clear algebraic working.

(3)
Show mark scheme
Marking pointMarkNotes
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.

15IGME-059-02-01Quadratic equations and inequalities Paper FEasy3 marks

Solve the equation below by factorising.
\(x^2 + 7x + 12 = 0\)

(a)

Factorise \(x^2 + 7x + 12\)

(2)
(b)

Hence solve \(x^2 + 7x + 12 = 0\)

(1)
Show mark scheme
Marking pointMarkNotes
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.

16IGME-059-02-02Quadratic equations and inequalities Paper FEasy5 marks

This question is about linear inequalities.

(a)

Solve \(5x + 3 > 2x - 9\)

(2)
(b)

n is an integer.
List all the values of n that satisfy \(-3 \le 2n + 1 < 9\)

(3)
Show mark scheme
Marking pointMarkNotes
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 <.

17IGME-059-02-03Quadratic equations and inequalities Paper FEasy4 marks

Solve each of these equations.

(a)

\(x^2 - 6x = 0\)

(2)
(b)

\(2x^2 = 72\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

18IGME-059-03-01Direct and inverse proportion and rearranging formulae Paper FEasy4 marks

A formula used for a moving object is
\(v = u + at\)

(a)

Make t the subject of the formula.

(2)
(b)

Find the value of t when v = 25, u = 4 and a = 3.5

(2)
Show mark scheme
Marking pointMarkNotes
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.

19IGME-059-03-02Direct and inverse proportion and rearranging formulae Paper FEasy4 marks

The area, A cm2, of a trapezium with parallel sides a cm and b cm and height h cm is given by
\(A = \frac{(a+b)h}{2}\)

(a)

Make h the subject of the formula.

(2)
(b)

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.

(2)
Show mark scheme
Marking pointMarkNotes
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.

20IGME-059-03-04Direct and inverse proportion and rearranging formulae Paper FEasy4 marks

8 identical machines can complete a printing job in 15 hours.
The time taken is inversely proportional to the number of machines used.

(a)

How long would the same job take using 6 of these machines?

(2)
(b)

The job must be finished in 5 hours.
How many machines are needed?

(2)
Show mark scheme
Marking pointMarkNotes
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.

21IGME-059-00-01Algebraic manipulation and factorising Paper FEasy3 marks

Answer each part.

(a)

Simplify 5x + 3y − 2x + y

(1)
(b)

Expand 4(2x − 5)

(1)
(c)

Factorise 6x + 15

(1)
Show mark scheme
Marking pointMarkNotes
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.

22IGME-059-00-02Algebraic manipulation and factorising Paper FEasy4 marks

Simplify each expression.

(a)

m4 × m3

(1)
(b)

(p5)2

(1)
(c)

\(\frac{12x^6y^3}{4x^2y}\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

23IGME-059-00-03Algebraic manipulation and factorising Paper FEasy5 marks

Answer each part.

(a)

Expand and simplify (x + 6)(x − 4)

(2)
(b)

Factorise x2 − 9x + 20

(2)
(c)

Hence solve x2 − 9x + 20 = 0

(1)
Show mark scheme
Marking pointMarkNotes
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.

24IGME-059-00-04Algebraic manipulation and factorising Paper FEasy4 marks

Answer each part.

(a)

Factorise fully 8x2y − 12xy2

(2)
(b)

Expand and simplify 3(2x + 1) − 2(x − 4)

(2)
Show mark scheme
Marking pointMarkNotes
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.

25IGME-059-00-05Algebraic manipulation and factorising Paper FEasy4 marks

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.

(a)

Work out the value of v when u = 4.5, a = 9.8 and t = 3

(2)
(b)

Make t the subject of the formula v = u + at

(2)
Show mark scheme
Marking pointMarkNotes
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.

26IGME-059-03-03Direct and inverse proportion and rearranging formulae Paper FEasy4 marks

The cost of notebooks is directly proportional to the number of notebooks bought.
6 notebooks cost £8.70

(a)

Work out the cost of 15 notebooks.

(2)
(b)

Priya spends exactly £30.45 on these notebooks.
How many notebooks does she buy?

(2)
Show mark scheme
Marking pointMarkNotes
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.

27IGME-059-02-04Quadratic equations and inequalities Paper FEasy5 marks

A rectangular rug has a width of x metres and a length of (x + 4) metres.
The area of the rug is 45 m2.

(a)

Show that \(x^2 + 4x - 45 = 0\)

(2)
(b)

Work out the perimeter of the rug.

(3)
Show mark scheme
Marking pointMarkNotes
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.

28IGME-059-03-05Direct and inverse proportion and rearranging formulae Paper FEasy3 marks

y is directly proportional to x.
When x = 12, y = 42

(a)

Find a formula for y in terms of x.

(2)
(b)

Find the value of x when y = 91

(1)
Show mark scheme
Marking pointMarkNotes
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.

29IGM-2-N1-072.2 Algebraic manipulation Paper FMedium5 marks

A rectangular vegetable garden is (x + 5) metres long and (x − 2) metres wide, where x > 2.

(a)

Show that the area of the garden, in m2, is x2 + 3x − 10

(2)
(b)

The owner increases both the length and the width by 1 metre.
Show that the area increases by (2x + 4) m2.

(3)
Show mark scheme
Marking pointMarkNotes
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.

30IGM-2-N1-082.2 Algebraic manipulation Paper FMedium5 marks

Factorising can make some calculations much quicker.

(a)

Factorise x2 − 11x + 28

(2)
(b)

Factorise y2 − 144

(1)
(c)

Use the identity a2 − b2 = (a − b)(a + b) to work out 622 − 382 without a calculator.

(2)
Show mark scheme
Marking pointMarkNotes
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.

31IGM-2-N2-032.4 Linear equations Paper FMedium1 mark

Nikos thinks of a number x. He finds that
(x + 4)/3 − (x − 2)/5 = 2
What is the value of x?

Show mark scheme
Marking pointMarkNotes
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
✓ 1Answer 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.

32IGM-2-N2-042.4 Linear equations Paper FMedium1 mark

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?

Show mark scheme
Marking pointMarkNotes
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°.)
✓ 1Answer 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.

33IGM-2-N2-062.3 Expressions and formulae Paper FMedium6 marks

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.

(a)

Work out the value of F when C = −15

(2)
(b)

Make C the subject of the formula.

(2)
(c)

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.

(2)
Show mark scheme
Marking pointMarkNotes
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).

34IGM-2-N2-082.4 Linear equations Paper FMedium5 marks

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.

(a)

Write down an expression, in terms of x, for her father's age in 12 years' time.

(1)
(b)

Form an equation in x and solve it to find Ana's age now.

(3)
(c)

Work out the total of their ages now.

(1)
Show mark scheme
Marking pointMarkNotes
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.

35IGM-2-N2-092.4 Linear equations Paper FMedium5 marks

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.

(a)

Show that the perimeter of the rectangle is (6x + 6) m.

(1)
(b)

Form an equation and solve it to find the value of x.

(3)
(c)

Work out the area of the rectangular bed.

(1)
Show mark scheme
Marking pointMarkNotes
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.

36IGM-2-N2-102.4 Linear equations Paper FMedium6 marks

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.

(a)

Write down an expression, in pounds, for the monthly cost of Plan A.

(1)
(b)

Form an equation and solve it to find the number of minutes for which the two plans cost the same.

(3)
(c)

Lena makes 420 minutes of calls in a month. Which plan is cheaper for her, and by how much?

(2)
Show mark scheme
Marking pointMarkNotes
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.

37IGM-2-N3-032.6 Simultaneous linear equations Paper FMedium1 mark

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?

Show mark scheme
Marking pointMarkNotes
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.
✓ 1Answer C

Answer: C

Exam tip: Re-read what the question actually asks for. Here you need a + c, not just one of the prices.

38IGM-2-N3-092.6 Simultaneous linear equations Paper FMedium5 marks

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.

(a)

Write down two equations in f and r.

(2)
(b)

Solve your equations to find the value of f and the value of r.

(2)
(c)

In May, Sofia uses 15 GB. Calculate how much she pays.

(1)
Show mark scheme
Marking pointMarkNotes
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.

39IGM-2-N3-102.6 Simultaneous linear equations Paper FMedium5 marks

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.

(a)

Write down two equations in x and y.

(2)
(b)

Solve the equations to find how many baskets of each type the team scored.

(3)
Show mark scheme
Marking pointMarkNotes
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.

40IGM-2-N4-042.7 Quadratic equations Paper FMedium5 marks

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.

(a)

Show that x2 + 5x − 84 = 0

(2)
(b)

Solve x2 + 5x − 84 = 0 and hence find the perimeter of the print. Give a reason for rejecting any value of x.

(3)
Show mark scheme
Marking pointMarkNotes
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).

41IGM-2-N4-052.7 Quadratic equations Paper FMedium4 marks

Two consecutive positive odd numbers have a product of 323.
Let the smaller number be n.

(a)

Write down an equation in n and rearrange it into the form n2 + bn + c = 0

(1)
(b)

Solve your equation to find the two odd numbers.

(3)
Show mark scheme
Marking pointMarkNotes
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.

42IGM-2-N5-032.8 Inequalities Paper FMedium1 mark

n is an integer such that
−7 < 3n − 1 ≤ 14
How many possible values of n are there?

Show mark scheme
Marking pointMarkNotes
Add 1: −6 < 3n ≤ 15
Divide by 3: −2 < n ≤ 5
Integers: −1, 0, 1, 2, 3, 4, 5, so there are 7 values.
✓ 1Answer 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.

43IGM-2-N5-052.8 Inequalities Paper FMedium4 marks

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.

(a)

Write down an inequality, in terms of t, that is true when Plan A costs less than Plan B. Work in pounds.

(1)
(b)

Solve your inequality.

(2)
(c)

Find the least number of texts in a month for which Plan A is cheaper.

(1)
Show mark scheme
Marking pointMarkNotes
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.

44IGM-2-N5-062.8 Inequalities Paper FMedium5 marks

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.

(a)

Show that x < 19/3

(2)
(b)

Given also that x is an integer and that 3x − 2 is the longest side, find all the possible values of x.

(3)
Show mark scheme
Marking pointMarkNotes
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.

45IGME-059-01-07Linear and simultaneous equations Paper FMedium6 marks

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 £\(c\).

(a)

Write down two equations in \(a\) and \(c\).

(2)
(b)

Solve your equations to find the cost of an adult ticket and the cost of a child ticket.

(3)
(c)

Work out the total cost of 4 adult tickets and 3 child tickets.

(1)
Show mark scheme
Marking pointMarkNotes
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.

46IGME-059-01-08Linear and simultaneous equations Paper FMedium5 marks

\(P = \frac{3(k - 2m)}{5}\)

(a)

Make \(m\) the subject of the formula.

(3)
(b)

Find the value of \(m\) when \(k = 12\) and \(P = 4.2\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

47IGME-059-01-10Linear and simultaneous equations Paper FMedium4 marks

Line L has equation \(y = 3x - 5\)
Line M has equation \(2y + x = 18\)

(a)

Use algebra to find the coordinates of the point where L and M intersect.

(3)
(b)

Show that the point \((-2, 10)\) lies on line M.

(1)
Show mark scheme
Marking pointMarkNotes
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.

48IGME-059-01-11Linear and simultaneous equations Paper FMedium4 marks

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.

(a)

Show clear algebraic working.

(4)
Show mark scheme
Marking pointMarkNotes
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, \(x + 12\).

49IGME-059-03-08Direct and inverse proportion and rearranging formulae Paper FMedium5 marks

The kinetic energy, E joules, of an object of mass m kg moving with speed v m/s is given by
\(E = \frac{1}{2}mv^2\)

(a)

Make v the subject of the formula.

(3)
(b)

Find the speed of an object of mass 12 kg with kinetic energy 540 J.
Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
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.

50IGME-059-00-11Algebraic manipulation and factorising Paper FMedium5 marks

Answer each part.

(a)

Show that (n + 3)2 − (n − 3)2 ≡ 12n

(3)
(b)

Hence work out the value of 532 − 472

(2)
Show mark scheme
Marking pointMarkNotes
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.

51IGME-059-02-11Quadratic equations and inequalities Paper FMedium6 marks

A ball is kicked straight up into the air.
Its height, h metres, after t seconds is given by the formula
\(h = 20t - 5t^2\)

(a)

Work out the height of the ball when t = 1.5

(2)
(b)

Find the two values of t for which the height of the ball is 15 m.

(3)
(c)

For how many seconds is the ball higher than 15 m?

(1)
Show mark scheme
Marking pointMarkNotes
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.

52IGME-059-01-06Linear and simultaneous equations Paper FMedium3 marks

Solve \(\frac{x}{4} + \frac{x - 1}{3} = 5\)
Give your answer as a fraction.

(a)

Show clear algebraic working.

(3)
Show mark scheme
Marking pointMarkNotes
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.

53IGME-059-00-09Algebraic manipulation and factorising Paper FMedium5 marks

Answer each part.

(a)

Make w the subject of P = 2l + 2w

(2)
(b)

Make x the subject of y = \(\frac{3x-2}{5}\)

(2)
(c)

Use your answer to the previous part to find x when y = 8

(1)
Show mark scheme
Marking pointMarkNotes
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.

54IGME-059-01-09Linear and simultaneous equations Paper FMedium4 marks

Solve the simultaneous equations
\(5x + 3y = 7.5\)
\(2x - 4y = 16\)

(a)

Show clear algebraic working.

(4)
Show mark scheme
Marking pointMarkNotes
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: \(2(3) - 4(-2.5) = 16\).

55IGME-059-01-13Linear and simultaneous equations Paper FHard4 marks

Solve \(\frac{3x - 2}{4} - \frac{2x + 5}{6} = \frac{x - 1}{3}\)

(a)

Show clear algebraic working.

(4)
Show mark scheme
Marking pointMarkNotes
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 \(-4x + 10\).

56IGME-059-01-14Linear and simultaneous equations Paper FHard6 marks

A straight line has equation \(y = mx + c\)
The line passes through the points \((2, 11)\) and \((-1, 2)\).

(a)

Form and solve a pair of simultaneous equations to find the values of \(m\) and \(c\).

(3)
(b)

The line meets the line with equation \(2x + y = 30\) at the point P.
Find the coordinates of P.

(3)
Show mark scheme
Marking pointMarkNotes
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: \(9 = 3m\).

57IGME-059-02-12Quadratic equations and inequalities Paper FHard6 marks

A right-angled triangle has two shorter sides of length x cm and (x + 7) cm.
Its hypotenuse has length (2x + 1) cm.

(a)

Show that \(x^2 - 5x - 24 = 0\)

(3)
(b)

Work out the area of the triangle.

(3)
Show mark scheme
Marking pointMarkNotes
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.

58IGME-059-02-05Quadratic equations and inequalities Paper HEasy4 marks

This question is about quadratic inequalities.

(a)

Solve \(x^2 > 49\)

(2)
(b)

n is an integer.
Find all the values of n such that \(n^2 < 10\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

59IGM-2-N1-022.1 Indices Paper HMedium1 mark

Find the value of 27−2/3

Show mark scheme
Marking pointMarkNotes
271/3 = 3, so 272/3 = 32 = 9.
The negative index means the reciprocal: 27−2/3 = 1/9
✓ 1Answer 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.

60IGM-2-N1-062.1 Indices Paper HMedium6 marks

Answer each part without a calculator. Show your working clearly.

(a)

Find the value of 163/4

(2)
(b)

Simplify fully (4x6)−1/2

(2)
(c)

Find the value of n such that 9n = 27

(2)
Show mark scheme
Marking pointMarkNotes
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.

61IGM-2-N2-072.3 Expressions and formulae Paper HMedium4 marks

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

(a)

Make v the subject of the formula. Give your answer for a positive value of v.

(2)
(b)

A ball of mass 0.45 kg carries 360 joules of energy. Find its speed.

(2)
Show mark scheme
Marking pointMarkNotes
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.

62IGM-2-N3-012.5 Proportion Paper HMedium1 mark

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?

Show mark scheme
Marking pointMarkNotes
d = kv2. 18 = k × 302 gives k = 18 ÷ 900 = 0.02.
When v = 50: d = 0.02 × 502 = 0.02 × 2500 = 50.0 m.
✓ 1Answer 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.

63IGM-2-N3-022.5 Proportion Paper HMedium1 mark

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.

Show mark scheme
Marking pointMarkNotes
P = k ÷ √T. 6 = k ÷ √16 = k ÷ 4, so k = 24.
When P = 8: 8 = 24 ÷ √T, so √T = 3 and T = 9.
✓ 1Answer 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.

64IGM-2-N3-052.5 Proportion Paper HMedium5 marks

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.

(a)

Find a formula for m in terms of r.

(2)
(b)

Calculate the mass of a ball bearing of radius 4.5 cm. Give your answer to 3 significant figures.

(1)
(c)

Another ball bearing has a mass of 2160 g. Find its radius.

(2)
Show mark scheme
Marking pointMarkNotes
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.

65IGM-2-N3-062.5 Proportion Paper HMedium5 marks

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.

(a)

Find a formula for F in terms of d.

(2)
(b)

Calculate F when d = 12.

(1)
(c)

Find the distance d at which the force is 20 N.

(2)
Show mark scheme
Marking pointMarkNotes
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.

66IGM-2-N4-022.7 Quadratic equations Paper HMedium1 mark

Solve 3x2 + 5x − 4 = 0, giving your solutions correct to 3 significant figures.

Show mark scheme
Marking pointMarkNotes
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.)
✓ 1Answer 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.

67IGM-2-N4-032.7 Quadratic equations Paper HMedium1 mark

Leo solves the equation x2 + 8x + 3 = 0 by completing the square.
Which of these is the correct exact solution?

Show mark scheme
Marking pointMarkNotes
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.
✓ 1Answer 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.

68IGM-2-N4-102.7 Quadratic equations Paper HMedium6 marks

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

(a)

Write x2 − 10x + 18 in the form (x − a)2 − b, where a and b are integers.

(2)
(b)

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.

(2)
(c)

Write down the price that gives the greatest daily profit and the value of this greatest profit.

(2)
Show mark scheme
Marking pointMarkNotes
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).

69IGM-2-N5-042.8 Inequalities Paper HMedium1 mark

Solve the inequality x2 − 3x − 28 ≤ 0

Show mark scheme
Marking pointMarkNotes
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.
✓ 1Answer 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.

70IGM-2-N5-092.2 Algebraic manipulation Paper HMedium4 marks

In a number game, a player chooses an integer n and works out (2n + 3)2 − (2n − 1)2.

(a)

Show that (2n + 3)2 − (2n − 1)2 simplifies to 16n + 8.

(2)
(b)

Hence prove that the result is always a multiple of 8 but never a multiple of 16.

(2)
Show mark scheme
Marking pointMarkNotes
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.

71IGME-059-02-07Quadratic equations and inequalities Paper HMedium4 marks

This question is about the expression \(2x^2 - 7x - 15\)

(a)

Factorise \(2x^2 - 7x - 15\)

(2)
(b)

Hence solve \(2x^2 - 7x - 15 \ge 0\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

72IGME-059-02-08Quadratic equations and inequalities Paper HMedium5 marks

\(y = x^2 - 8x + 3\)

(a)

Write \(x^2 - 8x + 3\) in the form \((x - a)^2 - b\), where a and b are integers.

(2)
(b)

Write down the coordinates of the turning point of the curve \(y = x^2 - 8x + 3\)

(1)
(c)

Hence solve \(x^2 - 8x + 3 = 0\)
Give your solutions in the form \(p \pm \sqrt{q}\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

73IGME-059-02-10Quadratic equations and inequalities Paper HMedium4 marks

Solve the equation below.
\(\frac{3}{x - 1} + \frac{2}{x + 2} = 1\)
Give your solutions correct to 3 significant figures.

(a)

Solve the equation. Show clear algebraic working.

(4)
Show mark scheme
Marking pointMarkNotes
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.

74IGME-059-03-06Direct and inverse proportion and rearranging formulae Paper HMedium6 marks

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

(a)

Find a formula for d in terms of t.

(3)
(b)

Find the distance fallen when t = 3.5

(1)
(c)

Find the time taken for the stone to fall 100 metres.
Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
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.

75IGME-059-03-07Direct and inverse proportion and rearranging formulae Paper HMedium5 marks

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

(a)

Find a formula for I in terms of d.

(3)
(b)

Find the distance from the lamp at which the intensity is 4 lux.

(2)
Show mark scheme
Marking pointMarkNotes
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).

76IGME-059-03-10Direct and inverse proportion and rearranging formulae Paper HMedium6 marks

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

(a)

Find a formula for T in terms of L.

(3)
(b)

Find the length of a pendulum for which T = 3

(2)
(c)

The length of a pendulum is multiplied by 4.
Describe the effect on T.

(1)
Show mark scheme
Marking pointMarkNotes
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.

77IGME-059-03-11Direct and inverse proportion and rearranging formulae Paper HMedium5 marks

For a lens, the focal length f cm, the object distance u cm and the image distance v cm are related by
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)

(a)

Make v the subject of the formula.
Give your answer as a single fraction.

(3)
(b)

Find v when f = 12 and u = 30

(2)
Show mark scheme
Marking pointMarkNotes
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.

78IGME-059-00-06Algebraic manipulation and factorising Paper HMedium3 marks

Answer each part.

(a)

Factorise 2x2 + 7x − 15

(2)
(b)

Hence solve 2x2 + 7x − 15 = 0

(1)
Show mark scheme
Marking pointMarkNotes
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.

79IGME-059-00-07Algebraic manipulation and factorising Paper HMedium3 marks

Expand and simplify (x + 3)(2x − 1)(x − 4)

(a)

Show all your working.

(3)
Show mark scheme
Marking pointMarkNotes
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.

80IGME-059-00-08Algebraic manipulation and factorising Paper HMedium3 marks

Simplify fully \(\frac{x^2-16}{x^2+x-12}\)

(a)

Show clear algebraic working.

(3)
Show mark scheme
Marking pointMarkNotes
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.

81IGME-059-00-10Algebraic manipulation and factorising Paper HMedium5 marks

Answer each part.

(a)

Make a the subject of 3a + b = ac − 2

(3)
(b)

Find the value of a when b = 5.4 and c = 7.5
Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
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.

82IGME-059-03-09Direct and inverse proportion and rearranging formulae Paper HMedium5 marks

\(a = \frac{3b + 2}{b - 4}\)

(a)

Make b the subject of the formula.

(4)
(b)

Find the value of b when a = 5

(1)
Show mark scheme
Marking pointMarkNotes
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.

83IGME-059-02-09Quadratic equations and inequalities Paper HMedium6 marks

The line with equation \(y = 2x - 5\) meets the curve with equation \(y = x^2 - 3x + 1\) at the points A and B.

(a)

Find the coordinates of A and B.

(4)
(b)

Find the exact length of AB.

(2)
Show mark scheme
Marking pointMarkNotes
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.

84IGME-059-02-06Quadratic equations and inequalities Paper HMedium4 marks

A number x satisfies \(x(3x - 5) = 4\)

(a)

Show that \(3x^2 - 5x - 4 = 0\)

(1)
(b)

Solve \(3x^2 - 5x - 4 = 0\)
Give your solutions correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
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.

85IGM-2-N1-052.2 Algebraic manipulation Paper HHard1 mark

Simplify fully  (x2 − 9) / (2x2 + 5x − 3)

Show mark scheme
Marking pointMarkNotes
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)
✓ 1Answer 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.

86IGM-2-N1-092.2 Algebraic manipulation Paper HHard6 marks

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).

(a)

Write 2x2 − 12x + 23 in the form a(x + b)2 + c, where a, b and c are integers.

(3)
(b)

Hence write down the minimum daily running cost, in euros, and the number of hours for which it occurs.

(2)
(c)

Explain why the model never gives a negative cost for any value of x.

(1)
Show mark scheme
Marking pointMarkNotes
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.

87IGM-2-N1-102.2 Algebraic manipulation Paper HHard6 marks

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.

(a)

Show that the time on the flat road minus the time on the hill, in hours, is
(x − 7) / ((x + 2)(x − 1))

(3)
(b)

Simplify fully
[(x2 + 4x + 4) / (x2 − 4)] ÷ [(3x + 6) / (x − 2)]

(3)
Show mark scheme
Marking pointMarkNotes
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.

88IGM-2-N2-022.3 Expressions and formulae Paper HHard1 mark

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?

Show mark scheme
Marking pointMarkNotes
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)
✓ 1Answer 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.

89IGM-2-N3-072.5 Proportion Paper HHard6 marks

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.

(a)

Find a formula for D in terms of h.

(2)
(b)

Find the distance to the horizon when the observer is 49 m above sea level.

(1)
(c)

Find the height the observer must be above sea level to see a horizon 30 km away. Give your answer to 3 significant figures.

(2)
(d)

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.

(1)
Show mark scheme
Marking pointMarkNotes
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.

90IGM-2-N3-082.6 Simultaneous linear equations Paper HHard5 marks

Show your working clearly. Answers without working may not gain full marks.

(a)

Solve the simultaneous equations
4x + 3y = 5
6x − 5y = 36

(3)
(b)

Hence solve the simultaneous equations
8u + 3(v + 1) = 5
12u − 5(v + 1) = 36

(2)
Show mark scheme
Marking pointMarkNotes
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.

91IGM-2-N4-062.7 Quadratic equations Paper HHard6 marks

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.

(a)

Show that v2 + 6v − 216 = 0

(3)
(b)

Solve the equation to find v, and hence find the total time for the round trip.

(3)
Show mark scheme
Marking pointMarkNotes
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.

92IGM-2-N4-072.7 Quadratic equations Paper HHard6 marks

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.

(a)

Find the coordinates of the two points where the walkway meets the edge of the fountain. Show your working clearly.

(4)
(b)

Calculate the length of walkway that lies inside the fountain. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
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.

93IGM-2-N4-082.7 Quadratic equations Paper HHard6 marks

An engineer models the curved underside of a footbridge by y = x2 − 4x + 1 and a straight support beam by y = 2x − 7.

(a)

Find the coordinates of the points where the beam meets the curve. Show your working clearly.

(4)
(b)

A second beam is planned along y = 2x − 10. Show that this beam does not meet the curve.

(2)
Show mark scheme
Marking pointMarkNotes
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.

94IGM-2-N4-092.7 Quadratic equations Paper HHard6 marks

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

(a)

Use the quadratic formula to find the time at which the ball reaches the ground. Give your answer correct to 3 significant figures.

(3)
(b)

Find, correct to 3 significant figures, the length of time for which the ball is more than 6 m above the ground.

(3)
Show mark scheme
Marking pointMarkNotes
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.

95IGM-2-N5-072.8 Inequalities Paper HHard5 marks

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.

(a)

Show that v2 + 20v − 800 < 0

(1)
(b)

Solve v2 + 20v − 800 < 0. Show your working clearly.

(3)
(c)

Hence state the range of speeds of the van for which it can stop within the required distance. Give a reason for your answer.

(1)
Show mark scheme
Marking pointMarkNotes
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.

96IGM-2-N5-082.8 Inequalities Paper HHard5 marks

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.

(a)

The lines x = 2, y = 1 and x + y = 8 form a triangle. Find the coordinates of its three vertices.

(2)
(b)

Find the number of possible combinations (x, y).

(2)
(c)

Each row of tomatoes earns £30 and each row of beans earns £60. Find the greatest possible total earnings.

(1)
Show mark scheme
Marking pointMarkNotes
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.

97IGM-2-N5-102.2 Algebraic manipulation Paper HHard5 marks

A tile designer studies patterns built from whole numbers.

(a)

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.

(2)
(b)

Hence explain why the product of two consecutive even numbers, plus 1, is always the square of an odd number.

(1)
(c)

Prove that the sum of the squares of any two consecutive integers is always odd.

(2)
Show mark scheme
Marking pointMarkNotes
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.

98IGME-059-02-13Quadratic equations and inequalities Paper HHard5 marks

This question is about the inequalities \(x^2 - 2x - 15 < 0\) and \(2x - 1 \ge 3\)

(a)

Solve \(x^2 - 2x - 15 < 0\)

(3)
(b)

Find all the integers that satisfy both \(x^2 - 2x - 15 < 0\) and \(2x - 1 \ge 3\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

99IGME-059-03-12Direct and inverse proportion and rearranging formulae Paper HHard6 marks

y is directly proportional to the cube of x.

(a)

x is increased by 20%.
Find the percentage increase in y.

(3)
(b)

Find the percentage decrease in x that would make y halve.
Give your answer correct to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
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.

100IGME-059-03-13Direct and inverse proportion and rearranging formulae Paper HHard6 marks

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

(a)

Find a formula for a in terms of c.

(4)
(b)

Find the positive value of c when a = 8

(2)
Show mark scheme
Marking pointMarkNotes
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.

101IGME-059-00-12Algebraic manipulation and factorising Paper HHard6 marks

Answer each part.

(a)

Simplify fully \(\frac{2x^2-5x-3}{4x^2-1} \div \frac{x^2-9}{6x-3}\)

(4)
(b)

Hence solve \(\frac{2x^2-5x-3}{4x^2-1} \div \frac{x^2-9}{6x-3} = \frac{3}{5}\)

(2)
Show mark scheme
Marking pointMarkNotes
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.

102IGME-059-03-14Direct and inverse proportion and rearranging formulae Paper HHard5 marks

Here is a formula.
\(k = \sqrt{\frac{m+n}{m-n}}\)

(a)

Make m the subject of the formula.

(4)
(b)

Find the value of m when k = 3 and n = 4

(1)
Show mark scheme
Marking pointMarkNotes
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.

103IGME-059-01-12Linear and simultaneous equations Paper HHard7 marks

The line with equation \(y = x + 1\) meets the circle with equation \(x^2 + y^2 = 25\) at the points A and B.

(a)

Use algebra to find the coordinates of A and B.

(5)
(b)

Find the length of AB. Give your answer correct to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
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.

104IGME-059-00-14Algebraic manipulation and factorising Paper HHard5 marks

(ax + 5)(3x − b) ≡ 12x2 + cx − 15, where a, b and c are integers.

(a)

Find the values of a, b and c.

(3)
(b)

Hence factorise fully 12x2 + cx − 15, using your value of c.

(2)
Show mark scheme
Marking pointMarkNotes
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.

105IGME-059-00-13Algebraic manipulation and factorising Paper HHard6 marks

Answer each part.

(a)

Write 2x2 − 12x + 23 in the form a(x + b)2 + c, where a, b and c are integers.

(3)
(b)

Hence write down the minimum value of 2x2 − 12x + 23 and the value of x at which it occurs.

(2)
(c)

Explain why the equation 2x2 − 12x + 23 = 0 has no real solutions.

(1)
Show mark scheme
Marking pointMarkNotes
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.