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Cambridge IGCSE Mathematics 0580 · Core and Extended · Topic 2

Algebra and graphs

295 free, high-quality past-paper-style questions on Algebra and graphs for Cambridge IGCSE Maths, each with a full mark scheme. Covers: Expanding, factorising and algebraic fractions; Quadratic equations and inequalities; Linear and simultaneous equations and rearranging formulae; 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.1,889 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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My progress

+ Add to test✓ In your test1IGMC-042-00-01Expanding, factorising and algebraic fractions Paper CEasy5 marks

No calculator

Answer all parts.

(a)

Expand $4(3x - 5)$.

(1)

(b)

Expand and simplify $3(2y + 1) - 2(y - 4)$.

(2)

(c)

Factorise completely $6a^2 - 9ab$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
12x − 20 B1	✓ 1
Part (b)
M1 for 6y + 3 − 2y + 8 seen, with the sign of +8 correct	✓ 1
A1 4y + 11	✓ 1
Part (c)
3a(2a − 3b) B2	✓ 2
[B1 for a correct partial factorisation such as a(6a − 9b) or 3(2a² − 3ab)]	—

Common error: In part (b), −2 × −4 = +8. Writing −8 is the most frequent slip when a negative number multiplies a bracket.

+ Add to test✓ In your test2IGMC-042-00-05Expanding, factorising and algebraic fractions Paper CEasy4 marks

Calculator

$E = 2x(x + 4) - x(3 - x)$

(a)

Expand and simplify E.

(2)

(b)

Factorise your answer to the first part completely.

(1)

(c)

Find the value of E when x = 2.6.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2x² + 8x − 3x + x² seen	✓ 1
A1 3x² + 5x	✓ 1
Part (b)
B1 x(3x + 5)	✓ 1
Part (c)
B1 33.28	✓ 1

Common error: −x × −x = +x². Getting this sign wrong gives x² + 5x, and that leads to a wrong value of E in the final part.

+ Add to test✓ In your test3IGMC-042-02-02Quadratic equations and inequalities Paper CEasy4 marks

No calculator

(a)

Solve 3x² = 48.

(2)

(b)

Solve x² − 9x = 0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x² = 16	✓ 1
A1 x = 4 or x = −4	✓ 1
Part (b)
M1 x(x − 9) = 0	✓ 1
A1 x = 0 or x = 9	✓ 1

Common error: Dividing both sides of x² − 9x = 0 by x loses the solution x = 0. Factorise instead of dividing by the variable.

+ Add to test✓ In your test4IGMC-042-02-04Quadratic equations and inequalities Paper CEasy5 marks

No calculator

(a)

Solve the inequality 4x + 3 > 2x − 9.

(2)

(b)

Find all the integer values of n that satisfy −3 < 2n + 1 ≤ 7.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2x > −12	✓ 1
A1 x > −6	✓ 1
Part (b)
M1 −4 < 2n ≤ 6	✓ 1
A1 −2 < n ≤ 3	✓ 1
B1 −1, 0, 1, 2, 3	✓ 1

Common error: Check the end values. −2 is excluded because the inequality is strict (<), but 3 is included because of ≤.

+ Add to test✓ In your test5IGMC-042-02-12Quadratic equations and inequalities Paper CEasy4 marks

No calculator

(a)

Expand and simplify (x + 6)(x − 2).

(2)

(b)

Solve the equation (x + 6)(x − 2) = x² + 20.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 three of the four terms x², −2x, 6x, −12 correct	✓ 1
A1 x² + 4x − 12	✓ 1
Part (b)
M1 4x − 12 = 20	✓ 1
A1 x = 8	✓ 1

Exam tip: After expanding, the x² terms on both sides cancel. This leaves a linear equation, so you do not need to factorise.

+ Add to test✓ In your test6IGMC-042-01-01Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

No calculator

Solve these equations.

(a)

$5x - 7 = 18$

(2)

(b)

$3(2y + 1) = 2y + 19$

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 5x = 25	✓ 1
A1 for x = 5	✓ 1
Part (b)
M1 for expanding the bracket correctly: 6y + 3 = 2y + 19	✓ 1
M1 for collecting the y terms and the numbers on opposite sides: 4y = 16	✓ 1
A1 for y = 4	✓ 1

Common error: When expanding 3(2y + 1), multiply both terms by 3. Writing 6y + 1 loses the method mark straight away.

+ Add to test✓ In your test7IGMC-042-01-02Linear and simultaneous equations and rearranging formulae Paper CEasy6 marks

Calculator

Rearranging formulae.

(a)

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

(2)

(b)

Make r the subject of the formula $A = \pi r^2$.

(2)

(c)

A circle has area 50 cm². Use your answer to part (b) to find its radius.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for v − u = at	✓ 1
A1 for $t = \frac{v-u}{a}$	✓ 1
Part (b)
M1 for $r^2 = \frac{A}{\pi}$	✓ 1
A1 for $r = \sqrt{\frac{A}{\pi}}$	✓ 1
Part (c)
M1 for $\sqrt{\frac{50}{\pi}}$	✓ 1
A1 for 3.99 cm	✓ 1

Exam tip: Keep the full calculator value (3.98942…) and round only at the end, to 3 significant figures.

+ Add to test✓ In your test8IGMC-042-01-03Linear and simultaneous equations and rearranging formulae Paper CEasy3 marks

No calculator

Solve the simultaneous equations.
$3x + 2y = 16$
$5x - 2y = 8$

(a)

Find the values of x and y. You must show all your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for a correct method to eliminate one variable, e.g. adding the equations to get 8x = 24, or a correct substitution	✓ 1
A1 for x = 3	✓ 1
A1 for y = 3.5 or $\frac{7}{2}$	✓ 1

Exam tip: The y coefficients already have opposite signs, so adding the equations removes y straight away. Then substitute x = 3 back into either equation and check the pair in the other one.

+ Add to test✓ In your test9IGMC-042-01-04Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

No calculator

An adult ticket for a museum costs $x. A child ticket costs $4 less than an adult ticket. Two adult tickets and three child tickets cost $63 in total.

(a)

Show that $5x - 12 = 63$.

(2)

(b)

Find the cost of one child ticket.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for child ticket = x − 4	✓ 1
B1 for 2x + 3(x − 4) = 63, simplified correctly to 5x − 12 = 63	✓ 1
Part (b)
M1 for 5x = 75	✓ 1
A1 for x = 15	✓ 1
A1 for child ticket = $11	✓ 1

Common error: Read the question carefully. Solving gives x = 15, but that is the adult price. The final answer is the child price, $11.

+ Add to test✓ In your test10IGMC-042-01-05Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

No calculator

Solve these equations involving fractions.

(a)

$\frac{x+4}{3} = 2x - 7$

(3)

(b)

$\frac{x}{2} + \frac{x}{5} = 14$

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for x + 4 = 3(2x − 7)	✓ 1
M1 for 6x − 21 = x + 4, leading to 5x = 25	✓ 1
A1 for x = 5	✓ 1
Part (b)
M1 for multiplying every term by 10: 5x + 2x = 140	✓ 1
A1 for x = 20	✓ 1

Common error: When you multiply both sides by 3, the whole right-hand side must be multiplied: 3(2x − 7), not 6x − 7.

+ Add to test✓ In your test11IGMC-042-03-01Sequences and graphs of functions Paper CEasy7 marks

No calculator

The first four terms of a sequence are
7, 11, 15, 19, ...

(a)

Write down the next term.

(1)

(b)

Find an expression, in terms of n, for the nth term.

(2)

(c)

Find the 50th term.

(2)

(d)

Show that 150 is not a term of this sequence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 23	✓ 1
Part (b)
B2 4n + 3	✓ 2
(B1 for 4n + k, where k ≠ 3)	—
Part (c)
M1 for 4 × 50 + 3	✓ 1
A1 203	✓ 1
Part (d)
M1 for 4n + 3 = 150 leading to n = 36.75 or $\frac{147}{4}$, OR for stating that every term is odd (odd start, even common difference), OR for 150 − 7 = 143 not being a multiple of 4	✓ 1
A1 correct conclusion: n is not a whole number / 150 is even / 143 is not divisible by 4, so 150 is not a term	✓ 1

Exam tip: The number multiplying n is the common difference. Always check your expression by substituting n = 1 and n = 2.

+ Add to test✓ In your test12IGMC-042-03-02Sequences and graphs of functions Paper CEasy4 marks

No calculator

The first four terms of a sequence are
40, 34, 28, 22, ...

(a)

Find an expression, in terms of n, for the nth term.

(2)

(b)

Find the value of the first term in the sequence that is negative.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B2 46 − 6n	✓ 2
(B1 for k − 6n or −6n + k, where k ≠ 46)	—
Part (b)
M1 for 46 − 6n < 0 or continuing the sequence 16, 10, 4, ...	✓ 1
A1 −2	✓ 1

Common error: In a decreasing sequence the coefficient of n is negative. Writing 6n + 46 gives an increasing sequence.

+ Add to test✓ In your test13IGMC-042-03-03Sequences and graphs of functions Paper CEasy6 marks

Calculator

The table shows some values of y = x² − 3x − 2.

x: −2, −1, 0, 1, 2, 3, 4, 5
y: 8, p, −2, −4, q, −2, r, 8

(a)

Find the values of p, q and r.

(3)

(b)

Write down the equation of the line of symmetry of the graph of y = x² − 3x − 2.

(1)

(c)

Find the minimum value of y.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 p = 2	✓ 1
B1 q = −4	✓ 1
B1 r = 2	✓ 1
Part (b)
B1 x = 1.5 (accept x = $\frac{3}{2}$)	✓ 1
Part (c)
M1 substituting x = 1.5: 1.5² − 3 × 1.5 − 2, or completing the square to get (x − 1.5)² − 4.25	✓ 1
A1 −4.25 (accept $-\frac{17}{4}$)	✓ 1

Exam tip: Equal y-values in a table of a quadratic sit either side of the line of symmetry; the line passes halfway between their x-values.

+ Add to test✓ In your test14IGMC-042-03-04Sequences and graphs of functions Paper CEasy6 marks

No calculator

A pattern of shapes is made from matchsticks.
Pattern 1 uses 5 matchsticks, Pattern 2 uses 9 matchsticks and Pattern 3 uses 13 matchsticks.
The number of matchsticks increases by the same amount each time.

(a)

Find the number of matchsticks in Pattern 6.

(1)

(b)

Find an expression for the number of matchsticks in Pattern n.

(2)

(c)

Rana has 150 matchsticks.
She makes the largest pattern she can.
Find the pattern number and the number of matchsticks she has left over.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 25	✓ 1
Part (b)
B2 4n + 1	✓ 2
(B1 for 4n + k, where k ≠ 1)	—
Part (c)
M1 for 4n + 1 ≤ 150 or (150 − 1) ÷ 4 = 37.25	✓ 1
A1 Pattern 37	✓ 1
A1 1 matchstick left over	✓ 1

Common error: Rounding 37.25 up to 38 gives a pattern that needs 153 matchsticks, which is more than Rana has. Round down when resources are limited.

+ Add to test✓ In your test15IGMC3-2-08c-01Expanding, factorising and algebraic fractions Paper CEasy5 marks

No calculator

This question is about algebraic expressions.

(a)

Expand and simplify 3(2x − 5) − 4(x − 2).

(2)

(b)

Factorise completely 12a²b − 18ab².

(2)

(c)

Find the value of the expression in part (a) when x = 4.5.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6x − 15 − 4x + 8 (at least one bracket correct)	✓ 1
A1 2x − 7	✓ 1
Part (b)
B1 common factor 6ab (or 6, a or b only for partial credit towards the next mark)	✓ 1
B1 6ab(2a − 3b)	✓ 1
Part (c)
B1 2	✓ 1

Common error: When removing a bracket with a negative sign in front, change the sign of every term inside it.

+ Add to test✓ In your test16IGMC3-2-08c-02Quadratic equations and inequalities Paper CEasy5 marks

Calculator

A taxi company charges €3.20 plus €1.45 for each kilometre travelled. The fare is €C for a journey of d km, so C = 3.20 + 1.45d.

(a)

Calculate the fare for a journey of 8 km.

(1)

(b)

Rearrange the formula to make d the subject.

(2)

(c)

A passenger pays €22.05. Calculate the distance travelled.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 €14.80	✓ 1
Part (b)
M1 C − 3.20 = 1.45d	✓ 1
A1 d = (C − 3.20) ÷ 1.45	✓ 1
Part (c)
M1 (22.05 − 3.20) ÷ 1.45	✓ 1
A1 13 km	✓ 1

Exam tip: Subtract the fixed charge before dividing by the rate per kilometre.

+ Add to test✓ In your test17IGMC3-2-08c-03Sequences and graphs of functions Paper CEasy5 marks

No calculator

The first four terms of a sequence are 7, 11, 15, 19.

(a)

Find an expression for the nth term.

(2)

(b)

Show that 100 is not a term of the sequence.

(2)

(c)

Find the 30th term.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 4 or 4n seen	✓ 1
A1 4n + 3	✓ 1
Part (b)
M1 4n + 3 = 100	✓ 1
A1 n = 24.25, which is not an integer, so 100 is not a term	✓ 1
Part (c)
B1 123	✓ 1

Common error: Check the nth term by substituting n = 1 to see that it gives the first term.

+ Add to test✓ In your test18IGMC3-2-08c-04Linear and simultaneous equations and rearranging formulae Paper CEasy6 marks

Calculator

At a cinema, 2 adult tickets and 3 child tickets cost €37. 3 adult tickets and 2 child tickets cost €40.50. An adult ticket costs €a and a child ticket costs €c.

(a)

Write down two equations in a and c.

(2)

(b)

Solve the equations to find a and c.

(3)

(c)

Find the cost of 1 adult ticket and 4 child tickets.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2a + 3c = 37	✓ 1
B1 3a + 2c = 40.5	✓ 1
Part (b)
M1 multiply to match coefficients of one letter	✓ 1
M1 subtract (or add) to eliminate it	✓ 1
A1 a = 9.50 and c = 6	✓ 1
Part (c)
B1 €33.50	✓ 1

Exam tip: Substitute your answers back into both original equations to check them.

+ Add to test✓ In your test19IGMC3-2-08c-05Quadratic equations and inequalities Paper CEasy5 marks

No calculator

This question is about inequalities.

(a)

Solve 5x − 7 < 18.

(2)

(b)

n is an integer such that −3 ⩽ 2n + 1 < 9. List all the possible values of n.

(2)

(c)

Find the sum of the values listed in (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5x < 25	✓ 1
A1 x < 5	✓ 1
Part (b)
M1 −4 ⩽ 2n < 8 or −2 ⩽ n < 4	✓ 1
A1 −2, −1, 0, 1, 2, 3 (all six values)	✓ 1
Part (c)
B1 3	✓ 1

Common error: A strict inequality (<) excludes the end value, while ⩽ includes it.

+ Add to test✓ In your test20IGMC3-2-08r5-01Expanding, factorising and algebraic fractions Paper CEasy4 marks

No calculator

An archaeologist records the number of pottery fragments found in two adjacent grid squares using algebraic expressions.

(a)

Expand and simplify 4(3x + 2) − 5(x − 1).

(2)

(b)

Factorise completely 6x²y − 9xy².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12x + 8 or −5x + 5 seen	✓ 1
A1 7x + 13	✓ 1
Part (b)
B1 a correct common factor taken out, for example 3 or xy	✓ 1
B1 3xy(2x − 3y)	✓ 1

Common error: Factorise completely means taking out every common factor, numbers and letters, not just one of them.

+ Add to test✓ In your test21IGMC3-2-08r5-02Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

Calculator

A cycling shop hires out bicycles. The total cost is a fixed fee of €12 plus €3.50 for each hour of hire. The total cost for h hours is €C.

(a)

Write a formula for C in terms of h.

(1)

(b)

Rearrange your formula to make h the subject.

(2)

(c)

Maria pays €40 for her hire. Find how many hours she hired the bicycle for.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 C = 12 + 3.5h	✓ 1
Part (b)
M1 subtract 12 from both sides, C − 12 = 3.5h	✓ 1
A1 h = (C − 12) ÷ 3.5	✓ 1
Part (c)
M1 substitute C = 40 into their rearranged formula, (40 − 12) ÷ 3.5	✓ 1
A1 8 hours	✓ 1

Exam tip: Check a rearranged formula by substituting a simple pair of values back into the original.

+ Add to test✓ In your test22IGMC3-2-08r5-11Sequences and graphs of functions Paper CEasy5 marks

No calculator

The cross-section of an excavated ditch is modelled by y = x² − 2x − 3, where x is the horizontal distance in metres and y is the height in metres relative to ground level (negative means below ground). A table of values is given.
x: −2, −1, 0, 1, 2, 3, 4
y: 5, 0, p, −4, q, 0, 5

(a)

Find the values of p and q.

(2)

(b)

Use the table to write down the values of x where the ditch meets ground level.

(1)

(c)

Write down the equation of the line of symmetry of the ditch.

(1)

(d)

Write down the width of the ditch at ground level.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 p = −3	✓ 1
B1 q = −3	✓ 1
Part (b)
B1 x = −1 and x = 3	✓ 1
Part (c)
B1 x = 1	✓ 1
Part (d)
B1 3 − (−1) = 4 m	✓ 1

Common error: Ground level is where y = 0; do not confuse the x-values there with the lowest point of the curve.

+ Add to test✓ In your test23IGMC3-2-08r9-01Expanding, factorising and algebraic fractions Paper CEasy4 marks

No calculator

Answer both parts, showing your working.

(a)

Expand and simplify $3(2x-5)-4(x-2)$.

(2)

(b)

Factorise completely $12ab-18a^2$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands both brackets: $6x-15-4x+8$	✓ 1
A1 $2x-7$	✓ 1
Part (b)
B1 $6a$ identified as the common factor	✓ 1
B1 $6a(2b-3a)$	✓ 1

Common error: When multiplying a bracket by a negative number, change the sign of every term inside it.

+ Add to test✓ In your test24IGMC3-2-08r9-02Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

Calculator

A cycle shop hires out bikes. Plan A costs £6 plus £3 for each hour. The total cost is £$C$ for $h$ hours, so $C=6+3h$.

(a)

Rearrange $C=6+3h$ to make $h$ the subject.

(2)

(b)

Plan B costs £12 plus £1.50 for each hour. Find the number of hours for which both plans cost the same.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $C-6=3h$	✓ 1
A1 $h=\frac{C-6}{3}$	✓ 1
Part (b)
M1 forms the equation $6+3h=12+1.5h$	✓ 1
M1 collects terms: $1.5h=6$	✓ 1
A1 $h=4$	✓ 1

Exam tip: Check an equation answer by substituting it into both sides: plan A gives £18 and plan B gives £18.

+ Add to test✓ In your test25IGMC3-2-08r9-05Linear and simultaneous equations and rearranging formulae Paper CEasy4 marks

Calculator

A cyclist moving at $u$ m/s accelerates at $a$ m/s² for $t$ seconds and reaches a speed of $v$ m/s, where $v=u+at$.

(a)

Rearrange the formula to make $t$ the subject.

(2)

(b)

Find $t$ when $u=3.0$, $v=11$ and $a=0.80$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $v-u=at$	✓ 1
A1 $t=\frac{v-u}{a}$	✓ 1
Part (b)
M1 substitutes: $t=\frac{11-3.0}{0.80}$	✓ 1
A1 $t=10$ s	✓ 1

Exam tip: Rearrange first and substitute afterwards; it is easier to check and to follow.

+ Add to test✓ In your test26IGMC3-2-08r10-01Functions and differentiation Paper CEasy5 marks

No calculator

A cycle shop normally charges £C for h hours of hire, where $C = 5h + 12$. On a promotion day, the shop charges £$(3x + 11)$ for x hours of hire, while a rival shop charges £$(7x - 5)$ for the same x hours.

(a)

Find the value of x for which the two shops charge the same amount.

(2)

(b)

Make h the subject of $C = 5h + 12$.

(2)

(c)

Use your answer to (b) to find the number of hours of hire that cost £37 at the normal rate.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 collect terms, e.g. $4x = 16$	✓ 1
A1 $x = 4$	✓ 1
Part (b)
M1 subtract 12 from both sides	✓ 1
A1 $h = \frac{C-12}{5}$	✓ 1
Part (c)
A1 5 hours	✓ 1

Exam tip: Do the same operation to both sides, one step at a time, and check by substituting back.

+ Add to test✓ In your test27IGMC3-2-08r10-02Sequences and graphs of functions Paper CEasy5 marks

Calculator

A data centre installs new server racks in batches. The numbers of servers in the first four batches are 12, 19, 26, 33, and the pattern continues in the same way.

(a)

Find an expression for the number of servers in the nth batch.

(2)

(b)

Find the number of servers in the 20th batch.

(1)

(c)

One batch contains 187 servers. Find the number of that batch.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 7 identified	✓ 1
A1 $7n + 5$	✓ 1
Part (b)
A1 145	✓ 1
Part (c)
M1 $7n + 5 = 187$	✓ 1
A1 $n = 26$	✓ 1

Common error: The nth term starts from the difference times n, then you adjust: check it by testing n = 1.

+ Add to test✓ In your test28IGMC3-2-08r10-10Functions and differentiation Paper CEasy4 marks

No calculator

Two cooling units in a data centre remove heat at rates of $\frac{x + 3}{2}$ kW and $\frac{2x - 1}{3}$ kW per cabinet. The two rates are equal.

(a)

Form an equation and solve it to find x.

(3)

(b)

Find the rate at which each unit removes heat.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{x+3}{2} = \frac{2x-1}{3}$ and multiply both sides by 6	✓ 1
M1 $3x + 9 = 4x - 2$	✓ 1
A1 $x = 11$	✓ 1
Part (b)
A1 7 kW	✓ 1

Common error: When clearing fractions, multiply every term by the common denominator, including terms without a fraction.

+ Add to test✓ In your test29IGMC3-2-08r10-14Sequences and graphs of functions Paper CEasy4 marks

Calculator

On the first four days of a dig, the team catalogues 5, 10, 20 and 40 items, and the pattern continues.

(a)

Write down the next two terms of the sequence.

(1)

(b)

Find the number of items catalogued on day 10.

(2)

(c)

Explain why this is not a linear sequence.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 80 and 160	✓ 1
Part (b)
M1 $5 \times 2^9$	✓ 1
A1 2560	✓ 1
Part (c)
B1 the differences (5, 10, 20, ...) are not constant	✓ 1

Common error: Doubling each time means multiplying by a constant ratio, so the 10th term needs nine multiplications, not ten.

+ Add to test✓ In your test30IGMC3-2-10r10-01Linear and simultaneous equations and rearranging formulae Paper CEasy6 marks

Calculator

A sports scientist estimates the maximum heart rate H beats per minute of an adult aged a years using H = 208 − 0.7a.

(a)

Calculate H for an athlete aged 40.

(2)

(b)

An athlete has H = 173. Find the athlete's age.

(2)

(c)

Rearrange the formula to make a the subject.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 208 − 0.7 × 40	✓ 1
A1 180	✓ 1
Part (b)
M1 0.7a = 208 − 173 (= 35)	✓ 1
A1 a = 50	✓ 1
Part (c)
M1 0.7a = 208 − H	✓ 1
A1 a = (208 − H) ÷ 0.7	✓ 1

Common error: When rearranging, do the same operation to every term on both sides; do not divide only part of the right-hand side.

+ Add to test✓ In your test31IGMC3-2-10r10-02Linear and simultaneous equations and rearranging formulae Paper CEasy7 marks

Calculator

At a railway station, 3 adult tickets and 2 child tickets cost £47.00 in total. 1 adult ticket and 4 child tickets cost £31.00 in total. An adult ticket costs £a and a child ticket costs £c.

(a)

Write down two simultaneous equations in a and c.

(2)

(b)

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

(3)

(c)

Find the cost of 2 adult tickets and 5 child tickets.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3a + 2c = 47	✓ 1
B1 a + 4c = 31	✓ 1
Part (b)
M1 eliminate one variable, e.g. a = 31 − 4c substituted into the first equation	✓ 1
A1 c = 4.60	✓ 1
A1 a = 12.60	✓ 1
Part (c)
M1 2 × 12.60 + 5 × 4.60	✓ 1
A1 £48.20	✓ 1

Exam tip: Substitute your answers back into both original equations to check them.

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Calculator

An Earth-observation satellite takes 12 images on day 1, 19 images on day 2, 26 images on day 3, and continues the pattern. The number of images on day n follows a linear sequence.

(a)

Write down the number of images on day 4 and on day 5.

(2)

(b)

Find an expression for the number of images on day n.

(2)

(c)

On day d the satellite takes 313 images. Find d.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 33	✓ 1
B1 40	✓ 1
Part (b)
M1 common difference 7, so 7n + …	✓ 1
A1 7n + 5	✓ 1
Part (c)
M1 7d + 5 = 313	✓ 1
M1 7d = 308	✓ 1
A1 d = 44	✓ 1

Exam tip: Test your nth term formula by substituting n = 1, 2 and 3 to see if it reproduces the given terms.

+ Add to test✓ In your test33IGMC3-2-10r10-04Quadratic equations and inequalities Paper CEasy7 marks

No calculator

A rectangular training pitch has width x metres and length (x + 5) metres. Its area is 84 m².

Diagram not to scale

(a)

Show that x² + 5x − 84 = 0.

(2)

(b)

Solve x² + 5x − 84 = 0 by factorising, and state the width of the pitch.

(3)

(c)

Find the perimeter of the pitch.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x(x + 5) = 84	✓ 1
A1 x² + 5x = 84, so x² + 5x − 84 = 0	✓ 1
Part (b)
M1 (x + 12)(x − 7)	✓ 1
A1 x = −12 or x = 7	✓ 1
B1 width = 7 m since a length cannot be negative	✓ 1
Part (c)
M1 2 × (7 + 12)	✓ 1
A1 38 m	✓ 1

Common error: Quadratic equations often give a negative solution; check whether it makes sense for a length.

+ Add to test✓ In your test34IGMC3-2-10r11-01Sequences and graphs of functions Paper CEasy5 marks

No calculator

The power output of a space probe's battery, in watts, at the start of each hour is 120, 114, 108, 102, ...

(a)

Write down the next two terms.

(1)

(b)

Find an expression for the nth term.

(2)

(c)

Find the power at the first hour in which it is below 40 W.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 96 and 90	✓ 1
Part (b)
M1 common difference −6 used as −6n	✓ 1
A1 126 − 6n	✓ 1
Part (c)
M1 126 − 6n < 40 so n > 14.3, hence n = 15	✓ 1
A1 36 W	✓ 1

Common error: The nth term of a linear sequence is d×n plus a constant; the constant is the first term minus d.

+ Add to test✓ In your test35IGMC3-2-10r11-02Linear and simultaneous equations and rearranging formulae Paper CEasy6 marks

No calculator

A café buys green coffee beans at £6.40 per kilogram and pays a fixed delivery charge of £15 for each order.

(a)

Write a formula for the total cost £C of an order of x kg.

(1)

(b)

An order costs £79 in total. Find x.

(2)

(c)

The roaster shows temperature F in °F, where F = 9C/5 + 32 and C is in °C. Make C the subject of the formula.

(2)

(d)

Find the temperature in °C when the display shows 392 °F.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 C = 6.4x + 15	✓ 1
Part (b)
M1 6.4x = 79 − 15 = 64	✓ 1
A1 x = 10	✓ 1
Part (c)
M1 F − 32 = 9C/5	✓ 1
A1 C = 5(F − 32)/9	✓ 1
Part (d)
B1 200 °C	✓ 1

Exam tip: Undo operations in the reverse order: remove the added constant before the multiplier.

+ Add to test✓ In your test36IGMC3-2-10r11-03Expanding, factorising and algebraic fractions Paper CEasy6 marks

No calculator

A bridge engineer writes expressions for the areas of steel plates.

(a)

Factorise fully 15x² − 10x.

(2)

(b)

Expand and simplify (2x + 3)(x − 4).

(2)

(c)

Factorise x² − 10x + 21.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 one correct common factor (5 or x)	✓ 1
A1 5x(3x − 2)	✓ 1
Part (b)
M1 2x² − 8x + 3x − 12	✓ 1
A1 2x² − 5x − 12	✓ 1
Part (c)
M1 (x − a)(x − b) with ab = 21	✓ 1
A1 (x − 3)(x − 7)	✓ 1

Exam tip: Check a factorisation by expanding it again.

+ Add to test✓ In your test37IGMC3-2-10r11-14Expanding, factorising and algebraic fractions Paper CEasy4 marks

No calculator

A bridge may carry a total mass of at most 100 tonnes. A crane of mass 5 tonnes carries n steel beams, each of mass 12 tonnes, onto the bridge.

(a)

Write an inequality in n and solve it.

(3)

(b)

Write down the greatest number of beams the crane can carry at once.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12n + 5 ≤ 100	✓ 1
M1 12n ≤ 95	✓ 1
A1 n ≤ 7 11/12	✓ 1
Part (b)
B1 7	✓ 1

Common error: The answer must be a whole number of beams, so round down, not to the nearest integer.

+ Add to test✓ In your test38IGMC3-2-08r16-01Expanding, factorising and algebraic fractions Paper CEasy6 marks

No calculator

A team of archaeologists records the dimensions of their trenches using algebra.

(a)

Expand and simplify $4(3x-2)-3(x-5)$.

(2)

(b)

Factorise completely $6x^2-9x$.

(2)

(c)

Expand and simplify $(x+4)(x-3)$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands at least one bracket correctly, for example $12x-8$ or $-3x+15$	✓ 1
A1 $9x+7$	✓ 1
Part (b)
B1 takes out a correct common factor, for example $3$ or $x$	✓ 1
B1 $3x(2x-3)$	✓ 1
Part (c)
M1 expands to four terms $x^2-3x+4x-12$	✓ 1
A1 $x^2+x-12$	✓ 1

Common error: When expanding $-3(x-5)$, the second term becomes $+15$, not $-15$.

+ Add to test✓ In your test39IGMC3-2-08r16-05Expanding, factorising and algebraic fractions Paper CEasy5 marks

Calculator

A cyclist trains each day. On day $n$ she rides $(3n+2)$ kilometres.

(a)

Write down the distances for the first three days.

(1)

(b)

Find the distance on day 20.

(1)

(c)

On which day does she first ride more than 100 km? Find the distance she rides on that day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5, 8, 11	✓ 1
Part (b)
B1 62 km	✓ 1
Part (c)
M1 $3n+2>100$, or solves $3n+2=100$ to get $n=32.67\ldots$	✓ 1
A1 day 33	✓ 1
A1 101 km	✓ 1

Common error: When the solution of an inequality is not a whole number of days, round up to the next whole day, not to the nearest.

+ Add to test✓ In your test40IGMC3-2-08r16-10Functions and differentiation Paper CEasy6 marks

Calculator

A cyclist increases speed uniformly. Her speed $v$ m/s after $t$ seconds is given by $v=u+at$, where $u$ is her starting speed and $a$ is her acceleration in m/s$^2$.

(a)

Find $v$ when $u=2.5$, $a=0.6$ and $t=8$.

(2)

(b)

Make $t$ the subject of $v=u+at$.

(2)

(c)

Use your answer to (b) to find $t$ when $v=10.0$, $u=2.5$ and $a=0.6$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $2.5+0.6\times 8$	✓ 1
A1 7.3 m/s	✓ 1
Part (b)
M1 $v-u=at$	✓ 1
A1 $t=\dfrac{v-u}{a}$	✓ 1
Part (c)
M1 $\dfrac{10.0-2.5}{0.6}$	✓ 1
A1 12.5 s	✓ 1

Common error: Subtract $u$ from both sides first and then divide by $a$; dividing only $at$ by $a$ gives the wrong rearrangement.

+ Add to test✓ In your test41IGMC3-2-08r16-12Functions and differentiation Paper CEasy7 marks

No calculator

A straight line $L$ has equation $y=2x-6$.

(a)

Write down the gradient of $L$ and the coordinates of the point where $L$ crosses the $y$-axis.

(2)

(b)

Find the coordinates of the point where $L$ crosses the $x$-axis.

(2)

(c)

A second line is parallel to $L$ and passes through $(0,4)$. Find the $x$-coordinate of the point where this line crosses the $x$-axis.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 gradient 2	✓ 1
B1 $(0,-6)$	✓ 1
Part (b)
M1 sets $y=0$ and solves $2x-6=0$	✓ 1
A1 $(3,0)$	✓ 1
Part (c)
B1 the equation is $y=2x+4$	✓ 1
M1 sets $y=0$	✓ 1
A1 $x=-2$	✓ 1

Exam tip: Parallel lines have the same gradient, so keep the 2 and change only the intercept.

+ Add to test✓ In your test42IGMC3-2-08r17-01Functions and differentiation Paper CEasy6 marks

No calculator

The temperature T °C at a height h km on a mountain is given by T = 18 − 6h.

(a)

Find T when h = 2.5.

(2)

(b)

Find the height at which the temperature is −12 °C.

(2)

(c)

Rearrange the formula to make h the subject.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute h = 2.5, 18 − 6 × 2.5	✓ 1
A1 T = 3	✓ 1
Part (b)
M1 −12 = 18 − 6h	✓ 1
A1 h = 5 (km)	✓ 1
Part (c)
M1 6h = 18 − T (or equivalent first step)	✓ 1
A1 h = (18 − T)/6	✓ 1

Common error: When rearranging, deal with the whole of the other side: h = (18 − T)/6, not 18 − T/6.

+ Add to test✓ In your test43IGMC3-2-08r17-04Expanding, factorising and algebraic fractions Paper CEasy6 marks

No calculator

(a)

Expand and simplify 4(3x + 2) − 3(2x − 5).

(2)

(b)

Factorise completely 12x² − 18x.

(2)

(c)

Factorise completely 15pq + 10p².

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expand both brackets: 12x + 8 − 6x + 15 (at least three terms correct)	✓ 1
A1 6x + 23	✓ 1
Part (b)
M1 partial factorisation, e.g. 6(2x² − 3x) or 3x(4x − 6)	✓ 1
A1 6x(2x − 3)	✓ 1
Part (c)
M1 partial factorisation, e.g. 5(3pq + 2p²) or p(15q + 10p)	✓ 1
A1 5p(3q + 2p)	✓ 1

Common error: In 4(3x + 2) − 3(2x − 5) the second bracket becomes −6x + 15; the sign of the last term changes.

+ Add to test✓ In your test44IGMC3-2-08r17-06Sequences and graphs of functions Paper CEasy6 marks

Calculator

A climber's altitude A metres is plotted against the time t hours since leaving camp. The graph is a straight line through the points (0, 1800) and (4, 2600).

Diagram not to scale

(a)

Calculate the gradient of the line.

(2)

(b)

Write down the equation of the line in the form A = mt + c.

(1)

(c)

Find the altitude of the climber after 5.5 hours.

(2)

(d)

State what the gradient represents.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (2600 − 1800)/(4 − 0)	✓ 1
A1 200	✓ 1
Part (b)
B1 A = 200t + 1800 (ECF from (a))	✓ 1
Part (c)
M1 200 × 5.5 + 1800	✓ 1
A1 2900 m	✓ 1
Part (d)
B1 the climber gains 200 m of altitude every hour (the rate of climbing)	✓ 1

Exam tip: The gradient of a graph of altitude against time is a rate, so give its units: metres per hour.

+ Add to test✓ In your test45IGMC3-2-08r17-08Functions and differentiation Paper CEasy6 marks

No calculator

A baking tray for a food science class is a trapezium shape. Its area A cm² is given by A = ½(a + b)h, where a cm and b cm are the lengths of the parallel sides and h cm is the distance between them.

Diagram not to scale

(a)

Find A when a = 24, b = 36 and h = 15.

(2)

(b)

Rearrange the formula to make h the subject.

(2)

(c)

Find h when a = 18, b = 30 and A = 216.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (24 + 36) × 15	✓ 1
A1 450	✓ 1
Part (b)
M1 2A = (a + b)h	✓ 1
A1 h = 2A/(a + b)	✓ 1
Part (c)
M1 2 × 216/(18 + 30)	✓ 1
A1 9	✓ 1

Exam tip: Substitute into your rearranged formula, not the original, and check that the answer is sensible for the shape.

+ Add to test✓ In your test46IGMC3-2-10r22-01Sequences and graphs of functions Paper CEasy6 marks

No calculator

The number of jars of honey collected from a beekeeper's hives in the first four weeks are 12, 17, 22, 27, and this pattern continues.

(a)

Find an expression for the $n$th term of the sequence.

(2)

(b)

Show that 150 is not a term of the sequence.

(2)

(c)

Which term of the sequence is 107?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 5, so $5n+\dots$	✓ 1
A1 $5n+7$	✓ 1
Part (b)
M1 $5n+7=150$	✓ 1
A1 $n=28.6$, which is not an integer	✓ 1
Part (c)
M1 $5n+7=107$	✓ 1
A1 20th term	✓ 1

Exam tip: To test whether a number is in a sequence, solve nth term = number and check that n is a positive whole number.

+ Add to test✓ In your test47IGMC3-2-10r22-04Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

Calculator

During a test run a Formula 1 car's fuel mass $m$ kg is recorded against the number of laps $t$. The graph is a straight line through the points $(0,110)$ and $(25,70)$.

Diagram not to scale

(a)

Calculate the gradient of the line.

(2)

(b)

Write down the equation of the line.

(1)

(c)

Find the number of laps after which the fuel mass is 30 kg.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{70-110}{25-0}$	✓ 1
A1 $-1.6$	✓ 1
Part (b)
B1 $m=110-1.6t$ (ECF from (a))	✓ 1
Part (c)
M1 $110-1.6t=30$	✓ 1
A1 50 laps	✓ 1

Exam tip: The y-intercept gives the starting value and the gradient gives the rate of change; state both with units when interpreting.

+ Add to test✓ In your test48IGMC3-2-08r24-01Expanding, factorising and algebraic fractions Paper CEasy6 marks

No calculator

Simplify and factorise the following expressions.

(a)

Expand and simplify $4(3x - 2) - 5(x - 3)$.

(2)

(b)

Factorise completely $12pq^2 - 18p^2q$.

(2)

(c)

Simplify $\dfrac{10a^2b}{15ab^2}$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expand both brackets: $12x - 8 - 5x + 15$	✓ 1
A1 $7x + 7$	✓ 1
Part (b)
M1 partial factorisation, e.g. $6p(2q^2 - 3pq)$ or $3pq(4q - 6p)$	✓ 1
A1 $6pq(2q - 3p)$	✓ 1
Part (c)
M1 cancel a common factor, e.g. $\frac{2a^2b}{3ab^2}$ or $\frac{10a}{15b}$	✓ 1
A1 $\dfrac{2a}{3b}$	✓ 1

Common error: Factorise completely: take out the highest common factor, including the letters, in one step.

+ Add to test✓ In your test49IGMC3-2-08r24-02Functions and differentiation Paper CEasy6 marks

Calculator

On a mountain, the air temperature $T$ °C at a height of $h$ metres above base camp is modelled by $T = 15 - 0.006h$.

(a)

Calculate the temperature at a height of 1800 m above base camp.

(2)

(b)

Rearrange the formula to make $h$ the subject.

(2)

(c)

Find the height above base camp at which the temperature is −6 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute $h = 1800$: $15 - 0.006 \times 1800$	✓ 1
A1 4.2 °C	✓ 1
Part (b)
M1 isolate the $h$ term: $0.006h = 15 - T$	✓ 1
A1 $h = \dfrac{15 - T}{0.006}$	✓ 1
Part (c)
M1 $h = \dfrac{15 - (-6)}{0.006}$	✓ 1
A1 3500 m	✓ 1

Exam tip: When you substitute a negative value, put it in brackets so that $15 - (-6)$ becomes $15 + 6$.

+ Add to test✓ In your test50IGMC3-2-08r24-03Sequences and graphs of functions Paper CEasy6 marks

No calculator

A telephone mast has cable segments whose lengths, in metres, are 120, 135, 150, 165, ... The lengths form an arithmetic sequence.

(a)

Find an expression for the length of the $n$th segment.

(2)

(b)

Show that no segment is exactly 400 m long.

(2)

(c)

Which segment is exactly 300 m long?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 common difference 15	✓ 1
B1 $15n + 105$	✓ 1
Part (b)
M1 $15n + 105 = 400$, so $n = \frac{295}{15} = \frac{59}{3}$	✓ 1
A1 $n$ is not a whole number, so there is no such segment	✓ 1
Part (c)
M1 $15n + 105 = 300$	✓ 1
A1 13th segment	✓ 1

Exam tip: To test whether a number is in a sequence, set the nth term equal to it: the term exists only if $n$ is a positive whole number.

+ Add to test✓ In your test51IGMC3-2-08r24-04Sequences and graphs of functions Paper CEasy4 marks

Calculator

A climber starts with an oxygen cylinder holding 1800 litres. She uses 150 litres per hour, and the team's safety rule is that at least 300 litres must always be left in the cylinder.

(a)

Write an inequality for the number of hours $h$ she can use the cylinder.

(2)

(b)

Solve the inequality and write down the greatest number of hours she can use the cylinder.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $1800 - 150h$	✓ 1
B1 $1800 - 150h \geqslant 300$	✓ 1
Part (b)
M1 $150h \leqslant 1500$	✓ 1
A1 $h \leqslant 10$, so 10 hours	✓ 1

Common error: Keep the inequality sign the same when dividing by a positive number; it reverses only when you divide or multiply by a negative.

+ Add to test✓ In your test52IGMC3-2-09r25-01Quadratic equations and inequalities Paper CEasy4 marks

Calculator

A sports scientist models a sprinter's speed using the formula $v = u + at$, where $u$ is the starting speed in m/s, $v$ is the speed in m/s after $t$ seconds and $a$ is the constant acceleration in m/s².

(a)

Make $a$ the subject of the formula.

(2)

(b)

A sprinter has $u = 1.2$, $v = 9.2$ and $t = 4.0$. Calculate the acceleration.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $v - u = at$	✓ 1
A1 $a = \frac{v-u}{t}$	✓ 1
Part (b)
M1 substitutes into $a = \frac{9.2-1.2}{4.0}$	✓ 1
A1 $a = 2$ m/s²	✓ 1

Common error: When rearranging, do the same operation to every term on both sides: subtracting $u$ from only one side gives $a = v - u/t$.

+ Add to test✓ In your test53IGMC3-2-09r25-02Linear and simultaneous equations and rearranging formulae Paper CEasy7 marks

Calculator

On a regional rail line, a family of 3 adults and 2 children pays £47 in total for tickets. A group of 1 adult and 4 children pays £39 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 the equations to find $a$ and $c$.

(3)

(c)

Find the total cost of tickets for 2 adults and 3 children.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $3a + 2c = 47$	✓ 1
B1 $a + 4c = 39$	✓ 1
Part (b)
M1 multiplies or rearranges to make one pair of coefficients equal (for example $3a + 12c = 117$)	✓ 1
A1 $c = 7$	✓ 1
A1 $a = 11$	✓ 1
Part (c)
M1 $2 \times 11 + 3 \times 7$ (ECF from (b))	✓ 1
A1 £43	✓ 1

Exam tip: Check your solution by substituting both values back into the original equations, not only the one you used last.

+ Add to test✓ In your test54IGMC3-2-09r25-03Expanding, factorising and algebraic fractions Paper CEasy4 marks

No calculator

A rectangular solar panel on a satellite has length $(x+4)$ m and width $(x-7)$ m, where $x > 7$.

Diagram not to scale

(a)

Expand and simplify $(x+4)(x-7)$ to find an expression for the area of the panel.

(2)

(b)

Factorise completely $8p^2 - 12p$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands to four terms ($x^2 - 7x + 4x - 28$)	✓ 1
A1 $x^2 - 3x - 28$	✓ 1
Part (b)
B1 takes out a common factor ($4$, $p$ or $2p$)	✓ 1
A1 $4p(2p - 3)$	✓ 1

Common error: Factorising completely means no common factor is left inside the bracket: $2p(4p-6)$ is not fully factorised.

+ Add to test✓ In your test55IGMC3-2-09r25-06Functions and differentiation Paper CEasy6 marks

Calculator

After a training run, an athlete's heart rate $H$ beats per minute is modelled by $H = 168 - 12t$, where $t$ is the time in minutes after stopping and $0 \le t \le 10$.

(a)

State what the value $-12$ represents in this model.

(1)

(b)

Find $t$ when $H = 96$.

(2)

(c)

Solve $168 - 12t < 120$ and explain what the answer tells you about the heart rate.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the heart rate falls by 12 beats per minute for every minute of recovery (gradient / rate of change)	✓ 1
Part (b)
M1 $168 - 12t = 96$	✓ 1
A1 $t = 6$ minutes	✓ 1
Part (c)
M1 $-12t < -48$	✓ 1
A1 $t > 4$ (inequality reversed when dividing by $-12$)	✓ 1
B1 the heart rate is below 120 beats per minute for more than 4 minutes after stopping	✓ 1

Common error: Dividing both sides of an inequality by a negative number reverses the inequality sign.

+ Add to test✓ In your test56IGMC3-2-12r27-01Expanding, factorising and algebraic fractions Paper CEasy6 marks

No calculator

Answer all parts of this question.

(a)

Expand and simplify 3(2x − 5) − 4(x − 3).

(2)

(b)

Factorise completely 12x²y − 18xy².

(2)

(c)

Factorise x² + x − 30.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6x − 15 and −4x + 12 (attempt to expand both brackets)	✓ 1
A1 2x − 3	✓ 1
Part (b)
B1 a correct common factor taken outside brackets (e.g. 2, 3, x or y or a product of these)	✓ 1
B1 6xy(2x − 3y)	✓ 1
Part (c)
M1 (x + a)(x + b) with ab = −30	✓ 1
A1 (x + 6)(x − 5)	✓ 1

Exam tip: “Factorise completely” means no common factor may be left inside the brackets, so check that the terms in the bracket share no further factor.

+ Add to test✓ In your test57IGMC3-2-12r27-02Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

Calculator

A marine biology team hires a boat to survey a reef. The cost, £C, for a group of n divers is given by C = 120 + 35n.

(a)

Work out the cost for a group of 6 divers.

(1)

(b)

Rearrange the formula to make n the subject.

(2)

(c)

A group is charged £540. Work out the number of divers in the group.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 £330	✓ 1
Part (b)
M1 C − 120 = 35n	✓ 1
A1 n = (C − 120) ÷ 35	✓ 1
Part (c)
M1 (540 − 120) ÷ 35	✓ 1
A1 12 divers	✓ 1

Common error: Undo the operations in reverse order: subtract the fixed charge first, then divide by the cost per diver; dividing the whole of C by 35 first is a common slip.

+ Add to test✓ In your test58IGMC3-2-12r27-03Sequences and graphs of functions Paper CEasy7 marks

Calculator

In a warehouse, pallets are stacked in rows. The first four rows contain 5, 8, 11 and 14 pallets, and the pattern continues.

(a)

Write down the number of pallets in the fifth and sixth rows.

(1)

(b)

Find an expression for the number of pallets in the nth row.

(2)

(c)

One row contains exactly 80 pallets. Find the number of this row.

(2)

(d)

Explain why no row can contain exactly 100 pallets.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 17 and 20	✓ 1
Part (b)
B1 3n	✓ 1
B1 3n + 2	✓ 1
Part (c)
M1 3n + 2 = 80	✓ 1
A1 26	✓ 1
Part (d)
M1 3n + 2 = 100 gives n = 98 ÷ 3	✓ 1
A1 32.67… is not a whole number, so there is no such row	✓ 1

Exam tip: For a linear sequence, the common difference gives the coefficient of n; test it by substituting n = 1 into your expression and comparing with the first term.

+ Add to test✓ In your test59IGMC3-2-12r27-04Functions and differentiation Paper CEasy6 marks

Calculator

The area, A m², of a trapezium-shaped window pane is given by A = ½(a + b)h, where a and b are the lengths of the parallel sides in metres and h is the distance between them in metres.

Diagram not to scale

(a)

Find A when a = 1.2, b = 0.8 and h = 1.5.

(2)

(b)

Rearrange the formula to make h the subject.

(2)

(c)

A different pane has A = 3.6, a = 2.4 and b = 1.2. Find h.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × (1.2 + 0.8) × 1.5	✓ 1
A1 1.5	✓ 1
Part (b)
M1 2A = (a + b)h	✓ 1
A1 h = 2A ÷ (a + b)	✓ 1
Part (c)
M1 2 × 3.6 ÷ (2.4 + 1.2)	✓ 1
A1 2 m	✓ 1

Common error: When you multiply both sides by 2 you must keep (a + b) as a single bracket when you divide, not divide by a alone.

+ Add to test✓ In your test60IGMC3-2-11r28-01Expanding, factorising and algebraic fractions Paper CEasy5 marks

No calculator

A shop manager works with algebraic price expressions.

(a)

Expand and simplify $3(2x-5)-4(x-2)$.

(2)

(b)

Factorise completely $12ab^2-18a^2b$.

(2)

(c)

Factorise $y^2-49$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to expand both brackets: $6x-15-4x+8$	✓ 1
A1 $2x-7$	✓ 1
Part (b)
B1 partial factorisation, e.g. $6(2ab^2-3a^2b)$ or $ab(12b-18a)$	✓ 1
B1 $6ab(2b-3a)$	✓ 1
Part (c)
B1 $(y-7)(y+7)$	✓ 1

Exam tip: Check a factorised answer by expanding it again; you should get back the original expression.

+ Add to test✓ In your test61IGMC3-2-11r28-02Quadratic equations and inequalities Paper CEasy6 marks

Calculator

A shop's monthly cost £$C$ for stocking $n$ items of one product is given by $C=F+pn$, where £$F$ is a fixed charge and £$p$ is the cost of each item.

(a)

Solve $4(2x+1)-3=5x+10$.

(2)

(b)

Make $n$ the subject of $C=F+pn$.

(2)

(c)

Find $n$ when $C=60$, $F=12$ and $p=0.80$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expand and collect: $8x+1=5x+10$, so $3x=9$	✓ 1
A1 $x=3$	✓ 1
Part (b)
M1 $C-F=pn$	✓ 1
A1 $n=\dfrac{C-F}{p}$	✓ 1
Part (c)
M1 $\dfrac{60-12}{0.80}$ (ECF from (b))	✓ 1
A1 $n=60$	✓ 1

Common error: When rearranging, subtract the fixed term $F$ from both sides before dividing by $p$; dividing only $pn$ by $p$ gives the wrong formula.

+ Add to test✓ In your test62IGMC3-2-11r28-03Sequences and graphs of functions Paper CEasy6 marks

Calculator

A solar installation firm fits panels to new homes. The crew fits 12 panels in week 1, 17 panels in week 2 and 22 panels in week 3, and the pattern continues in the same way.

(a)

Write down the number of panels fitted in week 4 and in week 5.

(2)

(b)

Find an expression for the number of panels fitted in week $n$.

(2)

(c)

In week $k$ the crew fits 112 panels. Find $k$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 27	✓ 1
B1 32	✓ 1
Part (b)
M1 common difference 5, so $5n+\dots$	✓ 1
A1 $5n+7$	✓ 1
Part (c)
M1 $5k+7=112$ (ECF from (b))	✓ 1
A1 $k=21$	✓ 1

Exam tip: Test your nth term by substituting $n=1$ and $n=2$; it must reproduce the first two terms.

+ Add to test✓ In your test63IGMC3-2-11r28-04Functions and differentiation Paper CEasy7 marks

Calculator

A courier charges a fixed fee plus a rate per kilometre, so the charge £$y$ for a delivery of $x$ km lies on a straight line. A 2 km delivery costs £7 and a 6 km delivery costs £19.

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line in the form $y=mx+c$.

(2)

(c)

Find the charge for a 15 km delivery.

(2)

(d)

State what the gradient represents in this context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\dfrac{19-7}{6-2}$	✓ 1
A1 3	✓ 1
Part (b)
M1 substitute a point into $y=3x+c$, e.g. $7=3(2)+c$	✓ 1
A1 $y=3x+1$	✓ 1
Part (c)
M1 $3\times15+1$ (ECF from (b))	✓ 1
A1 £46	✓ 1
Part (d)
B1 the charge per kilometre (£3 for each extra km)	✓ 1

Exam tip: Give the meaning of a gradient or intercept using the units of the problem, for example pounds per kilometre.

+ Add to test✓ In your test64IGMC3-2-12r28-01Expanding, factorising and algebraic fractions Paper CEasy6 marks

No calculator

An archaeology team marks out a rectangular trench of length $(x+5)$ metres and width $(x-2)$ metres.

Diagram not to scale

(a)

Expand and simplify the expression for the area of the trench.

(2)

(b)

Find the area when $x=6$.

(2)

(c)

Factorise $x^2-3x-28$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 four terms: $x^2-2x+5x-10$	✓ 1
A1 $x^2+3x-10$	✓ 1
Part (b)
M1 $6^2+3(6)-10$ or $11\times4$	✓ 1
A1 44 m²	✓ 1
Part (c)
M1 two numbers with product $-28$ and sum $-3$	✓ 1
A1 $(x-7)(x+4)$	✓ 1

Exam tip: Check an expansion by substituting a simple value such as $x=1$ into both the factorised and expanded forms.

+ Add to test✓ In your test65IGMC3-2-12r28-02Linear and simultaneous equations and rearranging formulae Paper CEasy6 marks

Calculator

A cycle-hire company charges £$C$ to hire a bike for $h$ hours, where $C=15+6h$.

(a)

Solve $7x-4=3x+20$.

(2)

(b)

Make $h$ the subject of $C=15+6h$.

(2)

(c)

A customer pays £63. For how many hours was the bike hired?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $4x=24$	✓ 1
A1 $x=6$	✓ 1
Part (b)
M1 $C-15=6h$	✓ 1
A1 $h=\dfrac{C-15}{6}$	✓ 1
Part (c)
M1 $\dfrac{63-15}{6}$ (ECF from (b))	✓ 1
A1 8 hours	✓ 1

Common error: Subtract the fixed charge from both sides before dividing by 6; dividing only $15$ by $6$ is a frequent slip.

+ Add to test✓ In your test66IGMC3-2-12r28-03Sequences and graphs of functions Paper CEasy6 marks

Calculator

A data centre has 200 free storage slots at the start of week 1. The numbers of free slots at the start of weeks 1, 2, 3, 4 are 200, 188, 176, 164, and the pattern continues.

(a)

Write down the number of free slots at the start of weeks 5 and 6.

(2)

(b)

Find an expression for the number of free slots at the start of week $n$.

(2)

(c)

In which week are there exactly 8 free slots?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 152	✓ 1
B1 140	✓ 1
Part (b)
M1 common difference $-12$, so $-12n+\dots$	✓ 1
A1 $212-12n$	✓ 1
Part (c)
M1 $212-12n=8$ (ECF from (b))	✓ 1
A1 week 17	✓ 1

Exam tip: For a decreasing linear sequence the coefficient of $n$ is the common difference, which is negative.

+ Add to test✓ In your test67IGMC3-2-12r28-04Sequences and graphs of functions Paper CEasy6 marks

Calculator

A data centre upgrade takes $t$ hours when $n$ technicians work together at the same rate, where $t=\dfrac{36}{n}$.

(a)

Find the values of $t$ when $n=2$, $n=4$ and $n=9$.

(3)

(b)

Describe how $t$ changes as $n$ increases.

(1)

(c)

Find the number of technicians needed to complete the upgrade in 3 hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18	✓ 1
B1 9	✓ 1
B1 4	✓ 1
Part (b)
B1 $t$ decreases as $n$ increases	✓ 1
Part (c)
M1 $\dfrac{36}{n}=3$ so $n=\dfrac{36}{3}$	✓ 1
A1 12 technicians	✓ 1

Common error: In $t=\dfrac{36}{n}$ you divide 36 by $n$; for a given time, divide 36 by the time to find $n$.

+ Add to test✓ In your test68IGMC3-2-10r29-01Expanding, factorising and algebraic fractions Paper CEasy6 marks

No calculator

Answer all parts, showing your working.

(a)

Expand and simplify 4(3x + 2) − 5(x − 1).

(2)

(b)

Factorise fully 15xy + 9y.

(2)

(c)

Factorise x² − 5x − 14.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands both brackets correctly: 12x + 8 − 5x + 5	✓ 1
A1 7x + 13	✓ 1
Part (b)
B1 3 or y taken out as a common factor	✓ 1
B1 3y(5x + 3)	✓ 1
Part (c)
B1 (x − 7)	✓ 1
B1 (x + 2)	✓ 1

Common error: When expanding a bracket with a minus sign in front, change the sign of every term inside it.

+ Add to test✓ In your test69IGMC3-2-10r29-02Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

Calculator

A beekeeper sells honey in small jars and large jars. Three small jars and two large jars cost £15.60 in total. Five small jars and one large jar cost £16.20 in total.

(a)

Let a small jar cost £x and a large jar cost £y. Write down two equations in x and y.

(2)

(b)

Solve the equations to find the cost of one small jar and the cost of one large jar.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3x + 2y = 15.6	✓ 1
B1 5x + y = 16.2	✓ 1
Part (b)
M1 eliminates one variable, e.g. y = 16.2 − 5x substituted: 3x + 32.4 − 10x = 15.6	✓ 1
A1 small jar £2.40	✓ 1
A1 large jar £4.20	✓ 1

Exam tip: Substitute your answers back into both original equations to check them.

+ Add to test✓ In your test70IGMC3-2-10r29-09Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

Calculator

A hospital orders boxes of gloves. Each box costs £14 and the delivery charge is £12 for the whole order. The hospital has a budget of £152.

(a)

The hospital orders n boxes. Write down an inequality in n for the total cost.

(1)

(b)

Solve your inequality and find the greatest number of boxes the hospital can order.

(2)

(c)

Solve 3(x + 2) > 2x + 11.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 14n + 12 ≤ 152	✓ 1
Part (b)
M1 14n ≤ 140	✓ 1
A1 n ≤ 10 so greatest number is 10	✓ 1
Part (c)
M1 3x + 6 > 2x + 11	✓ 1
A1 x > 5	✓ 1

Exam tip: Check the final answer of an inequality by testing a value inside the range and one outside it.

+ Add to test✓ In your test71IGMC3-2-10r29-14Linear and simultaneous equations and rearranging formulae Paper CEasy5 marks

Calculator

A hospital has three wards. Each ward has (2x − 1) empty beds. A fourth unit has (x + 17) empty beds. The total number of empty beds in the three wards is equal to the number of empty beds in the fourth unit.

(a)

Write down an equation in x.

(1)

(b)

Solve your equation to find x.

(3)

(c)

Find the number of empty beds in each of the three wards.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3(2x − 1) = x + 17	✓ 1
Part (b)
M1 6x − 3 = x + 17	✓ 1
M1 5x = 20	✓ 1
A1 x = 4	✓ 1
Part (c)
B1 2 × 4 − 1 = 7	✓ 1

Common error: Multiply the bracket by 3 on every term: 3(2x − 1) is 6x − 3, not 6x − 1.

+ Add to test✓ In your test72IGMC3-2-10r35-01Expanding, factorising and algebraic fractions Paper CEasy4 marks

No calculator

An airline engineer works with algebraic expressions.

(a)

Expand and simplify $3(2x-5)-4(x-2)$.

(2)

(b)

Factorise completely $12a^2b-18ab^2$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands both brackets: $6x-15-4x+8$	✓ 1
A1 $2x-7$	✓ 1
Part (b)
B1 partial factorisation, e.g. $6a(2ab-3b^2)$ or $6ab(\ldots)$	✓ 1
B1 $6ab(2a-3b)$	✓ 1

Common error: 'Factorise completely' means taking out the highest common factor, including every common letter.

+ Add to test✓ In your test73IGMC3-2-10r35-02Quadratic equations and inequalities Paper CEasy5 marks

Calculator

A farmer buys x sacks of seed at €14 each and pays €86 for fencing wire. The total cost is €240.

(a)

Write down an equation in x.

(1)

(b)

Solve your equation to find the number of sacks.

(2)

(c)

The farmer's total cost, €$C$, for $n$ sacks is given by $C=14n+86$. Make $n$ the subject of the formula.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $14x+86=240$	✓ 1
Part (b)
M1 $14x=240-86=154$	✓ 1
A1 $x=11$	✓ 1
Part (c)
M1 $C-86=14n$	✓ 1
A1 $n=\dfrac{C-86}{14}$	✓ 1

Exam tip: Check your answer by substituting it back: 14 × 11 + 86 = 240.

+ Add to test✓ In your test74IGMC3-2-10r35-14Functions and differentiation Paper CEasy6 marks

Calculator

The cost, €y, of irrigating a crop is a linear function of the area, x hectares. The line passes through (2, 5) and (6, 17).

Diagram not to scale

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line.

(2)

(c)

Find the value of x when $y=26$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\dfrac{17-5}{6-2}$	✓ 1
A1 3	✓ 1
Part (b)
M1 substitutes a point into $y=3x+c$	✓ 1
A1 $y=3x-1$	✓ 1
Part (c)
M1 $3x-1=26$	✓ 1
A1 $x=9$	✓ 1

Exam tip: Use the second point to check the equation you found: $3\times6-1=17$.

+ Add to test✓ In your test75IGMC3-2-12r46-01Quadratic equations and inequalities Paper CEasy5 marks

Calculator

A tailor buys 3 m of silk at £x per metre and 2 m of lining at £4.50 per metre. The total cost is £27.

(a)

Write an equation in x and solve it to find the cost of silk per metre.

(3)

(b)

The tailor then buys 5 m of silk and 4 m of lining at the same prices. Calculate the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $3x+2\times4.5=27$ (or $3x+9=27$)	✓ 1
A1 $3x=18$	✓ 1
A1 $x=6$, so £6 per metre	✓ 1
Part (b)
M1 $5\times6+4\times4.5$	✓ 1
A1 £48	✓ 1

Exam tip: Define the letter you use in an equation, then check your answer by substituting it back into the original situation.

+ Add to test✓ In your test76IGMC3-2-12r46-02Sequences and graphs of functions Paper CEasy6 marks

A fisheries student counts the fish in a series of tanks. The numbers of fish in tanks 1, 2, 3 and 4 are 12, 19, 26 and 33.

(a)

Write down the number of fish in tank 5.

(1)

(b)

Find an expression for the number of fish in tank n.

(2)

(c)

The first tank number in which there are more than 200 fish is tank k. Find k and the number of fish in tank k.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 40	✓ 1
Part (b)
M1 common difference 7 so $7n$	✓ 1
A1 $7n+5$	✓ 1
Part (c)
M1 $7n+5>200$ or $7n+5=200$ solved	✓ 1
A1 $k=28$ (since $n>27.86$)	✓ 1
A1 201 fish	✓ 1

Common error: A solution to 7n + 5 = 200 is not a whole number, so round up to the next whole tank to get the first value above 200.

+ Add to test✓ In your test77IGMC3-2-12r46-03Expanding, factorising and algebraic fractions Paper CEasy5 marks

No calculator

A robot's base plate is a rectangle with length (x + 4) cm and width (x − 7) cm.

Diagram not to scale

(a)

Expand and simplify (x + 4)(x − 7).

(2)

(b)

Factorise x² − 5x − 24.

(2)

(c)

Find the area of the plate when x = 10.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least three of four terms correct, $x^2-7x+4x-28$	✓ 1
A1 $x^2-3x-28$	✓ 1
Part (b)
M1 $(x+a)(x+b)$ with $ab=-24$ or $a+b=-5$	✓ 1
A1 $(x-8)(x+3)$	✓ 1
Part (c)
B1 $42$ cm² (from $14\times3$ or by substituting into the answer to (a))	✓ 1

Common error: When expanding two brackets, take care with the sign of the middle terms: +4x and −7x combine to −3x.

+ Add to test✓ In your test78IGMC3-2-12r46-07Linear and simultaneous equations and rearranging formulae Paper CEasy4 marks

Calculator

The volume of a conical feed hopper at a fish farm is given by V = ¹/₃πr²h, where r is the base radius and h is the height.

Cross-section of the conical hopper. Diagram not to scale

(a)

Rearrange the formula to make h the subject.

(2)

(b)

A hopper has V = 0.85 m³ and r = 0.60 m. Calculate h to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplies both sides by 3 and divides by $\pi r^2$	✓ 1
A1 $h=\frac{3V}{\pi r^2}$	✓ 1
Part (b)
M1 $\frac{3\times0.85}{\pi\times0.60^2}$	✓ 1
A1 2.25 m	✓ 1

Exam tip: Perform the same operation on both sides of the formula, one step at a time, and check units at the end.

+ Add to test✓ In your test79IGMC3-2-11r48-02Linear and simultaneous equations and rearranging formulae Paper CEasy6 marks

Calculator

A library charges a fine of £F for a book returned d days late, where F = 0.5 + 0.15d.

(a)

Work out the fine for a book returned 12 days late.

(2)

(b)

Rearrange the formula to make d the subject.

(2)

(c)

A fine is £4.25. Find the number of days the book was late.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.5 + 0.15 × 12	✓ 1
A1 £2.30	✓ 1
Part (b)
M1 F − 0.5 = 0.15d	✓ 1
A1 d = (F − 0.5) ÷ 0.15	✓ 1
Part (c)
M1 (4.25 − 0.5) ÷ 0.15	✓ 1
A1 25	✓ 1

Common error: When rearranging, subtract the fixed amount before dividing; dividing only 0.15d by 0.15 leaves the formula wrong.

+ Add to test✓ In your test80IGMC3-2-11r48-07Sequences and graphs of functions Paper CEasy7 marks

Calculator

The table shows some values of y = x² − 2x − 3.

x: −2, −1, 0, 1, 2, 3, 4
y: p, 0, −3, q, −3, r, 5

(a)

Find the values of p, q and r.

(3)

(b)

Use the table to write down the two solutions of x² − 2x − 3 = 0.

(2)

(c)

The curve is symmetrical. Write down the equation of its line of symmetry.

(1)

(d)

Write down the minimum value of y.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 p = 5	✓ 1
B1 q = −4	✓ 1
B1 r = 0	✓ 1
Part (b)
B1 x = −1	✓ 1
B1 x = 3	✓ 1
Part (c)
B1 x = 1	✓ 1
Part (d)
B1 −4	✓ 1

Exam tip: Solutions of an equation equal to zero are the x-values where y = 0 in the table.

+ Add to test✓ In your test81IGMC3-2-11r48-14Linear and simultaneous equations and rearranging formulae Paper CEasy7 marks

No calculator

Solve each equation.

(a)

5x − 3 = 2x + 12

(2)

(b)

(x + 4) ÷ 3 = 7

(2)

(c)

3(2x − 1) = 4(x + 3)

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3x = 15	✓ 1
A1 x = 5	✓ 1
Part (b)
M1 x + 4 = 21	✓ 1
A1 x = 17	✓ 1
Part (c)
M1 6x − 3 = 4x + 12	✓ 1
M1 2x = 15	✓ 1
A1 x = 7.5	✓ 1

Common error: Multiply every term inside a bracket, including the negative number: 3(2x − 1) gives 6x − 3, not 6x − 1.

+ Add to test✓ In your test82IGMC-042-01-07Linear and simultaneous equations and rearranging formulae Paper CMedium5 marks

Calculator

At a café, 3 coffees and 2 muffins cost $13.30.
5 coffees and 3 muffins cost $21.10.
A coffee costs $c and a muffin costs $m.

(a)

Write down two equations in c and m.

(2)

(b)

Solve your equations to find the cost of one coffee and the cost of one muffin.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for 3c + 2m = 13.30	✓ 1
B1 for 5c + 3m = 21.10	✓ 1
Part (b)
M1 for multiplying to give equal coefficients and eliminating one variable, e.g. 10c + 6m = 42.20 minus 9c + 6m = 39.90	✓ 1
A1 for c = 2.30, so a coffee costs $2.30	✓ 1
A1 for m = 3.20, so a muffin costs $3.20	✓ 1

Exam tip: Money answers should be written to 2 decimal places, e.g. $3.20, not $3.2.

+ Add to test✓ In your test83IGMC-042-01-10Linear and simultaneous equations and rearranging formulae Paper CMedium4 marks

No calculator

The three angles of a triangle are (2x + 10)°, (3x − 5)° and (x + 25)°.

(a)

Write down an equation in x and solve it.

(3)

(b)

Find the size of the largest angle.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (2x + 10) + (3x − 5) + (x + 25) = 180	✓ 1
A1 for 6x + 30 = 180	✓ 1
A1 for x = 25	✓ 1
Part (b)
B1 for 70°	✓ 1

Exam tip: Check your answer by working out all three angles (60°, 70°, 50°) and confirming that they add up to 180°.

+ Add to test✓ In your test84IGMC-042-01-12Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

No calculator

A straight line L passes through the points (−2, 7) and (4, −5).

Diagram not to scale

(a)

Find the equation of L in the form y = mx + c.

(3)

(b)

Find the coordinates of the point where L meets the line y = x − 6.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{-5-7}{4-(-2)}$	✓ 1
A1 for gradient = −2	✓ 1
A1 for y = −2x + 3	✓ 1
Part (b)
M1 for −2x + 3 = x − 6	✓ 1
A1 for x = 3	✓ 1
A1 for (3, −3)	✓ 1

Exam tip: Substitute one of the given points into your equation to check it. For example, at x = −2: −2(−2) + 3 = 7 ✓.

+ Add to test✓ In your test85IGMC-042-03-06Sequences and graphs of functions Paper CMedium5 marks

No calculator

The first five terms of a sequence are
4, 7, 12, 19, 28, ...

(a)

Write down the next term.

(1)

(b)

Find an expression, in terms of n, for the nth term.

(2)

(c)

Which term of the sequence is equal to 403?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 39	✓ 1
Part (b)
B2 n² + 3	✓ 2
(B1 for n² seen in an expression, or for second differences of 2 identified)	—
Part (c)
M1 for n² + 3 = 403 leading to n² = 400	✓ 1
A1 20th term	✓ 1

Exam tip: When the first differences increase by 2 each time, compare the sequence with the square numbers 1, 4, 9, 16, ... and look at what has been added.

+ Add to test✓ In your test86IGMC-042-03-09Sequences and graphs of functions Paper CMedium8 marks

Calculator

The graphs of y = $\frac{12}{x}$ and y = x + 1 intersect at two points, A and B.

Diagram not to scale

(a)

The point P lies on the curve y = $\frac{12}{x}$ and has x-coordinate 2.5. Find the y-coordinate of P.

(1)

(b)

Show that the x-coordinates of A and B satisfy the equation x² + x − 12 = 0.

(2)

(c)

Find the coordinates of A and B.

(3)

(d)

Calculate the length of AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4.8	✓ 1
Part (b)
M1 for $\frac{12}{x}$ = x + 1	✓ 1
A1 for 12 = x² + x leading to x² + x − 12 = 0 with no errors	✓ 1
Part (c)
M1 for (x + 4)(x − 3) = 0, or correct substitution into the quadratic formula, or equivalent valid method	✓ 1
A1 x = 3 and x = −4	✓ 1
A1 (3, 4) and (−4, −3), accepted with either point labelled A	✓ 1
Part (d)
M1 for $\sqrt{(3-(-4))^2 + (4-(-3))^2}$ or $\sqrt{7^2 + 7^2}$	✓ 1
A1 9.90, also accepting 9.9, 9.899… or $7\sqrt{2}$	✓ 1

Exam tip: In a 'show that' part, write down every step. Show the multiplication of both sides by x clearly before giving the final rearranged equation.

+ Add to test✓ In your test87IGMC3-2-08c-06Functions and differentiation Paper CMedium6 marks

Calculator

The formula to convert a temperature in degrees Celsius (C) to degrees Fahrenheit (F) is F = 1.8C + 32.

(a)

Rearrange the formula to make C the subject.

(2)

(b)

A sensor reads 104 °F. Calculate the temperature in °C.

(2)

(c)

At one temperature the Celsius and Fahrenheit readings are equal. Find this temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 F − 32 = 1.8C	✓ 1
A1 C = (F − 32) ÷ 1.8	✓ 1
Part (b)
M1 (104 − 32) ÷ 1.8	✓ 1
A1 40 °C	✓ 1
Part (c)
M1 C = 1.8C + 32	✓ 1
A1 −40	✓ 1

Exam tip: When both readings are equal, replace F by C in the formula and solve for one letter.

+ Add to test✓ In your test88IGMC3-2-08c-07Linear and simultaneous equations and rearranging formulae Paper CMedium5 marks

Calculator

The depth of water in a tank, h cm, is measured t hours after filling starts. The graph of h against t is a straight line passing through (2, 5) and (6, 13).

Diagram not to scale

(a)

Calculate the gradient of the line.

(2)

(b)

Find the equation of the line in the form h = mt + c.

(2)

(c)

Find the depth of water after 10 hours.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (13 − 5) ÷ (6 − 2)	✓ 1
A1 2	✓ 1
Part (b)
M1 5 = 2 × 2 + c or equivalent	✓ 1
A1 h = 2t + 1	✓ 1
Part (c)
B1 21 cm (ECF from (b))	✓ 1

Common error: Subtract the coordinates in the same order in the numerator and the denominator.

+ Add to test✓ In your test89IGMC3-2-08c-08Quadratic equations and inequalities Paper CMedium5 marks

No calculator

This question is about equations.

(a)

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

(2)

(b)

Solve $\frac{2x+1}{3} = \frac{x-4}{2}$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4x + 12 − x + 2	✓ 1
A1 3x + 14	✓ 1
Part (b)
M1 multiply both sides by 6 (or cross-multiply)	✓ 1
M1 4x + 2 = 3x − 12	✓ 1
A1 x = -14	✓ 1

Exam tip: Multiply every term by the common denominator to clear the fractions in one step.

+ Add to test✓ In your test90IGMC3-2-08r5-04Sequences and graphs of functions Paper CMedium6 marks

Calculator

The power P kW used by a rack in a data centre depends on the number of servers n installed. The relationship is linear. A rack with 2 servers uses 9 kW and a rack with 6 servers uses 21 kW.

(a)

Calculate the gradient of the line relating P to n.

(2)

(b)

State what the gradient represents in this situation.

(1)

(c)

Find the equation of the line, giving P in terms of n.

(2)

(d)

Find the power used by a rack with 15 servers.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (21 − 9) ÷ (6 − 2)	✓ 1
A1 3	✓ 1
Part (b)
B1 each extra server uses 3 kW more power	✓ 1
Part (c)
M1 P = 3n + c with a point substituted, for example 9 = 3 × 2 + c	✓ 1
A1 P = 3n + 3	✓ 1
Part (d)
B1 3 × 15 + 3 = 48 kW (ECF from (c))	✓ 1

Exam tip: In a context question, the gradient is the change in the output for each one-unit increase in the input; include the units in your statement.

+ Add to test✓ In your test91IGMC3-2-08r5-05Linear and simultaneous equations and rearranging formulae Paper CMedium7 marks

Calculator

At a cycling shop, 3 helmets and 2 locks cost €97. Also, 2 helmets and 5 locks cost €127. Let a helmet cost €h and a lock cost €l.

(a)

Write down two equations in h and l.

(2)

(b)

Solve the equations to find h and l.

(4)

(c)

Find the total cost of 4 helmets and 4 locks.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3h + 2l = 97	✓ 1
B1 2h + 5l = 127	✓ 1
Part (b)
M1 multiply to make the coefficients of one variable equal, for example ×2 and ×3	✓ 1
M1 add or subtract the equations to eliminate that variable	✓ 1
A1 h = 21	✓ 1
A1 l = 17	✓ 1
Part (c)
B1 4 × 21 + 4 × 17 = €152 (ECF from (b))	✓ 1

Common error: After finding the first unknown, substitute into an original equation and check the answer against the other equation.

+ Add to test✓ In your test92IGMC3-2-08r5-09Quadratic equations and inequalities Paper CMedium7 marks

Calculator

A data-centre room can use at most 20 kW of electrical power. Cooling equipment uses a fixed 4 kW, and each server uses 0.35 kW. There are n servers in the room.

(a)

Write down an inequality in n for the power limit.

(1)

(b)

Solve your inequality.

(2)

(c)

Write down the greatest number of servers that can be used.

(1)

(d)

Find the number of integers x that satisfy −3 < 2x + 1 ≤ 9.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4 + 0.35n ≤ 20	✓ 1
Part (b)
M1 0.35n ≤ 16	✓ 1
A1 n ≤ 45.7	✓ 1
Part (c)
B1 45	✓ 1
Part (d)
M1 subtract 1 from every part, −4 < 2x ≤ 8	✓ 1
A1 −2 < x ≤ 4	✓ 1
A1 6 integers (−1, 0, 1, 2, 3, 4)	✓ 1

Common error: For a double inequality, do the same operation to every part, and watch whether each end is included.

+ Add to test✓ In your test93IGMC3-2-08r5-10Sequences and graphs of functions Paper CMedium8 marks

No calculator

An archaeologist numbers batches of pottery using a rule. The first four batch codes form the sequence
5, 7, 11, 19, ...
Each term is found by doubling the previous term and subtracting 3.
A second sequence is arithmetic. Its 4th term is 17 and its 7th term is 32.

(a)

Find the next two terms of the first sequence.

(2)

(b)

Find the common difference of the second sequence.

(2)

(c)

Find an expression for the nth term of the second sequence.

(2)

(d)

Which term of the second sequence is equal to 112?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 35	✓ 1
B1 67	✓ 1
Part (b)
M1 (32 − 17) ÷ 3	✓ 1
A1 5	✓ 1
Part (c)
M1 5n + c with a term substituted, for example 17 = 5 × 4 + c	✓ 1
A1 5n − 3	✓ 1
Part (d)
M1 5n − 3 = 112	✓ 1
A1 n = 23	✓ 1

Exam tip: A term-to-term rule needs the previous term each time; an nth-term formula lets you jump straight to any term.

+ Add to test✓ In your test94IGMC3-2-08r5-14Expanding, factorising and algebraic fractions Paper CMedium6 marks

Calculator

Solve each of the following.

(a)

(x + 1) ÷ 3 + (x − 2) ÷ 4 = 8

(3)

(b)

(2x − 1) ÷ 3 > x − 2

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiply through by 12 or use a common denominator	✓ 1
M1 4(x + 1) + 3(x − 2) = 96	✓ 1
A1 x = 14	✓ 1
Part (b)
M1 2x − 1 > 3(x − 2)	✓ 1
M1 collect terms, 5 > x	✓ 1
A1 x < 5	✓ 1

Common error: When you multiply an inequality by a positive number the sign stays the same; check your final range with one test value.

+ Add to test✓ In your test95IGMC3-2-08r9-03Linear and simultaneous equations and rearranging formulae Paper CMedium5 marks

No calculator

A data centre has two types of server rack. Each type-A rack uses $a$ kW and each type-B rack uses $b$ kW. Three type-A racks and two type-B racks use a total of 19 kW. Two type-A racks and five type-B racks use a total of 31 kW.

(a)

Write down two equations in $a$ and $b$.

(2)

(b)

Solve the equations to find $a$ and $b$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $3a+2b=19$	✓ 1
B1 $2a+5b=31$	✓ 1
Part (b)
M1 multiplies to equalise a coefficient, e.g. $15a+10b=95$ and $4a+10b=62$	✓ 1
A1 $a=3$	✓ 1
A1 $b=5$	✓ 1

Common error: When subtracting two equations, subtract every term, including the right-hand sides.

+ Add to test✓ In your test96IGMC3-2-08r9-06Quadratic equations and inequalities Paper CMedium5 marks

No calculator

An archaeologist digs a rectangular trench. The trench is $x$ metres wide and its length is 3 m more than its width. Its area is 28 m².

Diagram not to scale

(a)

Show that $x^2+3x-28=0$.

(2)

(b)

Solve the equation by factorising and write down the width of the trench.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 length $=x+3$	✓ 1
M1 $x(x+3)=28$ and expands to $x^2+3x-28=0$	✓ 1
Part (b)
M1 $(x+7)(x-4)=0$	✓ 1
A1 $x=-7$ or $x=4$	✓ 1
R1 rejects $-7$ because a width cannot be negative, so the width is 4 m	✓ 1

Common error: A quadratic from a context usually gives two roots; always check each against the situation.

+ Add to test✓ In your test97IGMC3-2-08r9-08Sequences and graphs of functions Paper CMedium5 marks

Calculator

A cyclist's training plan has the following weekly distances: week 1 is 30 km, week 2 is 36 km, week 3 is 42 km, and the pattern continues as an arithmetic sequence.

(a)

Find an expression for the distance, in km, in week $n$.

(2)

(b)

Find the distance in week 12.

(1)

(c)

Find the first week in which the distance is more than 150 km.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference is 6	✓ 1
A1 $24+6n$	✓ 1
Part (b)
A1 96 km	✓ 1
Part (c)
M1 $24+6n>150$, so $n>21$	✓ 1
A1 week 22	✓ 1

Exam tip: With 'more than', test the boundary value: week 21 gives exactly 150 km, which is not more than 150.

+ Add to test✓ In your test98IGMC3-2-08r9-13Expanding, factorising and algebraic fractions Paper CMedium6 marks

No calculator

Answer all three parts.

(a)

Expand and simplify $(x+4)(x-6)$.

(2)

(b)

Factorise $x^2-5x-14$.

(2)

(c)

Factorise completely $2x^2-18$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 four terms: $x^2-6x+4x-24$	✓ 1
A1 $x^2-2x-24$	✓ 1
Part (b)
M1 $(x+a)(x+b)$ with $ab=-14$ or $a+b=-5$	✓ 1
A1 $(x-7)(x+2)$	✓ 1
Part (c)
B1 $2(x^2-9)$	✓ 1
B1 $2(x-3)(x+3)$	✓ 1

Exam tip: Always take out a common factor first; it often reveals a difference of two squares.

+ Add to test✓ In your test99IGMC3-2-08r9-14Functions and differentiation Paper CMedium5 marks

Calculator

A potter's bowl found at an archaeological site has a cross-section modelled by the curve $y=x^2-4x-5$, where $x$ and $y$ are in centimetres.

(a)

Find the value of $y$ when $x=-2$ and when $x=3$.

(2)

(b)

Find the $x$-coordinates of the points where the curve crosses the $x$-axis.

(2)

(c)

Write down the equation of the line of symmetry of the curve.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $y=7$ when $x=-2$	✓ 1
B1 $y=-8$ when $x=3$	✓ 1
Part (b)
M1 solves $(x-5)(x+1)=0$	✓ 1
A1 $x=-1$ and $x=5$	✓ 1
Part (c)
B1 $x=2$ (midway between $-1$ and $5$)	✓ 1

Exam tip: The line of symmetry of a quadratic lies halfway between its two roots.

+ Add to test✓ In your test100IGMC3-2-08r10-03Sequences and graphs of functions Paper CMedium7 marks

Calculator

A cyclist rides at a steady speed along a straight road. Her distance s km from the start of the road is related to the time t hours by a straight-line graph. At $t = 0.5$ she is 12 km from the start and at $t = 2.5$ she is 52 km from the start.

Diagram not to scale

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line in the form $s = mt + c$.

(2)

(c)

State what the value 2 in your equation represents.

(1)

(d)

Find the time at which she is 100 km from the start.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{52-12}{2.5-0.5}$	✓ 1
A1 20	✓ 1
Part (b)
M1 substitute a point into $s = 20t + c$	✓ 1
A1 $s = 20t + 2$	✓ 1
Part (c)
B1 her distance from the start of the road at $t = 0$ is 2 km	✓ 1
Part (d)
M1 $100 = 20t + 2$	✓ 1
A1 $t = 4.9$ hours	✓ 1

Exam tip: The gradient of a distance-time line is the speed; the intercept is the starting distance.

+ Add to test✓ In your test101IGMC3-2-08r10-04Quadratic equations and inequalities Paper CMedium6 marks

No calculator

A rectangular plot on an archaeological site has length $(3x - 5)$ m and width $(x + 4)$ m.

Diagram not to scale

(a)

Write down and simplify an expression for the perimeter of the plot.

(2)

(b)

The perimeter is 62 m. Find the value of x.

(2)

(c)

Find the area of the plot.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $2(3x - 5) + 2(x + 4)$	✓ 1
A1 $8x - 2$	✓ 1
Part (b)
M1 $8x - 2 = 62$	✓ 1
A1 $x = 8$	✓ 1
Part (c)
M1 length 19 and width 12 found	✓ 1
A1 228 m²	✓ 1

Common error: Multiply both terms in a bracket by the number outside it, including the sign.

+ Add to test✓ In your test102IGMC3-2-08r10-11Functions and differentiation Paper CMedium6 marks

Calculator

An archaeologist models the cross-section of a ditch by $y = x^2 - 4x - 1$, where x m is the horizontal distance and y m is the height of the ditch profile relative to ground level. Some values are shown in the table.

x: −1, 0, 1, 2, 3, 4, 5
y: 4, −1, p, q, r, −1, 4

(a)

Find the values of p, q and r.

(2)

(b)

Write down the equation of the line of symmetry of the graph of $y = x^2 - 4x - 1$.

(1)

(c)

Write down the coordinates of the lowest point of the graph.

(1)

(d)

The profile meets ground level where $y = 0$. State how many such points there are and give a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 two of the three values correct	✓ 1
B1 $p = -4$, $q = -5$, $r = -4$	✓ 1
Part (b)
B1 $x = 2$	✓ 1
Part (c)
B1 $(2, -5)$	✓ 1
Part (d)
B1 two	✓ 1
B1 the graph is a U-shape with its lowest point below the x-axis, so it crosses the axis twice	✓ 1

Common error: Use the symmetry of the table values to check your answers: equal y-values sit equally either side of the line of symmetry.

+ Add to test✓ In your test103IGMC3-2-08r10-13Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

Calculator

A cyclist's kinetic energy is $E = \frac{1}{2}mv^2$ and the force on the bike is $F = ma + R$, where E is in joules, m in kg, v in m/s, F and R in newtons and a in m/s².

(a)

Make m the subject of $E = \frac{1}{2}mv^2$.

(2)

(b)

A cyclist and bike have kinetic energy 1350 J at a speed of 6.0 m/s. Find their total mass.

(2)

(c)

Make a the subject of $F = ma + R$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiply both sides by 2	✓ 1
A1 $m = \frac{2E}{v^2}$	✓ 1
Part (b)
M1 $\frac{2 \times 1350}{6.0^2}$	✓ 1
A1 75 kg	✓ 1
Part (c)
M1 subtract R from both sides	✓ 1
A1 $a = \frac{F - R}{m}$	✓ 1

Exam tip: Undo operations in the reverse order to how they were applied to the subject.

+ Add to test✓ In your test104IGMC3-2-10r10-05Quadratic equations and inequalities Paper CMedium5 marks

Calculator

An astronomy club has a budget for hiring a telescope.

(a)

Solve 4 − 3x ≥ 19.

(2)

(b)

The telescope costs a €45 booking fee plus €12 for each night. The club can spend at most €189. Write down an inequality for n, the number of nights, and find the greatest number of nights the club can hire it.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −3x ≥ 15	✓ 1
A1 x ≤ −5	✓ 1
Part (b)
B1 45 + 12n ≤ 189	✓ 1
M1 12n ≤ 144	✓ 1
A1 n ≤ 12, so the greatest number of nights is 12	✓ 1

Common error: Reverse the inequality sign whenever you multiply or divide both sides by a negative number.

+ Add to test✓ In your test105IGMC3-2-10r10-06Quadratic equations and inequalities Paper CMedium6 marks

Calculator

A rail-link shuttle charges a fixed fee plus a rate per kilometre. The charge €C for a journey of d km is a straight-line relation. A 2 km journey costs €7 and a 6 km journey costs €19.

(a)

Find the charge per kilometre.

(2)

(b)

Find the fixed fee.

(2)

(c)

Find the charge for a journey of 15 km.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (19 − 7) ÷ (6 − 2)	✓ 1
A1 €3	✓ 1
Part (b)
M1 7 = 3 × 2 + c	✓ 1
A1 €1	✓ 1
Part (c)
M1 3 × 15 + 1	✓ 1
A1 €46	✓ 1

Exam tip: Gradient = change in y ÷ change in x; use one given point to find the intercept.

+ Add to test✓ In your test106IGMC3-2-10r10-09Quadratic equations and inequalities Paper CMedium6 marks

Calculator

A train journey lasts 4 hours and covers 330 km in total. The train travels at 90 km/h for t hours and then at 60 km/h for the rest of the journey.

(a)

Write down, in terms of t, the time spent at 60 km/h.

(1)

(b)

Form an equation in t and solve it.

(3)

(c)

Calculate the average speed for the whole journey.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4 − t	✓ 1
Part (b)
M1 90t + 60(4 − t) = 330	✓ 1
M1 30t = 90	✓ 1
A1 t = 3	✓ 1
Part (c)
M1 330 ÷ 4	✓ 1
A1 82.5 km/h	✓ 1

Exam tip: Distance = speed × time for each stage; add the stages to compare with the total.

+ Add to test✓ In your test107IGMC3-2-10r10-10Expanding, factorising and algebraic fractions Paper CMedium6 marks

Calculator

The roof of an observatory carries a rectangular solar panel of width (2x + 3) m and length (x − 5) m, where x > 5.

Diagram not to scale

(a)

Expand and simplify (2x + 3)(x − 5).

(2)

(b)

Find the area of the panel when x = 9.

(2)

(c)

Factorise completely 8st − 12t.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2x² − 10x + 3x − 15	✓ 1
A1 2x² − 7x − 15	✓ 1
Part (b)
M1 (2 × 9 + 3) × (9 − 5) = 21 × 4	✓ 1
A1 84 m²	✓ 1
Part (c)
M1 common factor 4t or 2t or 4	✓ 1
A1 4t(2s − 3)	✓ 1

Common error: When expanding two brackets, multiply every term in the first by every term in the second, and take care with the negative signs.

+ Add to test✓ In your test108IGMC3-2-10r11-04Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

No calculator

A probe carries three type A sensors and two type B sensors with total mass 17.5 kg. A second probe carries two type A sensors and five type B sensors with total mass 30 kg. Let each type A sensor have mass a kg and each type B sensor have mass b kg.

(a)

Write down two equations in a and b.

(2)

(b)

Solve the equations to find a and b.

(3)

(c)

Find the total mass of four type A sensors and three type B sensors.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3a + 2b = 17.5	✓ 1
B1 2a + 5b = 30	✓ 1
Part (b)
M1 multiply to match one coefficient, e.g. 6a + 4b = 35 and 6a + 15b = 90	✓ 1
A1 b = 5	✓ 1
A1 a = 2.5	✓ 1
Part (c)
B1 4 × 2.5 + 3 × 5 = 25 kg (ECF from (b))	✓ 1

Common error: After finding one unknown, substitute into an original equation and check the other equation as well.

+ Add to test✓ In your test109IGMC3-2-10r11-05Expanding, factorising and algebraic fractions Paper CMedium6 marks

No calculator

A rectangular deck plan on a bridge has length (x + 2) metres and width (x − 5) metres. Its area is 18 m².

Diagram not to scale

(a)

Show that x² − 3x − 28 = 0.

(2)

(b)

Solve the equation by factorising and find the length of the deck.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (x + 2)(x − 5) = x² − 3x − 10	✓ 1
A1 x² − 3x − 10 = 18, so x² − 3x − 28 = 0	✓ 1
Part (b)
M1 attempt to factorise	✓ 1
A1 (x − 7)(x + 4) = 0	✓ 1
A1 x = 7 (x = −4 rejected since the width must be positive)	✓ 1
B1 length = 9 m	✓ 1

Exam tip: Reject any root that makes a length negative, and say why.

+ Add to test✓ In your test110IGMC3-2-10r11-06Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

No calculator

A coffee roaster's thermometer shows that the temperature of the beans is 61 °C at 2 minutes and 169 °C at 8 minutes after loading. Assume the temperature T °C rises linearly with time t minutes.

(a)

Find the gradient of the graph of T against t, with its units.

(2)

(b)

Find the equation of the line in the form T = mt + c.

(2)

(c)

Find the time when the temperature reaches 205 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (169 − 61) ÷ (8 − 2)	✓ 1
A1 18 °C per minute	✓ 1
Part (b)
M1 61 = 18 × 2 + c	✓ 1
A1 T = 18t + 25	✓ 1
Part (c)
M1 18t + 25 = 205	✓ 1
A1 10 minutes	✓ 1

Common error: Keep the units with the gradient: it is a change in temperature per minute.

+ Add to test✓ In your test111IGMC3-2-10r11-13Sequences and graphs of functions Paper CMedium4 marks

No calculator

The number of panels in a satellite dish after 1, 2, 3 and 4 rings are built is 4, 9, 16, 25, ...

(a)

Write down the next two terms.

(1)

(b)

Write down an expression for the nth term.

(1)

(c)

The dish has 144 panels. Find the number of rings.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 36 and 49	✓ 1
Part (b)
B1 (n + 1)²	✓ 1
Part (c)
M1 (n + 1)² = 144 so n + 1 = 12	✓ 1
A1 n = 11	✓ 1

Common error: Square numbers start at 1, 4, 9, so this pattern is shifted by one term.

+ Add to test✓ In your test112IGMC3-2-08r16-02Linear and simultaneous equations and rearranging formulae Paper CMedium7 marks

Calculator

Shop A hires out bicycles for a fixed charge of £8 plus £2.50 for each hour. The cost, $C$ pounds, of hiring a bicycle for $h$ hours is $C=8+2.5h$. Shop B charges £3 plus £3.50 for each hour.

(a)

A cyclist pays £23 at Shop A. Find the number of hours for which the bicycle was hired.

(2)

(b)

Rearrange $C=8+2.5h$ to make $h$ the subject.

(2)

(c)

For one particular hire time the two shops charge the same amount. Find this number of hours and the cost.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $8+2.5h=23$ or $23-8=15$	✓ 1
A1 $h=6$	✓ 1
Part (b)
M1 subtracts 8 from both sides, $C-8=2.5h$	✓ 1
A1 $h=\dfrac{C-8}{2.5}$	✓ 1
Part (c)
M1 forms the equation $8+2.5h=3+3.5h$	✓ 1
A1 $h=5$	✓ 1
A1 £20.50	✓ 1

Exam tip: Write the cost for each shop as an expression in the same variable before you try to compare them.

+ Add to test✓ In your test113IGMC3-2-08r16-03Quadratic equations and inequalities Paper CMedium7 marks

Calculator

An archaeological trench is a rectangle of width $x$ metres and length $(x+5)$ metres. Its area is $36\ \text{m}^2$.

Diagram not to scale

(a)

Show that $x^2+5x-36=0$.

(2)

(b)

Solve $x^2+5x-36=0$ by factorising, and state the width of the trench.

(3)

(c)

Calculate the perimeter of the trench.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 area $=x(x+5)=36$	✓ 1
B1 expands to $x^2+5x=36$ and rearranges to $x^2+5x-36=0$	✓ 1
Part (b)
M1 $(x+9)(x-4)=0$	✓ 1
A1 $x=4$ or $x=-9$	✓ 1
B1 width is 4 m, since $x=-9$ is rejected as a length cannot be negative	✓ 1
Part (c)
M1 length $=4+5=9$ m and perimeter $=2(4+9)$	✓ 1
A1 26 m	✓ 1

Common error: Always reject any solution that gives a negative length, and say why.

+ Add to test✓ In your test114IGMC3-2-08r16-06Functions and differentiation Paper CMedium7 marks

Calculator

A curve has equation $y=x^2-4x-5$.

(a)

Find the value of $y$ when $x=-2$ and when $x=3$.

(2)

(b)

Find the values of $x$ where the curve crosses the $x$-axis.

(2)

(c)

Write down the $y$-coordinate of the point where the curve crosses the $y$-axis.

(1)

(d)

The curve is symmetrical about a vertical line and has a lowest point on this line. Find the equation of the line of symmetry and the minimum value of $y$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $y=7$ when $x=-2$	✓ 1
B1 $y=-8$ when $x=3$	✓ 1
Part (b)
M1 sets $y=0$ and factorises, $(x-5)(x+1)=0$	✓ 1
A1 $x=-1$ and $x=5$	✓ 1
Part (c)
B1 $-5$	✓ 1
Part (d)
B1 $x=2$, halfway between $-1$ and $5$	✓ 1
B1 minimum value is $-9$ (from $y=2^2-4(2)-5$)	✓ 1

Exam tip: The line of symmetry of a quadratic curve passes halfway between its two $x$-intercepts.

+ Add to test✓ In your test115IGMC3-2-08r16-13Sequences and graphs of functions Paper CMedium6 marks

Calculator

Archaeologists dig four trenches in a row. The numbers of pottery fragments found are 7, 12, 17, 22 in the first four trenches.

(a)

Find the $n$th term of this sequence.

(2)

(b)

Show that 150 is not a term of this sequence.

(2)

(c)

A second sequence begins 3, 6, 12, 24, and continues by multiplying each term by the same number. Find the 10th term of this sequence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 5, so $5n$ appears	✓ 1
A1 $5n+2$	✓ 1
Part (b)
M1 $5n+2=150$ gives $n=29.6$	✓ 1
R1 $n$ is not a whole number, so 150 is not a term	✓ 1
Part (c)
M1 the ratio is 2, so the 10th term is $3\times 2^9$	✓ 1
A1 1536	✓ 1

Common error: A number is a term of a sequence only if the value of $n$ you find is a positive whole number.

+ Add to test✓ In your test116IGMC3-2-08r17-02Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

Calculator

A food science lab buys flour and sugar. 3 bags of flour and 2 bags of sugar cost £8.10. 1 bag of flour and 4 bags of sugar cost £5.70. Let a bag of flour cost £x and a bag of sugar cost £y.

(a)

Write down two equations in x and y.

(2)

(b)

Solve the equations to find the cost of one bag of flour and the cost of one bag of sugar.

(3)

(c)

Find the cost of 2 bags of flour and 1 bag of sugar.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3x + 2y = 8.10	✓ 1
B1 x + 4y = 5.70	✓ 1
Part (b)
M1 a correct method to eliminate one variable, e.g. 3x + 12y = 17.10	✓ 1
A1 y = 0.90 (sugar £0.90)	✓ 1
A1 x = 2.10 (flour £2.10)	✓ 1
Part (c)
B1 £5.10 (ECF from (b))	✓ 1

Exam tip: Check your pair of answers in both of the original equations before moving on.

+ Add to test✓ In your test117IGMC3-2-08r17-05Sequences and graphs of functions Paper CMedium5 marks

Calculator

The altitudes of the camps on a mountain route form a sequence: Camp 1 is at 1800 m, Camp 2 at 2150 m, Camp 3 at 2500 m and Camp 4 at 2850 m.

(a)

Write down the altitude of Camp 5.

(1)

(b)

Find an expression for the altitude of Camp n, in metres.

(2)

(c)

Camp n is at an altitude of 4600 m. Find n.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3200 m	✓ 1
Part (b)
M1 common difference of 350 used, 350n	✓ 1
A1 350n + 1450	✓ 1
Part (c)
M1 350n + 1450 = 4600 (ECF from (b))	✓ 1
A1 n = 9	✓ 1

Exam tip: Check your nth term by substituting n = 1 and n = 2 into it.

+ Add to test✓ In your test118IGMC3-2-08r17-07Quadratic equations and inequalities Paper CMedium6 marks

No calculator

(a)

Solve 5x − 8 > 3x + 6.

(2)

(b)

Solve −7 < 2x − 3 ≤ 9.

(3)

(c)

Find the number of integers that satisfy the inequality in (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 collect terms: 2x > 14	✓ 1
A1 x > 7	✓ 1
Part (b)
M1 add 3 to all three parts: −4 < 2x ≤ 12	✓ 1
M1 divide all parts by 2	✓ 1
A1 −2 < x ≤ 6	✓ 1
Part (c)
B1 8 (the integers −1, 0, 1, 2, 3, 4, 5, 6)	✓ 1

Common error: −2 is not included because the left inequality is strict, but 6 is included because the right one is ≤.

+ Add to test✓ In your test119IGMC3-2-08r17-09Functions and differentiation Paper CMedium5 marks

Calculator

The height y metres (relative to a base camp) of a mountain path at a horizontal distance x hundred metres from the start is given by y = x² − 4x − 5, for −1 ≤ x ≤ 5.
Some values are shown in the table.
x: −1, 0, 1, 2, 3, 4, 5
y: 0, −5, p, −9, q, −5, 0

(a)

Find the values of p and q.

(2)

(b)

Write down the coordinates of the point where the curve crosses the y-axis.

(1)

(c)

Write down the equation of the line of symmetry of the curve.

(1)

(d)

Write down the minimum value of y.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 p = −8	✓ 1
B1 q = −8	✓ 1
Part (b)
B1 (0, −5)	✓ 1
Part (c)
B1 x = 2	✓ 1
Part (d)
B1 −9	✓ 1

Exam tip: A quadratic curve is symmetrical: the values in the table repeat either side of the line of symmetry, which gives a check on p and q.

+ Add to test✓ In your test120IGMC3-2-08r17-10Sequences and graphs of functions Paper CMedium7 marks

Calculator

A telecoms company offers two monthly plans. Plan A costs £15 plus 5p for each minute of calls. Plan B costs £9 plus 8p for each minute of calls. A customer makes x minutes of calls in a month.

(a)

Write down an expression, in pounds, for the cost of each plan.

(2)

(b)

Find the number of minutes for which the two plans cost the same.

(3)

(c)

A customer makes 350 minutes of calls. Which plan is cheaper and by how much?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 Plan A: 15 + 0.05x	✓ 1
B1 Plan B: 9 + 0.08x	✓ 1
Part (b)
M1 15 + 0.05x = 9 + 0.08x (ECF from (a))	✓ 1
M1 collect terms: 6 = 0.03x	✓ 1
A1 x = 200	✓ 1
Part (c)
M1 evaluate both costs: 32.50 and 37.00 (or find the difference)	✓ 1
A1 Plan A is cheaper by £4.50	✓ 1

Common error: Keep all the costs in pounds: 5p is £0.05 per minute, not £5.

+ Add to test✓ In your test121IGMC3-2-10r22-05Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

Calculator

A beekeeper sells honey in small jars and large jars. Three small jars and two large jars cost £22.50. Five small jars and one large jar cost £20.00.

(a)

Find the price of one small jar and the price of one large jar by solving simultaneous equations.

(3)

(b)

Find the cost of 4 small jars and 3 large jars.

(1)

(c)

The beekeeper gives a 20% discount on the total in part (b). Calculate the discounted cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 form $3s+2l=22.5$ and $5s+l=20$ and eliminate one variable, e.g. $10s+2l=40$	✓ 1
A1 $s=2.50$	✓ 1
A1 $l=7.50$	✓ 1
Part (b)
B1 £32.50 (ECF from (a))	✓ 1
Part (c)
M1 $32.50\times0.8$	✓ 1
A1 £26.00	✓ 1

Exam tip: Check your solution by substituting both values back into the other equation.

+ Add to test✓ In your test122IGMC3-2-10r22-08Quadratic equations and inequalities Paper CMedium6 marks

No calculator

Solve these inequalities.

(a)

Solve $5-2x<17$.

(2)

(b)

Find all the integers $n$ that satisfy $-3\le2n+1<9$.

(3)

(c)

Find the sum of these integers.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $-2x<12$	✓ 1
A1 $x>-6$ (the sign reverses when dividing by $-2$)	✓ 1
Part (b)
M1 $-4\le2n<8$	✓ 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: Dividing or multiplying an inequality by a negative number reverses the inequality sign.

+ Add to test✓ In your test123IGMC3-2-08r24-05Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

No calculator

A straight line passes through the points A(2, 7) and B(6, 19).

Diagram not to scale

(a)

Find the gradient of the line.

(2)

(b)

Find the equation of the line in the form $y = mx + c$.

(2)

(c)

The point P$(k, 25)$ lies on the line. Find $k$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\dfrac{19 - 7}{6 - 2}$	✓ 1
A1 3	✓ 1
Part (b)
M1 $y - 7 = 3(x - 2)$ or $7 = 3 \times 2 + c$ (ECF from (a))	✓ 1
A1 $y = 3x + 1$	✓ 1
Part (c)
M1 $25 = 3k + 1$	✓ 1
A1 $k = 8$	✓ 1

Exam tip: Check your equation by substituting the other given point: $3 \times 6 + 1 = 19$.

+ Add to test✓ In your test124IGMC3-2-08r24-06Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

Calculator

In food science, the heat energy $Q$ (joules) needed to warm a mass $m$ kg of liquid from temperature $T_1$ °C to $T_2$ °C is given by $Q = mc(T_2 - T_1)$, where $c$ J/(kg °C) is a constant for the liquid. For water $c = 4200$.

(a)

Rearrange the formula to make $T_2$ the subject.

(2)

(b)

0.50 kg of water at 15 °C receives 84 000 J. Calculate the final temperature.

(2)

(c)

0.75 kg of water receives 63 000 J and its final temperature is 60 °C. Calculate its starting temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 divide by $mc$: $\dfrac{Q}{mc} = T_2 - T_1$	✓ 1
A1 $T_2 = \dfrac{Q}{mc} + T_1$	✓ 1
Part (b)
M1 $\dfrac{84000}{0.50 \times 4200} + 15$	✓ 1
A1 55 °C	✓ 1
Part (c)
M1 $T_1 = 60 - \dfrac{63000}{0.75 \times 4200}$	✓ 1
A1 40 °C	✓ 1

Common error: $T_1$ is subtracted inside the bracket, so when you rearrange for $T_1$ you must subtract $\frac{Q}{mc}$ from $T_2$, not add it.

+ Add to test✓ In your test125IGMC3-2-08r24-07Linear and simultaneous equations and rearranging formulae Paper CMedium7 marks

Calculator

A mobile phone company offers two monthly plans. Plan X costs a fixed £8 plus £0.04 per minute of calls. Plan Y costs a fixed £5 plus £0.07 per minute of calls. In one month a customer makes $m$ minutes of calls.

(a)

Write an expression, in pounds, for the cost of each plan.

(2)

(b)

Find the number of minutes for which the two plans cost the same.

(3)

(c)

A customer makes 150 minutes of calls in a month. Work out how much cheaper the cheaper plan is than the other plan.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 Plan X: $8 + 0.04m$	✓ 1
B1 Plan Y: $5 + 0.07m$	✓ 1
Part (b)
M1 $8 + 0.04m = 5 + 0.07m$	✓ 1
M1 $3 = 0.03m$	✓ 1
A1 $m = 100$	✓ 1
Part (c)
M1 Plan X £14.00 and Plan Y £15.50	✓ 1
A1 £1.50	✓ 1

Exam tip: In a 'same cost' problem, write the two expressions equal to each other and solve for the unknown.

+ Add to test✓ In your test126IGMC3-2-08r24-08Sequences and graphs of functions Paper CMedium6 marks

Calculator

The strength $S$ of a mobile signal at a distance of $x$ km from a mast is modelled, for $0 \leqslant x \leqslant 5$, by $S = 4x - x^2 + 5$, and the model can be extended to other values of $x$. Some values are given: $x = 0$: $S = 5$; $x = 1$: $S = 8$; $x = 3$: $S = 8$; $x = 4$: $S = 5$.

(a)

Calculate $S$ when $x = 2$ and when $x = 5$.

(2)

(b)

The graph of $S$ against $x$ is symmetrical. Use the table of values to find the equation of its line of symmetry.

(2)

(c)

The model gives $S = -7$ when $x = -2$. Use symmetry to write down another value of $x$ for which $S = -7$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $S = 9$ at $x = 2$	✓ 1
B1 $S = 0$ at $x = 5$	✓ 1
Part (b)
M1 equal values of $S$ at $x = 1$ and $x = 3$ (or at 0 and 4), so the midpoint is 2	✓ 1
A1 $x = 2$	✓ 1
Part (c)
M1 reflect in the line $x = 2$: $2 \times 2 - (-2)$ (ECF from (b))	✓ 1
A1 $x = 6$	✓ 1

Exam tip: A line of symmetry lies midway between two points with the same $y$-value: add the $x$-values and halve.

+ Add to test✓ In your test127IGMC3-2-08r24-09Sequences and graphs of functions Paper CMedium6 marks

Calculator

A food scientist counts the number of yeast cells, in thousands, in a culture at the start of each hour. The first four counts are 2, 6, 18, 54.

(a)

Write down the rule for finding the next term.

(1)

(b)

Find the next two terms.

(2)

(c)

Explain why the $n$th term of this sequence cannot be written as $an + b$.

(1)

(d)

Find the first term that is greater than 1000.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Multiply the previous term by 3	✓ 1
Part (b)
B1 162	✓ 1
B1 486	✓ 1
Part (c)
The differences (4, 12, 36) are not constant, so it is not linear	✓ 1
Part (d)
M1 multiply 486 by 3 (ECF from (b))	✓ 1
A1 1458	✓ 1

Common error: A sequence that multiplies by a fixed number each time is not linear: its differences keep growing.

+ Add to test✓ In your test128IGMC3-2-09r25-04Sequences and graphs of functions Paper CMedium6 marks

Calculator

The number of seats in each row of a stadium stand forms a sequence. Row 1 has 14 seats, row 2 has 19 seats, row 3 has 24 seats and row 4 has 29 seats.

(a)

Find an expression for the number of seats in row $n$.

(2)

(b)

Show that no row can have exactly 200 seats.

(2)

(c)

The last row has 99 seats. Find the number of rows in the stand.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $5n$ (common difference 5)	✓ 1
B1 $5n + 9$	✓ 1
Part (b)
M1 $5n + 9 = 200$ leading to $n = 38.2$	✓ 1
A1 $n$ is not a whole number, so 200 is not in the sequence	✓ 1
Part (c)
M1 $5n + 9 = 99$ or $\frac{99 - 9}{5}$	✓ 1
A1 18 rows	✓ 1

Exam tip: To show a number is not in a sequence, solve for $n$ and state clearly that the result is not a positive whole number.

+ Add to test✓ In your test129IGMC3-2-09r25-05Quadratic equations and inequalities Paper CMedium7 marks

Calculator

A rail freight yard is a rectangle. Its length is $(x + 5)$ metres and its width is $x$ metres. The area of the yard is 336 m².

Diagram not to scale

(a)

Show that $x^2 + 5x - 336 = 0$.

(2)

(b)

Solve $x^2 + 5x - 336 = 0$ by factorising.

(3)

(c)

Find the perimeter of the yard.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x(x+5) = 336$	✓ 1
A1 expands to $x^2 + 5x = 336$ and rearranges to the given equation	✓ 1
Part (b)
M1 attempts to factorise with factors of $-336$ differing by 5	✓ 1
A1 $(x+21)(x-16) = 0$	✓ 1
A1 $x = 16$ chosen, and $x = -21$ rejected because a width cannot be negative	✓ 1
Part (c)
M1 $2(16 + 21)$ or $2(x + x + 5)$ with $x = 16$ (ECF from (b))	✓ 1
A1 74 m	✓ 1

Common error: Both roots satisfy the equation, but only one fits the context; always say why you reject the other.

+ Add to test✓ In your test130IGMC3-2-09r25-07Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

Calculator

The kinetic energy $E$ in joules of a moving rail wagon of mass $m$ kg travelling at speed $v$ m/s is given by $E = \frac{1}{2}mv^2$.

(a)

Make $m$ the subject of the formula.

(2)

(b)

Make $v$ the subject of the formula.

(2)

(c)

A wagon of mass 8000 kg has kinetic energy 450 000 J. Calculate its speed, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $2E = mv^2$	✓ 1
A1 $m = \frac{2E}{v^2}$	✓ 1
Part (b)
M1 $v^2 = \frac{2E}{m}$	✓ 1
A1 $v = \sqrt{\frac{2E}{m}}$	✓ 1
Part (c)
M1 $v = \sqrt{\frac{2 \times 450\,000}{8000}} = \sqrt{112.5}$	✓ 1
A1 10.6 m/s	✓ 1

Exam tip: Square root the whole fraction $\frac{2E}{m}$; taking the square root of only the numerator is a frequent slip.

+ Add to test✓ In your test131IGMC3-2-09r25-08Expanding, factorising and algebraic fractions Paper CMedium6 marks

No calculator

Solve these.

(a)

$\frac{2x-1}{3} + \frac{x+4}{2} = 11$

(4)

(b)

$5 - 2x \ge 17$

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplies every term by 6 to get $2(2x-1) + 3(x+4) = 66$	✓ 1
M1 expands brackets correctly ($4x - 2 + 3x + 12$)	✓ 1
M1 collects terms to get $7x = 56$	✓ 1
A1 $x = 8$	✓ 1
Part (b)
M1 $-2x \ge 12$	✓ 1
A1 $x \le -6$	✓ 1

Exam tip: When clearing fractions, multiply every term, including the one on the right-hand side, by the common denominator.

+ Add to test✓ In your test132IGMC3-2-09r25-09Expanding, factorising and algebraic fractions Paper CMedium7 marks

No calculator

(a)

Expand and simplify $(2x-3)^2 - (x+1)(x-4)$.

(3)

(b)

Factorise $x^2 - 5x - 24$.

(2)

(c)

Hence solve $x^2 - 5x - 24 = 0$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands $(2x-3)^2$ to $4x^2 - 12x + 9$	✓ 1
M1 expands $(x+1)(x-4)$ to $x^2 - 3x - 4$	✓ 1
A1 $3x^2 - 9x + 13$	✓ 1
Part (b)
M1 attempts two brackets with numbers whose product is $-24$	✓ 1
A1 $(x-8)(x+3)$	✓ 1
Part (c)
B1 $x = 8$	✓ 1
B1 $x = -3$	✓ 1

Common error: Subtracting a product: put the second expansion in brackets first, otherwise the signs of $-3x$ and $-4$ are lost.

+ Add to test✓ In your test133IGMC3-2-12r27-05Quadratic equations and inequalities Paper CMedium7 marks

No calculator

An architect designs a rectangular courtyard. The width of the courtyard is x metres and its length is (x + 5) metres. The area of the courtyard is 84 m².

Diagram not to scale

(a)

Show that x² + 5x − 84 = 0.

(2)

(b)

Solve x² + 5x − 84 = 0 by factorising, and state the width of the courtyard.

(3)

(c)

Work out the perimeter of the courtyard.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x(x + 5) = 84	✓ 1
A1 x² + 5x = 84, rearranged to x² + 5x − 84 = 0	✓ 1
Part (b)
M1 (x + 12)(x − 7)	✓ 1
A1 x = −12 and x = 7	✓ 1
B1 width = 7 m, as −12 is rejected because a length cannot be negative	✓ 1
Part (c)
M1 2 × (7 + 12) (ECF from (b))	✓ 1
A1 38 m	✓ 1

Common error: A quadratic gives two solutions, but in a context you must reject any that make no physical sense, such as a negative length.

+ Add to test✓ In your test134IGMC3-2-12r27-08Sequences and graphs of functions Paper CMedium6 marks

Calculator

A beam in a roof frame is modelled as a straight line L on a coordinate grid, with lengths in metres. The line L passes through the points A(1, 4) and B(5, 12).

Diagram not to scale

(a)

Find the gradient of L.

(1)

(b)

Find the equation of L in the form y = mx + c.

(2)

(c)

A second beam lies on a line parallel to L and passing through the point (0, -3). Write down the equation of this line.

(1)

(d)

Find the x-coordinate of the point where L crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2	✓ 1
Part (b)
M1 y − 4 = 2(x − 1) or substituting a point into y = 2x + c	✓ 1
A1 y = 2x + 2	✓ 1
Part (c)
B1 y = 2x - 3	✓ 1
Part (d)
M1 0 = 2x + 2	✓ 1
A1 x = -1	✓ 1

Common error: Parallel lines have the same gradient but a different y-intercept; do not change the gradient when you write the new equation.

+ Add to test✓ In your test135IGMC3-2-12r27-13Functions and differentiation Paper CMedium6 marks

No calculator

A curve has equation y = x² − 4x − 5.

(a)

Solve x² − 4x − 5 = 0 to find where the curve crosses the x-axis.

(3)

(b)

Write down the equation of the line of symmetry of the curve.

(1)

(c)

Find the coordinates of the lowest point of the curve.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (x + 1)(x − 5) = 0	✓ 1
A1 x = -1	✓ 1
A1 x = 5	✓ 1
Part (b)
B1 x = 2 (midway between the roots)	✓ 1
Part (c)
B1 x = 2 (ECF from (b))	✓ 1
B1 y = 2² − 4 × 2 − 5 = -9, so the point is (2, -9)	✓ 1

Exam tip: The lowest point of a symmetrical curve lies on the line of symmetry, so find x first and then substitute it into the equation to get y.

+ Add to test✓ In your test136IGMC3-2-11r28-05Linear and simultaneous equations and rearranging formulae Paper CMedium7 marks

Calculator

A café sells coffees and pastries. 3 coffees and 2 pastries cost £11.40. 2 coffees and 3 pastries cost £10.10. A coffee costs £$c$ and a pastry costs £$p$.

(a)

Write down two equations in $c$ and $p$.

(2)

(b)

Solve the equations simultaneously.

(3)

(c)

Find the cost of 4 coffees and 5 pastries.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $3c+2p=11.40$	✓ 1
B1 $2c+3p=10.10$	✓ 1
Part (b)
M1 match coefficients, e.g. $6c+4p=22.80$ and $6c+9p=30.30$ (or equivalent substitution)	✓ 1
A1 $p=1.50$	✓ 1
A1 $c=2.80$	✓ 1
Part (c)
M1 $4\times2.80+5\times1.50$ (ECF from (b))	✓ 1
A1 £18.70	✓ 1

Exam tip: Substitute your answers back into both original equations to check them before moving on.

+ Add to test✓ In your test137IGMC3-2-11r28-06Quadratic equations and inequalities Paper CMedium7 marks

Calculator

A rectangular field for a solar farm has width $x$ metres. Its length is 20 metres more than its width and its area is 4800 m².

Diagram not to scale

(a)

Show that $x^2+20x-4800=0$.

(2)

(b)

Solve the equation by factorising, and state the width of the field.

(3)

(c)

Find the perimeter of the field.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x(x+20)=4800$	✓ 1
A1 expands and rearranges to the given equation	✓ 1
Part (b)
M1 attempt to factorise	✓ 1
A1 $(x+80)(x-60)=0$	✓ 1
A1 width $60$ m, with $x=-80$ rejected because a width cannot be negative	✓ 1
Part (c)
M1 $2(60+80)$ (ECF from (b))	✓ 1
A1 280 m	✓ 1

Common error: A quadratic can give a negative solution that makes no sense for a length; always say why you reject it.

+ Add to test✓ In your test138IGMC3-2-11r28-07Expanding, factorising and algebraic fractions Paper CMedium5 marks

Calculator

(a)

Solve $7-3x>22$.

(2)

(b)

The integer $n$ satisfies $-5\le 3n+1<13$. List all possible values of $n$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $-3x>15$	✓ 1
A1 $x<-5$	✓ 1
Part (b)
M1 subtract 1 from each part: $-6\le3n<12$	✓ 1
A1 $-2\le n<4$	✓ 1
A1 $-2,-1,0,1,2,3$	✓ 1

Common error: When you multiply or divide an inequality by a negative number, the inequality sign must be reversed.

+ Add to test✓ In your test139IGMC3-2-11r28-08Functions and differentiation Paper CMedium5 marks

Calculator

A curve has equation $y=x^2-2x-3$. It is a smooth U-shaped curve with a minimum point.

(a)

Find the value of $y$ when $x=-2$ and when $x=2$.

(2)

(b)

Find the $x$-coordinates of the points where the curve crosses the $x$-axis.

(2)

(c)

Write down the equation of the line of symmetry of the curve.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 5	✓ 1
B1 $-3$	✓ 1
Part (b)
M1 $(x-3)(x+1)=0$	✓ 1
A1 $x=3$ and $x=-1$	✓ 1
Part (c)
B1 $x=1$ (halfway between the roots)	✓ 1

Exam tip: The line of symmetry of a parabola lies halfway between its two $x$-intercepts.

+ Add to test✓ In your test140IGMC3-2-12r28-05Linear and simultaneous equations and rearranging formulae Paper CMedium5 marks

Calculator

A cycling club buys tickets for 28 people to visit a velodrome. An adult ticket costs £12 and a child ticket costs £7. The total cost is £276. There are $a$ adults and $c$ children.

(a)

Write down two equations in $a$ and $c$.

(2)

(b)

Solve the equations simultaneously to find the number of adults.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $a+c=28$	✓ 1
B1 $12a+7c=276$	✓ 1
Part (b)
M1 eliminate a variable, e.g. $12(28-c)+7c=276$	✓ 1
A1 $c=12$	✓ 1
A1 $a=16$	✓ 1

Exam tip: Check by substituting both values back into the cost equation: $12\times16+7\times12=276$.

+ Add to test✓ In your test141IGMC3-2-12r28-07Quadratic equations and inequalities Paper CMedium7 marks

Calculator

A data centre converts a temperature of $F$ degrees Fahrenheit to $C$ degrees Celsius using the formula $F=\dfrac{9C}{5}+32$.

(a)

Make $C$ the subject of the formula.

(2)

(b)

Find $C$ when $F=77$.

(2)

(c)

The safe limit for a server room is 27 °C. The room reads 82 °F. Calculate the temperature in °C, correct to 1 decimal place, and state whether it is above the limit.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $F-32=\dfrac{9C}{5}$	✓ 1
A1 $C=\dfrac{5(F-32)}{9}$	✓ 1
Part (b)
M1 $\dfrac{5(77-32)}{9}$	✓ 1
A1 25	✓ 1
Part (c)
M1 $\dfrac{5(82-32)}{9}$	✓ 1
A1 27.8 °C	✓ 1
B1 above the limit because $27.8>27$	✓ 1

Exam tip: Subtract 32 from the whole Fahrenheit value before multiplying by 5 and dividing by 9.

+ Add to test✓ In your test142IGMC3-2-12r28-08Sequences and graphs of functions Paper CMedium5 marks

Calculator

The $n$th term of a sequence of server rack numbers is $7n-4$.

(a)

Find the 10th term.

(1)

(b)

Show that 150 is a term of the sequence and state which term it is.

(2)

(c)

Find the smallest term of the sequence that is greater than 300.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 66	✓ 1
Part (b)
M1 $7n-4=150$	✓ 1
A1 $n=22$, a whole number, so 150 is the 22nd term	✓ 1
Part (c)
M1 $7n-4>300$, so $n>43.4\ldots$	✓ 1
A1 $n=44$ gives 304	✓ 1

Common error: A number is a term of a sequence only if the value of $n$ that gives it is a whole number.

+ Add to test✓ In your test143IGMC3-2-10r29-03Quadratic equations and inequalities Paper CMedium5 marks

Calculator

A Formula 1 race engineer uses the formula t = d/v + p to estimate the time t seconds taken for a section of a race. Here d is the distance in metres, v is the speed in m/s and p is the time in seconds lost in the pit lane.

(a)

Rearrange the formula to make v the subject.

(3)

(b)

Find v when d = 4800, t = 205 and p = 25. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 t − p = d/v	✓ 1
M1 multiplies by v and divides by (t − p)	✓ 1
A1 v = d/(t − p)	✓ 1
Part (b)
M1 4800 ÷ (205 − 25) = 4800 ÷ 180	✓ 1
A1 26.7 m/s	✓ 1

Common error: Subtract p from both sides first; dividing only part of the right-hand side by v gives a wrong formula.

+ Add to test✓ In your test144IGMC3-2-10r29-04Sequences and graphs of functions Paper CMedium5 marks

Calculator

The numbers of beds on the first four floors of a hospital building are 12, 19, 26, 33, and this pattern continues in the same way for higher floors.

(a)

Find an expression for the number of beds on floor n.

(2)

(b)

Find the number of beds on floor 20.

(1)

(c)

A manager says that one floor has exactly 200 beds. Show that this cannot be correct.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 7n	✓ 1
B1 7n + 5	✓ 1
Part (b)
B1 145	✓ 1
Part (c)
M1 7n + 5 = 200 so n = 195/7	✓ 1
A1 n = 27.857… (3 d.p.), which is not a whole number, so 200 is not a term	✓ 1

Exam tip: To test whether a number is in a sequence, put it equal to the nth term and check that n is a positive whole number.

+ Add to test✓ In your test145IGMC3-2-10r29-05Quadratic equations and inequalities Paper CMedium6 marks

No calculator

A rectangular area for beehives has a width of w metres. Its length is 3 metres more than its width. The area is 40 m².

Diagram not to scale

(a)

Show that w² + 3w − 40 = 0.

(2)

(b)

Solve w² + 3w − 40 = 0 by factorising and state the width of the area.

(3)

(c)

Find the perimeter of the area.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 w(w + 3) = 40	✓ 1
A1 w² + 3w = 40 so w² + 3w − 40 = 0	✓ 1
Part (b)
M1 (w + 8)(w − 5) = 0	✓ 1
A1 w = 5 or w = −8	✓ 1
B1 width = 5 m since a length cannot be negative	✓ 1
Part (c)
B1 2(5 + 8) = 26 m	✓ 1

Common error: Always reject a solution that makes no sense in the context, such as a negative length.

+ Add to test✓ In your test146IGMC3-2-10r29-06Sequences and graphs of functions Paper CMedium6 marks

Calculator

A graph has equation y = x² − 4x − 5.

(a)

Factorise x² − 4x − 5.

(2)

(b)

Hence solve x² − 4x − 5 = 0.

(2)

(c)

Write down the coordinates of the points where the graph crosses the axes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 (x − 5)	✓ 1
B1 (x + 1)	✓ 1
Part (b)
A1 x = 5	✓ 1
A1 x = −1 (ECF from (a))	✓ 1
Part (c)
B1 (−1, 0) and (5, 0)	✓ 1
B1 (0, −5)	✓ 1

Exam tip: The y-intercept of a curve is found by putting x = 0; the x-intercepts come from putting y = 0.

+ Add to test✓ In your test147IGMC3-2-10r29-13Sequences and graphs of functions Paper CMedium6 marks

Calculator

During a race simulation, the mass m kg of fuel left in a Formula 1 car decreases steadily with the number of laps n. After 2 laps there are 95 kg of fuel and after 6 laps there are 79 kg.

(a)

Find the gradient of the straight line through (2, 95) and (6, 79).

(2)

(b)

Find the equation of the line in the form m = an + b.

(2)

(c)

Find the number of laps after which 15 kg of fuel remain.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (79 − 95)/(6 − 2)	✓ 1
A1 −4	✓ 1
Part (b)
M1 95 = −4 × 2 + b	✓ 1
A1 m = −4n + 103	✓ 1
Part (c)
M1 −4n + 103 = 15	✓ 1
A1 n = 22	✓ 1

Exam tip: Give the gradient its meaning as well: here the car uses 4 kg of fuel each lap.

+ Add to test✓ In your test148IGMC3-2-10r35-03Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

No calculator

Solve the simultaneous equations.
$2x+3y=13$
$5x-3y=1$

(a)

Solve for $x$ and $y$.

(4)

(b)

Rearrange $2x+3y=13$ to make $y$ the subject.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 eliminates $y$ (or $x$) correctly, e.g. adds the equations to get $7x=14$	✓ 1
A1 $x=2$	✓ 1
M1 substitutes to find the other variable	✓ 1
A1 $y=3$	✓ 1
Part (b)
M1 $3y=13-2x$	✓ 1
A1 $y=\dfrac{13-2x}{3}$	✓ 1

Exam tip: Substitute both values into the equation you did not use to solve, to check them.

+ Add to test✓ In your test149IGMC3-2-10r35-04Sequences and graphs of functions Paper CMedium8 marks

Calculator

A music teacher sets a practice plan. In week 1 a student practises for 15 minutes, in week 2 for 19 minutes, in week 3 for 23 minutes, in week 4 for 27 minutes, and so on. The times form an arithmetic sequence.

(a)

Write down the practice times in weeks 5 and 6.

(2)

(b)

Find an expression for the practice time in week n.

(2)

(c)

In which week does the student first practise for exactly 99 minutes?

(2)

(d)

Find the practice time in the first week in which it exceeds 120 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 31 minutes	✓ 1
B1 35 minutes	✓ 1
Part (b)
M1 common difference 4 used, e.g. $4n+\ldots$	✓ 1
A1 $4n+11$ minutes	✓ 1
Part (c)
M1 $4n+11=99$	✓ 1
A1 week 22	✓ 1
Part (d)
M1 $4n+11>120$ gives $n>27.25$, so $n=28$	✓ 1
A1 123 minutes	✓ 1

Common error: Because $n$ must be a whole number of weeks, round up when solving an inequality like $n>27.25$.

+ Add to test✓ In your test150IGMC3-2-10r35-05Quadratic equations and inequalities Paper CMedium7 marks

Calculator

A rectangular field has width x metres and length (x + 6) metres. The area of the field is 616 m².

Diagram not to scale

(a)

Show that $x^2+6x-616=0$.

(2)

(b)

Solve $x^2+6x-616=0$ and state the width of the field.

(3)

(c)

The farmer puts a fence all the way round the field, costing €3.50 per metre. Calculate the total cost of the fence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x(x+6)=616$	✓ 1
A1 expands and rearranges to $x^2+6x-616=0$ (no errors shown)	✓ 1
Part (b)
M1 $(x+28)(x-22)$ or correct use of the quadratic formula	✓ 1
A1 $x=22$ or $x=-28$	✓ 1
B1 rejects $x=-28$ because a length cannot be negative, width = 22 m	✓ 1
Part (c)
M1 perimeter $=2(22+28)=100$ m	✓ 1
A1 €350	✓ 1

Exam tip: Always reject a root that makes no sense in the context, and say why.

+ Add to test✓ In your test151IGMC3-2-10r35-09Functions and differentiation Paper CMedium6 marks

Calculator

A curve has equation $y=x^2-4x-5$.

(a)

Find the values of $x$ where the curve crosses the $x$-axis.

(2)

(b)

Write down the $y$-coordinate of the point where the curve crosses the $y$-axis.

(1)

(c)

Write down the equation of the line of symmetry of the curve.

(1)

(d)

Find the minimum value of $y$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $(x-5)(x+1)=0$ or correct use of the formula	✓ 1
A1 $x=5$ and $x=-1$	✓ 1
Part (b)
B1 $y=-5$	✓ 1
Part (c)
B1 $x=2$	✓ 1
Part (d)
M1 substitutes $x=2$ into the equation	✓ 1
A1 $y=-9$	✓ 1

Exam tip: The line of symmetry lies half-way between the $x$-intercepts: $(5+(-1))/2=2$.

+ Add to test✓ In your test152IGMC3-2-12r46-04Linear and simultaneous equations and rearranging formulae Paper CMedium6 marks

Calculator

A boutique sells shirts at €s each and scarves at €c each. On Monday it sells 3 shirts and 2 scarves for a total of €78. On Tuesday, at the same prices, it sells 2 shirts and 5 scarves for a total of €96.

(a)

Write down two equations in s and c.

(2)

(b)

Solve the equations to find the price of a shirt and the price of a scarf.

(3)

(c)

Calculate the cost of 4 shirts and 1 scarf.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $3s+2c=78$	✓ 1
B1 $2s+5c=96$	✓ 1
Part (b)
M1 correct elimination of one variable (for example, multiplying to equalise coefficients and subtracting)	✓ 1
A1 $s=18$	✓ 1
A1 $c=12$	✓ 1
Part (c)
B1 €84 (ECF from (b))	✓ 1

Exam tip: Check simultaneous solutions by substituting both values into the equation you did not use for elimination.

+ Add to test✓ In your test153IGMC3-2-12r46-12Functions and differentiation Paper CMedium6 marks

Calculator

A robot moves along a straight line that passes through the points (2, 5) and (6, 13), where the units are metres.

Diagram not to scale

(a)

Calculate the gradient of the line.

(2)

(b)

Find the equation of the line in the form y = mx + c.

(2)

(c)

A second robot moves along a line parallel to the first line that passes through (0, −3). Find the value of x where this second line crosses the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{13-5}{6-2}$	✓ 1
A1 2	✓ 1
Part (b)
M1 substitutes a point and the gradient, such as $5=2\times2+c$	✓ 1
A1 $y=2x+1$	✓ 1
Part (c)
M1 equation $y=2x-3$ with $y=0$	✓ 1
A1 $x=1.5$	✓ 1

Common error: Parallel lines have the same gradient, so only the intercept changes.

+ Add to test✓ In your test154IGMC3-2-11r48-01Sequences and graphs of functions Paper CMedium7 marks

Calculator

A wind farm is built in stages. The numbers of turbines installed in stages 1, 2, 3 and 4 are 6, 10, 14 and 18. The pattern continues in the same way.

(a)

Write down the number of turbines installed in stage 5 and in stage 6.

(2)

(b)

Find an expression for the number of turbines installed in stage n.

(2)

(c)

Explain why exactly 95 turbines cannot be installed in any stage.

(1)

(d)

One stage installs 102 turbines. Find the number of that stage.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 22	✓ 1
B1 26	✓ 1
Part (b)
M1 common difference 4 used, e.g. 4n + ...	✓ 1
A1 4n + 2	✓ 1
Part (c)
B1 every term 4n + 2 is even but 95 is odd (or 4n + 2 = 95 gives n = 23.25, not a whole number)	✓ 1
Part (d)
M1 4n + 2 = 102 (ECF from (b))	✓ 1
A1 n = 25	✓ 1

Exam tip: To show a number is not in a linear sequence, either use a property such as odd/even or solve for n and show it is not a whole number.

+ Add to test✓ In your test155IGMC3-2-11r48-12Quadratic equations and inequalities Paper CMedium7 marks

No calculator

(a)

Solve 3x − 7 < 14.

(2)

(b)

List all the integers n that satisfy −2 ≤ 2n + 1 < 9.

(3)

(c)

A wind farm must produce at least 40 MW. Each turbine produces 2.5 MW. There are n turbines. Write an inequality in n and use it to find the smallest possible number of turbines.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3x < 21	✓ 1
A1 x < 7	✓ 1
Part (b)
M1 subtracts 1 throughout, −3 ≤ 2n < 8	✓ 1
A1 −1.5 ≤ n < 4	✓ 1
A1 −1, 0, 1, 2, 3	✓ 1
Part (c)
B1 2.5n ≥ 40	✓ 1
B1 16 turbines	✓ 1

Common error: When listing integers, check the end-points: ≤ includes the end value and < does not.

+ Add to test✓ In your test156IGMC-042-01-14Linear and simultaneous equations and rearranging formulae Paper CHard6 marks

No calculator

A fraction is $\frac{n}{d}$.
When 1 is added to both the numerator and the denominator, the fraction becomes $\frac{2}{3}$.
When 1 is subtracted from both the numerator and the denominator, the fraction becomes $\frac{1}{2}$.

(a)

Show that $3n - 2d = -1$.

(2)

(b)

Form a second equation and solve the two equations to find the fraction $\frac{n}{d}$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{n+1}{d+1} = \frac{2}{3}$ leading to 3(n + 1) = 2(d + 1)	✓ 1
A1 for 3n + 3 = 2d + 2, rearranged correctly to 3n − 2d = −1	✓ 1
Part (b)
M1 for $\frac{n-1}{d-1} = \frac{1}{2}$ leading to 2(n − 1) = d − 1	✓ 1
A1 for 2n − d = 1	✓ 1
M1 for a correct method to solve the two equations simultaneously, e.g. substituting d = 2n − 1 into 3n − 2d = −1	✓ 1
A1 for $\frac{3}{5}$	✓ 1

Exam tip: Check your fraction against both conditions: $\frac{3+1}{5+1} = \frac{4}{6} = \frac{2}{3}$ and $\frac{3-1}{5-1} = \frac{2}{4} = \frac{1}{2}$.

+ Add to test✓ In your test157IGMC3-2-08c-09Sequences and graphs of functions Paper CHard7 marks

Calculator

A solar installer lays panels in rows. Row 1 has 8 panels, row 2 has 11 panels, row 3 has 14 panels, and the pattern continues. A second array has 5n − 4 panels in row n.

(a)

Find an expression for the number of panels in row n of the first array.

(2)

(b)

A row in the first array has 59 panels. Find the row number.

(2)

(c)

Find the first row in which the second array has more panels than the first array. Write down the number of panels in that row of the second array.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 common difference 3	✓ 1
A1 3n + 5	✓ 1
Part (b)
M1 3n + 5 = 59	✓ 1
A1 n = 18	✓ 1
Part (c)
M1 5n − 4 > 3n + 5	✓ 1
A1 n > 4.5, so row 5	✓ 1
A1 21 panels (5 × 5 − 4)	✓ 1

Common error: Test your answer in both formulas: row 4 gives 16 and 17 panels, so row 4 is too early.

+ Add to test✓ In your test158IGMC3-2-12r27-14Linear and simultaneous equations and rearranging formulae Paper CHard9 marks

Calculator

A courier company, Company A, charges a fixed fee of £f plus £r for each kilogram of the parcel. It charges £13.40 for a 4 kg parcel and £21.40 for a 9 kg parcel. A second company, Company B, charges £2.10 per kilogram with no fixed fee.

(a)

Find the values of f and r.

(4)

(b)

Find the mass of parcel for which both companies charge the same.

(3)

(c)

For a 20 kg parcel, work out how much cheaper Company A is than Company B.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 f + 4r = 13.40 and f + 9r = 21.40	✓ 1
M1 subtracting to eliminate f: 5r = 8.00	✓ 1
A1 r = 1.60	✓ 1
A1 f = 7.00	✓ 1
Part (b)
M1 7 + 1.6w = 2.1w	✓ 1
M1 7 = 0.5w	✓ 1
A1 14 kg	✓ 1
Part (c)
M1 Company A: 7 + 1.6 × 20 = £39.00 and Company B: 2.1 × 20 = £42.00	✓ 1
A1 £3.00	✓ 1

Exam tip: To compare two charges, set up an expression for each in terms of the same variable, then either equate them or substitute the same value into both.

+ Add to test✓ In your test159IGMC3-2-11r28-14Linear and simultaneous equations and rearranging formulae Paper CHard8 marks

Calculator

At a market, Stall A sells scarves at £19 each. Its costs are £180 hire fee plus £7 for each scarf sold. Profit is income minus costs. Stall A sells $x$ scarves.

Diagram not to scale

(a)

Show that the profit of Stall A is £$(12x-180)$.

(2)

(b)

Stall A needs a profit of at least £500. Find the least number of scarves it must sell.

(3)

(c)

The profit of Stall B is £$(15x-270)$ when it sells $x$ items. Find the value of $x$ for which both stalls make the same profit, and find that profit.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 income $19x$ minus costs $(180+7x)$	✓ 1
A1 $12x-180$	✓ 1
Part (b)
M1 $12x-180\ge500$	✓ 1
A1 $x\ge56.67\ldots$	✓ 1
A1 57 (a whole number of scarves)	✓ 1
Part (c)
M1 $12x-180=15x-270$	✓ 1
A1 $x=30$	✓ 1
A1 profit £180	✓ 1

Common error: A 'least number' answer must be a whole number of items; round up, even when the decimal is below 0.5.

+ Add to test✓ In your test160IGMC3-2-12r28-14Functions and differentiation Paper CHard10 marks

Calculator

Cyclist A leaves a village at 09:00 and rides along a straight road at a constant 18 km/h. Cyclist B leaves the same village 30 minutes later and rides in the same direction at a constant 24 km/h. Let $t$ be the time in hours after 09:00.

(a)

For $t\ge0.5$, write down expressions, in km, for the distance of each cyclist from the village.

(2)

(b)

Find the value of $t$ when B catches A.

(3)

(c)

Find the distance from the village at which B catches A.

(1)

(d)

In fact, A stops to rest at the roadside from 10:00 to 10:30 and does not move during this time. Show that B passes A while A is resting, and find the time after 09:00 at which this happens.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 A: $18t$	✓ 1
B1 B: $24(t-0.5)$	✓ 1
Part (b)
M1 $18t=24(t-0.5)$	✓ 1
M1 $18t=24t-12$, so $6t=12$	✓ 1
A1 $t=2$ (11:00)	✓ 1
Part (c)
B1 36 km	✓ 1
Part (d)
M1 A is 18 km from the village throughout the rest ($18\times1$)	✓ 1
M1 $24(t-0.5)=18$	✓ 1
A1 $t=1.25$ hours	✓ 1
B1 1.25 is between 1 and 1.5, so B reaches A during the rest, at 10:15	✓ 1

Exam tip: Write each cyclist's distance as a function of time on the same clock, then equate them.

+ Add to test✓ In your test161IGMC3-2-12r46-14Quadratic equations and inequalities Paper CHard7 marks

Calculator

A rectangular fish pond has a length that is 5 m longer than its width. The area of the pond is 84 m². Let the width be w metres.

Diagram not to scale

(a)

Show that w² + 5w − 84 = 0.

(2)

(b)

Solve the equation by factorising and state the width of the pond.

(3)

(c)

Fencing costs £18.50 per metre. Calculate the cost of fencing the whole perimeter of the pond.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $w(w+5)=84$	✓ 1
A1 expands and rearranges to the given equation	✓ 1
Part (b)
M1 $(w+12)(w-7)=0$	✓ 1
A1 $w=7$ or $w=-12$	✓ 1
B1 width is 7 m because a length cannot be negative	✓ 1
Part (c)
M1 perimeter $2\times(7+12)=38$ m and $38\times18.50$	✓ 1
A1 £703	✓ 1

Common error: Always check solutions in the context of the question; a negative length must be rejected.

+ Add to test✓ In your test162IGMC3-2-11r48-13Sequences and graphs of functions Paper CHard8 marks

Calculator

In a library display, the number of books in row n follows a linear sequence. Row 4 holds 17 books and row 10 holds 41 books.

(a)

Find the common difference of the sequence.

(2)

(b)

Find the nth term.

(2)

(c)

Which row holds 101 books?

(2)

(d)

Work out the total number of books in rows 1 to 4.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (41 − 17) ÷ (10 − 4)	✓ 1
A1 4	✓ 1
Part (b)
M1 first term 17 − 3 × 4 = 5 (or 4n + c with 17 substituted)	✓ 1
A1 4n + 1	✓ 1
Part (c)
M1 4n + 1 = 101 (ECF from (b))	✓ 1
A1 row 25	✓ 1
Part (d)
M1 5 + 9 + 13 + 17	✓ 1
A1 44	✓ 1

Exam tip: Use the difference between two known terms divided by the gap in positions to find the common difference.

+ Add to test✓ In your test163IGMC-042-00-02Expanding, factorising and algebraic fractions Paper EEasy6 marks

No calculator

Answer all parts.

(a)

Expand and simplify $(x + 5)(x - 3)$.

(2)

(b)

Expand and simplify $(2x - 1)^2$.

(2)

(c)

Factorise $x^2 - 7x + 12$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for at least 3 of the 4 terms x², −3x, +5x, −15 correct	✓ 1
A1 x² + 2x − 15	✓ 1
Part (b)
M1 for (2x − 1)(2x − 1) expanded with at least 3 of 4 terms correct	✓ 1
A1 4x² − 4x + 1	✓ 1
Part (c)
(x − 3)(x − 4) B2	✓ 2
[B1 for two brackets whose product gives two of the three terms correctly]	—

Common error: $(2x-1)^2$ is not $4x^2 + 1$. Write the bracket out twice and multiply every term so the middle term −4x is not lost.

+ Add to test✓ In your test164IGMC-042-00-03Expanding, factorising and algebraic fractions Paper EEasy5 marks

No calculator

Answer all parts.

(a)

Factorise $x^2 - 49$.

(1)

(b)

Factorise completely $5x^2 - 20$.

(2)

(c)

Use factorisation to work out the exact value of $83^2 - 17^2$. Show your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(x − 7)(x + 7) B1	✓ 1
Part (b)
M1 for 5(x² − 4) seen	✓ 1
A1 5(x − 2)(x + 2)	✓ 1
Part (c)
M1 for (83 − 17)(83 + 17) or 66 × 100	✓ 1
A1 6600	✓ 1

Exam tip: "Factorise completely" means take out the common factor first, then check whether what is left is a difference of two squares.

+ Add to test✓ In your test165IGMC-042-00-04Expanding, factorising and algebraic fractions Paper EEasy4 marks

No calculator

Simplify fully.

(a)

$\frac{12x^3y}{4xy^2}$

(2)

(b)

$\frac{x}{3} + \frac{x}{5}$

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 for two of: coefficient 3, x² in numerator, y in denominator	✓ 1
B1 $\frac{3x^2}{y}$ fully correct	✓ 1
Part (b)
M1 for $\frac{5x}{15} + \frac{3x}{15}$ or common denominator 15 used correctly	✓ 1
A1 $\frac{8x}{15}$	✓ 1

Common error: Adding the numerators and the denominators to get $\frac{2x}{8}$ is wrong. Fractions can only be added once they share a common denominator.

+ Add to test✓ In your test166IGMC-042-02-05Quadratic equations and inequalities Paper EEasy3 marks

No calculator

(a)

Show that x = 3 is a solution of 2x² − 5x − 3 = 0.

(1)

(b)

Find the other solution of the equation.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2 × 9 − 15 − 3 = 0 shown	✓ 1
Part (b)
M1 (x − 3)(2x + 1) = 0	✓ 1
A1 x = $-\frac{1}{2}$	✓ 1

Exam tip: If one solution is x = 3, then (x − 3) must be a factor. This gives you one bracket and makes the factorising quicker.

+ Add to test✓ In your test167IGMC-042-04-01Functions and differentiation Paper EEasy4 marks

No calculator

The functions f and g are defined by
f(x) = 3x − 5
g(x) = x² + 1

(a)

Find f(4).

(1)

(b)

Find g(−3).

(1)

(c)

Find fg(2).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 7	✓ 1
Part (b)
B1: 10	✓ 1
Part (c)
M1: g(2) = 2² + 1 = 5	✓ 1
A1: fg(2) = f(5) = 3(5) − 5 = 10	✓ 1

Exam tip: In fg(2), work from the inside out. Apply g first, then put that answer into f.

+ Add to test✓ In your test168IGMC-042-04-02Functions and differentiation Paper EEasy5 marks

No calculator

The function f is defined by f(x) = $\frac{x+7}{2}$.

(a)

Find f(5).

(1)

(b)

Find f⁻¹(x).

(2)

(c)

Solve the equation f(x) = f⁻¹(x).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 6	✓ 1
Part (b)
M1: write y = (x + 7)/2 and rearrange to x = 2y − 7, or swap x and y and rearrange	✓ 1
A1: f⁻¹(x) = 2x − 7	✓ 1
Part (c)
M1: (x + 7)/2 = 2x − 7 leading to x + 7 = 4x − 14	✓ 1
A1: x = 7	✓ 1

Common error: f⁻¹(x) is the inverse function, not the reciprocal. Do not write 2/(x + 7).

+ Add to test✓ In your test169IGMC-042-04-03Functions and differentiation Paper EEasy4 marks

No calculator

A curve has equation y = 4x³ − 6x² + 2x − 9.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the gradient of the curve at the point where x = 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: at least two of the terms 12x², −12x, +2 correct	✓ 1
A1: dy/dx = 12x² − 12x + 2, fully correct with the constant removed	✓ 1
Part (b)
M1: substitute x = 2 into their dy/dx: 12(4) − 12(2) + 2	✓ 1
A1: 26	✓ 1

Common error: A constant term such as −9 differentiates to 0. It does not stay in the derivative.

+ Add to test✓ In your test170IGMC-042-04-04Functions and differentiation Paper EEasy5 marks

No calculator

A curve has equation y = x² − 8x + 3.

(a)

Find $\frac{dy}{dx}$.

(1)

(b)

Find the coordinates of the turning point of the curve.

(3)

(c)

State whether the turning point is a maximum or a minimum. Give a reason.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 2x − 8	✓ 1
Part (b)
M1: set 2x − 8 = 0	✓ 1
A1: x = 4	✓ 1
A1: y = 16 − 32 + 3 = −13, giving the point (4, −13)	✓ 1
Part (c)
B1: minimum, with a valid reason, e.g. the coefficient of x² is positive, or d²y/dx² = 2 > 0	✓ 1

Exam tip: After finding x at a turning point, always substitute back into the original equation, not into dy/dx, to find y.

+ Add to test✓ In your test171IGMC-042-04-05Functions and differentiation Paper EEasy6 marks

No calculator

The function f is defined by f(x) = 5 − 2x.

(a)

Find the value of x when f(x) = 11.

(2)

(b)

Find f⁻¹(x).

(2)

(c)

Find ff(x), giving your answer in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 5 − 2x = 11	✓ 1
A1: x = −3	✓ 1
Part (b)
M1: y = 5 − 2x rearranged to x = (5 − y)/2, or equivalent	✓ 1
A1: f⁻¹(x) = (5 − x)/2, or equivalent	✓ 1
Part (c)
M1: 5 − 2(5 − 2x)	✓ 1
A1: 4x − 5	✓ 1

Common error: ff(x) means f applied twice, not [f(x)]². Substitute the whole expression 5 − 2x in place of x.

+ Add to test✓ In your test172IGMC-042-00-06Expanding, factorising and algebraic fractions Paper EMedium5 marks

No calculator

Answer all parts.

(a)

Factorise $2x^2 + 7x - 15$.

(2)

(b)

Hence solve $2x^2 + 7x - 15 = 0$.

(1)

(c)

Factorise completely $12x^2 - 27$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(2x − 3)(x + 5) B2	✓ 2
[B1 for (2x + 3)(x − 5), or for brackets (2x + a)(x + b) with ab = −15 or with 2b + a = 7]	—
Part (b)
x = $\frac{3}{2}$ (or 1.5) and x = −5, both required B1	✓ 1
Part (c)
M1 for 3(4x² − 9), or for a correct partial factorisation such as (6x − 9)(2x + 3) or (2x − 3)(6x + 9); implied by a correct final answer	✓ 1
A1 3(2x − 3)(2x + 3)	✓ 1

Exam tip: Always check a factorisation by expanding it again. For 'factorise completely', take out any common numerical factor first, then look for a difference of two squares.

+ Add to test✓ In your test173IGMC-042-00-07Expanding, factorising and algebraic fractions Paper EMedium5 marks

No calculator

Simplify fully.

(a)

$\frac{x^2 - 9}{x^2 + x - 6}$

(3)

(b)

$\frac{3x + 6}{x^2 - 4}$

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 (x − 3)(x + 3)	✓ 1
B1 (x + 3)(x − 2)	✓ 1
B1 $\frac{x-3}{x-2}$	✓ 1
Part (b)
M1 for $\frac{3(x+2)}{(x-2)(x+2)}$	✓ 1
A1 $\frac{3}{x-2}$	✓ 1

Common error: You can only cancel whole factors, never single terms. Crossing out the x² in the numerator and denominator earns no marks.

+ Add to test✓ In your test174IGMC-042-00-08Expanding, factorising and algebraic fractions Paper EMedium4 marks

No calculator

Answer both parts.

(a)

Write $\frac{3}{x+2} - \frac{2}{x-1}$ as a single fraction in its simplest form.

(3)

(b)

Hence solve $\frac{3}{x+2} - \frac{2}{x-1} = 0$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for common denominator (x + 2)(x − 1)	✓ 1
M1 for numerator 3(x − 1) − 2(x + 2)	✓ 1
A1 $\frac{x-7}{(x+2)(x-1)}$	✓ 1
Part (b)
x = 7 B1	✓ 1

Common error: −2(x + 2) = −2x − 4. Losing the bracket and writing −2x + 4 gives the wrong numerator.

+ Add to test✓ In your test175IGMC-042-00-09Expanding, factorising and algebraic fractions Paper EMedium5 marks

No calculator

Answer both parts.

(a)

Expand and simplify $(x + 2)(x - 3)(2x + 1)$.

(3)

(b)

Factorise $(x - 4)^2 - 3(x - 4)$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for any two brackets correctly expanded, e.g. x² − x − 6	✓ 1
M1 for multiplying their quadratic by the remaining bracket with at least 4 of 6 terms correct	✓ 1
A1 2x³ − x² − 13x − 6	✓ 1
Part (b)
M1 for (x − 4)[(x − 4) − 3], or for correct expansion to x² − 11x + 28	✓ 1
A1 (x − 4)(x − 7)	✓ 1

Exam tip: In the factorising part, treat (x − 4) as a single common factor. There is no need to expand first.

+ Add to test✓ In your test176IGMC-042-00-10Expanding, factorising and algebraic fractions Paper EMedium7 marks

Calculator

A rectangle has length (2x + 3) cm and width (x − 1) cm, where x > 1.
A square of side (x − 1) cm is cut from the rectangle.

Diagram not to scale

(a)

Show that the area of the rectangle is (2x² + x − 3) cm².

(2)

(b)

Show that the area remaining after the square is removed is (x² + 3x − 4) cm².

(2)

(c)

The remaining area is 30 cm². Find the value of x. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for (2x + 3)(x − 1) = 2x² − 2x + 3x − 3	✓ 1
A1 2x² + x − 3 with no errors seen	✓ 1
Part (b)
M1 for 2x² + x − 3 − (x² − 2x + 1)	✓ 1
A1 x² + 3x − 4 with brackets handled correctly	✓ 1
Part (c)
M1 for x² + 3x − 34 = 0	✓ 1
M1 for $\frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-34)}}{2}$ or better, e.g. $\frac{-3 \pm \sqrt{145}}{2}$	✓ 1
A1 4.52 only, negative root rejected	✓ 1

Common error: Subtracting x² − 2x + 1 without a bracket gives −x² − 2x + 1, which is wrong. Keep the bracket so that every sign changes.

+ Add to test✓ In your test177IGMC-042-00-11Expanding, factorising and algebraic fractions Paper EMedium7 marks

No calculator

Answer all parts.

(a)

Factorise $xy + 3x - 2y - 6$.

(2)

(b)

Factorise completely $x^4 - 16$.

(2)

(c)

Simplify fully $\frac{ax + bx - 2a - 2b}{x^2 - 4}$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for x(y + 3) − 2(y + 3)	✓ 1
A1 (x − 2)(y + 3)	✓ 1
Part (b)
M1 for (x² − 4)(x² + 4)	✓ 1
A1 (x − 2)(x + 2)(x² + 4)	✓ 1
Part (c)
B1 numerator as (a + b)(x − 2)	✓ 1
B1 denominator as (x − 2)(x + 2)	✓ 1
B1 $\frac{a+b}{x+2}$	✓ 1

Exam tip: When grouping, the two brackets must match exactly before you take them out as a common factor. Check this before writing your final line.

+ Add to test✓ In your test178IGMC-042-02-03Quadratic equations and inequalities Paper EMedium4 marks

Calculator

Solve the equation 2x² − 5x − 4 = 0.
Show all your working and give your answers correct to 2 decimal places.

(a)

Solve 2x² − 5x − 4 = 0, giving your answers correct to 2 decimal places.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
√((−5)² − 4(2)(−4)) or √57 seen, or completing the square to (x − $\frac{5}{4}$)² = $\frac{57}{16}$	✓ 1
$x = \frac{-(-5) \pm \sqrt{57}}{2\times 2}$, or $x = \frac{5}{4} \pm \sqrt{\frac{57}{16}}$	✓ 1
x = 3.14	✓ 1
x = −0.64	✓ 1

Exam tip: Write the formula, or your completed square, with the values substituted before using your calculator. If the final answers are wrong, the method marks can still be earned from this line.

+ Add to test✓ In your test179IGMC-042-02-07Quadratic equations and inequalities Paper EMedium5 marks

No calculator

This question is about the quadratic expression x² − x − 12.

(a)

Factorise x² − x − 12.

(2)

(b)

Hence solve the inequality x² − x − 12 < 0.

(2)

(c)

Write down the number of integers that satisfy x² − x − 12 < 0.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (x + a)(x + b) with ab = −12 and a + b = −1, or a pair of brackets giving the correct x² and constant terms	✓ 1
A1 (x − 4)(x + 3)	✓ 1
Part (b)
M1 critical values −3 and 4 identified	✓ 1
A1 −3 < x < 4	✓ 1
Part (c)
B1 6 (the integers −2, −1, 0, 1, 2, 3)	✓ 1

Common error: Writing x 4 gives the region where the expression is positive. A '< 0' quadratic with a positive x² term is negative between its roots.

+ Add to test✓ In your test180IGMC-042-02-08Quadratic equations and inequalities Paper EMedium7 marks

No calculator

A rectangle has length (x + 4) cm and width (x − 1) cm. The area of the rectangle is 36 cm².

Diagram not to scale

(a)

Show that x² + 3x − 40 = 0.

(2)

(b)

Solve x² + 3x − 40 = 0 and explain why only one solution is valid.

(3)

(c)

Find the perimeter of the rectangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (x + 4)(x − 1) = 36	✓ 1
A1 x² + 3x − 4 = 36 leading to x² + 3x − 40 = 0 with no errors	✓ 1
Part (b)
M1 (x + 8)(x − 5) = 0	✓ 1
A1 x = 5 or x = −8	✓ 1
B1 x = −8 rejected because the width x − 1 (and length) would be negative	✓ 1
Part (c)
M1 2 × (9 + 4)	✓ 1
A1 26 cm	✓ 1

Exam tip: In a 'show that' part, write every line of working down to the given equation. A result that appears without working earns no marks.

+ Add to test✓ In your test181IGMC-042-02-10Quadratic equations and inequalities Paper EMedium5 marks

No calculator

(a)

Show that $\frac{3}{x+1} + \frac{2}{x} = 2$ can be written as 2x² − 3x − 2 = 0.

(3)

(b)

Solve 2x² − 3x − 2 = 0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplying every term by x(x + 1): 3x + 2(x + 1) = 2x(x + 1)	✓ 1
M1 5x + 2 = 2x² + 2x	✓ 1
A1 rearranged to 2x² − 3x − 2 = 0 with no errors	✓ 1
Part (b)
M1 (2x + 1)(x − 2) = 0	✓ 1
A1 x = 2 or x = $-\frac{1}{2}$	✓ 1

Common error: When clearing fractions, the 2 on the right-hand side must also be multiplied by x(x + 1). Leaving it as 2 is a frequent slip.

+ Add to test✓ In your test182IGMC-042-01-11Linear and simultaneous equations and rearranging formulae Paper EMedium5 marks

No calculator

The lens formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.

(a)

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

(3)

(b)

Find v when f = 12 and u = 20.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$	✓ 1
M1 for combining into a single fraction: $\frac{1}{v} = \frac{u-f}{uf}$	✓ 1
A1 for $v = \frac{uf}{u-f}$	✓ 1
Part (b)
M1 for $\frac{20 \times 12}{20 - 12}$	✓ 1
A1 for 30	✓ 1

Common error: $\frac{1}{f} - \frac{1}{u}$ is not $\frac{1}{f-u}$. Use a common denominator uf before you take the reciprocal.

+ Add to test✓ In your test183IGMC-042-03-07Sequences and graphs of functions Paper EMedium6 marks

No calculator

The first five terms of a sequence are
2, 9, 20, 35, 54, ...

(a)

Find an expression, in terms of n, for the nth term.

(3)

(b)

Factorise your expression from part (a) and hence explain why no term after the first is a prime number.

(2)

(c)

Find the 30th term.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for second differences of 4 identified	✓ 1
M1 for 2n² seen in an expression	✓ 1
A1 2n² + n − 1	✓ 1
Part (b)
B1 (2n − 1)(n + 1)	✓ 1
B1 for n ≥ 2 both factors are at least 3, so every term is a product of two whole numbers greater than 1	✓ 1
Part (c)
B1 1829	✓ 1

Exam tip: Halve the second difference to get the coefficient of n², then subtract that part from each term to find the linear part.

+ Add to test✓ In your test184IGMC-042-03-08Functions and differentiation Paper EMedium8 marks

Calculator

A curve has equation y = x³ − 6x² + 9x + 1.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the coordinates of the two turning points.

(4)

(c)

Determine whether each turning point is a maximum or a minimum. Show how you decide.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B2 3x² − 12x + 9	✓ 2
(B1 for two of the three terms correct)	—
Part (b)
M1 for 3x² − 12x + 9 = 0	✓ 1
M1 for 3(x − 1)(x − 3) = 0 or x = 1 and x = 3 found	✓ 1
A1 (1, 5)	✓ 1
A1 (3, 1)	✓ 1
Part (c)
M1 for a valid method, e.g. $\frac{d^2y}{dx^2}$ = 6x − 12 evaluated at both points, or gradient values either side	✓ 1
A1 (1, 5) is a maximum and (3, 1) is a minimum	✓ 1

Common error: Finding the x-values of the turning points is not enough: substitute back into the equation of the curve, not into $\frac{dy}{dx}$, to find the y-coordinates.

+ Add to test✓ In your test185IGMC-042-03-10Functions and differentiation Paper EMedium7 marks

No calculator

A curve has equation y = 2x² − 5x + 3.

(a)

Find $\frac{dy}{dx}$.

(1)

(b)

Find the gradient of the curve at the point where x = 2.

(1)

(c)

The tangent to the curve at point P is parallel to the line y = 7x − 4.
Find the coordinates of P.

(3)

(d)

Find the equation of the tangent at P, giving your answer in the form y = mx + c.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4x − 5	✓ 1
Part (b)
B1 3	✓ 1
Part (c)
M1 for 4x − 5 = 7	✓ 1
A1 x = 3	✓ 1
A1 (3, 6)	✓ 1
Part (d)
M1 for 6 = 7 × 3 + c or y − 6 = 7(x − 3)	✓ 1
A1 y = 7x − 15	✓ 1

Common error: Parallel lines have equal gradients. Setting the curve's equation equal to 7x − 4 finds where the curve meets the line, not where its gradient is 7.

+ Add to test✓ In your test186IGMC-042-03-11Sequences and graphs of functions Paper EMedium6 marks

Calculator

The number of bacteria, y, in a sample after x hours is modelled by
y = 500 × 1.2^x.

(a)

Write down the number of bacteria at the start.

(1)

(b)

Find the number of bacteria after 6 hours.

(2)

(c)

Find the value of x when the number of bacteria is 2000. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 500	✓ 1
Part (b)
M1 for 500 × 1.2⁶	✓ 1
A1 1493 or 1490 (from 1492.99…)	✓ 1
Part (c)
M1 for 1.2^x = 4 or 500 × 1.2^x = 2000	✓ 1
M1 for a correct trial between 7 and 8, e.g. 500 × 1.2^7.6 = 1998.7, or for x = log 4 ÷ log 1.2	✓ 1
A1 7.60	✓ 1

Exam tip: In trial and improvement, test the value halfway between your two closest trials (here 7.605) to decide which way to round.

+ Add to test✓ In your test187IGMC-042-04-06Functions and differentiation Paper EMedium7 marks

No calculator

f(x) = 2x + 1
g(x) = x² − 3

(a)

Find gf(x), giving your answer in the form ax² + bx + c.

(2)

(b)

Solve gf(x) = 6.

(3)

(c)

Find f⁻¹(g(2)).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: (2x + 1)² − 3	✓ 1
A1: 4x² + 4x − 2	✓ 1
Part (b)
M1: 4x² + 4x − 8 = 0, or x² + x − 2 = 0	✓ 1
M1: correct factorisation (x + 2)(x − 1) = 0, or a correct method for solving	✓ 1
A1: x = 1 and x = −2	✓ 1
Part (c)
B1: g(2) = 1, or f⁻¹(x) = (x − 1)/2	✓ 1
B1: 0	✓ 1

Exam tip: When you expand (2x + 1)², write it as (2x + 1)(2x + 1). Squaring each term alone loses the middle term 4x.

+ Add to test✓ In your test188IGMC-042-04-07Functions and differentiation Paper EMedium8 marks

No calculator

A curve has equation y = x³ − 3x² − 9x + 5.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the coordinates of the two stationary points on the curve.

(4)

(c)

Determine the nature of each stationary point. Show your working.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: at least two terms correct	✓ 1
A1: 3x² − 6x − 9	✓ 1
Part (b)
M1: 3x² − 6x − 9 = 0	✓ 1
M1: 3(x − 3)(x + 1) = 0, or equivalent	✓ 1
A1: x = 3 and x = −1	✓ 1
A1: (3, −22) and (−1, 10)	✓ 1
Part (c)
M1: d²y/dx² = 6x − 6 evaluated at both points, or a valid gradient-sign test either side of each point	✓ 1
A1: (−1, 10) is a maximum and (3, −22) is a minimum	✓ 1

Exam tip: Take out the common factor 3 before factorising. The quadratic x² − 2x − 3 is much easier to handle.

+ Add to test✓ In your test189IGMC-042-04-08Functions and differentiation Paper EMedium7 marks

No calculator

A curve has equation y = 2x² − $\frac{8}{x}$.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the gradient of the curve at the point where x = 2.

(1)

(c)

Find the equation of the tangent to the curve at the point where x = 2. Give your answer in the form y = mx + c.

(3)

(d)

Find the x-coordinate of the point where this tangent crosses the x-axis.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 4x	✓ 1
B1: + 8/x², or + 8x⁻²	✓ 1
Part (b)
B1: 10	✓ 1
Part (c)
B1: y = 2(4) − 8/2 = 4 when x = 2	✓ 1
M1: y − 4 = 10(x − 2), or 4 = 10(2) + c	✓ 1
A1: y = 10x − 16	✓ 1
Part (d)
B1: 1.6	✓ 1

Common error: Rewrite −8/x as −8x⁻¹ before differentiating. Its derivative is +8x⁻². The sign changes because the power is negative.

+ Add to test✓ In your test190IGMC-042-04-09Functions and differentiation Paper EMedium6 marks

No calculator

The function f is defined by f(x) = $\frac{3x-2}{x+4}$, x ≠ −4.

(a)

Find f(1).

(1)

(b)

Find f⁻¹(x).

(3)

(c)

Solve f⁻¹(x) = 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 1/5	✓ 1
Part (b)
M1: y(x + 4) = 3x − 2, or x(y + 4) = 3y − 2 after swapping	✓ 1
M1: collect the terms in the subject on one side and factorise, e.g. x(y − 3) = −4y − 2	✓ 1
A1: f⁻¹(x) = (4x + 2)/(3 − x), or equivalent	✓ 1
Part (c)
M1: (4x + 2)/(3 − x) = 2 leading to 4x + 2 = 6 − 2x, or use f⁻¹(x) = 2 ⇔ x = f(2) = 4/6	✓ 1
A1: x = 2/3	✓ 1

Exam tip: To make x the subject when it appears twice, gather every x term on one side and then factorise x out.

+ Add to test✓ In your test191IGMC-042-04-10Functions and differentiation Paper EMedium7 marks

No calculator

A small robot moves along a straight track. Its distance, s metres, from a fixed point after t seconds is given by
s = t³ − 6t² + 15t.
The velocity v m/s is the rate of change of s, so v = $\frac{ds}{dt}$.

(a)

Find an expression for v in terms of t.

(2)

(b)

Find the velocity when t = 4.

(1)

(c)

Show that the robot is never at rest.

(2)

(d)

The acceleration is $\frac{dv}{dt}$. Find the velocity of the robot when its acceleration is zero.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: at least two terms correct	✓ 1
A1: v = 3t² − 12t + 15	✓ 1
Part (b)
B1: 15 m/s	✓ 1
Part (c)
M1: a valid method, e.g. discriminant (−12)² − 4(3)(15) = −36, or 3[(t − 2)² + 1]	✓ 1
A1: correct conclusion that v is never 0 (in fact v > 0 for all t), so the robot is never at rest	✓ 1
Part (d)
M1: 6t − 12 = 0 so t = 2	✓ 1
A1: v = 3 m/s	✓ 1

Exam tip: "At rest" means velocity = 0. A "show that" question needs a clear concluding statement.

+ Add to test✓ In your test192IGMC-042-04-11Functions and differentiation Paper EMedium7 marks

No calculator

A curve has equation y = ax² + bx, where a and b are constants.
The gradient of the curve is 4 when x = 1.
The gradient of the curve is 16 when x = 3.

(a)

Find the value of a and the value of b.

(4)

(b)

Find the coordinates of the point on the curve where the gradient is zero.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: dy/dx = 2ax + b	✓ 1
M1: 2a + b = 4 and 6a + b = 16	✓ 1
A1: a = 3	✓ 1
A1: b = −2	✓ 1
Part (b)
M1: 6x − 2 = 0	✓ 1
A1: x = 1/3	✓ 1
A1: y = 3(1/9) − 2(1/3) = −1/3, giving the point (1/3, −1/3)	✓ 1

Common error: The gradient conditions go into dy/dx, not into the equation of the curve. Substituting x = 1 and y = 4 into y = ax² + bx is wrong.

+ Add to test✓ In your test193IGMC3-2-08c-10Quadratic equations and inequalities Paper EMedium7 marks

No calculator

A rectangular garden has width x metres and length (x + 3) metres. Its area is 70 m².

Diagram not to scale

(a)

Show that x² + 3x − 70 = 0.

(2)

(b)

Solve the equation by factorising.

(3)

(c)

Find the perimeter of the garden.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x(x + 3) = 70	✓ 1
A1 x² + 3x − 70 = 0 shown	✓ 1
Part (b)
M1 (x + 10)(x − 7)	✓ 1
A1 x = −10 or x = 7	✓ 1
B1 reject x = −10 since a width cannot be negative	✓ 1
Part (c)
M1 length = 10 m (ECF from (b))	✓ 1
A1 34 m	✓ 1

Common error: After solving a quadratic in context, always check whether each root makes sense for the situation.

+ Add to test✓ In your test194IGMC3-2-08c-11Quadratic equations and inequalities Paper EMedium5 marks

Calculator

A ball is thrown upwards from a height of 1.5 m. Its height h metres after t seconds is h = 1.5 + 12t − 4.9t².

(a)

Calculate the height of the ball after 1 second.

(1)

(b)

Calculate the time taken for the ball to reach the ground. Show your working and give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 8.6 m	✓ 1
Part (b)
B1 4.9t² − 12t − 1.5 = 0	✓ 1
M1 correct use of the quadratic formula, e.g. t = (12 ± √(144 + 29.4)) ÷ 9.8	✓ 1
A1 t = 2.57 s	✓ 1
B1 other root t = −0.119 rejected as negative time	✓ 1

Exam tip: Set the height equal to zero for ground level and reject any negative time.

+ Add to test✓ In your test195IGMC3-2-08c-12Expanding, factorising and algebraic fractions Paper EMedium7 marks

No calculator

This question is about algebraic fractions.

(a)

Simplify $\frac{x^2-9}{x^2+5x+6}$.

(3)

(b)

Solve $\frac{1}{x-1} + \frac{2}{x+1} = 1$.

(4)

Show mark scheme

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 multiply through by (x − 1)(x + 1)	✓ 1
M1 (x + 1) + 2(x − 1) = x² − 1	✓ 1
A1 x² − 3x = 0	✓ 1
A1 x = 0 or x = 3	✓ 1

Common error: Cancel factors, never terms: x in the top and bottom of a fraction cannot be cancelled unless it is a factor of both.

+ Add to test✓ In your test196IGMC3-2-08r5-07Quadratic equations and inequalities Paper EMedium8 marks

No calculator

Consider the expression x² − 5x − 24.

(a)

Factorise x² − 5x − 24.

(2)

(b)

Solve x² − 5x − 24 = 0.

(1)

(c)

Solve the inequality x² − 5x − 24 ≤ 0.

(2)

(d)

Simplify (x² − 5x − 24) ÷ (x² − 64).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (x + a)(x + b) with ab = −24	✓ 1
A1 (x − 8)(x + 3)	✓ 1
Part (b)
B1 x = 8 and x = −3	✓ 1
Part (c)
M1 use of critical values −3 and 8 with the graph below the axis between them	✓ 1
A1 −3 ≤ x ≤ 8	✓ 1
Part (d)
B1 x² − 64 = (x − 8)(x + 8)	✓ 1
M1 cancel the common factor (x − 8)	✓ 1
A1 (x + 3) ÷ (x + 8)	✓ 1

Exam tip: For a quadratic inequality, find the roots first and then decide which region satisfies the sign; a quick sketch in words (U-shape, crossing points) helps.

+ Add to test✓ In your test197IGMC3-2-08r5-08Quadratic equations and inequalities Paper EMedium8 marks

Calculator

A rectangular data-centre floor has width x metres and length (x + 6) metres. Its area is 250 m².

Diagram not to scale

(a)

Show that x² + 6x − 250 = 0.

(2)

(b)

Solve x² + 6x − 250 = 0 using the quadratic formula. Give both answers correct to 2 decimal places.

(3)

(c)

State why only one of the solutions is a possible width.

(1)

(d)

Calculate the perimeter of the floor. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x(x + 6) = 250	✓ 1
A1 expand and rearrange to x² + 6x − 250 = 0	✓ 1
Part (b)
M1 correct substitution into the formula with a = 1, b = 6, c = −250	✓ 1
A1 x = 13.09	✓ 1
A1 x = −19.09	✓ 1
Part (c)
B1 a width cannot be negative	✓ 1
Part (d)
M1 2(13.09 + 19.09) or 2(2x + 6) with their positive x	✓ 1
A1 64.4 m	✓ 1

Exam tip: Keep unrounded values in your calculator for later parts, and round only the final answer to the accuracy asked.

+ Add to test✓ In your test198IGMC3-2-08r9-07Quadratic equations and inequalities Paper EMedium4 marks

No calculator

A data centre hall must contain $x$ server racks, where $x$ is a whole number satisfying $x^2-11x+24<0$ so that the cooling system works properly.

(a)

Solve the inequality $x^2-11x+24<0$.

(3)

(b)

Find the number of different whole-number values of $x$ that satisfy the inequality.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 critical values $x=3$ and $x=8$	✓ 1
M1 uses a sketch or sign test for an upward-opening parabola	✓ 1
A1 $3<x<8$	✓ 1
Part (b)
A1 4 values: 4, 5, 6 and 7	✓ 1

Common error: A less-than inequality for an upward-opening quadratic gives values between the roots, not outside them.

+ Add to test✓ In your test199IGMC3-2-08r9-10Functions and differentiation Paper EMedium7 marks

No calculator

The functions $f$ and $g$ are defined for all real $x$ by $f(x)=2x+3$ and $g(x)=x^2-1$.

(a)

Find $f(g(2))$.

(2)

(b)

Find $f^{-1}(x)$.

(2)

(c)

Solve $g(f(x))=24$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $g(2)=3$	✓ 1
A1 $f(3)=9$	✓ 1
Part (b)
M1 writes $y=2x+3$ and rearranges for $x$	✓ 1
A1 $f^{-1}(x)=\frac{x-3}{2}$	✓ 1
Part (c)
M1 $g(f(x))=(2x+3)^2-1$	✓ 1
M1 $(2x+3)^2=25$, so $2x+3=\pm5$	✓ 1
A1 $x=1$ or $x=-4$	✓ 1

Common error: $f(g(x))$ means apply $g$ first, then $f$; the order matters.

+ Add to test✓ In your test200IGMC3-2-08r10-06Linear and simultaneous equations and rearranging formulae Paper EMedium6 marks

No calculator

A cycle shop sells helmets at £h each and locks at £l each. Two helmets and three locks cost £97. One helmet and four locks cost £86.

(a)

Write down two equations in h and l.

(2)

(b)

Solve the equations to find h and l.

(3)

(c)

Find the cost of three helmets and two locks.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $2h + 3l = 97$	✓ 1
B1 $h + 4l = 86$	✓ 1
Part (b)
M1 eliminate one variable, e.g. double the second equation	✓ 1
A1 $l = 15$	✓ 1
A1 $h = 26$	✓ 1
Part (c)
A1 £108 (ECF)	✓ 1

Common error: After finding the first unknown, substitute into an original equation and check the other one as well.

+ Add to test✓ In your test201IGMC3-2-08r10-07Quadratic equations and inequalities Paper EMedium7 marks

Calculator

An archaeological excavation is a rectangle with length $(x + 4)$ m and width $(x - 1)$ m. Its area is 60 m².

Diagram not to scale

(a)

Show that $x^2 + 3x - 64 = 0$.

(2)

(b)

Solve $x^2 + 3x - 64 = 0$ using the quadratic formula. Give your answers correct to 2 decimal places.

(3)

(c)

Find the length of the excavation, correct to 2 decimal places. Use the unrounded value of x.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expand $(x + 4)(x - 1) = x^2 + 3x - 4$	✓ 1
A1 equate to 60 and rearrange to $x^2 + 3x - 64 = 0$	✓ 1
Part (b)
M1 correct substitution of $a = 1$, $b = 3$, $c = -64$ into the formula	✓ 1
A1 $x = 6.64$	✓ 1
A1 $x = -9.64$	✓ 1
Part (c)
B1 reject $x = -9.64$ because the width $x - 1$ must be positive	✓ 1
A1 10.64 m	✓ 1

Exam tip: A quadratic has two roots; use the context to decide which one is physically possible.

+ Add to test✓ In your test202IGMC3-2-08r10-08Quadratic equations and inequalities Paper EMedium5 marks

No calculator

A technician models the temperature margin M °C of a server room by $M = 24 + 2x - x^2$, where x is the time in hours after midnight and $0 \le x \le 12$.

(a)

Factorise $x^2 - 2x - 24$.

(2)

(b)

Hence find the values of x for which $M > 0$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 two numbers with product $-24$ and sum $-2$	✓ 1
A1 $(x - 6)(x + 4)$	✓ 1
Part (b)
M1 $M > 0$ becomes $x^2 - 2x - 24 < 0$	✓ 1
M1 critical values $x = 6$ and $x = -4$, with the region between them	✓ 1
A1 $0 \le x < 6$	✓ 1

Common error: Multiplying or dividing by a negative reverses an inequality; here rewrite so that the x² term is positive and flip the sign.

+ Add to test✓ In your test203IGMC3-2-10r10-07Quadratic equations and inequalities Paper EMedium8 marks

Calculator

A ball is thrown upwards by a sports scientist. Its height h metres above the ground after t seconds is h = −5t² + 18t + 1.2, for as long as the ball is in flight. Give final answers to 2 decimal places.

(a)

Show that the ball is at a height of 10 m when 5t² − 18t + 8.8 = 0.

(2)

(b)

Use the quadratic formula to find the two times at which the ball is at a height of 10 m.

(3)

(c)

Find the time at which the ball hits the ground.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 −5t² + 18t + 1.2 = 10	✓ 1
A1 rearranging to 5t² − 18t + 8.8 = 0	✓ 1
Part (b)
M1 t = (18 ± √(18² − 4 × 5 × 8.8)) ÷ (2 × 5)	✓ 1
A1 t = 0.58 s	✓ 1
A1 t = 3.02 s	✓ 1
Part (c)
M1 5t² − 18t − 1.2 = 0 with the formula, t = (18 ± √348) ÷ 10	✓ 1
A1 t = 3.67 s	✓ 1
B1 reject the negative root, −0.07, as the time is after the throw	✓ 1

Common error: Write the quadratic as ax² + bx + c = 0 with the right-hand side zero before reading off a, b and c, and keep the signs.

+ Add to test✓ In your test204IGMC3-2-10r10-08Expanding, factorising and algebraic fractions Paper EMedium8 marks

No calculator

Algebraic fractions.

(a)

Simplify fully (2x² − 18)/(x² + x − 12).

(3)

(b)

Write 3/(x + 2) − 2/(x − 1) as a single fraction in its simplest form.

(3)

(c)

Hence solve 3/(x + 2) − 2/(x − 1) = 0.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 numerator 2(x − 3)(x + 3)	✓ 1
M1 denominator (x + 4)(x − 3)	✓ 1
A1 2(x + 3)/(x + 4)	✓ 1
Part (b)
M1 common denominator (x + 2)(x − 1)	✓ 1
M1 3(x − 1) − 2(x + 2)	✓ 1
A1 (x − 7)/((x + 2)(x − 1))	✓ 1
Part (c)
M1 numerator x − 7 = 0	✓ 1
A1 x = 7	✓ 1

Common error: Only factors can be cancelled in a fraction, never terms; factorise numerator and denominator first.

+ Add to test✓ In your test205IGMC3-2-10r10-11Sequences and graphs of functions Paper EMedium7 marks

Calculator

Tiles in an observatory floor pattern follow the sequence 3, 10, 21, 36, …, and the pattern continues. The nth pattern uses T tiles.

(a)

Write down the next term.

(1)

(b)

Find the nth term.

(3)

(c)

Which pattern uses 300 tiles?

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 55	✓ 1
Part (b)
M1 differences 7, 11, 15 with a constant second difference of 4, so 2n²	✓ 1
M1 3 − 2 = 1 and 10 − 8 = 2, so the linear part is n	✓ 1
A1 2n² + n	✓ 1
Part (c)
M1 2n² + n = 300 so 2n² + n − 300 = 0	✓ 1
M1 (n − 12)(2n + 25) = 0	✓ 1
A1 n = 12 (reject −12.5)	✓ 1

Exam tip: For a quadratic sequence the nth term starts with half of the second difference multiplied by n².

+ Add to test✓ In your test206IGMC3-2-10r11-07Quadratic equations and inequalities Paper EMedium6 marks

Calculator

A lander descends towards a moon. Its height h metres above the ground, t seconds after the parachute opens, is modelled by h = 120 − 6t − 1.85t².

(a)

Calculate the height of the lander when t = 2.

(2)

(b)

Show that the lander reaches the ground when 1.85t² + 6t − 120 = 0.

(1)

(c)

Solve this equation using the quadratic formula to find the time of landing. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 120 − 6(2) − 1.85(2)²	✓ 1
A1 100.6 m	✓ 1
Part (b)
B1 set h = 0 and rearrange to 1.85t² + 6t − 120 = 0	✓ 1
Part (c)
M1 t = (−6 ± sqrt(6² − 4(1.85)(−120))) ÷ (2 × 1.85)	✓ 1
A1 (−6 + sqrt(924)) ÷ 3.7	✓ 1
A1 t = 6.59 s (negative root rejected)	✓ 1

Exam tip: Write the values of a, b and c, with their signs, before substituting into the quadratic formula.

+ Add to test✓ In your test207IGMC3-2-10r11-10Linear and simultaneous equations and rearranging formulae Paper EMedium5 marks

Calculator

A probe orbits Mars in a circular path of radius r metres, centre at the centre of Mars. Its speed v m/s satisfies v = sqrt(μ ÷ r), where μ = 4.28 × 10¹³ m³/s² for Mars.

(a)

Rearrange the formula to make r the subject.

(2)

(b)

The probe's speed is 3400 m/s. Calculate r. Give your answer to 3 significant figures.

(2)

(c)

Using the formula, state how v changes when r increases.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 square both sides: v² = μ ÷ r	✓ 1
A1 r = μ ÷ v²	✓ 1
Part (b)
M1 r = 4.28 × 10¹³ ÷ 3400²	✓ 1
A1 3.70 × 10⁶ m (3 700 000 m)	✓ 1
Part (c)
B1 v decreases, since r is in the denominator inside the square root	✓ 1

Common error: Squaring both sides is needed before you can move r out of the square root.

+ Add to test✓ In your test208IGMC3-2-10r11-11Quadratic equations and inequalities Paper EMedium6 marks

No calculator

The displacement s cm of the tip of a robotic sampling arm from its starting point, t seconds after it starts, is s = t³ − 9t² + 24t for 0 ≤ t ≤ 6.

(a)

Find ds/dt.

(2)

(b)

Find the two times when the arm is momentarily at rest.

(2)

(c)

Find the displacement of the arm at each of these times.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least two terms differentiated correctly	✓ 1
A1 3t² − 18t + 24	✓ 1
Part (b)
M1 3t² − 18t + 24 = 0, so t² − 6t + 8 = 0	✓ 1
A1 t = 2 and t = 4	✓ 1
Part (c)
B1 s(2) = 8 − 36 + 48 = 20 cm	✓ 1
B1 s(4) = 64 − 144 + 96 = 16 cm	✓ 1

Exam tip: Momentarily at rest means the rate of change of displacement is zero.

+ Add to test✓ In your test209IGMC3-2-08r16-04Quadratic equations and inequalities Paper EMedium7 marks

Calculator

A rectangular cable bay in a data centre has width $x$ metres and length $(3x-8)$ metres. Its floor area is $30\ \text{m}^2$.

Diagram not to scale

(a)

Show that $3x^2-8x-30=0$.

(2)

(b)

Solve $3x^2-8x-30=0$ using the quadratic formula. Give both answers correct to 2 decimal places.

(3)

(c)

State which solution is valid for the width and find the length of the bay. Use the unrounded value of $x$ and give the length correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $x(3x-8)=30$	✓ 1
B1 expands and rearranges to $3x^2-8x-30=0$	✓ 1
Part (b)
M1 substitutes $a=3$, $b=-8$, $c=-30$ into the quadratic formula	✓ 1
A1 $x=4.77$	✓ 1
A1 $x=-2.10$	✓ 1
Part (c)
B1 $x=4.77$ is valid as the width cannot be negative	✓ 1
A1 length $=3(4.7652\ldots)-8=6.30$ m	✓ 1

Exam tip: Keep the full calculator value of a root for later steps and round only the final answer.

+ Add to test✓ In your test210IGMC3-2-08r16-14Functions and differentiation Paper EMedium8 marks

Calculator

After a cooling fault, the temperature of a server is modelled by $T=20+12t-t^2$ degrees Celsius, where $t$ is the time in hours after the fault, for $0\le t\le 12$.

(a)

Find $\dfrac{dT}{dt}$ and the rate at which the temperature is changing when $t=2$.

(2)

(b)

Find the value of $t$ when $\dfrac{dT}{dt}=0$.

(2)

(c)

Show that the temperature is greatest at the time found in (b).

(2)

(d)

Find the greatest temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\dfrac{dT}{dt}=12-2t$	✓ 1
A1 8 °C per hour	✓ 1
Part (b)
M1 $12-2t=0$	✓ 1
A1 $t=6$	✓ 1
Part (c)
M1 gradient at $t=5$ is $2$ (positive) and at $t=7$ is $-2$ (negative)	✓ 1
R1 the gradient changes from positive to negative, so it is a maximum	✓ 1
Part (d)
M1 substitutes $t=6$ into $T$	✓ 1
A1 56 °C	✓ 1

Exam tip: Interpret $\dfrac{dT}{dt}$ as a rate with units: here degrees Celsius per hour.

+ Add to test✓ In your test211IGMC3-2-08r17-03Quadratic equations and inequalities Paper EMedium7 marks

No calculator

A rectangular telecoms panel has a length of (x + 4) metres and a width of x metres. Its area is 21 m².

Diagram not to scale

(a)

Show that x² + 4x − 21 = 0.

(2)

(b)

Solve x² + 4x − 21 = 0 by factorising and state the width of the panel.

(3)

(c)

Find the perimeter of the panel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x(x + 4) = 21	✓ 1
A1 x² + 4x = 21, so x² + 4x − 21 = 0 (shown)	✓ 1
Part (b)
M1 attempt to factorise	✓ 1
A1 (x + 7)(x − 3) = 0	✓ 1
A1 x = 3 (reject x = −7 because a width cannot be negative)	✓ 1
Part (c)
M1 2(3 + 7) (ECF from (b))	✓ 1
A1 20 m	✓ 1

Exam tip: Two roots do not always both fit the context: reject any root that gives a negative length.

+ Add to test✓ In your test212IGMC3-2-08r17-11Sequences and graphs of functions Paper EMedium6 marks

Calculator

A food factory stacks tins in a pattern. The numbers of tins in the first four layers are 5, 12, 23, 38. The number of tins in layer n follows a quadratic sequence.

(a)

Find the number of tins in layer 5.

(1)

(b)

Find an expression for the number of tins in layer n.

(3)

(c)

Find the number of tins in layer 20.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 57 (first differences 7, 11, 15, 19 so the next term is 38 + 19)	✓ 1
Part (b)
M1 second difference constant at 4, so the term is 2n²	✓ 1
M1 subtract 2n² from the terms: 3, 4, 5, 6, giving n + 2	✓ 1
A1 2n² + n + 2	✓ 1
Part (c)
M1 2 × 20² + 20 + 2 (ECF from (b))	✓ 1
A1 822	✓ 1

Exam tip: Check a quadratic nth term by substituting two different values of n, such as n = 1 and n = 4.

+ Add to test✓ In your test213IGMC3-2-10r22-09Expanding, factorising and algebraic fractions Paper EMedium8 marks

No calculator

Algebraic fractions.

(a)

Simplify $\frac{x^2-9}{x^2-x-6}$.

(3)

(b)

Write $\frac{2}{x-1}+\frac{2}{x+2}$ as a single fraction.

(2)

(c)

Hence find the positive value of $x$ for which $\frac{2}{x-1}+\frac{2}{x+2}=1$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 numerator $(x-3)(x+3)$	✓ 1
M1 denominator $(x-3)(x+2)$	✓ 1
A1 $\frac{x+3}{x+2}$	✓ 1
Part (b)
M1 common denominator $(x-1)(x+2)$	✓ 1
A1 $\frac{4x+2}{(x-1)(x+2)}$	✓ 1
Part (c)
M1 $4x+2=x^2+x-2$	✓ 1
M1 $x^2-3x-4=0$ and factorise or solve	✓ 1
A1 $x=4$ (the other solution is $x=-1$)	✓ 1

Exam tip: Factorise numerator and denominator fully before cancelling; only common factors can be cancelled.

+ Add to test✓ In your test214IGMC3-2-10r22-10Functions and differentiation Paper EMedium7 marks

No calculator

The functions $f$ and $g$ are defined by $f(x)=2x+3$ and $g(x)=x^2-1$.

(a)

Find $gf(x)$, giving your answer as an expanded expression.

(2)

(b)

Solve $gf(x)=0$.

(2)

(c)

Find $f^{-1}(x)$.

(2)

(d)

Find $f^{-1}(15)$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $(2x+3)^2-1$	✓ 1
A1 $4x^2+12x+8$	✓ 1
Part (b)
M1 $x^2+3x+2=0$ or $4(x+1)(x+2)=0$	✓ 1
A1 $x=-1$ and $x=-2$	✓ 1
Part (c)
M1 $y=2x+3$ rearranged to make $x$ the subject	✓ 1
A1 $f^{-1}(x)=\frac{x-3}{2}$	✓ 1
Part (d)
B1 6 (ECF from (c))	✓ 1

Common error: $gf(x)$ means apply $f$ first and then $g$; do not swap the order.

+ Add to test✓ In your test215IGMC3-2-10r22-11Functions and differentiation Paper EMedium7 marks

Calculator

The concentration of a drug in a patient's blood is $C$ mg per litre, $t$ hours after an injection, where $C=18t-3t^2$ for $0\le t\le6$.

(a)

Find $\frac{dC}{dt}$.

(2)

(b)

Find the time at which the concentration is greatest, and show that this is a maximum.

(3)

(c)

Calculate the greatest concentration.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 18	✓ 1
B1 $-6t$	✓ 1
Part (b)
M1 $18-6t=0$	✓ 1
A1 $t=3$ hours	✓ 1
B1 $\frac{d^2C}{dt^2}=-6<0$ or gradient changes from positive to negative	✓ 1
Part (c)
M1 $18\times3-3\times3^2$	✓ 1
A1 27 mg per litre	✓ 1

Exam tip: At a maximum or minimum the gradient is zero; check the type using the sign of the second derivative.

+ Add to test✓ In your test216IGMC3-2-08r24-10Linear and simultaneous equations and rearranging formulae Paper EMedium8 marks

Calculator

A mountaineering shop sells harnesses and helmets. Two harnesses and three helmets cost €270. Four harnesses and one helmet cost €240. Let a harness cost €$x$ and a helmet cost €$y$.

(a)

Write down two equations in $x$ and $y$.

(2)

(b)

Solve the equations to find the cost of one harness and the cost of one helmet.

(4)

(c)

Work out the cost of three harnesses and five helmets.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $2x + 3y = 270$	✓ 1
B1 $4x + y = 240$	✓ 1
Part (b)
M1 make the coefficients of one variable equal or substitute for one variable	✓ 1
M1 eliminate that variable	✓ 1
A1 $y = 60$, a helmet costs €60	✓ 1
A1 $x = 45$, a harness costs €45	✓ 1
Part (c)
M1 $3 \times 45 + 5 \times 60$ (ECF from (b))	✓ 1
A1 €435	✓ 1

Exam tip: Substitute your two answers back into both original equations to check them: $2 \times 45 + 3 \times 60 = 270$.

+ Add to test✓ In your test217IGMC3-2-08r24-11Expanding, factorising and algebraic fractions Paper EMedium5 marks

No calculator

A rectangular tent platform at base camp has width $x$ metres and length $(x + 3)$ metres. Its area is 70 m².

Diagram not to scale

(a)

Show that $x^2 + 3x - 70 = 0$.

(1)

(b)

Factorise $x^2 + 3x - 70$.

(2)

(c)

Solve the equation and write down the width of the platform.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x(x + 3) = 70$, so $x^2 + 3x = 70$ and $x^2 + 3x - 70 = 0$	✓ 1
Part (b)
M1 two brackets with numbers whose product is $-70$	✓ 1
A1 $(x + 10)(x - 7)$	✓ 1
Part (c)
A1 $x = 7$ or $x = -10$	✓ 1
R1 reject $-10$ because a width cannot be negative; width = 7 m	✓ 1

Common error: A quadratic gives two solutions: always check which one makes sense in the context.

+ Add to test✓ In your test218IGMC3-2-09r25-12Functions and differentiation Paper EMedium8 marks

Calculator

During the first 10 seconds of a sprint, the distance $s$ metres covered by an athlete is modelled by $s = 1.2t^2 - 0.04t^3$, where $t$ is the time in seconds and $0 \le t \le 10$.

(a)

Find an expression for the athlete's speed $v = \frac{ds}{dt}$.

(2)

(b)

Calculate the speed at $t = 4$.

(2)

(c)

Find the maximum speed during the 10 seconds and the time at which it occurs.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates (at least one term correct)	✓ 1
A1 $v = 2.4t - 0.12t^2$	✓ 1
Part (b)
M1 $2.4(4) - 0.12(4)^2$	✓ 1
A1 7.68 m/s	✓ 1
Part (c)
M1 differentiates $v$ to get $\frac{dv}{dt} = 2.4 - 0.24t$	✓ 1
A1 sets $\frac{dv}{dt} = 0$, giving $t = 10$ s (accept any valid method)	✓ 1
A1 $v = 2.4(10) - 0.12(10)^2$	✓ 1
A1 maximum speed 12 m/s	✓ 1

Exam tip: A maximum or minimum occurs where the derivative is zero, but check the answer lies inside the stated interval.

+ Add to test✓ In your test219IGMC3-2-09r25-13Sequences and graphs of functions Paper EMedium6 marks

Calculator

A rail platform has a row of lights. The first four terms of a sequence giving the number of lights in a design are 5, 12, 21, 32, ...

(a)

Find an expression for the $n$th term.

(3)

(b)

Find the position of the first term that is greater than 500.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 7, 9, 11 and second difference 2 identified	✓ 1
M1 uses $n^2$ as the leading term and examines the remainder ($4, 8, 12, 16$)	✓ 1
A1 $n^2 + 4n$	✓ 1
Part (b)
M1 tests or solves $n^2 + 4n > 500$ (for example finds the 20th term is 480)	✓ 1
A1 21st term is 525	✓ 1
A1 $n = 21$	✓ 1

Common error: For a quadratic sequence, the second difference equals $2a$, so the coefficient of $n^2$ is half of it, not the whole value.

+ Add to test✓ In your test220IGMC3-2-12r27-06Quadratic equations and inequalities Paper EMedium9 marks

Calculator

A dolphin leaves the water and its height, h metres, above the surface t seconds later is modelled by h = 7t − 4.9t², for the time it is in the air.

(a)

Find the two times at which the dolphin is 1.2 m above the surface. Give each answer to 3 significant figures.

(4)

(b)

Find the time at which the dolphin reaches its greatest height, and find this height.

(3)

(c)

Calculate how long the dolphin is more than 1.2 m above the surface. Use the unrounded times from (a). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.9t² − 7t + 1.2 = 0	✓ 1
M1 correct substitution into the quadratic formula, with sqrt(49 − 4 × 4.9 × 1.2) = sqrt(25.48)	✓ 1
A1 t = 0.199 s	✓ 1
A1 t = 1.23 s	✓ 1
Part (b)
M1 t = 7 ÷ (2 × 4.9) (or the midpoint of the two times where h = 0, or completing the square)	✓ 1
A1 t = 0.714 s	✓ 1
A1 h = 2.5 m	✓ 1
Part (c)
M1 1.229… − 0.199…	✓ 1
A1 1.03 s	✓ 1

Exam tip: The two solutions of a quadratic equation are symmetrical about the turning point, so the time of greatest height lies exactly half-way between them.

+ Add to test✓ In your test221IGMC3-2-12r27-09Sequences and graphs of functions Paper EMedium6 marks

Calculator

A marine biologist records the number of cells in an algae colony each day. The numbers, in thousands, for the first four days are 3, 6, 12 and 24. The pattern continues, with each term found by multiplying the previous term by the same number.

(a)

Write down the numbers for days 5 and 6.

(1)

(b)

Find the number of cells, in thousands, on day 10.

(2)

(c)

On which day does the number of cells first exceed 10 000 thousand? Give the number of cells, in thousands, on that day.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 48 and 96	✓ 1
Part (b)
M1 3 × 2⁹	✓ 1
A1 1536	✓ 1
Part (c)
M1 3 × 2ⁿ⁻¹ compared with 10 000 (or listing terms)	✓ 1
M1 day 12 is 3 × 2¹¹ = 6144 (not enough) and day 13 is 3 × 2¹²	✓ 1
A1 day 13 with 12288 thousand cells	✓ 1

Exam tip: For a sequence that doubles, the nth term is the first term multiplied by 2 raised to the power (n − 1), not n.

+ Add to test✓ In your test222IGMC3-2-11r28-09Expanding, factorising and algebraic fractions Paper EMedium9 marks

Calculator

(a)

Simplify $\dfrac{x^2-9}{x^2+5x+6}$.

(3)

(b)

Show that $\dfrac{2}{x+1}+\dfrac{3}{x-2}=2$ can be written as $2x^2-7x-3=0$.

(3)

(c)

Solve $2x^2-7x-3=0$. Give your answers correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factorise numerator: $(x-3)(x+3)$	✓ 1
M1 factorise denominator: $(x+2)(x+3)$	✓ 1
A1 $\dfrac{x-3}{x+2}$	✓ 1
Part (b)
M1 combine fractions: $\dfrac{2(x-2)+3(x+1)}{(x+1)(x-2)}$	✓ 1
M1 multiply both sides by $(x+1)(x-2)$: $5x-1=2(x^2-x-2)$	✓ 1
A1 rearranges to $2x^2-7x-3=0$	✓ 1
Part (c)
M1 use of formula: $x=\dfrac{7\pm\sqrt{49+24}}{4}$	✓ 1
A1 $x=3.89$	✓ 1
A1 $x=-0.386$	✓ 1

Exam tip: When a quadratic formula answer is asked to 3 significant figures, keep the unrounded square root in your calculator until the final step.

+ Add to test✓ In your test223IGMC3-2-11r28-10Functions and differentiation Paper EMedium9 marks

Calculator

A model for the number of new infections per day in a town gives $N$ hundred new cases $t$ weeks after an outbreak starts, where $N=2t^3-9t^2+12t+1$ for $0\le t\le3$.

(a)

Find $\dfrac{dN}{dt}$.

(2)

(b)

Find the values of $t$ for which $\dfrac{dN}{dt}=0$.

(3)

(c)

Determine the nature of each stationary point.

(3)

(d)

Find the value of $N$ at the local maximum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates at least two terms correctly	✓ 1
A1 $6t^2-18t+12$	✓ 1
Part (b)
M1 $6t^2-18t+12=0$, so $t^2-3t+2=0$	✓ 1
A1 $(t-1)(t-2)=0$	✓ 1
A1 $t=1$ and $t=2$	✓ 1
Part (c)
M1 $\dfrac{d^2N}{dt^2}=12t-18$	✓ 1
A1 at $t=1$ it is $-6<0$, a local maximum	✓ 1
A1 at $t=2$ it is $6>0$, a local minimum	✓ 1
Part (d)
A1 $N=2-9+12+1=6$ (hundred cases)	✓ 1

Common error: A positive second derivative means a minimum and a negative second derivative means a maximum; the two are easy to swap.

+ Add to test✓ In your test224IGMC3-2-11r28-11Sequences and graphs of functions Paper EMedium9 marks

Calculator

A model predicts the number of cases of a seasonal illness. On day 1 there are 64 cases, and on each later day the number of cases is 1.5 times the number on the previous day (64, 96, 144, ...).

(a)

Find an expression for the number of cases on day $n$.

(2)

(b)

Find the number of cases on day 7.

(2)

(c)

Find the first day on which the model gives more than 5000 cases.

(3)

(d)

Find the number of cases the model gives on day 12, correct to the nearest whole number.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 geometric sequence with first term 64 and common ratio 1.5	✓ 1
A1 $64\times1.5^{n-1}$	✓ 1
Part (b)
M1 $64\times1.5^{6}$ (ECF from (a))	✓ 1
A1 729	✓ 1
Part (c)
M1 $64\times1.5^{n-1}>5000$ (or $=5000$)	✓ 1
M1 valid method, e.g. $n-1>\dfrac{\log 78.125}{\log1.5}\approx10.7$, or compares day 11 ($3690.6\ldots$) with day 12 ($5535.8\ldots$)	✓ 1
A1 day 12	✓ 1
Part (d)
M1 $64\times1.5^{11}$	✓ 1
A1 5536	✓ 1

Common error: The $n$th term of a geometric sequence is $ar^{n-1}$, not $ar^{n}$; check with $n=1$ giving the first term.

+ Add to test✓ In your test225IGMC3-2-12r28-09Functions and differentiation Paper EMedium7 marks

Calculator

The height $h$ metres of a cycle path above its starting point, at a horizontal distance of $x$ hundred metres along it, is modelled by $h=x^3-6x^2+9x$ for $0\le x\le4$.

(a)

Find $\dfrac{dh}{dx}$.

(2)

(b)

Find the gradient of the path when $x=4$.

(2)

(c)

Show that there is only one value of $x$ for which the gradient is $-3$, and find it.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates at least one term correctly	✓ 1
A1 $3x^2-12x+9$	✓ 1
Part (b)
M1 $3(16)-12(4)+9$ (ECF from (a))	✓ 1
A1 9	✓ 1
Part (c)
M1 $3x^2-12x+9=-3$	✓ 1
M1 $3x^2-12x+12=0$, so $(x-2)^2=0$	✓ 1
A1 repeated root, so only $x=2$	✓ 1

Exam tip: A repeated root from a quadratic means the equation has exactly one solution; the discriminant is zero.

+ Add to test✓ In your test226IGMC3-2-12r28-10Functions and differentiation Paper EMedium7 marks

Calculator

The functions $f$ and $g$ are defined by $f(x)=3x+2$ and $g(x)=x^2-5$.

(a)

Find $gf(x)$ in the form $ax^2+bx+c$.

(2)

(b)

Find $f^{-1}(x)$.

(2)

(c)

Solve $gf(x)=0$. Give your answers correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $(3x+2)^2-5$	✓ 1
A1 $9x^2+12x-1$	✓ 1
Part (b)
M1 $y=3x+2$ rearranged to $x=\dfrac{y-2}{3}$	✓ 1
A1 $f^{-1}(x)=\dfrac{x-2}{3}$	✓ 1
Part (c)
M1 formula: $x=\dfrac{-12\pm\sqrt{144+36}}{18}$	✓ 1
A1 $x=0.0787$	✓ 1
A1 $x=-1.41$	✓ 1

Common error: In $gf(x)$ the function $f$ acts first; writing $f(g(x))$ instead gives a different expression.

+ Add to test✓ In your test227IGMC3-2-12r28-11Sequences and graphs of functions Paper EMedium9 marks

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Archaeologists count pottery fragments in successive soil layers. Layers 1, 2, 3 and 4 contain 5, 12, 23 and 38 fragments. The number in layer $n$ follows a quadratic sequence.

(a)

Find the number of fragments in layer 5.

(2)

(b)

Find an expression for the number of fragments in layer $n$.

(3)

(c)

Show that no layer contains exactly 100 fragments.

(2)

(d)

Find the number of fragments in layer 12.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences $7,11,15$ so next difference is 19	✓ 1
A1 57	✓ 1
Part (b)
M1 second difference 4, so the $n^2$ term is $2n^2$	✓ 1
M1 subtracting $2n^2$ leaves $3,4,5,6$, which is $n+2$	✓ 1
A1 $2n^2+n+2$	✓ 1
Part (c)
M1 $2n^2+n+2=100$, so $2n^2+n-98=0$ and discriminant $1+784=785$	✓ 1
A1 785 is not a perfect square ($n\approx6.75$), so $n$ is not a whole number	✓ 1
Part (d)
M1 $2(12)^2+12+2$ (ECF from (b))	✓ 1
A1 302	✓ 1

Exam tip: For a quadratic sequence, half the second difference gives the coefficient of $n^2$.

+ Add to test✓ In your test228IGMC3-2-10r29-10Quadratic equations and inequalities Paper EMedium7 marks

No calculator

The population P (in thousands) of bees in a hive after t weeks is modelled by P = 12t² − t³ for 0 ≤ t ≤ 12.

(a)

Find dP/dt.

(2)

(b)

Find the rate of change of the population when t = 2.

(1)

(c)

Find the value of t for which dP/dt = 0, where t > 0.

(2)

(d)

Find the population at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 24t	✓ 1
B1 −3t²	✓ 1
Part (b)
B1 24 × 2 − 3 × 4 = 36 thousand per week	✓ 1
Part (c)
M1 3t(8 − t) = 0	✓ 1
A1 t = 8	✓ 1
Part (d)
M1 12 × 8² − 8³ = 768 − 512	✓ 1
A1 256 thousand	✓ 1

Exam tip: At a stationary point set the derivative equal to zero, then substitute back into the original function (not the derivative) for the value.

+ Add to test✓ In your test229IGMC3-2-10r35-06Quadratic equations and inequalities Paper EMedium5 marks

Calculator

The height of a drone above a platform, in metres, is modelled by a function involving $3x^2-7x-5$.

(a)

Solve $3x^2-7x-5=0$. Give your answers correct to 2 decimal places.

(3)

(b)

Hence solve $3x^2-7x-5<0$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitutes correctly into the quadratic formula, $\dfrac{7\pm\sqrt{49+60}}{6}$	✓ 1
A1 $x=2.91$	✓ 1
A1 $x=-0.57$	✓ 1
Part (b)
M1 identifies the values between the two roots	✓ 1
A1 $-0.57<x<2.91$ (ECF from (a))	✓ 1

Common error: For '< 0' with a positive $x^2$ coefficient, the solution lies between the roots, not outside them.

+ Add to test✓ In your test230IGMC3-2-10r35-10Functions and differentiation Paper EMedium7 marks

Calculator

A model rocket is launched vertically. Its height h metres above the ground after t seconds is given by $h=40t-5t^2$, for $0\le t\le 8$.

(a)

Find $\dfrac{dh}{dt}$.

(2)

(b)

Find the speed of the rocket when $t=1.5$.

(2)

(c)

Find the maximum height reached by the rocket.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $40$	✓ 1
B1 $-10t$	✓ 1
Part (b)
M1 substitutes $t=1.5$ into the derivative	✓ 1
A1 25 m/s	✓ 1
Part (c)
M1 sets the derivative equal to 0	✓ 1
A1 $t=4$	✓ 1
A1 80 m	✓ 1

Exam tip: At a maximum point the gradient is zero, so set the derivative equal to 0.

+ Add to test✓ In your test231IGMC3-2-10r35-12Sequences and graphs of functions Paper EMedium7 marks

Calculator

A theatre designer arranges stage lights in patterns. The numbers of lights in the first four patterns are 5, 12, 23, 38.

(a)

Write down the number of lights in pattern 5.

(1)

(b)

Show that the number of lights in pattern n is $2n^2+n+2$.

(3)

(c)

One pattern uses 212 lights. Find the value of n for this pattern.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 57	✓ 1
Part (b)
M1 first differences 7, 11, 15 and second difference 4 found, so $2n^2$	✓ 1
M1 subtracts $2n^2$ from the terms (3, 4, 5, 6 ...) to find $n+2$	✓ 1
A1 $2n^2+n+2$ with a check, e.g. $n=3$: $18+3+2=23$	✓ 1
Part (c)
M1 $2n^2+n+2=212$ giving $2n^2+n-210=0$	✓ 1
M1 solves, e.g. $(2n+21)(n-10)=0$ or the formula	✓ 1
A1 $n=10$ (rejects $-10.5$)	✓ 1

Exam tip: For a quadratic sequence the second difference is $2a$, so halve it to get the coefficient of $n^2$.

+ Add to test✓ In your test232IGMC3-2-12r46-05Quadratic equations and inequalities Paper EMedium4 marks

Calculator

The position of a robotic arm along a rail is s = 3t² − 8t − 2 metres from a marker, where t ≥ 0 is the time in seconds after the arm starts moving. The arm passes the marker when s = 0.

(a)

Solve 3t² − 8t − 2 = 0. Show your working and give your answers correct to 3 significant figures.

(3)

(b)

State the time at which the arm passes the marker and give a reason for rejecting the other solution.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 correct substitution into the quadratic formula, $\frac{8\pm\sqrt{(-8)^2-4\times3\times(-2)}}{2\times3}$	✓ 1
A1 2.90	✓ 1
A1 −0.230	✓ 1
Part (b)
B1 2.90 s, since time cannot be negative ($t\ge0$)	✓ 1

Common error: Write the quadratic formula in full with brackets around negative values, otherwise the sign of b is often lost.

+ Add to test✓ In your test233IGMC3-2-12r46-06Expanding, factorising and algebraic fractions Paper EMedium5 marks

No calculator

A robot's sensor error is E = x² − 3x − 10, where x is an integer setting. The sensor is calibrated correctly when E < 0.

(a)

Factorise x² − 3x − 10.

(2)

(b)

Solve the inequality x² − 3x − 10 < 0.

(2)

(c)

How many positive integer settings give a correct calibration?

(1)

Show mark scheme

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 $-2$ and $5$ (ECF from (a))	✓ 1
A1 $-2<x<5$	✓ 1
Part (c)
B1 4 (the values 1, 2, 3, 4)	✓ 1

Common error: For a quadratic that is less than zero, the solution lies between the two roots, not outside them.

+ Add to test✓ In your test234IGMC3-2-12r46-09Sequences and graphs of functions Paper EMedium7 marks

The number of tiles a robot lays in pattern n follows the sequence 5, 12, 23, 38, 57, ...

(a)

Find the next term of the sequence.

(2)

(b)

Find an expression, in terms of n, for the nth term.

(3)

(c)

Find the number of tiles in pattern 15.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 first differences 7, 11, 15, 19 and second differences 4, so the next first difference is 23	✓ 1
A1 80	✓ 1
Part (b)
B1 $2n^2$ (half the second difference)	✓ 1
M1 subtracts $2n^2$ from the terms to leave 3, 4, 5, ...	✓ 1
A1 $2n^2+n+2$	✓ 1
Part (c)
M1 substitutes $n=15$, $2\times15^2+15+2$	✓ 1
A1 467	✓ 1

Exam tip: For a quadratic sequence, the coefficient of n squared is half of the constant second difference.

+ Add to test✓ In your test235IGMC3-2-11r48-05Quadratic equations and inequalities Paper EMedium8 marks

Calculator

A rectangular swimming pool has width w metres and length (w + 3) metres. Its area is 50 m².

Diagram not to scale

(a)

Show that w² + 3w − 50 = 0.

(2)

(b)

Solve w² + 3w − 50 = 0 using the quadratic formula. Give both answers correct to 2 decimal places.

(3)

(c)

Use the unrounded positive value of w to work out the perimeter of the pool. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 w(w + 3) = 50	✓ 1
A1 w² + 3w = 50, so w² + 3w − 50 = 0 shown	✓ 1
Part (b)
M1 correct substitution, w = (−3 ± √(3² − 4×1×(−50))) ÷ 2	✓ 1
A1 w = 5.73	✓ 1
A1 w = −8.73	✓ 1
Part (c)
B1 reject w = −8.73 as a width cannot be negative	✓ 1
M1 perimeter = 2(w + w + 3) = 4w + 6	✓ 1
A1 28.9 m	✓ 1

Common error: A negative solution of a quadratic from a length problem must be rejected explicitly, with a reason.

+ Add to test✓ In your test236IGMC3-2-11r48-08Functions and differentiation Paper EMedium8 marks

No calculator

The power output, P kilowatts, of a small wind turbine t hours after 06:00 is modelled by P = 2t³ − 9t² + 12t + 10, for 0 ≤ t ≤ 4.

(a)

Find dP/dt.

(2)

(b)

Find the values of t at which dP/dt = 0.

(3)

(c)

Show that the stationary point at t = 1 is a local maximum and find the power output at this point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at least one term differentiated correctly	✓ 1
A1 6t² − 18t + 12	✓ 1
Part (b)
M1 6t² − 18t + 12 = 0	✓ 1
M1 6(t − 1)(t − 2) = 0 or equivalent solving	✓ 1
A1 t = 1 and t = 2	✓ 1
Part (c)
M1 d²P/dt² = 12t − 18	✓ 1
A1 at t = 1 the value is −6 < 0, so a local maximum	✓ 1
A1 P = 2 − 9 + 12 + 10 = 15 kW	✓ 1

Common error: A negative second derivative means a maximum and a positive one a minimum; students often swap these.

+ Add to test✓ In your test237IGMC3-2-11r48-10Linear and simultaneous equations and rearranging formulae Paper EMedium6 marks

Calculator

The power P kilowatts produced by a model wind turbine with blades of diameter d metres, in a wind of speed v metres per second, is given by P = 0.3d²v³.

(a)

Work out P when d = 10 and v = 6.

(2)

(b)

Make v the subject of the formula.

(2)

(c)

A turbine with d = 12 produces P = 9331.2. Find the wind speed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.3 × 10² × 6³	✓ 1
A1 6480	✓ 1
Part (b)
M1 v³ = P ÷ (0.3d²)	✓ 1
A1 v = cube root of (P ÷ (0.3d²))	✓ 1
Part (c)
M1 v³ = 9331.2 ÷ (0.3 × 12²) = 216	✓ 1
A1 6 m/s	✓ 1

Common error: Take the cube root of the whole fraction; taking the cube root of only part of it gives a wrong answer.

+ Add to test✓ In your test238IGMC-042-00-12Expanding, factorising and algebraic fractions Paper EHard5 marks

No calculator

$\frac{2}{x-1} + \frac{3}{x+2} = 2$

(a)

Show that this equation simplifies to $2x^2 - 3x - 5 = 0$.

(3)

(b)

Solve $2x^2 - 3x - 5 = 0$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for 2(x + 2) + 3(x − 1) = 2(x − 1)(x + 2)	✓ 1
M1 for 5x + 1 = 2x² + 2x − 4	✓ 1
A1 correct rearrangement to 2x² − 3x − 5 = 0 with no errors	✓ 1
Part (b)
M1 for (2x − 5)(x + 1) = 0	✓ 1
A1 x = 2.5 and x = −1	✓ 1

Common error: When you multiply through by (x − 1)(x + 2), the 2 on the right-hand side must be multiplied as well. Leaving it as 2 is a common slip.

+ Add to test✓ In your test239IGMC-042-00-13Expanding, factorising and algebraic fractions Paper EHard8 marks

No calculator

Answer all parts.

(a)

Simplify fully $\frac{2x^2 + 5x - 3}{4x^2 - 1} \div \frac{x+3}{2x^2 + x}$.

(4)

(b)

Show that $\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}$.

(2)

(c)

Hence find the exact value of $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{9 \times 10}$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 2x² + 5x − 3 = (2x − 1)(x + 3)	✓ 1
B1 4x² − 1 = (2x − 1)(2x + 1) and 2x² + x = x(2x + 1)	✓ 1
M1 for inverting the second fraction and multiplying	✓ 1
A1 x	✓ 1
Part (b)
M1 for $\frac{(n+1) - n}{n(n+1)}$	✓ 1
A1 correct simplification to $\frac{1}{n(n+1)}$	✓ 1
Part (c)
M1 for writing the sum as $(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{9} - \frac{1}{10})$	✓ 1
A1 $\frac{9}{10}$	✓ 1

Exam tip: In part (c), write out the first two brackets and the last one. The middle terms cancel in pairs, leaving only the first term and the last term.

+ Add to test✓ In your test240IGMC-042-00-14Expanding, factorising and algebraic fractions Paper EHard6 marks

No calculator

Answer both parts.

(a)

$(3x + p)(x - 2) + q \equiv 3x^2 - x + 5$
Find the value of p and the value of q.

(3)

(b)

Two consecutive odd numbers can be written as 2n − 1 and 2n + 1, where n is an integer.
Prove that the difference between their squares is always a multiple of 8.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for expanding to 3x² + (p − 6)x − 2p + q, or for a correct equation from comparing coefficients, or for substituting x = 2 to give q = 12 − 2 + 5	✓ 1
A1 p = 5	✓ 1
A1 q = 15	✓ 1
Part (b)
M1 for (2n + 1)² − (2n − 1)²	✓ 1
M1 for correct expansion, 4n² + 4n + 1 − (4n² − 4n + 1), or for use of the difference of two squares giving (4n)(2)	✓ 1
A1 for 8n, with the conclusion that 8n is 8 × an integer and so is a multiple of 8	✓ 1

Exam tip: A proof needs a final statement. After reaching 8n, say that it is 8 × an integer and so is a multiple of 8.

+ Add to test✓ In your test241IGMC-042-02-11Quadratic equations and inequalities Paper EHard6 marks

Calculator

This question is about the quadratic expression 2x² + 3x − 7.

(a)

Solve 2x² + 3x − 7 = 0, giving your answers correct to 2 decimal places.

(3)

(b)

Hence solve the inequality 2x² + 3x − 7 ≥ 0.

(2)

(c)

Find how many integers do not satisfy 2x² + 3x − 7 ≥ 0.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Correct substitution into the quadratic formula, $\frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)}$, or an equivalent completing-the-square method	✓ 1
√65 (or 8.062…) seen	✓ 1
x = 1.27 and x = −2.77	✓ 1
Part (b)
Recognises that the solution lies outside the roots, because the coefficient of x² is positive. This may be implied by a correct final answer	✓ 1
x ≤ −2.77 or x ≥ 1.27. Accept exact forms, unrounded values, or values following through from the roots found in (a)	✓ 1
Part (c)
4 (the integers −2, −1, 0 and 1)	✓ 1

Exam tip: Sketch y = 2x² + 3x − 7 as a U-shape crossing the x-axis at the two roots. The curve is on or above the axis (≥ 0) in the two outer regions, so the integers that fail the inequality lie strictly between the roots.

+ Add to test✓ In your test242IGMC-042-02-13Quadratic equations and inequalities Paper EHard6 marks

No calculator

A car travels 240 km at an average speed of v km/h.
If the average speed had been 20 km/h faster, the journey would have taken 1 hour less.

(a)

Show that v² + 20v − 4800 = 0.

(3)

(b)

Solve the equation and find the time the journey takes at the original speed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 times written as $\frac{240}{v}$ and $\frac{240}{v+20}$	✓ 1
M1 $\frac{240}{v} - \frac{240}{v+20} = 1$	✓ 1
A1 240(v + 20) − 240v = v(v + 20) leading to v² + 20v − 4800 = 0 with no errors	✓ 1
Part (b)
M1 (v + 80)(v − 60) = 0	✓ 1
A1 v = 60, with v = −80 rejected	✓ 1
A1 4 hours	✓ 1

Common error: The slower speed gives the longer time, so the equation is $\frac{240}{v} - \frac{240}{v+20} = 1$. Reversing the subtraction gives the wrong sign.

+ Add to test✓ In your test243IGMC-042-02-14Quadratic equations and inequalities Paper EHard8 marks

Calculator

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

Diagram not to scale

(a)

Show that x² − 5x − 24 = 0.

(3)

(b)

Solve the equation and hence find the area of the triangle.

(3)

(c)

Calculate the smallest angle of the triangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x² + (x + 7)² = (2x + 1)²	✓ 1
M1 2x² + 14x + 49 = 4x² + 4x + 1	✓ 1
A1 2x² − 10x − 48 = 0 divided by 2 to give x² − 5x − 24 = 0 with no errors	✓ 1
Part (b)
M1 (x − 8)(x + 3) = 0	✓ 1
A1 x = 8, with x = −3 rejected	✓ 1
A1 area = ½ × 8 × 15 = 60 cm²	✓ 1
Part (c)
M1 tan θ = $\frac{8}{15}$ or sin θ = $\frac{8}{17}$	✓ 1
A1 28.1°	✓ 1

Exam tip: The smallest angle is always opposite the shortest side. Here that side is 8 cm, so use it as the opposite side in your trigonometry.

+ Add to test✓ In your test244IGMC-042-01-06Linear and simultaneous equations and rearranging formulae Paper EHard5 marks

No calculator

$y = \frac{2x+3}{x-1}$

(a)

Make x the subject of the formula.

(4)

(b)

Find the value of x when y = 5. Give your answer as a fraction.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for multiplying out: y(x − 1) = 2x + 3, i.e. xy − y = 2x + 3	✓ 1
M1 for collecting the x terms on one side, e.g. xy − 2x = y + 3 or 2x − xy = −y − 3	✓ 1
M1 for factorising out x, e.g. x(y − 2) = y + 3 or x(2 − y) = −y − 3	✓ 1
A1 for $x = \frac{y+3}{y-2}$, or an equivalent such as $x = \frac{-y-3}{2-y}$	✓ 1
Part (b)
B1 for $\frac{8}{3}$ (accept $2\frac{2}{3}$), from substituting into the answer to (a) or solving 5 = (2x + 3)/(x − 1) directly	✓ 1

Exam tip: When the new subject appears twice, collect every term containing x on one side, then take x out as a common factor.

+ Add to test✓ In your test245IGMC-042-01-13Linear and simultaneous equations and rearranging formulae Paper EHard6 marks

No calculator

Solve the simultaneous equations.
$y = 2x - 1$
$x^2 + y^2 = 13$

(a)

Find all the solutions. You must show all your working.

(6)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for substituting to get an equation in one variable: x² + (2x − 1)² = 13, or equivalent such as $\left(\frac{y+1}{2}\right)^2 + y^2 = 13$	✓ 1
M1 for correctly expanding the bracket, e.g. (2x − 1)² = 4x² − 4x + 1	✓ 1
A1 for 5x² − 4x − 12 = 0, or equivalent 5y² + 2y − 51 = 0	✓ 1
M1 for a correct method to solve their quadratic, e.g. (5x + 6)(x − 2) = 0 or (5y + 17)(y − 3) = 0 or the quadratic formula	✓ 1
A1 for x = 2 and x = $-\frac{6}{5}$	✓ 1
A1 for the correctly paired values: x = 2, y = 3 and x = $-\frac{6}{5}$, y = $-\frac{17}{5}$	✓ 1

Common error: (2x − 1)² is not 4x² + 1. Write it out as (2x − 1)(2x − 1) so that the middle term −4x is not lost.

+ Add to test✓ In your test246IGMC-042-03-12Sequences and graphs of functions Paper EHard8 marks

No calculator

Sequence A has nth term 5n − 2.
Sequence B has nth term n² + 2.

(a)

For some values of n, the nth term of sequence A is equal to the nth term of sequence B.
Find these values of n and the value of each of these terms.

(4)

(b)

Sequence C has first five terms
2, 10, 30, 68, 130, ...
Find an expression, in terms of n, for the nth term of sequence C.

(3)

(c)

Find the 10th term of sequence C.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for n² + 2 = 5n − 2	✓ 1
M1 for n² − 5n + 4 = 0 and (n − 1)(n − 4) = 0	✓ 1
A1 n = 1 and n = 4	✓ 1
A1 terms 3 and 18	✓ 1
Part (b)
M1 for third differences of 6 identified	✓ 1
M1 for n³ seen in an expression	✓ 1
A1 n³ + n	✓ 1
Part (c)
B1 1010	✓ 1

Exam tip: A constant third difference of 6 means the leading term is n³. Subtract the cubes 1, 8, 27, 64, 125 to reveal what remains.

+ Add to test✓ In your test247IGMC-042-03-13Sequences and graphs of functions Paper EHard8 marks

No calculator

A curve has equation y = x³ − 3x² − 9x + 5.

(a)

Find the range of values of x for which y is decreasing.

(4)

(b)

The tangent to the curve at the point where x = 0 meets the curve again at the point Q.
Find the coordinates of Q.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $\frac{dy}{dx}$ = 3x² − 6x − 9	✓ 1
M1 for 3x² − 6x − 9 < 0 or = 0	✓ 1
M1 for 3(x − 3)(x + 1) or x = −1 and 3 found	✓ 1
A1 −1 < x < 3	✓ 1
Part (b)
B1 tangent is y = 5 − 9x	✓ 1
M1 for x³ − 3x² − 9x + 5 = 5 − 9x	✓ 1
M1 for x²(x − 3) = 0 leading to x = 3	✓ 1
A1 (3, −22)	✓ 1

Common error: A curve is decreasing where the gradient is negative, so the answer is the interval between the roots of $\frac{dy}{dx}$ = 0, not the roots themselves.

+ Add to test✓ In your test248IGMC-042-03-14Functions and differentiation Paper EHard10 marks

Calculator

The curve y = ax² + bx + c passes through the points (0, −3), (1, 2) and (−2, 5).

(a)

Find the values of a, b and c.

(4)

(b)

Find the exact coordinates of the turning point of the curve.

(3)

(c)

Solve ax² + bx + c = 0, giving your answers correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 c = −3	✓ 1
M1 for a + b − 3 = 2 and 4a − 2b − 3 = 5	✓ 1
M1 for correct method to eliminate one variable	✓ 1
A1 a = 3 and b = 2	✓ 1
Part (b)
M1 for $\frac{dy}{dx}$ = 6x + 2 = 0 or completing the square	✓ 1
A1 x = $-\frac{1}{3}$	✓ 1
A1 y = $-\frac{10}{3}$	✓ 1
Part (c)
M1 for $\frac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-3)}}{2 \times 3}$	✓ 1
A1 0.721	✓ 1
A1 −1.39	✓ 1

Exam tip: Use the point with x = 0 first: it gives c immediately and reduces the problem to two simultaneous equations.

+ Add to test✓ In your test249IGMC-042-04-12Functions and differentiation Paper EHard9 marks

Calculator

A square sheet of card has side 30 cm. A square of side x cm is cut from each corner. The sides are then folded up to make an open box with a square base.

Net of the card: a square of side x cm is removed at each corner. Diagram not to scale

(a)

Show that the volume, V cm³, of the box is given by V = 4x³ − 120x² + 900x.

(2)

(b)

Find $\frac{dV}{dx}$.

(2)

(c)

Find the value of x that gives the maximum volume, and find this maximum volume.

(4)

(d)

Explain why the other solution of $\frac{dV}{dx}$ = 0 is not suitable.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: base side 30 − 2x and V = x(30 − 2x)²	✓ 1
A1: correct expansion shown, x(900 − 120x + 4x²) = 4x³ − 120x² + 900x	✓ 1
Part (b)
M1: at least two terms correct	✓ 1
A1: 12x² − 240x + 900	✓ 1
Part (c)
M1: 12x² − 240x + 900 = 0	✓ 1
M1: x² − 20x + 75 = 0 solved, e.g. (x − 5)(x − 15) = 0	✓ 1
A1: x = 5	✓ 1
A1: V = 2000 cm³	✓ 1
Part (d)
B1: x = 15 gives a base side of 30 − 2(15) = 0, so there is no box (the volume is zero)	✓ 1

Exam tip: In context problems, check each solution of dV/dx = 0 against the physical limits. Here x must satisfy 0 < x < 15.

+ Add to test✓ In your test250IGMC-042-04-13Functions and differentiation Paper EHard7 marks

Calculator

f(x) = 2x − 3
g(x) = $\frac{1}{x+1}$, x ≠ −1
h(x) = x²

(a)

Find hg(1).

(1)

(b)

Find g⁻¹(x).

(2)

(c)

Solve fg(x) = x. Give your answers correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 1/4	✓ 1
Part (b)
M1: y = 1/(x + 1) rearranged to x + 1 = 1/y, or equivalent	✓ 1
A1: g⁻¹(x) = (1 − x)/x, or 1/x − 1	✓ 1
Part (c)
M1: fg(x) = 2/(x + 1) − 3	✓ 1
M1: multiply through by (x + 1) and rearrange to x² + 4x + 1 = 0	✓ 1
M1: correct use of the quadratic formula or completing the square, e.g. x = −2 ± √3	✓ 1
A1: x = −0.268 and x = −3.73	✓ 1

Common error: When you multiply 2/(x + 1) − 3 = x by (x + 1), every term must be multiplied, including the −3 and the x on the right-hand side.

+ Add to test✓ In your test251IGMC-042-04-14Functions and differentiation Paper EHard8 marks

No calculator

A curve has equation y = x³ − 4x + 1.
At two points on the curve, the tangent is parallel to the line y = 8x + 3.

(a)

Find $\frac{dy}{dx}$.

(1)

(b)

Find the coordinates of the two points where the tangent is parallel to y = 8x + 3.

(4)

(c)

The two tangents cross the y-axis at the points P and Q. Find the distance PQ.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 3x² − 4	✓ 1
Part (b)
M1: 3x² − 4 = 8	✓ 1
A1: x = 2 and x = −2	✓ 1
M1: substitute at least one x value into the equation of the curve	✓ 1
A1: (2, 1) and (−2, 1)	✓ 1
Part (c)
M1: correct method for the equation of either tangent, e.g. y − 1 = 8(x − 2) or y − 1 = 8(x + 2), or equivalent	✓ 1
A1: y-intercepts −15 and 17 (both)	✓ 1
A1: PQ = 32	✓ 1

Exam tip: Parallel lines have equal gradients. Set dy/dx equal to 8, and remember that x² = 4 has two solutions.

+ Add to test✓ In your test252IGMC3-2-08c-13Functions and differentiation Paper EHard11 marks

Calculator

The height of a drone above the ground is h metres at time t seconds, where h = 2t³ − 15t² + 24t + 30 for 0 ⩽ t ⩽ 5.

(a)

Find $\frac{dh}{dt}$.

(2)

(b)

Find the values of t at which the drone is momentarily stationary.

(3)

(c)

Find $\frac{d^2h}{dt^2}$ and use it to show that the stationary point at t = 1 is a maximum.

(2)

(d)

Find the height of the drone at this maximum.

(1)

(e)

Find the equation of the tangent to the graph of h against t at t = 2.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (one correct term)	✓ 1
A1 6t² − 30t + 24	✓ 1
Part (b)
M1 6t² − 30t + 24 = 0	✓ 1
M1 t² − 5t + 4 = 0 and factorise (t − 1)(t − 4)	✓ 1
A1 t = 1 and t = 4	✓ 1
Part (c)
M1 12t − 30	✓ 1
A1 −18 < 0 at t = 1, so a maximum	✓ 1
Part (d)
B1 41 m	✓ 1
Part (e)
M1 h(2) = 34	✓ 1
M1 gradient 6(4) − 60 + 24 = −12	✓ 1
A1 h = −12t + 58	✓ 1

Common error: A stationary point has zero gradient; do not substitute into h when you need the gradient.

+ Add to test✓ In your test253IGMC3-2-08c-14Functions and differentiation Paper EHard8 marks

No calculator

In a graphics program, f(x) = 2x − 3 and g(x) = x² + 1.

(a)

Find fg(x) in its simplest form.

(2)

(b)

Find f⁻¹(x).

(2)

(c)

Solve gf(x) = 5.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2(x² + 1) − 3	✓ 1
A1 2x² − 1	✓ 1
Part (b)
M1 x = 2y − 3 (swap and rearrange)	✓ 1
A1 (x + 3) ÷ 2	✓ 1
Part (c)
M1 (2x − 3)² + 1 = 5 giving 4x² − 12x + 5 = 0	✓ 1
M1 (2x − 1)(2x − 5) = 0	✓ 1
A1 x = 0.5	✓ 1
A1 x = 2.5	✓ 1

Common error: gf(x) means apply f first, then g; reversing the order gives a different function.

+ Add to test✓ In your test254IGMC3-2-08r5-06Functions and differentiation Paper EHard7 marks

No calculator

A cyclist coasting downhill has final speed v after covering a distance s, with initial speed u and constant acceleration a.

(a)

Make s the subject of the formula v = √(u² + 2as).

(3)

(b)

Make x the subject of the formula y = (3x + 1) ÷ (x − 2).

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 square both sides, v² = u² + 2as	✓ 1
M1 subtract u² and isolate the term in s	✓ 1
A1 s = (v² − u²) ÷ (2a)	✓ 1
Part (b)
M1 multiply by (x − 2), y(x − 2) = 3x + 1	✓ 1
M1 expand and collect the x terms on one side, yx − 3x = 1 + 2y	✓ 1
M1 factorise, x(y − 3) = 1 + 2y	✓ 1
A1 x = (1 + 2y) ÷ (y − 3)	✓ 1

Common error: When the new subject appears twice, collect those terms on one side and factorise before dividing.

+ Add to test✓ In your test255IGMC3-2-08r5-12Sequences and graphs of functions Paper EHard9 marks

No calculator

The height y hundred metres of a cycle path above sea level, at a horizontal distance x hundred metres from the start, is modelled by y = x³ − 4x + 1, for 0 ≤ x ≤ 5.

(a)

Find dy/dx.

(2)

(b)

Find the gradient of the path where x = 2.

(1)

(c)

Find the value of y where x = 2.

(1)

(d)

Find the equation of the tangent to the curve at x = 2. Give your answer in the form y = mx + c.

(2)

(e)

Find the positive value of x where the gradient of the path is 23.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, at least one term correct	✓ 1
A1 3x² − 4	✓ 1
Part (b)
B1 3(4) − 4 = 8	✓ 1
Part (c)
B1 8 − 8 + 1 = 1	✓ 1
Part (d)
M1 y − 1 = 8(x − 2) (ECF from (b) and (c))	✓ 1
A1 y = 8x − 15	✓ 1
Part (e)
M1 3x² − 4 = 23	✓ 1
M1 x² = 9	✓ 1
A1 x = 3 (reject −3 as outside the range)	✓ 1

Common error: A tangent needs both the gradient and the point; use the gradient at the given x-value and the y-value on the curve there.

+ Add to test✓ In your test256IGMC3-2-08r5-13Functions and differentiation Paper EHard11 marks

No calculator

A survey drone moves along a straight rail at an archaeological site. Its displacement from the start is s metres at time t seconds, where s = 2t³ − 15t² + 36t, for 0 ≤ t ≤ 5.

(a)

Find an expression for the velocity v of the drone.

(3)

(b)

Find the times when the drone is instantaneously at rest.

(3)

(c)

Find the acceleration of the drone when t = 4.

(2)

(d)

The velocity is negative for 2 < t < 3, so the drone moves backwards in this interval. Find the distance it moves backwards.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 at least two terms correct	✓ 1
A1 v = 6t² − 30t + 36	✓ 1
Part (b)
M1 6t² − 30t + 36 = 0	✓ 1
M1 factorise or use the formula, 6(t − 2)(t − 3) = 0	✓ 1
A1 t = 2 and t = 3	✓ 1
Part (c)
M1 differentiate the velocity, a = 12t − 30	✓ 1
A1 18 m/s²	✓ 1
Part (d)
M1 s(2) = 16 − 60 + 72 = 28	✓ 1
M1 s(3) = 54 − 135 + 108 = 27	✓ 1
A1 1 m	✓ 1

Exam tip: Velocity is the derivative of displacement and acceleration is the derivative of velocity; zero velocity marks where the direction can change.

+ Add to test✓ In your test257IGMC3-2-08r9-04Expanding, factorising and algebraic fractions Paper EHard8 marks

Calculator

Answer both parts.

(a)

Simplify fully $\frac{2x^2-8}{x^2+x-6}$.

(3)

(b)

Solve $\frac{3}{x+1}+\frac{2}{x-2}=1$. Give your answers correct to 3 significant figures.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factorises the numerator: $2(x-2)(x+2)$	✓ 1
M1 factorises the denominator: $(x+3)(x-2)$	✓ 1
A1 $\frac{2(x+2)}{x+3}$	✓ 1
Part (b)
M1 multiplies through by $(x+1)(x-2)$	✓ 1
M1 $3(x-2)+2(x+1)=(x+1)(x-2)$	✓ 1
A1 $x^2-6x+2=0$	✓ 1
M1 correct use of the quadratic formula	✓ 1
A1 $x=5.65$ or $x=0.354$	✓ 1

Exam tip: Fully factorise before cancelling; you can only cancel factors, never single terms.

+ Add to test✓ In your test258IGMC3-2-08r9-11Functions and differentiation Paper EHard7 marks

Calculator

A cyclist accelerates from rest. Her speed $v$ m/s at time $t$ seconds is modelled by $v=3t-0.25t^2$ for $0\le t\le12$.

(a)

Find an expression for the acceleration, $\frac{dv}{dt}$.

(2)

(b)

Find the time at which her speed is greatest, and the greatest speed.

(3)

(c)

Find the acceleration when $t=10$ and explain what its sign shows about her motion.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates, with at least one term correct	✓ 1
A1 $3-0.5t$	✓ 1
Part (b)
M1 sets $3-0.5t=0$	✓ 1
A1 $t=6$ s	✓ 1
A1 greatest speed 9 m/s	✓ 1
Part (c)
M1 substitutes $t=10$ into $3-0.5t$, giving $-2$ m/s²	✓ 1
R1 the negative value shows she is slowing down (decelerating)	✓ 1

Exam tip: A maximum of a smooth function occurs where the gradient is zero; always go on to find the value at that point.

+ Add to test✓ In your test259IGMC3-2-08r9-12Functions and differentiation Paper EHard7 marks

No calculator

A survey models the height $y$ metres of a buried ridge above a datum at horizontal distance $x$ hundred metres by $y=x^3-6x^2+9x+2$ for $0\le x\le5$.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the coordinates of the two stationary points.

(3)

(c)

Use the second derivative to determine the nature of each stationary point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates at least one term correctly	✓ 1
A1 $3x^2-12x+9$	✓ 1
Part (b)
M1 sets $3x^2-12x+9=0$ and factorises as $3(x-1)(x-3)$	✓ 1
A1 $x=1$ and $x=3$	✓ 1
A1 $(1,6)$ and $(3,2)$	✓ 1
Part (c)
M1 $\frac{d^2y}{dx^2}=6x-12$	✓ 1
A1 at $x=1$ it is $-6<0$ so a maximum; at $x=3$ it is $6>0$ so a minimum	✓ 1

Common error: Substitute the $x$-value back into the original function, not the derivative, to find the $y$-coordinate.

+ Add to test✓ In your test260IGMC3-2-08r10-05Expanding, factorising and algebraic fractions Paper EHard5 marks

No calculator

The cooling efficiency ratio of a data-centre loop is modelled by the fraction $\frac{6x^2 - 7x - 5}{4x^2 - 1}$, where $x > 1$.

(a)

Factorise $6x^2 - 7x - 5$.

(2)

(b)

Hence simplify $\frac{6x^2 - 7x - 5}{4x^2 - 1}$ as far as possible.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 correct form of factors, e.g. $(3x \pm 5)(2x \pm 1)$	✓ 1
A1 $(3x - 5)(2x + 1)$	✓ 1
Part (b)
M1 factorise denominator as $(2x - 1)(2x + 1)$	✓ 1
M1 cancel the common factor $(2x + 1)$	✓ 1
A1 $\frac{3x - 5}{2x - 1}$	✓ 1

Exam tip: Factorise numerator and denominator fully before cancelling; never cancel individual terms.

+ Add to test✓ In your test261IGMC3-2-08r10-09Functions and differentiation Paper EHard10 marks

No calculator

The height h metres of a cycle track above its starting point, at a horizontal distance x hundred metres along the track, is modelled by $h = x^3 - 6x^2 + 9x + 2$, for $0 \le x \le 4$.

(a)

Find $\frac{dh}{dx}$.

(2)

(b)

Find the values of x at which $\frac{dh}{dx} = 0$.

(3)

(c)

Find the coordinates of the turning points and determine their nature.

(4)

(d)

Find the difference in height between the crest and the dip of the track.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate a power of x	✓ 1
A1 $3x^2 - 12x + 9$	✓ 1
Part (b)
M1 set $3x^2 - 12x + 9 = 0$	✓ 1
M1 factorise as $3(x - 1)(x - 3)$	✓ 1
A1 $x = 1$ and $x = 3$	✓ 1
Part (c)
B1 $(1, 6)$	✓ 1
B1 $(3, 2)$	✓ 1
M1 $\frac{d^2h}{dx^2} = 6x - 12$ evaluated at both points	✓ 1
A1 maximum at $x = 1$ (value $-6$) and minimum at $x = 3$ (value $6$)	✓ 1
Part (d)
A1 4 m (ECF)	✓ 1

Exam tip: Substitute x back into the original equation, not the derivative, to get the height of a turning point.

+ Add to test✓ In your test262IGMC3-2-08r10-12Functions and differentiation Paper EHard8 marks

No calculator

A planner models a cable route by the curve $y = x^2 - 2x - 3$ and a service road by the straight line $y = x + 7$ in a data-centre site plan.

Diagram not to scale

(a)

Show that the x-coordinates of the points where the route meets the road satisfy $x^2 - 3x - 10 = 0$.

(2)

(b)

Solve $x^2 - 3x - 10 = 0$.

(2)

(c)

Find the coordinates of the two points where the route meets the road.

(2)

(d)

Hence solve $x^2 - 2x - 3 > x + 7$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equate $x^2 - 2x - 3 = x + 7$	✓ 1
A1 rearrange to $x^2 - 3x - 10 = 0$	✓ 1
Part (b)
M1 $(x - 5)(x + 2) = 0$	✓ 1
A1 $x = 5$ and $x = -2$	✓ 1
Part (c)
B1 $(5, 12)$	✓ 1
B1 $(-2, 5)$	✓ 1
Part (d)
M1 U-shaped quadratic $x^2 - 3x - 10 > 0$ is positive outside the roots	✓ 1
A1 $x < -2$ or $x > 5$	✓ 1

Common error: For a 'greater than' inequality with a U-shaped quadratic, the solution is two separate regions, not one.

+ Add to test✓ In your test263IGMC3-2-10r10-12Functions and differentiation Paper EHard11 marks

Calculator

A model for the height h metres of a rail line above a reference level at a distance x km along the line is h = x³ − 6x² + 9x + 2, for 0 ≤ x ≤ 4.

(a)

Find dh/dx.

(2)

(b)

Find the values of x at which the line has a stationary point.

(3)

(c)

Find the height at each stationary point.

(3)

(d)

Use the second derivative to decide the nature of each stationary point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate a term correctly	✓ 1
A1 3x² − 12x + 9	✓ 1
Part (b)
M1 3x² − 12x + 9 = 0	✓ 1
M1 3(x − 1)(x − 3) = 0	✓ 1
A1 x = 1 and x = 3	✓ 1
Part (c)
M1 substitute x = 1 or x = 3	✓ 1
A1 h = 6 m at x = 1	✓ 1
A1 h = 2 m at x = 3	✓ 1
Part (d)
M1 d²h/dx² = 6x − 12	✓ 1
A1 at x = 1 the value is −6, so a maximum	✓ 1
A1 at x = 3 the value is 6, so a minimum	✓ 1

Common error: A stationary point has dh/dx = 0, but you must substitute back into the original function, not the derivative, to find the height.

+ Add to test✓ In your test264IGMC3-2-10r10-13Functions and differentiation Paper EHard8 marks

Calculator

The functions f and g are defined by f(x) = 4x − 3 and g(x) = x² + 2.

(a)

Find f⁻¹(x).

(2)

(b)

Show that gf(x) = 16x² − 24x + 11.

(3)

(c)

Solve gf(x) = 18.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 let y = 4x − 3 and make x the subject	✓ 1
A1 f⁻¹(x) = (x + 3)/4	✓ 1
Part (b)
M1 gf(x) = (4x − 3)² + 2	✓ 1
M1 expand (4x − 3)² = 16x² − 24x + 9	✓ 1
A1 16x² − 24x + 11	✓ 1
Part (c)
M1 16x² − 24x − 7 = 0	✓ 1
M1 (4x + 1)(4x − 7) = 0	✓ 1
A1 x = −1/4 (−0.25) or x = 7/4 (1.75)	✓ 1

Common error: In gf(x) the function f is applied first; do not apply g to x first.

+ Add to test✓ In your test265IGMC3-2-10r10-14Sequences and graphs of functions Paper EHard9 marks

Calculator

The net gain G units in a sprint-training session of n minutes is modelled by G = −n² + 14n − 24.

(a)

Factorise n² − 14n + 24.

(2)

(b)

Solve G > 0.

(3)

(c)

Write G in the form 25 − (n − 7)² and state the maximum gain and the time at which it occurs.

(3)

(d)

How many whole numbers of minutes give G > 0?

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 two numbers with product 24 and sum −14	✓ 1
A1 (n − 2)(n − 12)	✓ 1
Part (b)
M1 critical values n = 2 and n = 12	✓ 1
M1 sketch or sign test, since G > 0 means n² − 14n + 24 < 0	✓ 1
A1 2 < n < 12	✓ 1
Part (c)
B1 −(n − 7)² = −n² + 14n − 49	✓ 1
B1 G = 25 − (n − 7)²	✓ 1
B1 maximum gain 25 at n = 7	✓ 1
Part (d)
B1 n = 3, 4, …, 11, which is 9 values	✓ 1

Exam tip: Dividing by a negative reverses an inequality, so rewrite G > 0 as n² − 14n + 24 < 0 before factorising.

+ Add to test✓ In your test266IGMC3-2-10r11-08Quadratic equations and inequalities Paper EHard7 marks

No calculator

The central arch of a footbridge has height h metres above the deck at a horizontal distance x metres from one end, where h = 0.04x(40 − x) for 0 ≤ x ≤ 40.

(a)

Find the greatest height of the arch above the deck.

(2)

(b)

Show that the arch is more than 12 m above the deck when x² − 40x + 300 < 0.

(2)

(c)

Solve the inequality to find the values of x for which the arch is more than 12 m above the deck.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 greatest height at x = 20 by symmetry (midpoint of the roots 0 and 40)	✓ 1
A1 0.04 × 20 × 20 = 16 m	✓ 1
Part (b)
M1 0.04x(40 − x) > 12 so 40x − x² > 300	✓ 1
A1 x² − 40x + 300 < 0	✓ 1
Part (c)
M1 (x − 10)(x − 30)	✓ 1
A1 critical values 10 and 30	✓ 1
A1 10 < x < 30	✓ 1

Common error: For a positive x² term, the inequality with < 0 gives the values between the roots.

+ Add to test✓ In your test267IGMC3-2-10r11-09Expanding, factorising and algebraic fractions Paper EHard9 marks

Calculator

Simplify and solve algebraic fractions in these parts.

(a)

Simplify fully (x² + 5x + 6) ÷ (x² − 4).

(3)

(b)

Solve 3 ÷ (x − 1) + 2 ÷ (x + 2) = 1. Give your answers correct to 2 decimal places.

(6)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 numerator = (x + 2)(x + 3)	✓ 1
M1 denominator = (x − 2)(x + 2)	✓ 1
A1 (x + 3) ÷ (x − 2)	✓ 1
Part (b)
M1 combine over the common denominator (x − 1)(x + 2)	✓ 1
A1 (5x + 4) ÷ ((x − 1)(x + 2)) = 1	✓ 1
M1 5x + 4 = (x − 1)(x + 2)	✓ 1
A1 x² − 4x − 6 = 0	✓ 1
M1 quadratic formula (or completing the square)	✓ 1
A1 x = 5.16 or x = −1.16	✓ 1

Common error: Cancel only common factors, never common terms.

+ Add to test✓ In your test268IGMC3-2-10r11-12Linear and simultaneous equations and rearranging formulae Paper EHard9 marks

Calculator

The cost £C per kilogram of roasting a batch of x kg of beans is modelled by C = 0.5x + 200x⁻¹ for 5 ≤ x ≤ 50.

(a)

Calculate C when x = 10.

(1)

(b)

Find dC/dx.

(2)

(c)

Find the batch size for which the cost per kilogram is stationary.

(3)

(d)

Show that this stationary point is a minimum.

(2)

(e)

Find the minimum cost per kilogram.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 £25	✓ 1
Part (b)
M1 one term differentiated correctly	✓ 1
A1 0.5 − 200x⁻²	✓ 1
Part (c)
M1 0.5 − 200x⁻² = 0	✓ 1
M1 x² = 400	✓ 1
A1 x = 20 kg (−20 rejected)	✓ 1
Part (d)
M1 d²C/dx² = 400x⁻³	✓ 1
A1 at x = 20 this is 0.05 > 0, so a minimum	✓ 1
Part (e)
B1 0.5 × 20 + 200 ÷ 20 = £20	✓ 1

Exam tip: Use the second derivative to justify a minimum rather than just assuming it.

+ Add to test✓ In your test269IGMC3-2-08r16-07Expanding, factorising and algebraic fractions Paper EHard10 marks

No calculator

Give all answers in their simplest exact form.

(a)

Simplify $\dfrac{x^2-9}{x^2+5x+6}$.

(3)

(b)

Write $\dfrac{3}{x-1}+\dfrac{2}{x+2}$ as a single fraction, simplifying the numerator.

(3)

(c)

Solve $\dfrac{3}{x-1}+\dfrac{2}{x+2}=1$. Give the larger solution in the form $p+\sqrt{q}$, where $p$ and $q$ are integers.

(4)

Show mark scheme

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 $\dfrac{x-3}{x+2}$	✓ 1
Part (b)
M1 uses the common denominator $(x-1)(x+2)$	✓ 1
M1 numerator $3(x+2)+2(x-1)$	✓ 1
A1 $\dfrac{5x+4}{(x-1)(x+2)}$	✓ 1
Part (c)
M1 sets $5x+4=(x-1)(x+2)$	✓ 1
A1 $x^2-4x-6=0$	✓ 1
M1 uses the quadratic formula or completes the square, $(x-2)^2=10$	✓ 1
A1 $x=2+\sqrt{10}$	✓ 1

Exam tip: Check that each solution does not make a denominator zero; here $x=1$ and $x=-2$ are the values to avoid.

+ Add to test✓ In your test270IGMC3-2-08r16-08Functions and differentiation Paper EHard9 marks

Calculator

The height of a cycling route above a starting point is modelled by $y=2x^3-9x^2+12x+1$ for $0\le x\le 3$, where $x$ is the horizontal distance in kilometres and $y$ is the height in tens of metres.

(a)

Find $\dfrac{dy}{dx}$.

(2)

(b)

Find the coordinates of the two stationary points of the curve.

(3)

(c)

Use $\dfrac{d^2y}{dx^2}$ to determine the nature of each stationary point.

(2)

(d)

Find the gradient of the route at $x=0.5$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to differentiate, with at least one term correct	✓ 1
A1 $6x^2-18x+12$	✓ 1
Part (b)
M1 sets $\dfrac{dy}{dx}=0$, giving $6(x-1)(x-2)=0$	✓ 1
A1 $x=1$ and $x=2$	✓ 1
A1 $(1,6)$ and $(2,5)$	✓ 1
Part (c)
M1 $\dfrac{d^2y}{dx^2}=12x-18$ and evaluates it at $x=1$ and $x=2$	✓ 1
A1 $-6$ at $x=1$ so a maximum; $+6$ at $x=2$ so a minimum	✓ 1
Part (d)
M1 substitutes $x=0.5$ into $6x^2-18x+12$	✓ 1
A1 4.5	✓ 1

Common error: A stationary point needs both coordinates: after finding $x$, substitute back into $y$, not into $\dfrac{dy}{dx}$.

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Calculator

A data centre's net cooling saving is $S=6x-x^2-4$, where $S$ is in thousands of euros and $x$ is the number of cooling units in use. Treat $x$ as a real number in parts (a) to (c).

(a)

Write $S$ in the form $q-(x-p)^2$.

(2)

(b)

Write down the maximum value of $S$ and the value of $x$ at which it occurs.

(2)

(c)

Solve $S>0$, giving the end points of the interval in exact form.

(3)

(d)

The data centre can only use a whole number of units. How many different whole numbers of units give a positive saving?

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 completes the square, $(x-3)^2$ appears	✓ 1
A1 $S=5-(x-3)^2$	✓ 1
Part (b)
B1 maximum saving is 5 (thousand euros)	✓ 1
B1 at $x=3$	✓ 1
Part (c)
M1 $(x-3)^2<5$	✓ 1
M1 $-\sqrt5<x-3<\sqrt5$	✓ 1
A1 $3-\sqrt5<x<3+\sqrt5$	✓ 1
Part (d)
M1 evaluates the end points as $0.76\ldots$ and $5.24\ldots$	✓ 1
A1 5 (that is $x=1,2,3,4,5$)	✓ 1

Common error: For an inequality like $(x-3)^2<5$, the solution is an interval between two values, not just $x<\sqrt5+3$.

+ Add to test✓ In your test272IGMC3-2-08r17-12Expanding, factorising and algebraic fractions Paper EHard8 marks

No calculator

(a)

Simplify fully $\frac{x^2-16}{x^2-x-12}$.

(3)

(b)

Solve $\frac{1}{x-3}+\frac{2}{x-1}=1$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factorise the numerator: (x − 4)(x + 4)	✓ 1
M1 factorise the denominator: (x − 4)(x + 3)	✓ 1
A1 (x + 4)/(x + 3)	✓ 1
Part (b)
M1 multiply through by (x − 3)(x − 1) or use a common denominator	✓ 1
A1 (x − 1) + 2(x − 3) = (x − 3)(x − 1), i.e. 3x − 7 = x² − 4x + 3	✓ 1
A1 x² − 7x + 10 = 0	✓ 1
M1 factorise or solve the quadratic, e.g. (x − 2)(x − 5) = 0	✓ 1
A1 x = 2 and x = 5	✓ 1

Common error: Cancel only factors, never terms: in (x + 4)/(x + 3) nothing more can be cancelled.

+ Add to test✓ In your test273IGMC3-2-08r17-13Functions and differentiation Paper EHard10 marks

No calculator

On a mountain trail, the height h (in hundreds of metres) at a distance x km from the start is given by h = x³ − 6x² + 9x + 2, for 0 ≤ x ≤ 4.

(a)

Find dh/dx.

(2)

(b)

Find the values of x at the stationary points of the trail.

(3)

(c)

Find the coordinates of each stationary point and determine whether it is a maximum or a minimum.

(4)

(d)

Find the difference in height between the two stationary points.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, at least one term correct	✓ 1
A1 3x² − 12x + 9	✓ 1
Part (b)
M1 set dh/dx = 0, 3x² − 12x + 9 = 0 (ECF from (a))	✓ 1
M1 factorise or solve, 3(x − 1)(x − 3) = 0	✓ 1
A1 x = 1 and x = 3	✓ 1
Part (c)
B1 (1, 6)	✓ 1
B1 (3, 2)	✓ 1
M1 second derivative 6x − 12, evaluated at x = 1 and x = 3 (or a sign-change test)	✓ 1
A1 (1, 6) is a maximum and (3, 2) is a minimum	✓ 1
Part (d)
A1 4 hundred metres (400 m), ECF from (c)	✓ 1

Exam tip: To decide the nature of a stationary point, substitute into d²h/dx²: negative gives a maximum, positive gives a minimum.

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Calculator

On a map, with distances in kilometres, a telecoms cable follows the curve y = x² − 4x + 5. A straight ridge line has equation y = x + 1. The cable crosses the ridge line at two points.

Diagram not to scale

(a)

Show that the x-coordinates of the two points satisfy x² − 5x + 4 = 0.

(2)

(b)

Solve the equation to find the coordinates of the two points.

(3)

(c)

Find the distance between the two points. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x² − 4x + 5 = x + 1	✓ 1
A1 x² − 5x + 4 = 0 (shown)	✓ 1
Part (b)
M1 factorise or use the quadratic formula, (x − 1)(x − 4) = 0	✓ 1
A1 x = 1 and x = 4	✓ 1
A1 (1, 2) and (4, 5)	✓ 1
Part (c)
M1 √((4 − 1)² + (5 − 2)²) (ECF from (b))	✓ 1
A1 4.24 km	✓ 1

Common error: After finding the x-values, substitute into the linear equation to find y; using the curve also works but is slower and more error-prone.

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Calculator

A beekeeper builds a rectangular paddock for hives with a perimeter of 46 m and an area of 100 m². One side has length $x$ m.

Diagram not to scale

(a)

Show that $x^2-23x+100=0$.

(2)

(b)

Solve $x^2-23x+100=0$ using the quadratic formula. Give your answers correct to 2 decimal places.

(3)

(c)

Calculate the length of a diagonal of the paddock. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 other side $=23-x$	✓ 1
M1 $x(23-x)=100$ rearranged to $x^2-23x+100=0$	✓ 1
Part (b)
M1 $\frac{23\pm\sqrt{(-23)^2-4\times100}}{2}$	✓ 1
A1 17.18	✓ 1
A1 5.82	✓ 1
Part (c)
M1 $\sqrt{17.18^2+5.82^2}$ (or $\sqrt{529-200}$)	✓ 1
M1 correct evaluation 18.13... or 18.14...	✓ 1
A1 18.1 m	✓ 1

Common error: Write the formula with correct signs: the coefficient $-23$ gives $-(-23)=+23$.

+ Add to test✓ In your test276IGMC3-2-08r24-12Functions and differentiation Paper EHard10 marks

Calculator

The height $h$, in hundreds of metres above sea level, of a ridge route at a horizontal distance of $x$ km from its start is modelled by $h = 2x^3 - 9x^2 + 12x + 1$, for $0 \leqslant x \leqslant 3$.

(a)

Find $\dfrac{dh}{dx}$.

(2)

(b)

Find the values of $x$ at which the route has stationary points.

(3)

(c)

Determine the nature of each stationary point.

(3)

(d)

Calculate the difference in height, in metres, between the local maximum and the local minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least two terms correct)	✓ 1
A1 $6x^2 - 18x + 12$	✓ 1
Part (b)
M1 set $\dfrac{dh}{dx} = 0$	✓ 1
M1 $x^2 - 3x + 2 = 0$ and factorise or solve	✓ 1
A1 $x = 1$ and $x = 2$	✓ 1
Part (c)
M1 $\dfrac{d^2h}{dx^2} = 12x - 18$	✓ 1
A1 at $x = 1$ the second derivative is $-6 < 0$, so a local maximum	✓ 1
A1 at $x = 2$ the second derivative is $+6 > 0$, so a local minimum	✓ 1
Part (d)
M1 height 6 at $x = 1$ and height 5 at $x = 2$	✓ 1
A1 difference 1 hundred metres = 100 m	✓ 1

Exam tip: Give a reason for the nature of a stationary point: the sign of the second derivative, or the sign of the gradient either side.

+ Add to test✓ In your test277IGMC3-2-08r24-13Functions and differentiation Paper EHard9 marks

No calculator

The functions $f$ and $g$ are defined by $f(x) = 2x + 3$ and $g(x) = x^2$.

(a)

Find $f^{-1}(x)$.

(2)

(b)

Find $gf(x)$ and write it in the form $ax^2 + bx + c$.

(2)

(c)

Find $fg(x)$.

(1)

(d)

Solve $gf(x) = fg(x)$. Give your answers in exact form.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $y = 2x + 3$, swap and rearrange, or $x = 2y + 3$	✓ 1
A1 $f^{-1}(x) = \dfrac{x - 3}{2}$	✓ 1
Part (b)
M1 $(2x + 3)^2$	✓ 1
A1 $4x^2 + 12x + 9$	✓ 1
Part (c)
$2x^2 + 3$	✓ 1
Part (d)
M1 $4x^2 + 12x + 9 = 2x^2 + 3$ (ECF from (b) and (c))	✓ 1
M1 $x^2 + 6x + 3 = 0$	✓ 1
M1 use the quadratic formula or complete the square: $(x + 3)^2 = 6$	✓ 1
A1 $x = -3 \pm \sqrt{6}$	✓ 1

Common error: $gf(x)$ means apply $f$ first and then $g$, so $gf(x) = g(f(x))$, not $f(g(x))$.

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Calculator

A food scientist records the number of bacterial colonies, in hundreds, on the first four days of an experiment: 5, 12, 23, 38. The numbers follow a quadratic sequence.

(a)

Show that the second differences are constant.

(2)

(b)

Find an expression for the $n$th term.

(3)

(c)

The model continues to hold. Find the first term of the sequence that is greater than 500.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 first differences 7, 11, 15	✓ 1
B1 second differences 4, 4 (constant)	✓ 1
Part (b)
M1 coefficient of $n^2$ is $4 \div 2 = 2$	✓ 1
M1 subtracting $2n^2$ leaves 3, 4, 5, 6, i.e. $n + 2$	✓ 1
A1 $2n^2 + n + 2$	✓ 1
Part (c)
M1 $n = 15$ gives 467	✓ 1
M1 $n = 16$ gives 530	✓ 1
A1 530	✓ 1

Exam tip: For a quadratic sequence, the $n^2$ coefficient is half the constant second difference.

+ Add to test✓ In your test279IGMC3-2-09r25-11Sequences and graphs of functions Paper EHard8 marks

Calculator

A rail operator models its daily profit $P$ (in thousands of euros) as $P = 15 + 7x - 2x^2$, where $x$ is the number of thousands of passengers carried in a day and $x \ge 0$.

(a)

Solve $P = 0$.

(3)

(b)

Find the values of $x$, with $x \ge 0$, for which the operator makes a profit.

(2)

(c)

By completing the square, find the maximum daily profit in euros.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rearranges to $2x^2 - 7x - 15 = 0$	✓ 1
M1 factorises $(2x+3)(x-5)$	✓ 1
A1 $x = 5$ and $x = -1.5$	✓ 1
Part (b)
B1 sketch or reasoning: the graph of $P$ is a downward-opening curve, so $P > 0$ between the roots	✓ 1
A1 $0 \le x < 5$	✓ 1
Part (c)
M1 $-2(x - 1.75)^2$ obtained from $-2(x^2 - 3.5x)$	✓ 1
A1 $P = -2(x-1.75)^2 + 21.125$	✓ 1
A1 maximum profit €21 125	✓ 1

Exam tip: For a quadratic with a negative $x^2$ coefficient, the completed-square form gives the maximum value directly: the constant term.

+ Add to test✓ In your test280IGMC3-2-09r25-14Functions and differentiation Paper EHard8 marks

Calculator

On a map, one unit represents 1 km. A railway track follows the curve $y = x^2 - 4x + 5$. A straight service road follows the line $y = 2x - 3$. The track and the road meet at two points, $A$ and $B$.

Diagram not to scale

(a)

Show that the $x$-coordinates of $A$ and $B$ satisfy $x^2 - 6x + 8 = 0$.

(2)

(b)

Solve $x^2 - 6x + 8 = 0$.

(2)

(c)

Find the coordinates of $A$ and $B$.

(2)

(d)

Calculate the length of the service road between $A$ and $B$, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equates $x^2 - 4x + 5 = 2x - 3$	✓ 1
A1 rearranges to $x^2 - 6x + 8 = 0$	✓ 1
Part (b)
M1 factorises $(x-2)(x-4)$ or uses the formula	✓ 1
A1 $x = 2$ and $x = 4$	✓ 1
Part (c)
M1 substitutes into $y = 2x - 3$	✓ 1
A1 $(2, 1)$ and $(4, 5)$	✓ 1
Part (d)
M1 $\sqrt{(4-2)^2 + (5-1)^2}$ (ECF from (c))	✓ 1
A1 4.47 km	✓ 1

Exam tip: To find where a line meets a curve, set the two expressions for $y$ equal, then substitute back into the simpler (linear) equation.

+ Add to test✓ In your test281IGMC3-2-12r27-07Quadratic equations and inequalities Paper EHard9 marks

Calculator

A lorry makes a delivery of 120 km along a motorway at an average speed of x km/h. It returns along the same route at an average speed of (x + 20) km/h. The return journey takes 30 minutes less than the outward journey.

(a)

Show that x² + 20x − 4800 = 0.

(4)

(b)

Solve x² + 20x − 4800 = 0 and state the average speed on the outward journey.

(3)

(c)

Find the total time for the two journeys, in hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 outward time 120 ÷ x and return time 120 ÷ (x + 20)	✓ 1
M1 120 ÷ x − 120 ÷ (x + 20) = 0.5	✓ 1
M1 multiplying through by 2x(x + 20) to give 240(x + 20) − 240x = x(x + 20)	✓ 1
A1 4800 = x² + 20x, so x² + 20x − 4800 = 0	✓ 1
Part (b)
M1 (x + 80)(x − 60) or correct use of the quadratic formula	✓ 1
A1 x = 60 and x = −80	✓ 1
B1 outward speed 60 km/h, rejecting −80 as a speed cannot be negative	✓ 1
Part (c)
M1 120 ÷ 60 + 120 ÷ 80	✓ 1
A1 3.5 hours	✓ 1

Exam tip: In a “show that” question, every line of the algebra must appear: multiplying by the common denominator and expanding should be shown before the final equation.

+ Add to test✓ In your test282IGMC3-2-12r27-12Expanding, factorising and algebraic fractions Paper EHard11 marks

Calculator

The curved top of an arch is modelled by h = 4x − ¼x², for 0 ≤ x ≤ 16, where x metres is the horizontal distance from the left foot of the arch and h metres is the height above the ground.

(a)

Show that h can be written in the form h = 16 − ¼(x − 8)².

(3)

(b)

Write down the greatest height of the arch and the value of x at which it occurs.

(2)

(c)

A horizontal beam at height 12 m is fixed between two points on the arch. Find the length of the beam.

(3)

(d)

A delivery vehicle has a rectangular cross-section 9 m high and is centred on the line x = 8. Find the greatest width of the vehicle that can pass under the arch. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factorising −¼ from the two terms: −¼(x² − 16x)	✓ 1
M1 completing the square: −¼[(x − 8)² − 64]	✓ 1
A1 16 − ¼(x − 8)²	✓ 1
Part (b)
B1 16 m	✓ 1
B1 x = 8 m	✓ 1
Part (c)
M1 16 − ¼(x − 8)² = 12, so (x − 8)² = 16	✓ 1
M1 x = 4 and x = 12	✓ 1
A1 8 m	✓ 1
Part (d)
M1 16 − ¼(x − 8)² = 9, so (x − 8)² = 28	✓ 1
M1 x − 8 = ±sqrt(28) = ±5.29…	✓ 1
A1 width = 2 × 5.29… = 10.6 m	✓ 1

Exam tip: The completed-square form shows the turning point directly: the number outside the bracket is the greatest height and the number inside gives its position.

+ Add to test✓ In your test283IGMC3-2-11r28-12Functions and differentiation Paper EHard7 marks

No calculator

The functions $f$ and $g$ are defined by $f(x)=\dfrac{2x+3}{x-1}$ for $x\ne1$, and $g(x)=x^2-4$.

(a)

Find $f^{-1}(x)$.

(3)

(b)

Solve $gf(x)=0$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $y=\dfrac{2x+3}{x-1}$ and clear the fraction: $y(x-1)=2x+3$	✓ 1
M1 collect $x$ terms: $x(y-2)=y+3$	✓ 1
A1 $f^{-1}(x)=\dfrac{x+3}{x-2}$	✓ 1
Part (b)
M1 $gf(x)=\left(\dfrac{2x+3}{x-1}\right)^2-4$	✓ 1
M1 $f(x)=2$ or $f(x)=-2$	✓ 1
A1 $f(x)=2$ gives $2x+3=2x-2$, which has no solution	✓ 1
A1 $f(x)=-2$ gives $x=-\dfrac14$	✓ 1

Exam tip: In $gf(x)$ the function $f$ is applied first; write $g(f(x))$ in full before simplifying.

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A solar panel plot is a rectangle with sides $x$ m and $y$ m. Its perimeter is 34 m and its diagonal is 13 m.

(a)

Show that $x^2-17x+60=0$.

(3)

(b)

Solve the equation in (a).

(2)

(c)

Find the area of the plot.

(1)

(d)

Fencing for the whole perimeter costs £14.50 per metre. Find the total cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $x+y=17$	✓ 1
B1 $x^2+y^2=169$ (Pythagoras)	✓ 1
M1 substitute $y=17-x$ and simplify to $x^2-17x+60=0$	✓ 1
Part (b)
M1 $(x-5)(x-12)=0$	✓ 1
A1 $x=5$ and $x=12$	✓ 1
Part (c)
B1 $5\times12=60$ m²	✓ 1
Part (d)
M1 $34\times14.50$	✓ 1
A1 £493	✓ 1

Exam tip: A perimeter equation and a Pythagoras equation give a linear and a quadratic equation; substitute the linear one into the quadratic one.

+ Add to test✓ In your test285IGMC3-2-12r28-12Functions and differentiation Paper EHard10 marks

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A data centre builds a storage frame from metal rods with total length 120 m. The frame is a cuboid with a square base of side $x$ m and height $h$ m, using 8 rods of length $x$ m and 4 rods of length $h$ m.

Square-based cuboid frame. Diagram not to scale

(a)

Show that $h=30-2x$.

(2)

(b)

Show that the volume is $V=30x^2-2x^3$.

(1)

(c)

Find $\dfrac{dV}{dx}$ and the value of $x$ for which $V$ is stationary.

(3)

(d)

Show that this value of $x$ gives a maximum volume.

(2)

(e)

Find the maximum volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $8x+4h=120$	✓ 1
A1 rearranges to $h=30-2x$	✓ 1
Part (b)
B1 $V=x^2h=x^2(30-2x)$	✓ 1
Part (c)
M1 $\dfrac{dV}{dx}=60x-6x^2$	✓ 1
M1 set equal to 0: $6x(10-x)=0$	✓ 1
A1 $x=10$ (rejecting $x=0$)	✓ 1
Part (d)
M1 $\dfrac{d^2V}{dx^2}=60-12x$	✓ 1
A1 at $x=10$ it equals $-60<0$, so a maximum	✓ 1
Part (e)
M1 $10^2\times(30-20)$	✓ 1
A1 1000 m³	✓ 1

Exam tip: State the sign of the second derivative and what it means; a stationary value alone does not show it is a maximum.

+ Add to test✓ In your test286IGMC3-2-12r28-13Quadratic equations and inequalities Paper EHard9 marks

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An excavation area is a rectangle of length $(3x+1)$ m and width $(x+2)$ m, where $x>0$. The area must be at least 50 m² and the perimeter must be at most 62 m.

Diagram not to scale

(a)

Show that the area condition leads to $3x^2+7x-48\ge0$.

(2)

(b)

Solve $3x^2+7x-48\ge0$ for $x>0$.

(3)

(c)

Use the perimeter condition to find an upper limit for $x$.

(2)

(d)

Find the greatest possible area.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $(3x+1)(x+2)=3x^2+7x+2$	✓ 1
A1 $3x^2+7x+2\ge50$, so $3x^2+7x-48\ge0$	✓ 1
Part (b)
M1 $(3x+16)(x-3)$	✓ 1
A1 critical values $x=3$ and $x=-\dfrac{16}{3}$	✓ 1
A1 $x\ge3$	✓ 1
Part (c)
M1 $2(3x+1+x+2)=8x+6\le62$	✓ 1
A1 $x\le7$	✓ 1
Part (d)
M1 $(3\times7+1)(7+2)$ (ECF from (c))	✓ 1
A1 198 m²	✓ 1

Common error: For a quadratic inequality of the form $\ge0$ with a positive $x^2$ coefficient, the solution lies outside the roots; then apply the condition $x>0$.

+ Add to test✓ In your test287IGMC3-2-10r29-07Quadratic equations and inequalities Paper EHard7 marks

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A hospital garden bed is a rectangle with sides of length x metres and (2x − 7) metres. Its area is 5 m².

Diagram not to scale

(a)

Show that 2x² − 7x − 5 = 0.

(2)

(b)

Solve 2x² − 7x − 5 = 0 using the quadratic formula. Give both answers correct to 2 decimal places.

(3)

(c)

Reject one solution and find the length of the side (2x − 7), correct to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 x(2x − 7) = 5	✓ 1
A1 2x² − 7x − 5 = 0	✓ 1
Part (b)
M1 x = (7 ± √(49 + 40))/4 = (7 ± √89)/4	✓ 1
A1 x = 4.11	✓ 1
A1 x = −0.61	✓ 1
Part (c)
B1 reject −0.61 as a length cannot be negative	✓ 1
A1 2 × 4.1085… − 7 = 1.22 m (using 4.11 also gives 1.22 m)	✓ 1

Exam tip: Show the substitution into the quadratic formula before evaluating the root; method marks depend on it.

+ Add to test✓ In your test288IGMC3-2-10r29-08Expanding, factorising and algebraic fractions Paper EHard8 marks

No calculator

Answer all parts.

(a)

Simplify (x² − 9)/(x² + x − 12).

(3)

(b)

Write 3/(x − 2) + 2/(x + 1) as a single fraction in its simplest form.

(3)

(c)

Solve (x + 3)/(x + 4) = 2/3.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 numerator (x − 3)(x + 3)	✓ 1
M1 denominator (x + 4)(x − 3)	✓ 1
A1 (x + 3)/(x + 4)	✓ 1
Part (b)
M1 common denominator (x − 2)(x + 1)	✓ 1
M1 numerator 3(x + 1) + 2(x − 2)	✓ 1
A1 (5x − 1)/((x − 2)(x + 1))	✓ 1
Part (c)
M1 3(x + 3) = 2(x + 4)	✓ 1
A1 x = −1	✓ 1

Common error: Do not cancel terms across a sum, for example the x² in a numerator and denominator; factorise first and cancel whole brackets only.

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The height y metres, relative to the start line, of part of a Formula 1 circuit is modelled by y = x³ − 6x² + 5 for 0 ≤ x ≤ 5, where x is the distance in hundreds of metres along the circuit.

(a)

Find dy/dx.

(2)

(b)

Find the values of x at which the gradient of the circuit is zero.

(3)

(c)

Find the height of the circuit when x = 4.

(1)

(d)

Find the equation of the tangent to the curve at the point where x = 2.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3x²	✓ 1
B1 −12x	✓ 1
Part (b)
M1 3x² − 12x = 0	✓ 1
M1 3x(x − 4) = 0	✓ 1
A1 x = 0 and x = 4	✓ 1
Part (c)
B1 64 − 96 + 5 = −27 m	✓ 1
Part (d)
B1 gradient = 3(2)² − 12(2) = −12	✓ 1
B1 y = 8 − 24 + 5 = −11	✓ 1
A1 y = −12x + 13	✓ 1

Common error: The gradient of the tangent comes from dy/dx, but the point on the line needs the original equation of the curve.

+ Add to test✓ In your test290IGMC3-2-10r35-08Expanding, factorising and algebraic fractions Paper EHard6 marks

No calculator

Algebraic fractions.

(a)

Simplify $\dfrac{x^2-9}{x^2-x-12}$.

(3)

(b)

Write $\dfrac{2}{x+3}+\dfrac{5}{x-1}$ as a single fraction in its simplest form.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factorises numerator $(x-3)(x+3)$	✓ 1
M1 factorises denominator $(x-4)(x+3)$	✓ 1
A1 $\dfrac{x-3}{x-4}$	✓ 1
Part (b)
M1 common denominator $(x+3)(x-1)$	✓ 1
M1 numerator $2(x-1)+5(x+3)$	✓ 1
A1 $\dfrac{7x+13}{(x+3)(x-1)}$	✓ 1

Exam tip: Factorise fully before cancelling, and cancel only whole factors, never single terms.

+ Add to test✓ In your test291IGMC3-2-10r35-11Functions and differentiation Paper EHard8 marks

No calculator

A curve has equation $y=x^3-6x^2+9x+2$.

(a)

Find $\dfrac{dy}{dx}$.

(2)

(b)

Find the coordinates of the stationary points of the curve.

(4)

(c)

Determine the nature of each stationary point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $3x^2$ and $-12x$ correct	✓ 1
B1 $+9$ and constant term differentiated to 0, so $3x^2-12x+9$	✓ 1
Part (b)
M1 sets $\dfrac{dy}{dx}=0$	✓ 1
A1 $x=1$ and $x=3$	✓ 1
A1 $(1,6)$	✓ 1
A1 $(3,2)$	✓ 1
Part (c)
M1 finds $\dfrac{d^2y}{dx^2}=6x-12$ and substitutes (or uses a sign-change test)	✓ 1
A1 $(1,6)$ is a maximum and $(3,2)$ is a minimum	✓ 1

Common error: Finding the $x$-values is not enough: substitute back into the original equation to get the $y$-coordinates.

+ Add to test✓ In your test292IGMC3-2-10r35-13Quadratic equations and inequalities Paper EHard5 marks

Calculator

The frequency f hertz of a vibrating guitar string of length L metres, tension T newtons and mass per unit length μ kg/m is given by $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$.

(a)

Make T the subject of the formula.

(3)

(b)

A string has $L=0.65$, $\mu=0.0010$ and $f=440$. Calculate $T$, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplies by $2L$ to get $2Lf=\sqrt{T/\mu}$	✓ 1
M1 squares both sides	✓ 1
A1 $T=4L^2f^2\mu$	✓ 1
Part (b)
M1 substitutes into $4\times0.65^2\times440^2\times0.0010$	✓ 1
A1 $T=327$ N	✓ 1

Common error: When squaring both sides, square the whole of each side, including the factor $2L$.

+ Add to test✓ In your test293IGMC3-2-12r46-10Expanding, factorising and algebraic fractions Paper EHard7 marks

No calculator

A robotics engineer simplifies expressions in a control-system model.

(a)

Simplify fully $\frac{x^2-9}{x^2+5x+6}$.

(3)

(b)

Solve $\frac{3}{x+1}+\frac{2}{x-2}=2$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 factorises the numerator, $(x-3)(x+3)$	✓ 1
M1 factorises the denominator, $(x+2)(x+3)$	✓ 1
A1 $\frac{x-3}{x+2}$	✓ 1
Part (b)
M1 multiplies through by $(x+1)(x-2)$ or uses a common denominator	✓ 1
A1 $3(x-2)+2(x+1)=2(x+1)(x-2)$ (or equivalent)	✓ 1
M1 rearranges to $2x^2-7x=0$ (or equivalent quadratic)	✓ 1
A1 $x=0$ and $x=\frac72$	✓ 1

Common error: Cancel only factors, never terms: x² in the numerator cannot be cancelled with x² in the denominator before factorising.

+ Add to test✓ In your test294IGMC3-2-12r46-11Functions and differentiation Paper EHard9 marks

No calculator

A fisheries scientist models the stock P, in thousands of fish, after t years by P = t³ − 6t² + 9t + 2, for 0 ≤ t ≤ 4.

(a)

Find $\frac{dP}{dt}$.

(2)

(b)

Find the values of t at which the stock has a stationary value.

(3)

(c)

Find $\frac{d^2P}{dt^2}$ and use it to decide which stationary value is a maximum and which is a minimum.

(2)

(d)

Find the minimum value of P, in thousands of fish.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 $3t^2$	✓ 1
B1 $-12t+9$	✓ 1
Part (b)
M1 sets $3t^2-12t+9=0$ (ECF from (a))	✓ 1
M1 factorises, $3(t-1)(t-3)=0$	✓ 1
A1 $t=1$ and $t=3$	✓ 1
Part (c)
B1 $6t-12$	✓ 1
B1 $t=1$ gives $-6<0$ so a maximum, and $t=3$ gives $6>0$ so a minimum	✓ 1
Part (d)
M1 substitutes $t=3$, $27-54+27+2$	✓ 1
A1 2	✓ 1

Exam tip: Use the second derivative to justify the nature of a turning point; quote its sign at each value of t.

+ Add to test✓ In your test295IGMC3-2-11r48-09Functions and differentiation Paper EHard9 marks

No calculator

The functions f and g are defined by f(x) = $\frac{2x+1}{x-3}$ for x ≠ 3, and g(x) = x − 2.

(a)

Find f(5).

(1)

(b)

Find f⁻¹(x).

(3)

(c)

Show that gf(x) = $\frac{7}{x-3}$.

(3)

(d)

Solve gf(x) = 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 11/2 or 5.5	✓ 1
Part (b)
M1 y(x − 3) = 2x + 1 (after swapping or rearranging)	✓ 1
M1 collects x terms and factorises, x(y − 2) = 3y + 1	✓ 1
A1 f⁻¹(x) = (3x + 1) ÷ (x − 2)	✓ 1
Part (c)
M1 gf(x) = f(x) − 2	✓ 1
M1 common denominator, (2x + 1 − 2(x − 3)) ÷ (x − 3)	✓ 1
A1 numerator simplified to 7, giving 7 ÷ (x − 3) shown	✓ 1
Part (d)
M1 7 = 2(x − 3)	✓ 1
A1 x = 6.5	✓ 1

Exam tip: gf(x) means apply f first, then g; write g(f(x)) to avoid reversing the order.

Marking point	Mark	Notes
Part (a)
M1 expands both brackets: \(6x-15-4x+8\)	✓ 1
A1 \(2x-7\)	✓ 1
Part (b)
B1 \(6a\) identified as the common factor	✓ 1
B1 \(6a(2b-3a)\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(C-6=3h\)	✓ 1
A1 \(h=\frac{C-6}{3}\)	✓ 1
Part (b)
M1 forms the equation \(6+3h=12+1.5h\)	✓ 1
M1 collects terms: \(1.5h=6\)	✓ 1
A1 \(h=4\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(v-u=at\)	✓ 1
A1 \(t=\frac{v-u}{a}\)	✓ 1
Part (b)
M1 substitutes: \(t=\frac{11-3.0}{0.80}\)	✓ 1
A1 \(t=10\) s	✓ 1

Marking point	Mark	Notes
Part (a)
M1 collect terms, e.g. \(4x = 16\)	✓ 1
A1 \(x = 4\)	✓ 1
Part (b)
M1 subtract 12 from both sides	✓ 1
A1 \(h = \frac{C-12}{5}\)	✓ 1
Part (c)
A1 5 hours	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{x+3}{2} = \frac{2x-1}{3}\) and multiply both sides by 6	✓ 1
M1 \(3x + 9 = 4x - 2\)	✓ 1
A1 \(x = 11\)	✓ 1
Part (b)
A1 7 kW	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(2.5+0.6\times 8\)	✓ 1
A1 7.3 m/s	✓ 1
Part (b)
M1 \(v-u=at\)	✓ 1
A1 \(t=\dfrac{v-u}{a}\)	✓ 1
Part (c)
M1 \(\dfrac{10.0-2.5}{0.6}\)	✓ 1
A1 12.5 s	✓ 1

Marking point	Mark	Notes
Part (a)
M1 common difference 5, so \(5n+\dots\)	✓ 1
A1 \(5n+7\)	✓ 1
Part (b)
M1 \(5n+7=150\)	✓ 1
A1 \(n=28.6\), which is not an integer	✓ 1
Part (c)
M1 \(5n+7=107\)	✓ 1
A1 20th term	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{70-110}{25-0}\)	✓ 1
A1 \(-1.6\)	✓ 1
Part (b)
B1 \(m=110-1.6t\) (ECF from (a))	✓ 1
Part (c)
M1 \(110-1.6t=30\)	✓ 1
A1 50 laps	✓ 1

Marking point	Mark	Notes
Part (a)
M1 expand both brackets: \(12x - 8 - 5x + 15\)	✓ 1
A1 \(7x + 7\)	✓ 1
Part (b)
M1 partial factorisation, e.g. \(6p(2q^2 - 3pq)\) or \(3pq(4q - 6p)\)	✓ 1
A1 \(6pq(2q - 3p)\)	✓ 1
Part (c)
M1 cancel a common factor, e.g. \(\frac{2a^2b}{3ab^2}\) or \(\frac{10a}{15b}\)	✓ 1
A1 \(\dfrac{2a}{3b}\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 substitute \(h = 1800\): \(15 - 0.006 \times 1800\)	✓ 1
A1 4.2 °C	✓ 1
Part (b)
M1 isolate the \(h\) term: \(0.006h = 15 - T\)	✓ 1
A1 \(h = \dfrac{15 - T}{0.006}\)	✓ 1
Part (c)
M1 \(h = \dfrac{15 - (-6)}{0.006}\)	✓ 1
A1 3500 m	✓ 1

Marking point	Mark	Notes
Part (a)
B1 \(1800 - 150h\)	✓ 1
B1 \(1800 - 150h \geqslant 300\)	✓ 1
Part (b)
M1 \(150h \leqslant 1500\)	✓ 1
A1 \(h \leqslant 10\), so 10 hours	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(v - u = at\)	✓ 1
A1 \(a = \frac{v-u}{t}\)	✓ 1
Part (b)
M1 substitutes into \(a = \frac{9.2-1.2}{4.0}\)	✓ 1
A1 \(a = 2\) m/s²	✓ 1

Marking point	Mark	Notes
Part (a)
B1 \(3a + 2c = 47\)	✓ 1
B1 \(a + 4c = 39\)	✓ 1
Part (b)
M1 multiplies or rearranges to make one pair of coefficients equal (for example \(3a + 12c = 117\))	✓ 1
A1 \(c = 7\)	✓ 1
A1 \(a = 11\)	✓ 1
Part (c)
M1 \(2 \times 11 + 3 \times 7\) (ECF from (b))	✓ 1
A1 £43	✓ 1

Marking point	Mark	Notes
Part (a)
M1 expands to four terms (\(x^2 - 7x + 4x - 28\))	✓ 1
A1 \(x^2 - 3x - 28\)	✓ 1
Part (b)
B1 takes out a common factor (\(4\), \(p\) or \(2p\))	✓ 1
A1 \(4p(2p - 3)\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 attempt to expand both brackets: \(6x-15-4x+8\)	✓ 1
A1 \(2x-7\)	✓ 1
Part (b)
B1 partial factorisation, e.g. \(6(2ab^2-3a^2b)\) or \(ab(12b-18a)\)	✓ 1
B1 \(6ab(2b-3a)\)	✓ 1
Part (c)
B1 \((y-7)(y+7)\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 expand and collect: \(8x+1=5x+10\), so \(3x=9\)	✓ 1
A1 \(x=3\)	✓ 1
Part (b)
M1 \(C-F=pn\)	✓ 1
A1 \(n=\dfrac{C-F}{p}\)	✓ 1
Part (c)
M1 \(\dfrac{60-12}{0.80}\) (ECF from (b))	✓ 1
A1 \(n=60\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\dfrac{19-7}{6-2}\)	✓ 1
A1 3	✓ 1
Part (b)
M1 substitute a point into \(y=3x+c\), e.g. \(7=3(2)+c\)	✓ 1
A1 \(y=3x+1\)	✓ 1
Part (c)
M1 \(3\times15+1\) (ECF from (b))	✓ 1
A1 £46	✓ 1
Part (d)
B1 the charge per kilometre (£3 for each extra km)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 four terms: \(x^2-2x+5x-10\)	✓ 1
A1 \(x^2+3x-10\)	✓ 1
Part (b)
M1 \(6^2+3(6)-10\) or \(11\times4\)	✓ 1
A1 44 m²	✓ 1
Part (c)
M1 two numbers with product \(-28\) and sum \(-3\)	✓ 1
A1 \((x-7)(x+4)\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(4x=24\)	✓ 1
A1 \(x=6\)	✓ 1
Part (b)
M1 \(C-15=6h\)	✓ 1
A1 \(h=\dfrac{C-15}{6}\)	✓ 1
Part (c)
M1 \(\dfrac{63-15}{6}\) (ECF from (b))	✓ 1
A1 8 hours	✓ 1

Marking point	Mark	Notes
Part (a)
B1 \(14x+86=240\)	✓ 1
Part (b)
M1 \(14x=240-86=154\)	✓ 1
A1 \(x=11\)	✓ 1
Part (c)
M1 \(C-86=14n\)	✓ 1
A1 \(n=\dfrac{C-86}{14}\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\dfrac{17-5}{6-2}\)	✓ 1
A1 3	✓ 1
Part (b)
M1 substitutes a point into \(y=3x+c\)	✓ 1
A1 \(y=3x-1\)	✓ 1
Part (c)
M1 \(3x-1=26\)	✓ 1
A1 \(x=9\)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(3x+2\times4.5=27\) (or \(3x+9=27\))	✓ 1
A1 \(3x=18\)	✓ 1
A1 \(x=6\), so £6 per metre	✓ 1
Part (b)
M1 \(5\times6+4\times4.5\)	✓ 1
A1 £48	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{-5-7}{4-(-2)}\)	✓ 1
A1 for gradient = −2	✓ 1
A1 for y = −2x + 3	✓ 1
Part (b)
M1 for −2x + 3 = x − 6	✓ 1
A1 for x = 3	✓ 1
A1 for (3, −3)	✓ 1

Marking point	Mark	Notes
Part (a)
B1 \(3a+2b=19\)	✓ 1
B1 \(2a+5b=31\)	✓ 1
Part (b)
M1 multiplies to equalise a coefficient, e.g. \(15a+10b=95\) and \(4a+10b=62\)	✓ 1
A1 \(a=3\)	✓ 1
A1 \(b=5\)	✓ 1

Marking point	Mark	Notes
Part (a)
B1 length \(=x+3\)	✓ 1
M1 \(x(x+3)=28\) and expands to \(x^2+3x-28=0\)	✓ 1
Part (b)
M1 \((x+7)(x-4)=0\)	✓ 1
A1 \(x=-7\) or \(x=4\)	✓ 1
R1 rejects \(-7\) because a width cannot be negative, so the width is 4 m	✓ 1

Marking point	Mark	Notes
Part (a)
M1 four terms: \(x^2-6x+4x-24\)	✓ 1
A1 \(x^2-2x-24\)	✓ 1
Part (b)
M1 \((x+a)(x+b)\) with \(ab=-14\) or \(a+b=-5\)	✓ 1
A1 \((x-7)(x+2)\)	✓ 1
Part (c)
B1 \(2(x^2-9)\)	✓ 1
B1 \(2(x-3)(x+3)\)	✓ 1

Marking point	Mark	Notes
Part (a)
B1 \(y=7\) when \(x=-2\)	✓ 1
B1 \(y=-8\) when \(x=3\)	✓ 1
Part (b)
M1 solves \((x-5)(x+1)=0\)	✓ 1
A1 \(x=-1\) and \(x=5\)	✓ 1
Part (c)
B1 \(x=2\) (midway between \(-1\) and \(5\))	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(\frac{52-12}{2.5-0.5}\)	✓ 1
A1 20	✓ 1
Part (b)
M1 substitute a point into \(s = 20t + c\)	✓ 1
A1 \(s = 20t + 2\)	✓ 1
Part (c)
B1 her distance from the start of the road at \(t = 0\) is 2 km	✓ 1
Part (d)
M1 \(100 = 20t + 2\)	✓ 1
A1 \(t = 4.9\) hours	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(2(3x - 5) + 2(x + 4)\)	✓ 1
A1 \(8x - 2\)	✓ 1
Part (b)
M1 \(8x - 2 = 62\)	✓ 1
A1 \(x = 8\)	✓ 1
Part (c)
M1 length 19 and width 12 found	✓ 1
A1 228 m²	✓ 1

Marking point	Mark	Notes
Part (a)
M1 \(8+2.5h=23\) or \(23-8=15\)	✓ 1
A1 \(h=6\)	✓ 1
Part (b)
M1 subtracts 8 from both sides, \(C-8=2.5h\)	✓ 1
A1 \(h=\dfrac{C-8}{2.5}\)	✓ 1
Part (c)
M1 forms the equation \(8+2.5h=3+3.5h\)	✓ 1
A1 \(h=5\)	✓ 1
A1 £20.50	✓ 1