Substitution & rearranging formulae (changing the subject) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

$P = 3a - 2b$

(a) Work out the value of $P$ when $a = 5$ and $b = -4$ [2]

(b) Work out the value of $a$ when $P = 20$ and $b = 2$ [2]

(a) for correct substitution, e.g. $3 \times 5 - 2 \times (-4)$ or $15 - (-8)$ or $15 + 8$M1
$23$A1
(b) for a correct substitution or a correct first step to isolate $a$, e.g. $20 = 3a - 2 \times 2$ or $20 = 3a - 4$ or $3a = 24$ or $(20 + 4) \div 3$ or $a = \frac{24}{3}$M1
$8$A1
Answer: (a) $23$ (b) $8$
Examiner tip: Subtracting a negative is adding: $-2 \times (-4) = +8$, so $P = 15 + 8 = 23$. Writing $15 - 8 = 7$ is wrong; show the substitution $3 \times 5 - 2 \times (-4)$ so you can still earn the method mark if you slip.

2Grade 1–3FoundationNon-calculator2 marks

Make $t$ the subject of the formula $w = 5t - 9$ [2]

for a correct first step, e.g. $w + 9 = 5t$ or $\frac{w}{5} = t - \frac{9}{5}$M1
$t = \frac{w + 9}{5}$ oeA1
Answer: $t = \dfrac{w + 9}{5}$
Examiner tip: Undo in reverse order – add 9 first, then divide the WHOLE of $w + 9$ by 5 (writing $w + 9 \div 5$ loses the A mark).

3Grade 4–5HigherCalculator4 marks

$E = \frac{1}{2}mv^2$

(a) Work out the value of $E$ when $m = 4.8$ and $v = 6.5$ [2]

(b) Make $v$ the subject of the formula, where $v$ is positive. [2]

(a) for $0.5 \times 4.8 \times 6.5^2$ or $0.5 \times 4.8 \times 42.25$ or $2.4 \times 42.25$M1
$101.4$A1
(b) for a correct first step, e.g. $2E = mv^2$ or $\frac{E}{0.5m} = v^2$ or $v^2 = \frac{2E}{m}$M1
$v = \sqrt{\frac{2E}{m}}$ oeA1
Answer: (a) $101.4$ (b) $v = \sqrt{\dfrac{2E}{m}}$
Examiner tip: Only $v$ is squared – work out $6.5^2$ before multiplying, not $(4.8 \times 6.5)^2$.

4Grade 6–7HigherNon-calculator3 marks

The time, $T$ seconds, for one swing of a pendulum of length $L$ metres is given by

$$T = 2\pi\sqrt{\frac{L}{g}}$$

Make $L$ the subject of the formula. [3]

for dividing by $2\pi$: $\frac{T}{2\pi} = \sqrt{\frac{L}{g}}$, OR squaring both sides correctly: $T^2 = 4\pi^2 \frac{L}{g}$M1
for completing both of these steps correctly, e.g. $\frac{T^2}{4\pi^2} = \frac{L}{g}$M1
$L = \frac{gT^2}{4\pi^2}$ oeA1
Answer: $L = \dfrac{gT^2}{4\pi^2}$
Examiner tip: When you square $2\pi\sqrt{\dots}$ the $2\pi$ is squared too – $4\pi^2$, not $2\pi^2$.

5Grade 8–9HigherNon-calculator4 marks

Make $x$ the subject of $\;y = \dfrac{3x + 2}{x - 5}$ [4]

for multiplying both sides by $(x - 5)$ and expanding: $xy - 5y = 3x + 2$M1
for collecting the $x$ terms on one side and the other terms on the other side, e.g. $xy - 3x = 5y + 2$M1
for factorising out $x$: $x(y - 3) = 5y + 2$M1
$x = \frac{5y + 2}{y - 3}$ oe, e.g. $\frac{-5y - 2}{3 - y}$A1
Answer: $x = \dfrac{5y + 2}{y - 3}$
Examiner tip: When $x$ appears twice you MUST collect the $x$ terms and factorise – an answer with $x$ on both sides scores at most M2.

6Grade 4–5Calculator3 marks

$s = ut + \frac{1}{2}at^2$

Work out the value of $a$ when $s = 60$, $u = 4$ and $t = 5$ [3]

for correct substitution: $60 = 4 \times 5 + \frac{1}{2} \times a \times 5^2$M1
for simplifying and a correct step to isolate $a$, e.g. $40 = 12.5a$ or $a = \frac{60 - 20}{12.5}$M1
$3.2$A1
Answer: $a = 3.2$
Examiner tip: Only $t$ is squared: $\frac{1}{2} \times 5^2 = 12.5$, not $(\frac{1}{2} \times 5)^2 = 6.25$.

7Grade 8–9HigherNon-calculator3 marks

$$\frac{1}{m} + \frac{1}{n} = \frac{1}{f}$$

Make $m$ the subject of the formula. Give your answer as a single fraction. [3]

for isolating $\frac{1}{m}$: $\frac{1}{m} = \frac{1}{f} - \frac{1}{n}$, OR multiplying through by $mnf$: $nf + mf = mn$M1
for a single fraction $\frac{1}{m} = \frac{n - f}{fn}$, OR collecting and factorising: $m(n - f) = nf$M1
$m = \frac{fn}{n - f}$ oeA1
Answer: $m = \dfrac{fn}{n - f}$
Examiner tip: $\frac{1}{m} = \frac{1}{f} - \frac{1}{n}$ does NOT give $m = f - n$ – combine into one fraction first, then flip it.

8Grade 1–3FoundationNon-calculator4 marks

A recipe book uses this formula to work out the time, $T$ minutes, needed to roast a chicken of mass $w$ kg.

$$T = 40w + 20$$

(a) Complete the table. [2]

Mass, $w$ (kg)	1.5	3.5
Time, $T$ (minutes)

(b) Jamal roasts a chicken for exactly 3 hours, the time given by the formula.
Work out the mass of the chicken. [2]

(a) $80$ (for $w = 1.5$)B1
$160$ (for $w = 3.5$)B1
(b) for converting 3 hours to 180 minutes and starting to solve, e.g. $40w + 20 = 180$ or $180 - 20\,(= 160)$ or $160 \div 40$
Do not award for working with $T = 3$M1
$4$ (kg)A1
Answer: (a) $80$, $160$ (b) $4$ kg
Examiner tip: Change the 3 hours into 180 minutes before you substitute: $T$ is measured in minutes, so using $T = 3$ gives a negative mass.

9Grade 4–5Non-calculator3 marks

(a) Work out the value of $\dfrac{2x^2 - y}{x + y}$ when $x = -3$ and $y = 4$ [2]

(b) Leah says,

“When $a = -2$, the value of $5a^2$ is $100$”

Explain what Leah has done wrong. [1]

(a) for correct substitution into numerator or denominator, e.g. $2 \times (-3)^2 - 4$ or $2 \times 9 - 4$ or $14$ seen, or $-3 + 4$ or $1$ seen as denominatorM1
$14$A1
(b) explanation that she has squared $5 \times (-2)$ (i.e. found $(-10)^2$) instead of squaring only $a$, e.g. “she squared the 5 as well”; accept correct value $5 \times 4 = 20$ shown with a reasonC1
Answer: (a) $14$ (b) She squared $5a = -10$; only $a$ is squared, so $5a^2 = 5 \times 4 = 20$
Examiner tip: $(-3)^2 = 9$, not $-9$ – put negative numbers in brackets when you substitute, and remember an index applies only to the letter it is attached to.

10Grade 4–5Non-calculator5 marks

The diagram shows a trapezium with parallel sides of length $a$ and $b$ and perpendicular height $h$.

The area, $A$, of the trapezium is given by

$$A = \frac{1}{2}(a + b)h$$

(a) Make $b$ the subject of the formula. [3]

(b) A trapezium has area $35\text{ cm}^2$, height $5$ cm and one parallel side of length $4$ cm.
Work out the length of the other parallel side. [2]

(a) for a correct first step, e.g. $2A = (a + b)h$ or $\frac{A}{h} = \frac{1}{2}(a+b)$M1
for a correct second step, e.g. $\frac{2A}{h} = a + b$ or $2A = ah + bh$ then $bh = 2A - ah$M1
$b = \frac{2A}{h} - a$ oe, e.g. $b = \frac{2A - ah}{h}$; must start $b =$A1
(b) for substitution into a correct rearrangement or into the original formula, e.g. $\frac{2 \times 35}{5} - 4$ or $35 = \frac{1}{2}(4 + b) \times 5$ or $14 = 4 + b$M1
$10$ (cm)A1
Answer: (a) $b = \dfrac{2A}{h} - a$ (b) $10$ cm
Examiner tip: Undo the $\frac{1}{2}$ by doubling the whole of the other side ($2A$) first; a common slip is $b = \frac{A}{2h} - a$.

11Grade 4–5Calculator5 marks

On a clear day, a person whose eyes are $h$ metres above sea level can see a distance of $d$ kilometres out to sea, where

$$d = 3.57\sqrt{h}$$

(a) Mo stands on a beach with his eyes $1.8$ m above sea level.
Work out how far out to sea he can see. Give your answer correct to 3 significant figures. [2]

(b) From the top of a lighthouse, a keeper can see a distance of $25$ km.
Work out the height of the keeper's eyes above sea level. Give your answer to the nearest metre. [3]

(a) for $3.57 \times \sqrt{1.8}$ or $3.57 \times 1.34(16...)$M1
$4.79$ (km); accept $4.789...$ then rounded to $4.79$; do not accept $4.8$ aloneA1
(b) for a correct first step, e.g. $\sqrt{h} = \frac{25}{3.57}$ or $\sqrt{h} = 7.00(28...)$ or $625 = 3.57^2 h$M1
for a complete method, e.g. $\left(\frac{25}{3.57}\right)^2$ or $\frac{625}{12.7449}$ or $49.03...$ or $49.04$M1
$49$ (m)A1
Answer: (a) $4.79$ km (b) $49$ m
Examiner tip: To undo a square root you square – but only after dividing by $3.57$; squaring $25$ and then dividing by $3.57$ (not $3.57^2$) gives the wrong answer of $175$.

12Grade 6–7HigherNon-calculator3 marks

Make $k$ the subject of the formula

$$3(k - 2a) = ak + 5$$

[3]

for expanding the bracket and collecting the $k$ terms on one side and other terms on the other, e.g. $3k - ak = 5 + 6a$ or $ak - 3k = -6a - 5$M1
for factorising out $k$, e.g. $k(3 - a) = 5 + 6a$ or $k(a - 3) = -6a - 5$M1
$k = \frac{5 + 6a}{3 - a}$ oe, e.g. $k = \frac{-6a - 5}{a - 3}$A1
Answer: $k = \dfrac{6a + 5}{3 - a}$
Examiner tip: When the subject appears twice, get every $k$ term on the same side and factorise: $k(3 - a)$. Dividing by $3$ and by $a$ separately is not a valid step.

13Grade 6–7HigherCalculator5 marks

The volume, $V$, of a sphere of radius $r$ is given by $V = \dfrac{4}{3}\pi r^3$

(a) Make $r$ the subject of the formula. [2]

(b) A rubber ball is a sphere with volume $500\text{ cm}^3$.
Work out the diameter of the ball. Give your answer correct to 3 significant figures. [3]

(a) for a correct step to isolate $r^3$, e.g. $3V = 4\pi r^3$ or $r^3 = \frac{3V}{4\pi}$M1
$r = \sqrt[3]{\frac{3V}{4\pi}}$ oeA1
(b) for $r^3 = \frac{3 \times 500}{4\pi}$ oe or $r^3 = 119.3(66...)$M1
for $r = 4.92(37...)$ or $2 \times \sqrt[3]{\frac{1500}{4\pi}}$M1
$9.85$ (cm)A1
Answer: (a) $r = \sqrt[3]{\dfrac{3V}{4\pi}}$ (b) $9.85$ cm
Examiner tip: Read the question: it asks for the diameter, so double the radius $4.92$ cm – many students stop at the radius and lose the final mark.

14Grade 6–7HigherNon-calculator3 marks

Priya is asked to make $t$ the subject of the formula $s = \dfrac{t^2}{4} + c$, where $t$ is positive.

Here is her working.

$4s = t^2 + c$
$t^2 = 4s - c$
$t = \sqrt{4s - c}$

(a) Explain the mistake that Priya has made. [1]

(b) Make $t$ the subject of the formula correctly. [2]

(a) in the first line she has not multiplied $c$ by 4 (she should have $4s = t^2 + 4c$); accept “she only multiplied one term on the right by 4”C1
(b) for a correct step towards isolating $t^2$, e.g. $4s = t^2 + 4c$ or $s - c = \frac{t^2}{4}$ or $t^2 = 4s - 4c$M1
$t = \sqrt{4s - 4c}$ oe, e.g. $t = 2\sqrt{s - c}$ or $t = \sqrt{4(s - c)}$; ignore $\pm$A1
Answer: (a) She did not multiply $c$ by 4 (b) $t = \sqrt{4s - 4c}$ or $2\sqrt{s - c}$
Examiner tip: When you multiply both sides by 4, every term must be multiplied – it is safer to subtract $c$ first so there is only one term left to multiply.

15Grade 8–9HigherNon-calculator4 marks

Make $x$ the subject of the formula

$$b = \sqrt{\frac{x + a}{x}}$$

[4]

for squaring both sides correctly, $b^2 = \frac{x + a}{x}$M1
for removing the fraction, $b^2 x = x + a$ (ft their first step if it involves squaring)M1
for collecting $x$ terms and factorising, e.g. $x(b^2 - 1) = a$ or $x(1 - b^2) = -a$M1
$x = \frac{a}{b^2 - 1}$ oe, e.g. $x = \frac{-a}{1 - b^2}$A1
Answer: $x = \dfrac{a}{b^2 - 1}$
Examiner tip: After squaring and multiplying by $x$, the subject appears on both sides – collect the $x$ terms together and factorise rather than trying to divide by $x$.

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