Formulae: substitution, deriving & changing the subject – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator2 marks

$P = 5a - 2b^2$

Work out the value of $P$ when $a = -3$ and $b = 4$. [2]

correct substitution, e.g. $5 \times (-3) - 2 \times 4^2$ or $-15 - 32$ (allow $-15 - 2 \times 16$)M1
$-47$ caoA1
Answer: $-47$
Examiner tip: Square $b$ first and then multiply by 2: working out $(2 \times 4)^2 = 64$ is the most common error.

2Grade 1–3FoundationCalculator4 marks

A company hires out vans. The cost is £45 plus £0.30 for each mile driven.

A van is driven $m$ miles. The total cost is £$C$.

(a) Write down a formula for $C$ in terms of $m$. [2]

(b) Ravi pays £117 to hire a van. Use your formula to find how many miles he drove. [2]

(a) $45 + 0.3m$ oe seen (allow $0.30m + 45$, or $4500 + 30m$ if working in pence)B1
a fully correct formula $C = 45 + 0.3m$ oe (must include '$C =$'; B0 B0 for $C = 45 + 0.3 + m$ or $C = 45m + 0.3$)B1
(b) $117 - 45 = 72$ and $72 \div 0.3$ seen, or $117 = 45 + 0.3m$ rearranged to $0.3m = 72$M1
$240$ (miles)A1
Answer: (a) $C = 45 + 0.3m$ (b) 240 miles
Examiner tip: In (a) the answer must be a formula starting '$C =$' — an expression alone only gets 1 of the 2 marks.

3Grade 4–5FoundationCalculator2 marks

Make $r$ the subject of the formula

$$T = \frac{5r}{2} - 8$$

[2]

a correct first step, e.g. $T + 8 = \dfrac{5r}{2}$ or $2T = 5r - 16$M1
$r = \dfrac{2(T + 8)}{5}$ oe, e.g. $r = \dfrac{2T + 16}{5}$ or $r = 0.4T + 3.2$ (must have '$r =$')A1
Answer: $r = \dfrac{2(T+8)}{5}$
Examiner tip: Undo the operations in reverse order: deal with the $-8$ before the fraction, and multiply every term by 2 if you clear the fraction first.

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4Grade 4–5Calculator4 marks

The distance, $D$ miles, that a person can see to the horizon from a height of $h$ feet above sea level is given approximately by

$$D = \sqrt{\frac{3h}{2}}$$

(a) Work out the value of $D$ when $h = 20$. Give your answer correct to 3 significant figures. [2]

(b) Find the height $h$ from which a person can see 12 miles to the horizon. [2]

(a) $\sqrt{\dfrac{3 \times 20}{2}}$ or $\sqrt{30}$ seenM1
$5.48$ (awrt 5.48)A1
(b) $12^2 = \dfrac{3h}{2}$ or $144 = 1.5h$ or $h = \dfrac{2D^2}{3}$ seenM1
$96$ (feet)A1
Answer: (a) 5.48 miles (b) 96 feet
Examiner tip: In (b) square both sides first to remove the square root — dividing 12 by 1.5 before squaring is a common slip.

5Grade 4–5FoundationCalculator5 marks

The diagram shows a pentagon made from a rectangle and a triangle. All lengths are in centimetres.

Diagram NOT accurately drawn

The perimeter of the pentagon is $P$ cm.

(a) Show that $P = 6x + 8$ [3]

(b) The perimeter of the pentagon is 50 cm. Work out the value of $x$. [2]

(a) adds all five side lengths (and no internal side), e.g. $2x + (x + 3) + (x + 3) + (x + 1) + (x + 1)$M1
expands/collects correctly at least the $x$ terms ($6x$) or the number terms ($+8$)M1
completes to $P = 6x + 8$ with no errors seen (shown)A1
(b) $6x + 8 = 50$ and a correct first step, e.g. $6x = 42$M1
$x = 7$A1
Answer: (a) shown (b) $x = 7$
Examiner tip: Do not include the line where the triangle meets the rectangle — it is inside the shape, so it is not part of the perimeter.

6Grade 6–7HigherCalculator4 marks

Make $g$ the subject of the formula

$$h = \frac{5 - 2g}{3g + 1}$$

[4]

multiplies both sides by $3g + 1$ correctly: $h(3g + 1) = 5 - 2g$ or $3gh + h = 5 - 2g$M1
collects terms in $g$ on one side and other terms on the other, e.g. $3gh + 2g = 5 - h$ (ft their expansion with one error)M1
factorises $g$ out, e.g. $g(3h + 2) = 5 - h$M1
$g = \dfrac{5 - h}{3h + 2}$ oe, e.g. $g = \dfrac{h - 5}{-3h - 2}$A1
Answer: $g = \dfrac{5 - h}{3h + 2}$
Examiner tip: When the subject appears twice, gather every $g$ term on one side and factorise — you cannot 'divide by $g$' to finish.

7Grade 6–7HigherCalculator5 marks

$$F = \frac{5m}{d^2} - 2$$

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

(b) Find the positive value of $d$ when $m = 24$ and $F = 0.5$. Give your answer correct to 3 significant figures. [2]

(a) $F + 2 = \dfrac{5m}{d^2}$M1
$d^2 = \dfrac{5m}{F + 2}$ oeM1
$d = \sqrt{\dfrac{5m}{F + 2}}$ (allow $\pm$)A1
(b) $\sqrt{\dfrac{5 \times 24}{0.5 + 2}}$ or $\sqrt{48}$ seen (ft their (a) if it involves a square root)M1
$6.93$ (awrt 6.93)A1
Answer: (a) $d = \sqrt{\dfrac{5m}{F+2}}$ (b) 6.93
Examiner tip: Add the 2 before you deal with the fraction; and remember the last step is a square root, not dividing by 2.

8Grade 8–9HigherCalculator4 marks

Make $x$ the subject of

$$k = \sqrt{\frac{2x + p}{x}}$$

[4]

squares both sides: $k^2 = \dfrac{2x + p}{x}$M1
multiplies by $x$: $k^2x = 2x + p$M1
collects $x$ terms and factorises, e.g. $x(k^2 - 2) = p$M1
$x = \dfrac{p}{k^2 - 2}$ oe, e.g. $x = \dfrac{-p}{2 - k^2}$A1
Answer: $x = \dfrac{p}{k^2 - 2}$
Examiner tip: Square first to remove the root, then clear the fraction — only then can you collect the two $x$ terms and factorise.

9Grade 6–7HigherCalculator5 marks

For a lens, the object distance $u$ cm, the image distance $v$ cm and the focal length $f$ cm are connected by the formula

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

(a) Show that $v = \dfrac{fu}{u - f}$ [3]

(b) A lens has focal length 15 cm. An object is placed 40 cm from the lens. Work out the image distance. [2]

(a) isolates $\dfrac{1}{v}$: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ (or multiplies through by $fuv$: $uv = fv + fu$)M1
combines into a single fraction, $\dfrac{1}{v} = \dfrac{u - f}{fu}$ (or collects to $uv - fv = fu$, $v(u - f) = fu$)M1
completes to $v = \dfrac{fu}{u - f}$ with no errors (shown)A1
(b) $\dfrac{15 \times 40}{40 - 15}$ or $\dfrac{1}{15} - \dfrac{1}{40}$ $\left(= \dfrac{1}{24}\right)$M1
$24$ (cm)A1
Answer: (a) shown (b) 24 cm
Examiner tip: $\dfrac{1}{f} - \dfrac{1}{u}$ is NOT $\dfrac{1}{f - u}$ — use a common denominator before flipping the fraction.

10Grade 4–5Calculator4 marks

The area $A$ of a trapezium with parallel sides $a$ and $b$ and height $h$ is $A = \dfrac{h(a + b)}{2}$

Leila makes $b$ the subject. Her working is

$2A = h(a + b)$
$2A - h = a + b$
$b = 2A - h - a$

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

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

(a) explanation that $h$ multiplies the bracket so she should have divided by $h$, not subtracted $h$ (in line 2)C1
(b) $\dfrac{2A}{h} = a + b$ oeM1
$b = \dfrac{2A}{h} - a$ oe, e.g. $b = \dfrac{2A - ah}{h}$A1
(c) $8.5$B1
Answer: (a) She subtracted $h$ instead of dividing by $h$ (b) $b = \dfrac{2A}{h} - a$ (c) 8.5
Examiner tip: Check how a letter is attached before removing it: $h(a+b)$ means $h$ times the bracket, so divide.

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