Edexcel GCSE Maths June 2022 Paper 3 Higher answers

Question 1

Pythagoras and trigonometry2 marksGrade 4CalculatorOn both tiers

Use Pythagoras' theorem to find a shorter side of a right-angled triangle.

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Answer	7.5	2 marks

Worked solution Checked against the mark scheme

The hypotenuse is 8.5 cm.

$x^2 = 8.5^2 - 4^2 = 72.25 - 16 = 56.25$

$x = \sqrt{56.25} = 7.5$

Watch out: Adding the squares instead of subtracting, giving about 9.39.

Question 2

Formulae and changing the subject4 marksGrade 4CalculatorOn both tiers

Substitute a negative number into a formula with a square, then rearrange a linear formula.

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(a)	$T = 25$	2 marks
(b)	$p = \frac{d - 4}{3}$	2 marks

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(a) $(-3)^2 = 9$ , so $T = 4 \times 9 - 11 = 36 - 11 = 25$

(b) $d = 3p + 4$

Subtract 4: $d - 4 = 3p$

Divide by 3: $p = \frac{d - 4}{3}$

Watch out: In (a) squaring $4 \times -3$ to get 144, or using $-9$ for $(-3)^2$ .

Question 3

Ratio5 marksGrade 5CalculatorOn both tiers

Form and solve an equation for the counters three people hold, then write a ratio in the form 1 : p.

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Answer	1.5	5 marks

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Let Rick have $n$ . Selma has $2n$ , Tony has $2n - 6$ .

$n + 2n + 2n - 6 = 54 \Rightarrow 5n = 60 \Rightarrow n = 12$

Tony has $24 - 6 = 18$

$12 : 18 = 1 : 1.5$ , so $p = 1.5$

Watch out: Using Selma's counters instead of Tony's in the ratio.

Question 4

Percentage change and reverse percentages4 marksGrade 4CalculatorOn both tiers

Compare the cost of 15 rolls of wallpaper from two shops with different offers.

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Answer	Chic Decor (£180 compared with £184.80)	4 marks

Worked solution Checked against the mark scheme

Chic Decor: $15 \div 3 = 5$ lots of £36, so $5 \times 36 = £180$

Style Papers: $3 \times 70 = £210$ , then $210 \times 0.88 = £184.80$

£180 < £184.80, so buy from Chic Decor.

Watch out: Taking 12% off only one pack, or finding 12% but not subtracting it.

Question 5

Charts, graphs and time series2 marksGrade 4CalculatorOn both tiers

Spot two errors in a frequency polygon drawn from a grouped table.

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Answer	The vertical scale misses 40 (not evenly spaced); the last point is plotted at 50 instead of the midpoint 45	2 marks

Worked solution Checked against the mark scheme

Look at the frequency axis: it goes 30 then 50, so 40 is missing and the scale is not even.

Points should be at class midpoints (5, 15, 25, 35, 45). The last point is at (50, 5) but should be at (45, 5).

Watch out: Saying the polygon should start at the origin, which is not a mistake.

Question 6

Speed, density and pressure4 marksGrade 5CalculatorOn both tiers

Find a runner's total distance over two stages, then the speed another runner needs to cover it in a given time.

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Answer	10	4 marks

Worked solution Checked against the mark scheme

Stage 1: $6 \times \frac{15}{60} = 1.5$ miles

Stage 2: $9 \times \frac{40}{60} = 6$ miles

Total = 7.5 miles

Amy: $7.5 \div 0.75 = 10$ mph

Watch out: Using 45 hours instead of 0.75 hours, or averaging the two speeds.

Question 7

Area and perimeter3 marksGrade 5CalculatorOn both tiers

Show that the area of a shaded trapezium in a rectangle is a given quadratic expression in x.

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Answer	$A = 2x^2 + 20x$ shown	3 marks

Worked solution Checked against the mark scheme

$QU = (3x + 5) - 2x = x + 5$ and $RV = 5$ ; the height is $4x$ .

$A = \frac{1}{2}(QU + RV) \times 4x = \frac{1}{2}(x + 10) \times 4x$

$A = 2x(x + 10) = 2x^2 + 20x$

Watch out: Using $3x + 5$ for the top side QU instead of $x + 5$ .

Question 8

Real-life graphs and areas under graphs3 marksGrade 5Calculator

Find the gradient of a cost-against-units graph and interpret it in context.

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(a)	$0.14$	2 marks
(b)	The cost of each unit of electricity (£0.14, i.e. 14p per unit)	1 mark

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(a) The line passes through $(0, 0)$ and $(100, 14)$ .

Gradient $= \frac{14}{100} = 0.14$

(b) The gradient is pounds per unit, so it is the price of one unit of electricity: 14p.

Watch out: In (b), saying only that it shows how steep the line is, without linking it to cost per unit.

Question 9

Powers, roots and indices4 marksGrade 6Calculator

Simplify a square root of a quotient of powers of 10, and explain the error in raising a power to a power.

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(a)	$10^{60}$	3 marks
(b)	He should multiply the powers, $50\times2 = 100$ , giving $12^{100}$ , not square the 50	1 mark

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(a) $10^{150} \times 10^{90} = 10^{240}$

$10^{360} \div 10^{240} = 10^{120}$

$\sqrt{10^{120}} = 10^{60}$

(b) For a power of a power you multiply the indices: $(12^{50})^2 = 12^{100}$ . Liam squared 50 instead.

Watch out: In (a), halving only the numerator's power before dividing, or taking the square root of 120 instead of halving it.

Question 10

Compound interest, growth and decay3 marksGrade 6Calculator

Work backwards from a car's value after one percentage fall and two further equal percentage falls to find the original price.

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Answer	$£24\,000$	3 marks

Worked solution Checked against the mark scheme

Multiplier after 3 years: $0.875 \times 0.9 \times 0.9 = 0.70875$

Original value $= 17\,010 \div 0.70875 = £24\,000$

Watch out: Adding the percentages (12.5 + 10 + 10 = 32.5%) or increasing £17 010 by those percentages.

Question 11

Systematic listing and the product rule2 marksGrade 5Calculator

Count how many outfits can be made from given numbers of shirts, jeans and jackets.

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Answer	$240$	2 marks

Worked solution Checked against the mark scheme

Each choice can go with every other choice, so multiply:

$16 \times 5 \times 3 = 240$

Watch out: Adding the numbers to get 24.

Question 12

Pythagoras and trigonometry3 marksGrade 6Calculator

Use right-angled trigonometry in two connected triangles to find a length to 3 significant figures.

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Answer	$23.4$ cm	3 marks

Worked solution Checked against the mark scheme

In triangle ACD, $\tan 45^\circ = \frac{AC}{8}$ , so $AC = 8$ cm.

In triangle ABC, $AC$ is opposite the $20^\circ$ angle and $AB$ is the hypotenuse:

$AB = \frac{8}{\sin 20^\circ} = 23.39\ldots \approx 23.4$ cm

Watch out: Using tan 20° instead of sin 20°, which finds BC rather than AB.

Question 13

Vectors3 marksGrade 6Calculator

Find an unknown column vector from a vector equation involving a known vector.

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Answer	$\begin{pmatrix}-1\\4\end{pmatrix}$	3 marks

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$3\mathbf{a} = \begin{pmatrix}6\\-9\end{pmatrix}$

$2\mathbf{b} = 3\mathbf{a} - \begin{pmatrix}8\\-17\end{pmatrix} = \begin{pmatrix}-2\\8\end{pmatrix}$

$\mathbf{b} = \begin{pmatrix}-1\\4\end{pmatrix}$

Watch out: Sign error when rearranging, giving (1, −4).

Question 14

Factorising5 marksGrade 6Calculator

Fully factorise a difference of two squares with a common factor, and expand a product of three brackets into a cubic.

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(a)	$4(p-3)(p+3)$	2 marks
(b)	$6m^3 + 11m^2 - 57m - 20$	3 marks

Worked solution Checked against the mark scheme

(a) $4p^2 - 36 = 4(p^2 - 9) = 4(p - 3)(p + 3)$

(b) $(m + 4)(2m - 5) = 2m^2 + 3m - 20$

$(2m^2 + 3m - 20)(3m + 1) = 6m^3 + 2m^2 + 9m^2 + 3m - 60m - 20$

$= 6m^3 + 11m^2 - 57m - 20$

Watch out: In (a), stopping at 4(p² − 9) without factorising fully.

Question 15

Circle theorems3 marksGrade 7Calculator

Prove two triangles formed by intersecting chords in a circle are similar, giving reasons.

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Answer	Angle PQX = angle SRX (same segment); angle QPX = angle RSX (same segment); angle PXQ = angle SXR (vertically opposite); all three angles equal, so similar	3 marks

Worked solution Checked against the mark scheme

Angles $PQS$ and $PRS$ stand on the same arc $PS$ , so angle $PQX$ = angle $SRX$ (angles in the same segment).

Angles $QPR$ and $QSR$ stand on arc $QR$ , so angle $QPX$ = angle $RSX$ (angles in the same segment).

Angle $PXQ$ = angle $SXR$ (vertically opposite).

All three pairs of angles are equal, so the triangles are similar.

Watch out: Stating equal angles without naming the circle theorem.

Question 16

Error intervals and bounds3 marksGrade 7Calculator

Find the upper bound of a square-root formula using rounded values, to 3 significant figures.

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Answer	$17.4$	3 marks

Worked solution Checked against the mark scheme

Bounds: $6.75 \le e < 6.85$ and $0.045 \le f < 0.055$

For the largest $p$ , make the top big and the bottom small:

$p = \sqrt{\frac{2 \times 6.85}{0.045}} = \sqrt{304.44\ldots} = 17.448\ldots \approx 17.4$

Watch out: Dividing by the upper bound of f, which gives the smaller value.

Question 17

Histograms5 marksGrade 7Calculator

Draw a histogram from grouped data with unequal widths, then use a given histogram to express one class frequency in terms of another.

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(a)	Bars with frequency densities 90, 96, 44, 8 and 6 over the correct class widths	3 marks
(b)	$0.4n$	2 marks

Worked solution Checked against the mark scheme

(a) Frequency density = frequency ÷ class width:

$90\div1 = 90$ , $48\div0.5 = 96$ , $22\div0.5 = 44$ , $8\div1 = 8$ , $12\div2 = 6$

(b) On the grid, the 1–2 bar is 5 times as tall as the 3–5 bar, but the 3–5 bar is twice as wide.

Area ratio $= 5\times1 : 1\times2 = 5 : 2$

So the number is $\frac{2}{5}n = 0.4n$

Watch out: Drawing bar heights equal to the frequencies instead of the frequency densities.

Question 18

Pythagoras and trigonometry5 marksGrade 8Calculator

Find the angle between a line in a prism and its base, using a ratio, trapezium area, Pythagoras and trigonometry.

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Answer	$30.6^\circ$	5 marks

Worked solution Checked against the mark scheme

$TC = 14 \times \frac{3}{7} = 6$ cm

$TD = \sqrt{14^2 + 6^2} = \sqrt{232} = 15.23\ldots$ cm

Trapezium: $\frac{1}{2}(SD + 12)\times14 = 147 \Rightarrow SD = 9$ cm

Angle $= \tan^{-1}\left(\frac{9}{15.23\ldots}\right) = 30.6^\circ$

Watch out: Using TC or BT alone as the horizontal distance instead of TD.

Question 19

Algebraic fractions4 marksGrade 8Calculator

$Combine three algebraic fractions (including a whole number) over a common denominator to reach a single given form.$

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Answer	$\frac{-11x + 2}{x^2 - 4}$ , so $a = -11$ , $b = 2$	4 marks

Worked solution Checked against the mark scheme

Common denominator $(x + 2)(x - 2) = x^2 - 4$

$\frac{3x(x - 2) - (2x + 1)(x + 2) - (x^2 - 4)}{x^2 - 4}$

Numerator: $3x^2 - 6x - (2x^2 + 5x + 2) - x^2 + 4 = -11x + 2$

So the expression is $\frac{-11x + 2}{x^2 - 4}$

Watch out: Not distributing the minus sign across $(2x + 1)(x + 2)$ .

Question 20

Sequences and the nth term4 marksGrade 7Calculator

Use two consecutive years' values to find the constant in a recurrence relation, then predict two years ahead.

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Answer	$£44\,384$	4 marks

Worked solution Checked against the mark scheme

$29\,600 = 24\,000a + 800 \Rightarrow a = \frac{28\,800}{24\,000} = 1.2$

2020: $1.2 \times 29\,600 + 800 = 36\,320$

2021: $1.2 \times 36\,320 + 800 = £44\,384$

Watch out: Stopping at 2020 (£36 320).

Question 21

Tree diagrams and combined events4 marksGrade 8Calculator

Find the probability that three cards drawn without replacement from nine have an even total.

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Answer	$\frac{11}{21}$	4 marks

Worked solution Checked against the mark scheme

There are 4 even and 5 odd cards. The sum is even for EEE or one even and two odd.

EEE: $\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7} = \frac{24}{504}$

One even, two odd (3 orders): $3\times\frac{4}{9}\times\frac{5}{8}\times\frac{4}{7} = \frac{240}{504}$

Total $= \frac{264}{504} = \frac{11}{21}$

Watch out: Treating the draws as with replacement, giving 364/729.

Question 22

Simultaneous equations5 marksGrade 8Calculator

Find the intersection points of a straight line and a curve by substituting and solving a quadratic.

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Answer	$(-3, -11)$ and $(5.5, 6)$	5 marks

Worked solution Checked against the mark scheme

Substitute $y = 2x - 5$ : $(2x - 5)^2 = 6x^2 - 25x - 8$

$4x^2 - 20x + 25 = 6x^2 - 25x - 8$

$2x^2 - 5x - 33 = 0 \Rightarrow (2x - 11)(x + 3) = 0$

$x = 5.5 \Rightarrow y = 6$ ; $x = -3 \Rightarrow y = -11$

Watch out: Expanding $(2x - 5)^2$ as $4x^2 + 25$ .

June 2022 Paper 3 Higher

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