Edexcel GCSE Maths November 2024 Paper 2 Higher answers

Question 1

Calculating with integers, decimals and negatives2 marksGrade 3Calculator

$Use a calculator to evaluate the square root of a fraction whose top and bottom are both calculations, giving the full display.$

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Answer	$4.643069317$	2 marks

Worked solution Checked against the mark scheme

Top: $208.3 - 15.7 = 192.6$

Bottom: $5.694 + 1.8^2 = 5.694 + 3.24 = 8.934$

$192.6 \div 8.934 = 21.5580…$

$\sqrt{21.5580…} = 4.643069317$

Watch out: Not using brackets on the calculator, so only part of the numerator is divided.

Question 2

Scatter graphs and correlation3 marksGrade 3CalculatorOn both tiers

Describe the correlation on a scatter graph, draw a line of best fit and use it to estimate a value.

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(a)	Positive	1 mark
(b)	Line of best fit drawn	1 mark
(c)	About 28 m (26.5 to 29.5)	1 mark

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(a) Longer ships tend to be wider, so the correlation is positive.

(b) Draw one straight line through the middle of the crosses, with roughly as many points above as below.

(c) Go up from 194 m on the length axis to the line, then across: about 28 m.

Watch out: Joining the points dot to dot instead of drawing one straight line.

Question 3

Units and converting units3 marksGrade 5CalculatorOn both tiers

Decide which city's chocolate bar is better value using an exchange rate.

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Answer	Zurich (e.g. 200 g costs 4.00 francs in Zurich against 4.375 francs in London)	3 marks

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Change the London price to francs: $3.50 \times 1.25 = 4.375$ francs for 200 g.

Zurich price for 200 g: $7.20 \div 360 \times 200 = 4.00$ francs.

$4.00 < 4.375$ , so Zurich is better value.

Watch out: Dividing by 1.25 when changing pounds to francs.

Question 4

Venn diagrams and sets2 marksGrade 4CalculatorOn both tiers

Spot the two errors in a Venn diagram drawn from given sets.

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Answer	Remove the 17 from 'A only' so it is only in the overlap; put 1 in the region outside both circles	2 marks

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The universal set is 1, 3, 5, ..., 19.

17 is in both A and B, so it belongs only in the overlap; Tom also wrote it in A only.

1 is odd but in neither A nor B, so it must go inside the rectangle, outside both circles.

Watch out: Forgetting that 1 is an odd number between 0 and 20.

Question 5

Quadratic graphs4 marksGrade 4CalculatorOn both tiers

Complete a table of values for a quadratic and draw its graph.

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(a)	$5, -3, -4, 0, 5$ (for $x = -2, 0, 1, 3, 4$ )	2 marks
(b)	Smooth U-shaped curve through the points	2 marks

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(a) $x = -2$ : $4 + 4 - 3 = 5$ ; $x = 0$ : $-3$ ; $x = 1$ : $1 - 2 - 3 = -4$ ; $x = 3$ : $9 - 6 - 3 = 0$ ; $x = 4$ : $16 - 8 - 3 = 5$ .

(b) Plot $(-2, 5), (-1, 0), (0, -3), (1, -4), (2, -3), (3, 0), (4, 5)$ and join them with a smooth curve.

Watch out: Working out $(-2)^2$ as $-4$ , or joining the points with straight lines.

Question 6

Percentage change and reverse percentages4 marksGrade 5CalculatorOn both tiers

Compare two percentage increases to test a claim.

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Answer	Yes: $11.8\%$ is more than $7 \times 1.54\% = 10.8\%$	4 marks

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First class: $\frac{85 - 76}{76} \times 100 = 11.84\%$

Second class: $\frac{66 - 65}{65} \times 100 = 1.54\%$

$7 \times 1.54 = 10.77\%$ , and $11.84 > 10.77$ , so Filip is correct.

Watch out: Comparing the actual increases (9p and 1p) instead of percentages.

Question 7

Pythagoras and trigonometry5 marksGrade 5CalculatorOn both tiers

Use equal sides and a perimeter to find the side lengths of an isosceles triangle, then its area.

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Answer	240	5 marks

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$AC = BC$ : $2x + 11 = 4x - 4$ , so $2x = 15$ and $x = 7.5$ .

Sides: $AC = BC = 26$ , $AB = 20$ (perimeter $72$ ✓).

Height from $C$ to the midpoint of $AB$ : $\sqrt{26^2 - 10^2} = \sqrt{576} = 24$ .

Area $= \frac{1}{2} \times 20 \times 24 = 240\text{ cm}^2$

Watch out: Using 26 as the height instead of finding the perpendicular height.

Question 8

Standard form2 marksGrade 5CalculatorOn both tiers

Solve for an unknown in a standard form equation and give it in standard form.

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Answer	$3.125 \times 10^{7}$	2 marks

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$k = \frac{1.25 \times 10^{-12}}{4 \times 10^{-20}}$

$1.25 \div 4 = 0.3125$ and $10^{-12} \div 10^{-20} = 10^{8}$

$k = 0.3125 \times 10^{8} = 3.125 \times 10^{7}$

Watch out: Subtracting the powers the wrong way to get $10^{-32}$ or $10^{-8}$ .

Question 9

Angles in polygons3 marksGrade 5Calculator

Use the angles of two regular nine-sided polygons to prove a quadrilateral between them is not a square.

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Answer	Angle $ABC = 80°$ , not $90°$ , so ABCD is not a square	3 marks

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Exterior angle $= 360 \div 9 = 40°$ , so interior angle $= 140°$ .

Angles at point B add to 360°: $360 - 140 - 140 = 80°$ .

A square has all angles 90°, but angle $ABC = 80°$ , so ABCD is not a square.

Watch out: Using the angle sum of the nonagon (1260°) as if it were one angle.

Question 10

Simultaneous equations4 marksGrade 5Calculator

Solve a pair of linear simultaneous equations by elimination.

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Answer	$x = -2.5$ , $y = -6$	4 marks

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Multiply the first by 7 and the second by 5:

$28x - 35y = 140$ and $30x + 35y = -285$

Add: $58x = -145$ , so $x = -2.5$

$4(-2.5) - 5y = 20 \Rightarrow -5y = 30 \Rightarrow y = -6$

Watch out: Adding when the signs mean you should subtract, or a sign slip with the negatives.

Question 11

Transformations2 marksGrade 6Calculator

$Describe the single transformation (a fractional enlargement) that maps one triangle onto another on a grid.$

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Answer	Enlargement, scale factor $\frac{2}{3}$ , centre $(-4, -2)$	2 marks

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B's sides are $\frac{2}{3}$ of A's (height 6 → 4, width 3 → 2), so it is an enlargement with scale factor $\frac{2}{3}$ .

Join matching corners, e.g. $(5, 7)$ to $(2, 4)$ and $(5, 1)$ to $(2, 0)$ , and extend the lines: they meet at $(-4, -2)$ .

Watch out: Giving scale factor $\frac{3}{2}$ (mapping B to A) or omitting the centre.

Question 12

Speed, density and pressure5 marksGrade 6Calculator

Equate the masses of a steel cylinder and a brass cube using densities to find the cube's side length.

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Answer	$5.9$	5 marks

Worked solution Checked against the mark scheme

Cylinder volume $= \pi \times 3^2 \times 8 = 226.19…$ cm³

Mass $= 226.19… \times 7.86 = 1777.89…$ g

Cube volume $= 1777.89… \div 8.5 = 209.16…$ cm³

$h = \sqrt[3]{209.16…} = 5.936… \approx 5.9$

Watch out: Multiplying by the density to get the cube's volume instead of dividing.

Question 13

Direct and inverse proportion5 marksGrade 7Calculator

Judge whether a table shows direct proportion, then use a square-root proportion to find a value and sketch its graph.

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(a)	No: $y \div x$ is not constant (e.g. $4 \div 4 = 1$ but $8 \div 6 \neq 1$ ), and the graph would not pass through the origin	1 mark
(b(i))	$49$	3 marks
(b(ii))	Curve starting at the origin, increasing and getting less steep (shape of $w = k\sqrt{t}$ )	1 mark

Worked solution Checked against the mark scheme

(a) In direct proportion $y \div x$ is constant. Here $0 \div 2 = 0$ but $4 \div 4 = 1$ , so Nadia is wrong.

(b)(i) $w = k\sqrt{t}$ : $140 = k \times 8$ , so $k = 17.5$

$w = 17.5 \times \sqrt{7.84} = 17.5 \times 2.8 = 49$

(ii) Start at O and rise, curving so it gets less steep.

Watch out: Using $t$ instead of $\sqrt{t}$ and getting $k = 140 \div 64$ .

Question 14

Systematic listing and the product rule2 marksGrade 6Calculator

Count the total number of matches when every team in a league plays each other team several times.

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Answer	$180$	2 marks

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Number of pairs of teams: $\frac{10 \times 9}{2} = 45$

Each pair plays 4 times: $45 \times 4 = 180$

Watch out: Forgetting to halve and giving 360.

Question 15

Sequences and the nth term6 marksGrade 7Calculator

Find the nth term of a quadratic sequence, then use a recurrence rule with an unknown constant to find a later term.

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(a)	$5n^2 + 2n - 4$	3 marks
(b)	$1.2$	3 marks

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(a) Differences: 17, 27, 37, 47; second difference 10, so $5n^2$ .

Sequence − $5n^2$ : $-2, 0, 2, 4, 6$ , which is $2n - 4$ .

nth term $= 5n^2 + 2n - 4$

(b) $u_2 = 9k + k = 10k = 4$ , so $k = 0.4$

$u_3 = 0.4 \times 4 + 0.4 = 2$ , $u_4 = 0.4 \times 2 + 0.4 = 1.2$

Watch out: Using the second difference 10 as the coefficient of $n^2$ instead of halving it.

Question 16

Histograms4 marksGrade 7Calculator

Use frequency densities and given facts about the totals to fill in the missing bars of a histogram.

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Answer	Bar 5–10 at frequency density 2.8 and bar 20–30 at frequency density 1.4	4 marks

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Known frequencies: 0–5: 11; 10–20: $10 \times 1.5 = 15$ ; 30–50: $20 \times 0.3 = 6$

Left: $60 - 11 - 15 - 6 = 28$ , shared equally: 14 each.

5–10: $14 \div 5 = 2.8$ ; 20–30: $14 \div 10 = 1.4$

Watch out: Drawing bar heights as frequencies rather than frequency densities.

Question 17

Algebraic fractions3 marksGrade 7Calculator

$Combine three algebraic fractions into one and show the result is an integer over y.$

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Answer	$\frac{2}{y}$ (so $k = 2$ )	3 marks

Worked solution Checked against the mark scheme

Common denominator $10xy$ :

$\frac{6x - y}{10xy} + \frac{5y}{10xy} - \frac{2(2y - 7x)}{10xy}$

$= \frac{6x - y + 5y - 4y + 14x}{10xy} = \frac{20x}{10xy} = \frac{2}{y}$

Watch out: Not multiplying the minus sign through the bracket, getting $-14x$ instead of $+14x$ .

Question 18

Congruent triangles and geometric proof3 marksGrade 8Calculator

Prove that the line from the centre of a circle to a point equidistant from two others on the circle bisects the angle there.

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Answer	Triangles ABO and CBO are congruent (SSS), so angle ABO = angle CBO	3 marks

Worked solution Checked against the mark scheme

$BA = BC$ (given).

$OA = OC$ (radii of the circle).

$OB$ is common to both triangles.

So triangles $ABO$ and $CBO$ are congruent (SSS), and angle $ABO$ = angle $CBO$ : $OB$ bisects angle $ABC$ .

Watch out: Stating sides are equal without giving reasons, or assuming what must be proved.

Question 19

Error intervals and bounds5 marksGrade 8Calculator

Use upper and lower bounds of three rounded values to find a formula's value to a justified degree of accuracy.

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Answer	$60$ , because both bounds (58.4… and 60.7…) round to 60	5 marks

Worked solution Checked against the mark scheme

Bounds: $w$ : 432.5 to 437.5; $a$ : 9.75 to 9.85; $c$ : 2.45 to 2.55

Lower bound of $T = \frac{432.5}{9.85 - 2.45} = 58.44…$

Upper bound of $T = \frac{437.5}{9.75 - 2.55} = 60.76…$

Both round to 60 (1 sf), so $T = 60$ .

Watch out: Using the upper bound of c in the lower bound of T, forgetting that subtracting reverses which bound is needed.

Question 20

Pythagoras and trigonometry3 marksGrade 8Calculator

Use trigonometry in the vertical cross-section of a pyramid on an equilateral base to find the base's side length.

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Answer	$10.8$ cm	3 marks

Worked solution Checked against the mark scheme

In triangle $VFC$ : $FC = \frac{8}{\tan 52°} = 6.250…$

$MF : FC = 1 : 2$ , so $MC = 1.5 \times 6.250… = 9.375…$

$MC$ is the height of the equilateral triangle: side $= \frac{9.375…}{\sin 60°} = 10.8$ cm

Watch out: Taking FC as the full height MC of the triangle.

Question 21

Coordinates and straight-line graphs5 marksGrade 9Calculator

Find the possible values of k for a point R so that the angle at R, between lines to two given points, is a right angle.

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Answer	$k = -6$ or $k = 3.5$	5 marks

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Gradient $PR = \frac{k - 2}{k + 4}$ , gradient $QR = \frac{k + 9}{k - 6}$

Perpendicular: $\frac{k-2}{k+4} \times \frac{k+9}{k-6} = -1$

$(k-2)(k+9) = -(k+4)(k-6)$

$k^2 + 7k - 18 = -k^2 + 2k + 24 \Rightarrow 2k^2 + 5k - 42 = 0$

$(k + 6)(2k - 7) = 0$ , so $k = -6$ or $k = 3.5$

Watch out: Setting the gradients equal (parallel) instead of making their product −1.

Question 22

Tree diagrams and combined events5 marksGrade 8Calculator

$Given the fraction of red counters and the chance two picked without replacement match in colour, find how many yellow counters there are.$

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Answer	$26$	5 marks

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Let there be $5x$ counters: $3x$ red, $2x$ yellow.

$\frac{3x}{5x} \times \frac{3x-1}{5x-1} + \frac{2x}{5x} \times \frac{2x-1}{5x-1} = \frac{41}{80}$

$\frac{13x - 5}{5(5x-1)} = \frac{41}{80} \Rightarrow 1040x - 400 = 1025x - 205$

$15x = 195$ , $x = 13$ , so yellow $= 2 \times 13 = 26$

Watch out: Treating the picks as with replacement, so the equation has no solution.

November 2024 Paper 2 Higher

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