Edexcel GCSE Maths November 2023 Paper 2 Higher answers

Question 1

Expanding brackets and simplifying6 marksGrade 4Calculator

Expand and simplify a sum of two brackets, factorise a quadratic expression fully, and rearrange a linear formula.

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(a)	$13y - 1$	2 marks
(b)	$3x(2x + 5)$	2 marks
(c)	$g = \frac{f - 11}{3}$	2 marks

Worked solution Checked against the mark scheme

(a) $6y - 15 + 7y + 14 = 13y - 1$

(b) The highest common factor of $6x^2$ and $15x$ is $3x$ , so $6x^2 + 15x = 3x(2x + 5)$ .

(c) $f - 11 = 3g$ , so $g = \frac{f - 11}{3}$ .

Watch out: In (b), taking out only 3 or only x, so the answer is not fully factorised.

Question 2

Percentage change and reverse percentages4 marksGrade 5CalculatorOn both tiers

Work out the percentage profit from ticket sales after event costs.

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Answer	35%	4 marks

Worked solution Checked against the mark scheme

Income: $54 \times 7.50 = £405$

Costs: $100 + 120 + 80 = £300$

Profit: $405 - 300 = £105$

Percentage profit $= \frac{105}{300} \times 100 = 35\%$

Watch out: Dividing the profit by the income (405) instead of the costs.

Question 3

Compound interest, growth and decay2 marksGrade 5CalculatorOn both tiers

Find the value of an investment after two years of compound interest.

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Answer	£4811.20	2 marks

Worked solution Checked against the mark scheme

Multiplier for 3.4% interest: $1.034$

$4500 \times 1.034^2 = 4811.202$ , so £4811.20

Watch out: Using simple interest to get £4806.

Question 4

Forming and solving linear equations3 marksGrade 4CalculatorOn both tiers

Solve a linear equation with the unknown on both sides.

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Answer	$x = 11$	3 marks

Worked solution Checked against the mark scheme

Add $x$ to both sides: $6x - 14 = 52$

Add 14 to both sides: $6x = 66$

$x = 11$

Watch out: Subtracting x instead of adding it, giving 4x.

Question 5

Sharing in a ratio4 marksGrade 6CalculatorOn both tiers

Use a share in a three-way ratio to find the total, then compare with a new ratio to see how much one person passed to another.

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Answer	£21	4 marks

Worked solution Checked against the mark scheme

Debbie has 4 parts $= 120$ , so 1 part $= 30$ and the total is $9 \times 30 = 270$ .

At first Errol has $2 \times 30 = 60$ .

New ratio has $2 + 5 + 3 = 10$ parts, so 1 part $= 27$ and Errol now has $3 \times 27 = 81$ .

Chris gave Errol $81 - 60 = £21$

Watch out: Assuming Debbie still has £120 after the change.

Question 6

Bearings and scale drawings2 marksGrade 4CalculatorOn both tiers

Find a back bearing from a given bearing.

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Answer	327°	2 marks

Worked solution Checked against the mark scheme

The return bearing differs by $180°$ .

$147 + 180 = 327°$

Watch out: Subtracting from 360 to give 213°.

Question 7

Volume and surface area4 marksGrade 5CalculatorOn both tiers

Find the volume of a cylinder, convert it to litres and work out the filling time at a given rate.

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Answer	65 minutes	4 marks

Worked solution Checked against the mark scheme

Volume $= \pi \times 15^2 \times 43 = 30\,394.9\ldots\text{ cm}^3$

$1000\text{ cm}^3 = 1$ litre, so this is $30.3949\ldots$ litres.

Time $= 30.3949 \div 0.47 = 64.67\ldots$ , so 65 minutes

Watch out: Forgetting to convert cm³ to litres.

Question 8

Error intervals and bounds2 marksGrade 5Calculator

Write the error interval for a number rounded to 2 significant figures.

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Answer	$1.85 \le x < 1.95$	2 marks

Worked solution Checked against the mark scheme

Rounding to 2 significant figures here means to the nearest 0.1, so half of that is 0.05.

Lower bound: $1.9 - 0.05 = 1.85$

Upper bound: $1.9 + 0.05 = 1.95$

$1.85 \le x < 1.95$

Watch out: Using 1.94 or 1.949 as the upper bound with ≤.

Question 9

Expanding brackets and simplifying3 marksGrade 6Calculator

Expand the product of three linear brackets and simplify.

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Answer	$x^3 + 8x^2 + x - 42$	3 marks

Worked solution Checked against the mark scheme

$(x + 7)(x - 2) = x^2 + 5x - 14$

$(x^2 + 5x - 14)(x + 3) = x^3 + 3x^2 + 5x^2 + 15x - 14x - 42$

$= x^3 + 8x^2 + x - 42$

Watch out: Sign errors with the −2 and −14 terms.

Question 10

Tree diagrams and combined events5 marksGrade 5Calculator

Complete a probability tree for passing two tests where the second depends on the first, then find the probability of passing at least one.

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(a)	First test: fail 0.13. After a pass: pass 0.94, fail 0.06. After a fail: pass 0.73, fail 0.27	2 marks
(b)	$0.9649$	3 marks

Worked solution Checked against the mark scheme

(a) Each pair of branches adds up to 1: $1 - 0.87 = 0.13$ , $1 - 0.94 = 0.06$ , $1 - 0.73 = 0.27$ .

(b) The only way to pass neither is fail, fail: $0.13 \times 0.27 = 0.0351$

$P(\text{at least one}) = 1 - 0.0351 = 0.9649$

Watch out: Using 0.94 and 0.73 on the wrong branches.

Question 11

Direct and inverse proportion2 marksGrade 7Calculator

Match four proportionality relationships to sketches of their graphs.

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Answer	$y \propto x^2$ : D; $y \propto x$ : A; $y \propto \frac{1}{x}$ : B; $y \propto \sqrt{x}$ : C	2 marks

Worked solution Checked against the mark scheme

$y \propto x$ is a straight line through the origin: graph A.

$y \propto x^2$ curves upwards more and more steeply: graph D.

$y \propto \frac{1}{x}$ falls as $x$ grows and never reaches the axes: graph B.

$y \propto \sqrt{x}$ rises steeply at first and then flattens: graph C.

Watch out: Mixing up the square and square-root graphs (C and D).

Question 12

Pythagoras and trigonometry3 marksGrade 6Calculator

Find two midpoints from given coordinates and the distance between them.

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Answer	$6.4$	3 marks

Worked solution Checked against the mark scheme

$M = \left(\frac{7 + (-3)}{2}, \frac{13 + 21}{2}\right) = (2, 17)$

$N = \left(\frac{-3 + 15}{2}, \frac{21 + 23}{2}\right) = (6, 22)$

$MN = \sqrt{4^2 + 5^2} = \sqrt{41} = 6.403... \approx 6.4$

Watch out: Subtracting coordinates instead of adding them when finding a midpoint.

Question 13

Fractions, decimals and recurring decimals3 marksGrade 7Calculator

$Prove algebraically that a recurring decimal equals a given fraction.$

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Answer	Proof that $0.0\dot{7}2\dot{3} = \frac{241}{3330}$	3 marks

Worked solution Checked against the mark scheme

Let $x = 0.0723723...$

Then $10x = 0.723723...$ and $10\,000x = 723.723723...$

Subtract: $9990x = 723$

$x = \frac{723}{9990} = \frac{241}{3330}$ (dividing top and bottom by 3)

Watch out: Subtracting multiples whose recurring parts do not line up, so the decimals do not cancel.

Question 14

Direct and inverse proportion5 marksGrade 8Calculator

Combine a direct square proportion with an inverse proportion to find y for a given w.

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Answer	$y = 0.48$	5 marks

Worked solution Checked against the mark scheme

$y = kx^2$ : $3 = k \times 0.5^2$ , so $k = 12$ .

$x = \frac{j}{w}$ : $2 = \frac{j}{0.2}$ , so $j = 0.4$ .

When $w = 2$ : $x = \frac{0.4}{2} = 0.2$

$y = 12 \times 0.2^2 = 0.48$

Watch out: Treating the inverse proportion as x = jw.

Question 15

Histograms6 marksGrade 7Calculator

Complete a frequency table and histogram from each other, then estimate a median from a second histogram.

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(a)	$3$	1 mark
(b)	Bars of frequency density 4.5 for $19 < t \le 21$ and 1.4 for $21 < t \le 26$	2 marks
(c)	$18.5$ seconds	3 marks

Worked solution Checked against the mark scheme

(a) The 12–16 bar has frequency 10 and width 4, so its frequency density is 2.5. The 10–12 bar is 1.5 high, so its frequency is $1.5 \times 2 = 3$ .

(b) $9 \div 2 = 4.5$ and $7 \div 5 = 1.4$ , so draw bars of those heights.

(c) Frequencies: $4, 8, 16, 13, 2$ (total 43). The median is the 22nd value. After 16 s there are 12 students, so we need 10 more of the 16 in the 16–20 class: $16 + \frac{10}{16} \times 4 = 18.5$ seconds.

Watch out: Using bar heights as frequencies instead of frequency density × class width.

Question 16

Sine rule, cosine rule and area of a triangle4 marksGrade 7Calculator

In a trapezium, use the sine rule then the sine area formula to find the area of a triangle.

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Answer	$56.0$ cm²	4 marks

Worked solution Checked against the mark scheme

In triangle $ABD$ : $\frac{BD}{\sin 112°} = \frac{7.9}{\sin 42°}$ , so $BD = 10.946...$ cm.

AD is parallel to BC, so angle $DBC = 42°$ (alternate angles).

Area of $BCD = \frac{1}{2} \times 10.946... \times 15.3 \times \sin 42° = 56.03... \approx 56.0$ cm²

Watch out: Not spotting that angle DBC equals 42° by alternate angles.

Question 17

Iteration6 marksGrade 7Calculator

Show a cubic has a root between 1 and 2, rearrange it into an iterative form, and use the iteration twice.

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(a)	$f(1) = -3$ and $f(2) = 6$ ; sign change, so a root lies between 1 and 2	2 marks
(b)	$x^3 + 2x = 6 \Rightarrow x(x^2 + 2) = 6 \Rightarrow x = \frac{6}{x^2 + 2}$	1 mark
(c)	$1.4496$	3 marks

Worked solution Checked against the mark scheme

(a) $1^3 + 2(1) - 6 = -3$ and $2^3 + 2(2) - 6 = 6$ . The sign changes, so there is a solution between 1 and 2.

(b) $x^3 + 2x = 6$ , so $x(x^2 + 2) = 6$ , giving $x = \frac{6}{x^2 + 2}$ .

(c) $x_1 = \frac{6}{1.45^2 + 2} = 1.46252...$

$x_2 = \frac{6}{1.46252...^2 + 2} = 1.44963... \approx 1.4496$

Watch out: In (a), working out the values but not stating that the sign changes.

Question 18

Vectors4 marksGrade 9Calculator

Use vectors to prove that joining the midpoints of a quadrilateral's sides makes a parallelogram.

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Answer	Proof: $\overrightarrow{EH} = \overrightarrow{FG} = \mathbf{a} - \mathbf{b}$ , so EFGH is a parallelogram	4 marks

Worked solution Checked against the mark scheme

$\overrightarrow{EH} = -\mathbf{b} + \mathbf{a}$

$\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AD} + \overrightarrow{DC} = -2\mathbf{b} + 2\mathbf{a} + 2\mathbf{c}$

$\overrightarrow{FG} = \frac{1}{2}\overrightarrow{BC} + \overrightarrow{CG} = (-\mathbf{b} + \mathbf{a} + \mathbf{c}) - \mathbf{c} = \mathbf{a} - \mathbf{b}$

$\overrightarrow{EH} = \overrightarrow{FG}$ , so EH and FG are parallel and equal in length, and EFGH is a parallelogram.

Watch out: Forgetting that the full side vectors are twice a, b and c.

Question 19

Functions4 marksGrade 7Calculator

Evaluate a composite function at a value and find the inverse of a linear function.

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(a)	$17$	2 marks
(b)	$\frac{x + 1}{2}$	2 marks

Worked solution Checked against the mark scheme

(a) $f(-3) = (2(-3) + 3)^2 = (-3)^2 = 9$ , then $g(9) = 2 \times 9 - 1 = 17$ .

(b) Let $y = 2x - 1$ . Then $y + 1 = 2x$ , so $x = \frac{y + 1}{2}$ .

$g^{-1}(x) = \frac{x + 1}{2}$

Watch out: In (a), working out fg(−3) instead of gf(−3).

Question 20

Algebraic fractions4 marksGrade 9Calculator

$Add an algebraic fraction to the result of dividing by another algebraic fraction, giving one simplified fraction.$

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Answer	$\frac{6x + 37}{6x - 42}$	4 marks

Worked solution Checked against the mark scheme

$2x^2 - 6x - 56 = 2(x + 4)(x - 7)$ and $3x - 21 = 3(x - 7)$ .

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

$\frac{14}{3(x - 7)} + \frac{2x + 3}{2(x - 7)} = \frac{28 + 3(2x + 3)}{6(x - 7)} = \frac{6x + 37}{6x - 42}$

Watch out: Multiplying by the divisor instead of its reciprocal.

Question 21

Conditional probability4 marksGrade 8Calculator

Use a conditional probability from a Venn diagram to form an equation and find the total number of items.

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Answer	$98$	4 marks

Worked solution Checked against the mark scheme

Gold medals: $x + (5x - 3) = 6x - 3$

$\frac{x}{6x - 3} = \frac{2}{11}$ , so $11x = 12x - 6$ and $x = 6$ .

Total: $(7 \times 6 - 11) + 6 + (5 \times 6 - 3) + 34 = 31 + 6 + 27 + 34 = 98$

Watch out: Dividing by the total number of medals instead of the number of gold medals.

November 2023 Paper 2 Higher

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