Edexcel GCSE Maths November 2018 Paper 2 Higher answers

Question 1

Venn diagrams and sets6 marksGrade 4CalculatorOn both tiers

Place set elements in a three-set Venn diagram, then find a probability for an intersection.

Show answer

(a)	A only: 2, 10, 14; A∩B∩C: 8; B only: 6; B∩C only: 20; C only: 18, 22; outside: 4, 12, 16, 24	4 marks
(b)	$\frac{1}{12}$	2 marks

Worked solution Checked against the mark scheme

(a) $\mathcal{E} = \{2, 4, 6, ..., 24\}$ , 12 numbers.

8 is in all three sets. 20 is in B and C only. 6 is B only; 2, 10, 14 are A only; 18, 22 are C only.

4, 12, 16, 24 go outside the circles.

(b) $A \cap B = \{8\}$ , so P $= \frac{1}{12}$

Watch out: Forgetting to place the numbers that are in no set.

Question 2

Scatter graphs and correlation2 marksGrade 3CalculatorOn both tiers

Identify two faults in a student's scatter graph and line of best fit.

Show answer

Answer	The line of best fit is wrong (it does not follow the points); the height scale is not evenly spaced (150 is missing)	2 marks

Worked solution Checked against the mark scheme

Check the line: it should pass through the middle of the crosses with roughly equal numbers either side, which Sean's does not.

Check the axes: the height axis jumps from 140 to 160 in the same space as 160 to 170, so the scale is uneven.

Watch out: Saying there is no title or the axis should start at 0.

Question 3

Angle facts and parallel lines4 marksGrade 5CalculatorOn both tiers

Find an angle in a triangle between parallel lines, giving reasons.

Show answer

Answer	$60°$	4 marks

Worked solution Checked against the mark scheme

Angle $ABE = 180 - 110 = 70°$ (co-interior angles add up to 180°)

Angle $EBG = 180 - 70 - 35 = 75°$ (angles on a straight line add up to 180°)

Angle $BEG = 180 - 110 - 25 = 45°$ (angles on a straight line)

$x = 180 - 75 - 45 = 60°$ (angles in a triangle add up to 180°)

Watch out: Giving no reasons, or naming corresponding angles when they are alternate.

Question 4

Compound interest, growth and decay5 marksGrade 5CalculatorOn both tiers

Compare compound interest on two accounts over 3 years, then decide if a rate change in year 3 alters the result.

Show answer

(a)	Ben: £173.95 interest against Ali's £153.78	4 marks
(b)	No: Ben now gets about £182.52, still more than Ali	1 mark

Worked solution Checked against the mark scheme

(a) Ali: $2000 \times 1.025^3 = £2153.78$ , interest $£153.78$

Ben: $1600 \times 1.035^3 = £1773.95$ , interest $£173.95$

Ben gets more.

(b) Ben: $1600 \times 1.035^2 \times 1.04 = £1782.52$ , interest $£182.52$ . Still Ben, so no change.

Watch out: Using simple interest, or comparing final totals instead of interest.

Question 5

Area and perimeter5 marksGrade 5CalculatorOn both tiers

Find a trapezium floor's area, the paint needed and its cost, and decide if a budget is enough.

Show answer

Answer	No: he needs 10 tins costing £169.90	5 marks

Worked solution Checked against the mark scheme

Area $= \frac{1}{2}(10 + 16) \times 7 = 91$ m²

Paint $= 91 \div 2 = 45.5$ litres

Tins $= 45.5 \div 5 = 9.1$ , so 10 tins

Cost $= 10 \times 16.99 = £169.90 > £160$ , so No.

Watch out: Rounding 9.1 down to 9 tins.

Question 6

Coordinates and straight-line graphs3 marksGrade 5CalculatorOn both tiers

Find an unknown x-coordinate from two points and the gradient of the line joining them.

Show answer

Answer	7	3 marks

Worked solution Checked against the mark scheme

Gradient $= \frac{\text{change in } y}{\text{change in } x}$

Change in $y = 15 - 9 = 6$ , so change in $x = 6 \div 3 = 2$

$d = 5 + 2 = 7$

Watch out: Using change in x over change in y, giving 23.

Question 7

Standard form3 marksGrade 5Calculator

Write a small decimal in standard form, then evaluate a calculation with numbers in standard form to 3 significant figures.

Show answer

(a)	$8.623 \times 10^{-5}$	1 mark
(b)	$7.44 \times 10^{6}$	2 marks

Worked solution Checked against the mark scheme

(a) Move the decimal point 5 places: $8.623 \times 10^{-5}$ .

(b) $\frac{3200 + 0.051}{0.00043} = \frac{3200.051}{0.00043} = 7\,441\,979.07...$

To 3 s.f. in standard form: $7.44 \times 10^6$ .

Watch out: Only dividing the second term of the numerator by the denominator (no brackets on the calculator).

Question 8

Transformations3 marksGrade 6Calculator

Carry out two reflections on a triangle and describe the single transformation from the final image back to the original.

Show answer

Answer	Rotation $90°$ anticlockwise about $(-1, 1)$	3 marks

Worked solution Checked against the mark scheme

P has vertices $(-1, 3), (3, 3), (3, 5)$ .

Reflect in $y = -x$ : Q is $(-3, 1), (-3, -3), (-5, -3)$ .

Reflect in $x = -1$ : R is $(1, 1), (1, -3), (3, -3)$ .

Turning R through $90°$ anticlockwise about $(-1, 1)$ lands it exactly on P.

Watch out: Giving two transformations or leaving out the centre of rotation.

Question 9

Error intervals and bounds2 marksGrade 6Calculator

Write the error interval for a number that gives 7 when truncated to one digit.

Show answer

Answer	$7 \le N < 8$	2 marks

Worked solution Checked against the mark scheme

Truncating chops off the digits after the first one, so N can be 7 itself or anything up to (but not including) 8.

$7 \le N < 8$

Watch out: Treating it as rounding and writing $6.5 \le N < 7.5$ .

Question 10

Percentage change and reverse percentages5 marksGrade 6Calculator

Find the cost of mixing paint from tins in a given ratio and work out the percentage profit on each tin sold.

Show answer

Answer	35%	5 marks

Worked solution Checked against the mark scheme

Yellow $= \frac{2}{5} \times 50 = 20$ litres = 4 tins; blue $= 30$ litres = 3 tins.

Cost $= 4 \times 26 + 3 \times 48 = 104 + 144 = £248$ for 50 litres.

Cost per 10 litre tin $= 248 \div 5 = £49.60$ .

Profit $= 66.96 - 49.60 = £17.36$

$\frac{17.36}{49.60} \times 100 = 35\%$

Watch out: Dividing the profit by the selling price (66.96) instead of the cost.

Question 11

Systematic listing and the product rule3 marksGrade 6Calculator

Show the total number of ways to choose a meal from three menu combinations using the product rule.

Show answer

Answer	$135 + 120 + 1080 = 1335$	3 marks

Worked solution Checked against the mark scheme

Starter and main: $9 \times 15 = 135$

Main and dessert: $15 \times 8 = 120$

All three: $9 \times 15 \times 8 = 1080$

Total: $135 + 120 + 1080 = 1335$

Watch out: Adding the numbers of dishes instead of multiplying.

Question 12

Algebraic fractions6 marksGrade 8Calculator

$Simplify a product of algebraic fractions by factorising, and combine three algebraic fractions into one.$

Show answer

(a)	$\frac{4x - 6}{3x - 9}$	3 marks
(b)	$\frac{8 - x}{x(x+1)(x-2)}$	3 marks

Worked solution Checked against the mark scheme

(a) $\frac{(2x-3)(2x+3)}{3(2x+3)} \times \frac{2x}{x(x-3)} = \frac{2(2x-3)}{3(x-3)} = \frac{4x-6}{3x-9}$

(b) Common denominator $x(x+1)(x-2)$ .

Numerator: $3x(x-2) + x(x+1) - 4(x+1)(x-2)$
$= 3x^2 - 6x + x^2 + x - 4x^2 + 4x + 8 = 8 - x$

Answer: $\frac{8 - x}{x(x+1)(x-2)}$

Watch out: Sign errors when subtracting $4(x+1)(x-2)$ .

Question 13

Circles: parts, circumference and area3 marksGrade 6Calculator

Use a circle's circumference to find its diameter, then find the area of an equilateral triangle built on that diameter.

Show answer

Answer	$84.9$ cm²	3 marks

Worked solution Checked against the mark scheme

$\pi d = 44$ , so $d = \frac{44}{\pi} = 14.0056...$ cm

Area $= \frac{1}{2} d^2 \sin 60° = \frac{1}{2} \times 14.0056^2 \times \sin 60°$

$= 84.9$ cm² (3 s.f.)

Watch out: Using $2\pi r$ but then taking the triangle side as the radius.

Question 14

Cubic, reciprocal, exponential and trig graphs2 marksGrade 7Calculator

Sketch an exponential curve and mark where it meets the axes.

Show answer

Answer	Increasing exponential curve, above the x-axis for negative x, crossing the y-axis at $(0, 1)$	2 marks

Worked solution Checked against the mark scheme

$2^x$ is always positive, so the curve never touches the x-axis.

When $x = 0$ , $y = 2^0 = 1$ , so it crosses the y-axis at $(0, 1)$ .

It gets close to the x-axis on the left and rises ever more steeply on the right.

Watch out: Drawing the curve crossing the x-axis or passing through the origin.

Question 15

Equation of a circle1 markGrade 6Calculator

Find the radius of a circle from its equation.

Show answer

Answer	$6.5$	1 mark

Worked solution Checked against the mark scheme

$x^2 + y^2 = r^2$ , so $r^2 = 42.25$

$r = \sqrt{42.25} = 6.5$

Watch out: Halving 42.25 instead of square rooting.

Question 16

Tree diagrams and combined events4 marksGrade 6Calculator

Find the probability two people with replacement pick different colours, and use a probability to find the number of blue counters.

Show answer

(a)	$0.455$	2 marks
(b)	$42$	2 marks

Worked solution Checked against the mark scheme

(a) $P(\text{blue}) = 1 - 0.65 = 0.35$

$P(\text{different}) = 0.65 \times 0.35 + 0.35 \times 0.65 = 0.455$

(b) Total $= 78 \div 0.65 = 120$ counters, so blue $= 120 - 78 = 42$ .

Watch out: Only counting red then blue, giving 0.2275.

Question 17

Ratio5 marksGrade 8Calculator

Form and solve simultaneous equations from two ratio conditions to find the ratio of two numbers.

Show answer

Answer	$4 : 1$	5 marks

Worked solution Checked against the mark scheme

$\frac{p-5}{q-5} = 5$ gives $p = 5q - 20$

$\frac{p+20}{q+20} = \frac{5}{2}$ gives $2p + 40 = 5q + 100$

Substitute: $10q - 40 + 40 = 5q + 100$ , so $q = 20$ and $p = 80$ .

$p : q = 80 : 20 = 4 : 1$

Watch out: Giving the ratio the wrong way round as $1 : 4$ .

Question 18

Coordinates and straight-line graphs4 marksGrade 7Calculator

Find where a line through two points meets a line through the origin with a given gradient.

Show answer

Answer	$\left(-\frac{16}{5}, \frac{48}{5}\right)$	4 marks

Worked solution Checked against the mark scheme

Gradient of L1 $= \frac{2 - 6}{12 - 4} = -\frac{1}{2}$

$6 = -\frac{1}{2}(4) + c$ , so $c = 8$ : L1 is $y = -\frac{1}{2}x + 8$

L2 is $y = -3x$ .

$-\frac{1}{2}x + 8 = -3x \Rightarrow 2.5x = -8 \Rightarrow x = -3.2$ , $y = 9.6$

$P = (-3.2, 9.6)$

Watch out: Getting the gradient sign wrong for L1.

Question 19

Inequalities5 marksGrade 8Calculator

Solve a double inequality involving a squared term, giving both ranges of solutions.

Show answer

Answer	$9 < m < 11$ or $-11 < m < -9$	5 marks

Worked solution Checked against the mark scheme

Multiply by 4: $88 < m^2 + 7 < 128$

Subtract 7: $81 < m^2 < 121$

So $9 < |m| < 11$ .

$9 < m < 11$ or $-11 < m < -9$

Watch out: Forgetting the negative range.

Question 20

Volume and surface area5 marksGrade 8Calculator

Find the average density of a solid made from a frustum and a hemisphere with different densities.

Show answer

Answer	$3.75$ g/cm³	5 marks

Worked solution Checked against the mark scheme

Frustum $= \frac{1}{3}\pi(3.6)^2(6.4) - \frac{1}{3}\pi(1.8)^2(3.2) = 76.00$ cm³

Hemisphere $= \frac{2}{3}\pi(3.6)^3 = 97.72$ cm³

Mass $= 76.00 \times 2.4 + 97.72 \times 4.8 = 182.40 + 469.04 = 651.44$ g

Density $= \frac{651.44}{173.72} = 3.75$ g/cm³

Watch out: Averaging the two densities (3.6) instead of using total mass over total volume.

Question 21

Circle theorems4 marksGrade 9Calculator

Prove two angles in a triangle are equal using a cyclic quadrilateral and an isosceles triangle formed by radii.

Show answer

Answer	Proof that angle $CAB$ = angle $ABC$	4 marks

Worked solution Checked against the mark scheme

Let angle $CAB = x$ .

$AORC$ is cyclic, so angle $CRO = 180 - x$ (opposite angles of a cyclic quadrilateral add up to $180°$ ).

Angle $ORB = 180 - (180 - x) = x$ (angles on a straight line add up to $180°$ ).

$OR = OB$ (both radii), so triangle $ORB$ is isosceles and angle $OBR = x$ (base angles are equal).

So angle $ABC = x =$ angle $CAB$ .

Watch out: Stating results without the reasons, or using the wrong circle for the cyclic quadrilateral.

November 2018 Paper 2 Higher

Finished the paper?