Higher Tier - Grades 4-9

IGCSE Maths Higher Paper 4

Covers Number and Calculation, Fractions, Decimals and Percentages, Indices, Surds and Standard Form and 9 more.

13 questions - 60 marks - calculator allowed

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Questions

Question 1 [3 marks]

Linear Equations and Inequalities

Expand and simplify 3(2x - 5) + 4(x + 6).

Question 2 [3 marks]

Indices, Surds and Standard Form

A red blood cell has diameter 0.0000075 m. Write this number in standard form.

Question 3 [3 marks]

Linear Equations and Inequalities

Solve the inequality 15 - 3x > 6.

Question 4 [4 marks]

Fractions, Decimals and Percentages

Ryan mixes 3.6 litres of orange squash.

He pours 40% of it into a jug for a party, then shares 1/4 of what remains between two bottles.

How many millilitres of squash are left in the original container?

Question 5 [5 marks]

Ratio and Proportion

y is directly proportional to the square root of x.

When x = 16, y = 20.

Find y when x = 81.

Question 6 [5 marks]

Number and Calculation

Three warning lights flash every 18 seconds, 24 seconds and 30 seconds respectively.

All three lights flash together at 9:00 am. Find the next time, in seconds after 9:00 am, that all three lights flash together at the same moment.

Question 7 [5 marks]

Statistics and Probability

A biased coin has P(heads) = 0.65. A spinner has three equal sections labelled 1, 2 and 3.

The coin is tossed once and the spinner is spun once.

Find the probability of getting tails on the coin and an odd number on the spinner.

Question 8 [4 marks]

Mensuration and Vectors

OABC is a parallelogram, where O is the origin.

Vector OA = a and vector OC = c. M is the midpoint of AB.

Find vector OM in terms of a and c, giving your answer in its simplest form.

Question 9 [5 marks]

Sequences and Graphs

The first four terms of a quadratic sequence are 4, 11, 22, 37.

Find an expression for the nth term.

Question 10 [5 marks]

Quadratics and Simultaneous Equations

A rectangle has length (x + 9) cm and width (x - 3) cm. The area of the rectangle is 133 cm^2.

Find the value of x, given that x is positive.

Question 11 [6 marks]

Angles, Polygons and Circle Theorems

Two similar triangles, triangle ABC and triangle PQR, have area 18 cm^2 and 50 cm^2 respectively.

The length of side AB is 6 cm.

Find the length of the corresponding side PQ.

Question 12 [6 marks]

Pythagoras and Trigonometry

A square-based pyramid has a base of side 10 cm and a vertical height of 12 cm,

with its apex directly above the centre of the base.

Find the length of a sloping edge, and find the angle between a sloping edge and the base, to 1 decimal place.

Question 13 [6 marks]

Functions and Rates of Change

The displacement, s metres, of a particle at time t seconds is given by s = t^3 - 6t^2 + 9t.

Find the particle's velocity when t = 4, and find the value(s) of t at which the particle is momentarily at rest.

Model solutions

Mark scheme for Question 1 [3 marks]
Question 1[3 marks]
Answer or working	Marks
expanding to 6x - 15 + 4x + 24	M1
collecting like x terms and constants	M1
10x + 9	A1

Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or working	Marks
identifying the first significant figures as 7.5	M1
counting the place value shift to give power -6	M1
7.5 x 10^-6	A1
Final answer: 7.5 x 10^-6 m

Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or working	Marks
subtracting 15 from both sides to get -3x > -9	M1
dividing both sides by -3 and reversing the inequality	M1
x < 3	A1

Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or working	Marks
60% of 3.6 = 2.16 litres remaining after the jug is filled	M1
finding 1/4 of 2.16	M1
subtracting to leave 3/4 of 2.16	M1
1620 ml	A1

Mark scheme for Question 5 [5 marks]
Question 5[5 marks]
Answer or working	Marks
y = k sqrt(x)	M1
20 = k x 4, since sqrt(16) = 4	M1
k = 5	A1
substituting x = 81 into y = 5 sqrt(x)	M1
45	A1

Mark scheme for Question 6 [5 marks]
Question 6[5 marks]
Answer or working	Marks
expressing each number as a product of primes	M1
18 = 2 x 3^2, 24 = 2^3 x 3, 30 = 2 x 3 x 5	M1
identifying the highest power of each prime: 2^3, 3^2 and 5	M1
multiplying to get 8 x 9 x 5	M1
360 seconds	A1

Mark scheme for Question 7 [5 marks]
Question 7[5 marks]
Answer or working	Marks
P(tails) = 1 - 0.65 = 0.35	M1
P(odd number) = 2/3, since 1 and 3 are odd out of 3 equally likely sections	M1
using the AND rule: multiplying the two probabilities	M1
0.35 x 2/3	M1
7/30, or 0.233 (3 s.f.)	A1
Final answer: 7/30

Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or working	Marks
recognising vector AB = OC = c, since OABC is a parallelogram	M1
OM = OA + (1/2)AB	M1
substituting OA = a and AB = c	M1
OM = a + (1/2)c	A1

Mark scheme for Question 9 [5 marks]
Question 9[5 marks]
Answer or working	Marks
first differences 7, 11, 15	M1
second difference 4	M1
starting with 2n^2	M1
comparing with 2, 8, 18, 32 to get adjustment 2, 3, 4, 5	M1
2n^2 + n + 1	A1

Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or working	Marks
forming the equation (x + 9)(x - 3) = 133	M1
expanding to x^2 + 6x - 27 = 133	M1
rearranging to x^2 + 6x - 160 = 0	M1
factorising or using the quadratic formula to solve	M1
x = 10 (rejecting x = -16 since x must be positive)	A1
Final answer: x = 10

Mark scheme for Question 11 [6 marks]
Question 11[6 marks]
Answer or working	Marks
finding the area scale factor 50/18 = 25/9	M1
taking the square root to find the length scale factor	M1
length scale factor = 5/3	A1
PQ = AB x 5/3	M1
6 x 5/3	M1
10 cm	A1

Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or working	Marks
finding the diagonal of the base using Pythagoras: sqrt(10^2 + 10^2)	M1
halving the diagonal to get 7.07 cm, the horizontal distance from the centre to a corner	M1
using Pythagoras to find the sloping edge: sqrt(12^2 + 7.07^2)	M1
sloping edge = 13.9 cm (3 s.f.)	A1
using tan(angle) = height / horizontal distance = 12 / 7.07	M1
angle = 59.5 degrees (1 d.p.)	A1
Final answer: sloping edge = 13.9 cm, angle = 59.5 degrees

Mark scheme for Question 13 [6 marks]
Question 13[6 marks]
Answer or working	Marks
differentiating to get v = 3t^2 - 12t + 9	M1
substituting t = 4 into v	M1
velocity = 9 m/s	A1
setting v = 0 and dividing by 3 to get t^2 - 4t + 3 = 0	M1
factorising to (t - 1)(t - 3) = 0	M1
t = 1 and t = 3 seconds	A1
Final answer: velocity = 9 m/s; t = 1 and t = 3 seconds