Foundation Tier - Grades 1-5

IGCSE Maths Foundation Paper 4

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

15 questions - 60 marks - calculator allowed

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Questions

Question 1 [2 marks]

Angles, Polygons and Circle Theorems

Two parallel lines are crossed by a transversal. One angle formed is 118 degrees.

Find the size of its co-interior (allied) angle on the same side of the transversal.

Question 2 [4 marks]

Pythagoras and Trigonometry

In a right-angled triangle, the side opposite angle y is 9 cm and the side adjacent to angle y is 6.5 cm.

Find y to 1 decimal place.

Question 3 [4 marks]

Indices, Surds and Standard Form

Simplify (2m^4 n^2)^3, giving your answer as a single term.

Question 4 [4 marks]

Quadratics and Simultaneous Equations

A number x satisfies x(x + 2) = 24. Find the two possible values of x.

Question 5 [4 marks]

Sequences and Graphs

The first four terms of a sequence are 5, 9, 13, 17.

Find an expression for the nth term, then use it to determine whether 205 is a term of the sequence.

Question 6 [4 marks]

Statistics and Probability

The ages, in years, of 7 people in a choir are 34, 28, 41, 28, 55, 39, 28.

Find the median age, the mode, and state which of these two averages best represents a typical age, giving a reason.

Question 7 [4 marks]

Functions and Rates of Change

f(x) = 2x + 1 and g(x) = x^2 - 3.

Find fg(x) in terms of x, and hence find fg(4).

Question 8 [4 marks]

Ratio and Proportion

In a school, the ratio of Year 10 students to Year 11 students is 5:4.

The ratio of Year 11 students to Year 12 students is 2:3.

Find the ratio of Year 10 to Year 11 to Year 12 students, in its simplest form.

Question 9 [4 marks]

Linear Equations and Inequalities

Solve (3x - 2)/4 = (x + 5)/2.

Question 10 [4 marks]

Mensuration and Vectors

A prism has cross-sectional area 22 cm^2 and length 16 cm.

Find the volume of the prism.

Question 11 [4 marks]

Number and Calculation

Find the HCF of 168 and 180 by expressing each number as a product of its prime factors.

Question 12 [4 marks]

Sequences and Graphs

Find the distance between the points (-1, 2) and (5, -4), giving your answer as a simplified surd.

Question 13 [4 marks]

Mensuration and Vectors

A circular pond has radius 6.5 m.

Find its circumference to 1 decimal place.

Question 14 [5 marks]

Pythagoras and Trigonometry

A cuboid has length 8 cm, width 5 cm and height 6 cm.

Find the length of the diagonal that runs from one corner of the cuboid to the opposite corner, to 1 decimal place.

Question 15 [5 marks]

Fractions, Decimals and Percentages

Container A holds 5/6 of its capacity in juice.

Sam pours 2/5 of the juice in container A into container B, which was empty.

What fraction of container A's capacity is now in container B, and what fraction remains in container A?

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or working	Marks
using co-interior angles sum to 180 degrees	M1
62 degrees	A1

Mark scheme for Question 2 [4 marks]
Question 2[4 marks]
Answer or working	Marks
using tan y = opposite/adjacent	M1
tan y = 9/6.5	M1
y = tan^-1(9/6.5)	M1
54.2 degrees	A1

Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or working	Marks
cubing the coefficient: 2^3 = 8	M1
multiplying the power of m: 4 x 3 = 12	M1
multiplying the power of n: 2 x 3 = 6	M1
8m^12 n^6	A1

Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or working	Marks
rearranging to x^2 + 2x - 24 = 0	M1
factorising to (x + 6)(x - 4) = 0	M1
setting each bracket equal to zero	M1
x = -6 or x = 4	A1

Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or working	Marks
identifying the common difference 4	M1
comparing with 4n to find the adjustment +1, giving 4n + 1	M1
solving 4n + 1 = 205	M1
n = 51, so 205 is a term of the sequence	A1
Final answer: 4n + 1; yes, 205 is the 51st term

Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or working	Marks
arranging the ages in order: 28, 28, 28, 34, 39, 41, 55	M1
median = 34	A1
mode = 28	A1
stating the median, with a valid reason such as the mode being lower than most of the ages	B1
Final answer: median = 34, mode = 28; the median better represents a typical age

Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or working	Marks
substituting g(x) = x^2 - 3 into f(x) = 2x + 1	M1
fg(x) = 2(x^2 - 3) + 1	M1
fg(x) = 2x^2 - 5	A1
fg(4) = 27	A1
Final answer: fg(x) = 2x^2 - 5; fg(4) = 27

Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or working	Marks
scaling the second ratio so the Year 11 parts match: 2:3 becomes 4:6	M1
combining to get Year 10 : Year 11 : Year 12 = 5 : 4 : 6	M1
checking the ratio cannot be simplified further	M1
5:4:6	A1

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
cross-multiplying to get 2(3x - 2) = 4(x + 5)	M1
expanding to 6x - 4 = 4x + 20	M1
collecting terms to get 2x = 24	M1
x = 12	A1

Mark scheme for Question 10 [4 marks]
Question 10[4 marks]
Answer or working	Marks
volume = cross-sectional area x length	M1
22 x 16	M1
352	A1
cm^3	B1
Final answer: 352 cm^3

Mark scheme for Question 11 [4 marks]
Question 11[4 marks]
Answer or working	Marks
168 = 2^3 x 3 x 7	M1
180 = 2^2 x 3^2 x 5	M1
identifying the common prime factors 2^2 and 3	M1
HCF = 12	A1
Final answer: 12

Mark scheme for Question 12 [4 marks]
Question 12[4 marks]
Answer or working	Marks
finding the horizontal difference 5 - (-1) = 6	M1
finding the vertical difference -4 - 2 = -6	M1
using distance = sqrt(6^2 + 6^2) = sqrt(72)	M1
6 sqrt(2)	A1

Mark scheme for Question 13 [4 marks]
Question 13[4 marks]
Answer or working	Marks
using circumference = 2 pi r	M1
2 x pi x 6.5	M1
40.840... before rounding	M1
40.8 m	A1

Mark scheme for Question 14 [5 marks]
Question 14[5 marks]
Answer or working	Marks
using the 3D Pythagoras formula d = sqrt(l^2 + w^2 + h^2)	M1
8^2 + 5^2 + 6^2 = 125	M1
taking the square root of 125	M1
11.18033...	A1
11.2 cm	A1

Mark scheme for Question 15 [5 marks]
Question 15[5 marks]
Answer or working	Marks
multiplying 2/5 by 5/6 to find the amount poured out	M1
2/5 x 5/6 = 10/30 = 1/3	M1
1/3 of the capacity is now in container B	A1
subtracting 1/3 from 5/6, using a common denominator of 6	M1
1/2 of the capacity remains in container A	A1
Final answer: 1/3 in container B, 1/2 remains in container A