IGCSE Mathematics 0580 Paper 21, May/June 2025

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Question paper

0580/21 · May/June 2025

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Simplify. $7 c - 5 d + c + 3 d$

[2]

The diagram shows two parallel lines intersecting two straight lines. Find the values of $w$ , $x$ and $y$ .

[4]

Sally invests $1500 at 3% per year simple interest. Work out the total value of her investment at the end of 6 years.

[3]

Work out. $\frac{5}{6} - \frac{2}{3} \times \frac{3}{8}$

[3]

The interior angle of a regular polygon is $150^{\circ}$ . Find the number of sides of this polygon.

[2]

The line $x + y = 7$ is drawn on the grid.

On the grid, draw the line $y = 2 x + 1$ .

[2]

Use your graph to solve these simultaneous equations. $x + y = 7$ $y = 2 x + 1$

[1]

Write the recurring decimal $0.2 \dot{6}$ as a fraction. Give your answer in its simplest form.

[3]

$m = (\begin{array}{c} 11 \\ 5 \end{array}) n = (\begin{array}{c} 8 \\ - 3 \end{array})$

Find $2 m - n$ .

[2]

The vector $(\begin{array}{c} 5 \\ \sqrt{y} \end{array})$ has a magnitude of 7. Find the value of $y$ .

[2]

The table shows some information about the marks scored by a group of students in a test.

Test mark	4	5	8
Frequency	2	4	$n$

The mean mark is 6. Work out the value of $n$ .

[3]

$A$ , $B$ and $C$ are three points on a circle, centre $O$ . $D E$ is a tangent to the circle at $A$ . Angle $A C O = 35^{\circ}$ and angle $B C O = 40^{\circ}$ . Find

angle $A O C$

[1]

angle $A B C$

[1]

angle $D A C$

[1]

angle $O A B$ .

[1]

The diagram shows the graph of $y = f (x)$ and the point $P (- 2, 11)$ . The tangent from $P$ touches the graph of $y = f (x)$ at the point $(a, b)$ . The values of $a$ and $b$ are integers.

By drawing this tangent, find the value of $a$ and the value of $b$ .

[2]

Find the equation of the tangent. Give your answer in the form $y = m x + c$ .

[3]

The time spent on the internet by each of 120 adults is recorded for one day. The cumulative frequency diagram shows this information.

Use the cumulative frequency diagram to find an estimate of the interquartile range.

[2]

70% of the adults spent less than $k$ hours on the internet. Use the cumulative frequency diagram to find an estimate of the value of $k$ .

[2]

The diagram shows solid $A$ and solid $B$ . Solid $A$ is made from a hemisphere and a cone each with radius $6 cm$ . The cone has sloping edge $10 cm$ . Solid $B$ is a cylinder with radius $4 cm$ and height $h cm$ . The total surface area of solid $A$ is equal to the total surface area of solid $B$ .

Work out the value of $h$ .

[5]

Work out the height of solid $A$ .

[3]

$f (x) = 3 x - 4 g (x) = 4 x + 1$

Find $f (- 2)$ .

[1]

Find $f^{- 1} (x)$ .

[2]

$f g (x) = a x + b$ Find the value of $a$ , and the value of $b$ .

[2]

Simplify. $\frac{2}{f (x)} - \frac{5}{g (x)}$ Give your answer as a single fraction in terms of $x$ .

[3]

Expand and simplify. $(2 - \sqrt{5}) (1 - 3 \sqrt{5})$

[2]

Rationalise the denominator. Give your answer in its simplest form. $\frac{6}{\sqrt{10}}$

[2]

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

[3]

A bag contains 6 red marbles, 3 green marbles and 1 blue marble. Two marbles are picked at random from the bag with replacement. Find the probability that both marbles are green.

[2]

b(i)

Another bag contains 4 red counters and 2 yellow counters. Two counters are picked at random from this bag without replacement. Complete the tree diagram.

[2]

b(ii)

Find the probability that one of the two counters is yellow.

[3]

One day, Anya runs $12 km$ at a speed of $x km/h$ . The next day she walks $10 km$ at a speed of $(x - 4) km/h$ .

Write down an expression, in terms of $x$ , for the time she spends running.

[1]

Write down an expression, in terms of $x$ , for the time she spends walking.

[1]

The time Anya spends walking is 1 hour more than the time she spends running. Write an equation in terms of $x$ and show that it simplifies to $x^{2} - 2 x - 48 = 0$ .

[4]

Use factorisation to solve the equation $x^{2} - 2 x - 48 = 0$ .

[3]

Find the time Anya spends running.

[1]

Find the value of $27^{- \frac{2}{3}}$ .

[2]

Find the exact value of $x$ .

[4]

$O A B C$ is a rhombus and $O$ is the origin. The diagonals of the rhombus intersect at $P$ . $\vec{O P} = 2 m$ and $\vec{A P} = n$ .

a(i)

Find, in terms of $m$ and $n$ , in its simplest form $\vec{O A}$

[1]

a(ii)

$\vec{O C}$ .

[1]

$D$ is the point such that $\vec{A D} = 10 m - 3 n$ . Show that $O A D C$ is a trapezium.

[3]

A curve has equation $y = x^{n} + q x^{2} + 9 x$ . $\frac{d y}{d x} = 3 x^{2} - 12 x + 9$

Find the value of $n$ , and the value of $q$ .

[2]

Work out the coordinates of the turning points of the curve.

[4]

Simplify. $\frac{2 x^{2} + 10 x}{x^{2} - 25}$

[3]

Mark scheme

0580/21 · May/June 2025

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1 · 2 marks

2 marks
8c – 2d final answer

2 · 4 marks

4 marks
[w =] 158 [x =] 76 [y =] 82

3 · 3 marks

3 marks
1770

4 · 3 marks

3 marks
$\frac{7}{12}$ oe

5 · 2 marks

2 marks
12

6a · 2 marks

2 marks
Correct line drawn

6b · 1 mark

1 mark
[x =] 2, [y =] 5

7 · 3 marks

3 marks
$\frac{4}{15}$

8a · 2 marks

2 marks
$(\begin{array}{c} 14 \\ 13 \end{array})$

8b · 2 marks

2 marks
24

9 · 3 marks

3 marks
4 nfww

10a · 1 mark

1 mark
110

10b · 1 mark

1 mark
55

10c · 1 mark

1 mark
55

10d · 1 mark

1 mark
15

11a · 2 marks

2 marks
For correct ruled tangent and [a =] 1, [b =] 2

11b · 3 marks

3 marks
[y =] 5 – 3x

12a · 2 marks

2 marks
1.4

12b · 2 marks

2 marks
5.4

13a · 5 marks

5 marks
12.5 oe

13b · 3 marks

3 marks
14

14a · 1 mark

1 mark
–10

14b · 2 marks

2 marks
$\frac{x + 4}{3}$ oe

14c · 2 marks

2 marks
a = 12, b = –1

14d · 3 marks

3 marks
$\frac{22 - 7 x}{(3 x - 4) (4 x + 1)}$ or $\frac{22 - 7 x}{12 x^{2} - 13 x - 4}$ final answer

15a · 2 marks

2 marks
17 – 7 $\sqrt{5}$ final answer

15b · 2 marks

2 marks
$\frac{3 \sqrt{10}}{5}$ cao

16 · 3 marks

3 marks
3x^3 + 5x^2 – 34x – 24 final answer

17a · 2 marks

2 marks
$\frac{9}{100}$ oe

17b(i) · 2 marks

2 marks
$\frac{2}{5}$ $\frac{4}{5}$ $\frac{1}{5}$

17b(ii) · 3 marks

3 marks
$\frac{16}{30}$ oe

18a · 1 mark

1 mark
$\frac{12}{x}$

18b · 1 mark

1 mark
$\frac{10}{x - 4}$

18c · 4 marks

M1
their $\frac{10}{x - 4}$ – their $\frac{12}{x} = 1$ oe
M2
10x – 12x + 48 = x^2 – 4x
A1
Leading to 0 = x^2 – 2x – 48

18d · 3 marks

M2
(x + 6)(x – 8)
B1
–6, 8

18e · 1 mark

1 mark
1.5 or 1 $\frac{1}{2}$

19 · 2 marks

2 marks
$\frac{1}{9}$

20 · 4 marks

4 marks
6 $\sqrt{3}$ oe

21a(i) · 1 mark

1 mark
2m – n

21a(ii) · 1 mark

1 mark
2m + n

21b · 3 marks

M2
$\vec{C D} = 10 m - 5 n$
A1
$\vec{C D} = 5 \vec{O A}$ leading to CD is parallel to OA [∴ OACD is a trapezium]

22a · 2 marks

2 marks
[n =] 3, [q =] – 6

22b · 4 marks

4 marks
(1, 4) and (3, 0)

23 · 3 marks

3 marks
$\frac{2 x}{x - 5}$ final answer

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