IGCSE Mathematics 0580 Paper 23, May/June 2025

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0580/23 · May/June 2025

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The probability of picking a green pen from a box is $0.17$ . Find the probability of not picking a green pen from the box.

[1]

ABCD is a quadrilateral. CDE is a straight line. AFB is an isosceles triangle. Find the value of $x$ and the value of $y$ .

[4]

The diagram shows a pyramid $A B C D E$ with a square base. $M$ is the centre of the square base. $E$ is vertically above $M$ .

Write down the number of planes of symmetry of this pyramid.

[1]

Using two of the letters from $A$ , $B$ , $C$ , $D$ , $E$ and $M$ , complete the statement about the pyramid. The axis of rotational symmetry passes through the points ............... and ...............

[1]

The number of ice creams sold increases as the temperature rises. What type of correlation does this statement describe?

[1]

Describe fully the single transformation that maps triangle $A$ onto triangle $B$ .

[3]

b(i)

On the grid, draw the image of triangle $A$ after a reflection in the line $x = - 1$

[2]

b(ii)

On the grid, draw the image of triangle $A$ after a rotation $90^{\circ}$ clockwise, centre $(1, - 2)$ .

[2]

There are 15 giraffes in a group. The table gives information about the heights of the 15 giraffes. One giraffe has a height of $2.6 m$ No giraffe is shorter than $2.5 m$ The range of heights for the 15 giraffes is $2.3 m$ More than 3 giraffes have the same height The modal height for the giraffes is $3.9 m$ The stem-and-leaf diagram shows information about the height of 9 of these giraffes. Key: $4 ∣ 1$ represents a giraffe height of $4.1 m$ Use the information in the table to complete the stem-and-leaf diagram for the group of 15 giraffes.

[3]

Work out.

$\frac{5}{6} - \frac{7}{12}$ Give your answer as a fraction in its simplest form.

[2]

$1 \frac{1}{3} \div \frac{8}{15}$ Give your answer as a mixed number in its simplest form.

[3]

Write 42 as a product of its prime factors.

[2]

Find the highest common factor (HCF) of 84 and 70.

[2]

Solve. $5 x^{2} = 12 - 17 x$

[4]

$a x^{2} + a = b$ where $a$ and $b$ are integers. One solution of this equation is $x = 6$ . Write down the other solution.

[1]

Solve the simultaneous equations. $4 x - 5 y = 13$ $3 x - 2 y = 8$

[4]

Angela picks a number at random from the numbers 1, 2 and 3. She then picks a number at random from the numbers 4, 5 and 6. She adds the two numbers to find the total.

Complete the table to show the possible outcomes.

[2]

Given that the total is odd, find the probability that one of the numbers Angela picks is 3.

[2]

$v = (\begin{array}{c} 2 \\ - 3 \end{array})$ Find $5 v$ .

[1]

$H$ is the point $(- 3, 8)$ and $K$ is the point $(- 4, 0)$ . $\vec{H J} = (\begin{array}{c} 7 \\ - 2 \end{array})$ Find $∣ \vec{J K} ∣$ .

[4]

$A$ , $B$ , $C$ and $D$ are points on a circle. $E F$ is a tangent to the circle at $A$ . $A B$ is parallel to $D C$ .

Find angle $D C B$ , giving a geometrical reason.

[2]

Find angle $D B C$ .

[2]

Find the lowest common multiple (LCM) of $15 x y^{3}$ and $18 x^{4} y$ .

[2]

Simplify. $\sqrt{27} + \sqrt{12}$

[2]

$\frac{40 \sqrt{8}}{5 \sqrt{2}} = k$ , where $k$ is an integer. Find the value of $k$ .

[2]

Rationalise the denominator. $\frac{1}{3 - \sqrt{5}}$

[2]

Write $0.3 \dot{2} \dot{8}$ as a fraction in its simplest form.

[3]

Solid $A$ is mathematically similar to solid $B$ . The height of solid $A$ is $7 cm$ and its surface area is $60 {cm}^{2}$ . The surface area of solid $B$ is $540 {cm}^{2}$ . Calculate the height of solid $B$ .

[3]

Make $t$ the subject of the formula. $2 = \frac{m (1 - t)}{p t}$

[4]

Simplify. $\frac{7 x - x^{2}}{49 - x^{2}}$

[3]

The diagram shows a sector of a circle, centre $O$ . The radius of the circle is $6 cm$ . Calculate the length of the major arc $A B$ . Give your answer in its simplest form in terms of $π$ .

[3]

Differentiate $x^{3} - 3 x^{2} + 1$ .

[2]

Find the coordinates of the turning points of the graph of $y = x^{3} - 3 x^{2} + 1$ .

[4]

$f (x) = 2 x + 5 g (x) = x - 4 h (x) = 5^{x}$

Find $f (3)$ .

[1]

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

[2]

Solve $f g (x) = 25$ .

[3]

Find $x$ when $h^{- 1} (x) = 2$ .

[2]

Write down the value of $\cos 90^{\circ}$ .

[1]

The diagram shows two different right-angled triangles joined by a common side. Find $x$ in terms of $n$ .

[5]

a(i)

$A$ is the point $(a, 12)$ and $B$ is the point $(b, 27)$ . Find the $y$ -coordinate of the midpoint of $A B$ .

[1]

a(ii)

$A$ is the point $(a, 12)$ and $B$ is the point $(b, 27)$ . The line $A B$ has gradient 3. Find an expression for $a$ in terms of $b$ .

[3]

$D$ is the point $(22, 34)$ and $E$ is the point $(23, 39)$ . $D$ is the point on $C E$ such that $2 C E = 5 D E$ . Find the coordinates of $C$ .

[3]

Mark scheme

0580/23 · May/June 2025

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1 · 1 mark

1 mark
0.83 oe

2 · 4 marks

B2
x = 84
B2
y = 75

3a · 1 mark

1 mark
4

3b · 1 mark

1 mark
E and M

4 · 1 mark

1 mark
Positive

5a · 3 marks

B1
Enlargement
B1
[Scale factor] 3
B1
(1, 2)

5b(i) · 2 marks

2 marks
Triangle at (–6, 1), (–3, 3), (–4, 4)

5b(ii) · 2 marks

2 marks
Triangle at (4, –5), (6, –2), (7, –3)

6 · 3 marks

B1
Row 2: 5 6
B1
Row 3: 2 7 7 9 9 9 9
B1
Row 4: 1 1 4 5 7 8

7a · 2 marks

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

7b · 3 marks

3 marks
$2 \frac{1}{2}$ cao

8a · 2 marks

2 marks
2 × 3 × 7

8b · 2 marks

2 marks
14

9a · 4 marks

4 marks
$\frac{3}{5}$ or 0.6 and –4

9b · 1 mark

1 mark
–6

10 · 4 marks

A1
[x =] 2
A1
[y =] –1

11a · 2 marks

2 marks
Fully correct table

11b · 2 marks

2 marks
$\frac{2}{5}$ oe

12a · 1 mark

1 mark
$(\begin{array}{c} 10 \\ - 15 \end{array})$

12b · 4 marks

4 marks
10

13a · 2 marks

2 marks
120 and opposite angles of a cyclic quadrilateral sum to 180°

13b · 2 marks

2 marks
25

14 · 2 marks

2 marks
$90 x^{4} y^{3}$ final answer

15a · 2 marks

2 marks
$5 \sqrt{3}$

15b · 2 marks

2 marks
16

15c · 2 marks

2 marks
$\frac{3 + \sqrt{5}}{4}$

16 · 3 marks

3 marks
$\frac{65}{198}$ cao

17 · 3 marks

3 marks
21

18 · 4 marks

4 marks
$[t =] \frac{m}{2 p + m}$ final answer

19 · 3 marks

3 marks
$\frac{x}{7 + x}$ final answer

20 · 3 marks

3 marks
$\frac{32}{3} π$ or $10 \frac{2}{3} π$ cao

21a · 2 marks

2 marks
$3 x^{2} – 6 x$ or $3 x (x – 2)$

21b · 4 marks

4 marks
(0, 1) (2, –3)

22a · 1 mark

1 mark
11

22b · 2 marks

2 marks
$\frac{x - 5}{2}$

22c · 3 marks

3 marks
14

22d · 2 marks

2 marks
25

23a · 1 mark

1 mark
0

23b · 5 marks

5 marks
$[x =] \frac{4 n}{\sqrt{3}}$ or $\frac{4 n \sqrt{3}}{3}$ oe

24a(i) · 1 mark

1 mark
19.5

24a(ii) · 3 marks

3 marks
[a =] b – 5 oe

24b · 3 marks

3 marks
(20.5, 26.5) or ( $\frac{41}{2}$ , $\frac{53}{2}$ )

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