O Level Mathematics 4024 Paper 12, Oct/Nov 2025

Question paper and mark scheme100 marks23 questions

Solve it on ExamPal

Practise

Untimed. Write on the paper and check as you go.

Start

Exam

Timed, with a clock you set, like sitting the real paper.

Start

Opens full screen.

Download PDF Mark scheme

100%

Opening the question paper

Read as textThe questions and the mark scheme as plain text. Easier to read on a phone.

Question paper

4024/12 · Oct/Nov 2025

Download PDF

Ruth records the colour of each of 24 cars. Red Blue Silver Blue Silver Silver White Silver Red Silver Silver Blue Grey Grey Silver Red White Red Blue Grey Blue Silver Red Red

Complete the frequency table. You may use the tally column to help you.

[2]

Draw a bar chart to show the information in the table. Complete the scale on the frequency axis.

[3]

Work out.

0.78 - 0.2

[1]

$- 8 \times - 5$

[1]

ABCD is a parallelogram. $E$ is a point on $A B$ and $E C = B C$ . Angle $E C B = 40^{\circ}$ and angle $D E C = 60^{\circ}$ . Find the value of $x$ .

[4]

Petra thinks of a number. The number is a multiple of 7. Petra says: When I write my number correct to the nearest ten it is 30. Find the number that Petra thinks of.

[2]

Write these numbers in order of size, starting with the smallest. 0.8 $\frac{7}{9}$ $\frac{17}{20}$ 84.5%

[2]

Solve. $6 x - 3 = 4 x + 9$

[2]

Write down the inequality represented on the number line.

[1]

Solve the inequality. $x - 2 < 7$

[1]

Point $A$ and point $B$ are plotted on the grid.

Write down the coordinates of point $A$ .

[1]

The coordinates of point $C$ are $(p, - 1)$ . The gradient of the line $B C$ is $- 2$ . Find the value of $p$ .

[2]

$A B$ and $B C$ are two sides of the parallelogram $A B C D$ . Find the coordinates of point $D$ .

[2]

Sara has a bag containing 60 red counters, 72 blue counters and 36 green counters. Find the ratio of red counters : blue counters : green counters in the bag. Write the ratio in its simplest form.

[2]

Sunil has a bag containing yellow counters and white counters. The ratio of yellow counters to white counters is $5 : 8$ . There are 18 more white counters than yellow counters. Work out the number of yellow counters and the number of white counters.

[3]

Work out. $1 \frac{2}{5} \div \frac{3}{10}$ Give your answer as a mixed number in its simplest form.

[3]

Shape $A$ and shape $B$ are drawn on the grid.

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

[2]

Draw the image of shape $A$ after a rotation of $90^{\circ}$ clockwise about the origin.

[2]

Luca has these two fair spinners.

a(i)

Luca spins both spinners and adds the two numbers to find his score. Complete the sample space diagram to show all the possible scores.

[1]

a(ii)

Find the probability that Luca's score is greater than 10.

[2]

Luca spins spinner A twice. He now multiplies the two numbers to find his score. Find the probability that his score is greater than 30.

[2]

Thomas works in a shop. He is paid $12 for each hour he works. He is also paid a bonus of 5% of the value of the items he sells. One day, Thomas works for 5 hours. He is paid a total of $82 including his bonus. Work out the value of the items Thomas sells on this day.

[4]

$A$ , $B$ , $C$ and $D$ are points on a circle, centre $O$ . $E F$ is a tangent to the circle at $A$ . Angle $B A F = 65^{\circ}$ and angle $A O D = 80^{\circ}$ . Find angle $B C D$ . Give a geometrical property to explain each step of your working.

[4]

Complete the table for $y = x^{3} - 4 x^{2} + 12$ .

[1]

Draw the graph of $y = x^{3} - 4 x^{2} + 12$ for $- 2 ⩽ x ⩽ 4$ .

[4]

Use your graph to solve the equation $x^{3} - 4 x^{2} + 12 = 0$ .

[1]

By drawing a suitable line on the grid, find the roots of the equation $x^{3} - 4 x^{2} + x + 4 = 0$ .

[4]

Expand and simplify. $(3 + \sqrt{2}) (1 + 4 \sqrt{2})$

[2]

Rationalise the denominator. $\frac{1}{1 + \sqrt{7}}$

[2]

Triangle $A B C$ is similar to triangle $P Q R$ .

Calculate $P R$ .

[2]

The area of triangle $A B C$ is $16 {cm}^{2}$ . Calculate the area of triangle $P Q R$ .

[2]

The diagram shows a solid cylinder and a solid hemisphere. The radius of the cylinder is equal to the radius of the hemisphere. The volume of the cylinder is equal to the volume of the hemisphere. The height of the cylinder is $6 cm$ . Calculate the total surface area of the hemisphere. Give your answer in terms of $π$ .

[5]

Evaluate $16^{- \frac{3}{4}}$ .

[2]

Write $9^{2} \times 3^{- \frac{1}{2}}$ as a single power of 3.

[2]

$f (x) = a x + 3 g (x) = (x + b)^{2}$

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

[2]

$g f (x) = 9 x^{2} + 24 x + 16$ $a$ and $b$ are positive integers. Find the value of $a$ and the value of $b$ .

[3]

$O P Q R$ is a parallelogram. $\vec{O P} = a$ and $\vec{O R} = 2 b$ . $M$ is the midpoint of $Q R$ . $N$ is a point on $P Q$ and $P N : N Q = 1 : 3$ .

a(i)

Find, as simply as possible, in terms of $a$ and/or $b$ $\vec{M Q}$

[1]

a(ii)

$\vec{N M}$

[2]

Line $O P$ is extended to point $X$ . $M N X$ is a straight line. Find the position vector of $X$ .

[2]

Ade and Maha make cards. One day they each work for 4 hours making cards. Ade takes $x$ minutes to make one card.

Show that the number of cards Ade makes in 4 hours is given by $\frac{240}{x}$ .

[1]

Maha takes 2 minutes less than Ade to make one card. Maha and Ade make a total of 70 cards in 4 hours. Form an equation and show that it simplifies to $7 x^{2} - 62 x + 48 = 0$ .

[4]

Solve by factorisation. $7 x^{2} - 62 x + 48 = 0$

[3]

Work out the number of cards Maha makes in 1 hour.

[2]

The diagram shows a square-based pyramid. $X$ is the centre of the square base. Vertex $E$ is vertically above $X$ and $E X = 7 cm$ . The perpendicular height of triangle $B E C$ is $11 cm$ . Work out the length of the base of the pyramid. Give your answer as a surd in its simplest form.

[4]

Mark scheme

4024/12 · Oct/Nov 2025

Download PDF

1a · 2 marks

2 marks
| Blue | 5 | | Grey | 3 | | Red | 6 | | Silver | 8 | | White | 2 |

1b · 3 marks

3 marks
Correct bar chart with linear scale starting from 0 on frequency axis

2a · 1 mark

1 mark
0.58

2b · 1 mark

1 mark
40

3 · 4 marks

4 marks
20 nfww

4 · 2 marks

2 marks
28 cao

5 · 2 marks

2 marks
$\frac{7}{9} 0.8 84.5 % \frac{17}{20}$

6 · 2 marks

2 marks
[x = ] 6

7a · 1 mark

1 mark
x > 3 or 3 < x final answer

7b · 1 mark

1 mark
x < 9 or 9 > x final answer

8a · 1 mark

1 mark
(–2, 1)

8b · 2 marks

2 marks
[p = ] 6

8c · 2 marks

2 marks
(0, –3) or ((their p – 6), –3)

9a · 2 marks

2 marks
5 : 6 : 3

9b · 3 marks

3 marks
[yellow = ] 30 [white = ] 48

10 · 3 marks

3 marks
$4 \frac{2}{3}$ cao

11a · 2 marks

2 marks
Translation $(\begin{array}{c} - 5 \\ - 4 \end{array})$

11b · 2 marks

2 marks
Shape at (2, –1), (3, –1), (3, –2), (2, –3)

12a(i) · 1 mark

1 mark
3 5 7 9
3 5 7 9
5 7 9 11
7 9 11 13

3	5	7	9
3	5	7	9
5	7	9	11
7	9	11	13

12a(ii) · 2 marks

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

12b · 2 marks

2 marks
$\frac{6}{16}$ oe

13 · 4 marks

4 marks
440

14 · 4 marks

B2
[∠ BCD = ] 105°
B1
alternate segment theorem associated with ∠BCA
B1
Angle at the centre is twice the angle at the circumference associated with ∠ACD

15a · 1 mark

1 mark
12

15b · 4 marks

4 marks
Correct smooth curve

15c · 1 mark

1 mark
–1.7 to –1.3

15d · 4 marks

M2
Ruled line $y = 8 - x$
A2
–0.9 to –0.7 1.4 to 1.6 3.2 to 3.5

16a · 2 marks

2 marks
$11 + 13 \sqrt{2}$ final answer

16b · 2 marks

2 marks
$\frac{- 1 + \sqrt{7}}{6}$ final answer

17a · 2 marks

2 marks
$\frac{21}{2}$ oe

17b · 2 marks

2 marks
36 nfww

18 · 5 marks

5 marks
243π

19a · 2 marks

2 marks
$\frac{1}{8}$

19b · 2 marks

2 marks
$3^{\frac{7}{2}}$

20a · 2 marks

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

20b · 3 marks

3 marks
[a = ] 3 [b = ] 1

21a(i) · 1 mark

1 mark
$\frac{1}{2} a$

21a(ii) · 2 marks

2 marks
$\frac{3}{2} b - \frac{1}{2} a$ final answer

21b · 2 marks

2 marks
$\frac{7}{6} a$

22a · 1 mark

1 mark
$\frac{60 \times 4}{x}$ leading to $\frac{240}{x}$

22b · 4 marks

M1
$\frac{240}{x - 2} + \frac{240}{x} = 70$ oe
M1
$\frac{240 x}{x (x - 2)} + \frac{240 (x - 2)}{x (x - 2)} = 70$ or $240 x + 240 (x - 2) = 70 x (x - 2)$
M1
$240 x + 240 x - 480 = 70 x^{2} - 140 x$
A1
Leading to $7 x^{2} - 62 x + 48 = 0$

22c · 3 marks

M2
$(7 x - 6) (x - 8)$
B1
$\frac{6}{7}$ and 8

22d · 2 marks

2 marks
10

23 · 4 marks

4 marks
$12 \sqrt{2}$ cao

Questions

1
Question 15 marks
Mark scheme ›
2
Question 22 marks
Mark scheme ›
3
Question 34 marks
Mark scheme ›
4
Question 42 marks
Mark scheme ›
5
Question 52 marks
Mark scheme ›
6
Question 62 marks
Mark scheme ›
7
Question 72 marks
Mark scheme ›
8
Question 85 marks
Mark scheme ›
9
Question 95 marks
Mark scheme ›
10
Question 103 marks
Mark scheme ›
11
Question 114 marks
Mark scheme ›
12
Question 125 marks
Mark scheme ›
13
Question 134 marks
Mark scheme ›
14
Question 144 marks
Mark scheme ›
15
Question 1510 marks
Mark scheme ›
16
Question 164 marks
Mark scheme ›
17
Question 174 marks
Mark scheme ›
18
Question 185 marks
Mark scheme ›
19
Question 194 marks
Mark scheme ›
20
Question 205 marks
Mark scheme ›
21
Question 215 marks
Mark scheme ›
22
Question 2210 marks
Mark scheme ›
23
Question 234 marks
Mark scheme ›