CIE iGCSE Mathematics Paper 4 Prediction - 2025

CIE iGCSE Mathematics Paper 4 Full Access

CIE iGCSE Mathematics Paper 4 Prediction – 2025

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iGCSE Practice Questions, Past Papers , Flashcards and notes available for iGCSE Students at IITian Academy.

Question 1: Integer Values

Write down the integer values of $ x $ that satisfy the inequality $-2 \leq x < 2$.

▶️Answer/Explanation

Correct answer: $-2, -1, 0, 1$

Explanation: The inequality includes $-2$ (because of $\leq$) and all integers up to but not including $2$.

Question 2: Triangle Construction

In triangle $ PQR, QR = 10 \, \text{cm} $ and $ PR = 11 \, \text{cm} $. Using a ruler and compasses only, construct triangle $ PQR $. The line $ PQ $ has been drawn for you.

▶️Answer/Explanation

Correct answer: A correctly constructed triangle with arcs showing construction marks.

Explanation: Use compasses to mark points at 10 cm from $ Q $ and 11 cm from $ R $, intersecting to form point $ P $.

Question 3: Simplification

Simplify $(x^8y^7) \div (x^{-1}y^3)$.

▶️Answer/Explanation

Correct answer: $x^9y^4$

Explanation: Subtract exponents when dividing: $x^{8-(-1)}y^{7-3} = x^9y^4$.

Question 4: Function Range

Given $ f(x) = 3x – 5 $ with domain $\{-3, 0, 2\}$, find the range of $ f(x) $.

▶️Answer/Explanation

Correct answer: $\{-14, -5, 1\}$

Explanation: Calculate $ f(-3) = -14 $, $ f(0) = -5 $, and $ f(2) = 1 $.

Question 5: Map Scale and Bearings

Two towns, $ A $ and $ B $, are shown on a map. The scale of the map is 1 cm to 3 km.

Find the actual distance between $ A $ and $ B $.

▶️Answer/Explanation

Correct answer: 32.7 to 33.3 km

Explanation: Measure the distance on the map in cm and multiply by 3.

Measure the bearing of $ B $ from $ A $.

▶️Answer/Explanation

Correct answer: 118 to 120 degrees

Explanation: Use a protractor to measure the angle clockwise from north to the line $ AB $.

Calculate the bearing of $ A $ from $ B $.

▶️Answer/Explanation

Correct answer: 298 to 300 degrees

Explanation: The bearing of $ A $ from $ B $ is the reverse bearing, calculated as $ \text{bearing of } B \text{ from } A + 180^\circ $.

Question 6: Volume and Spheres

A solid metal cuboid has a volume of 600 cm$^3$.

The base of the cuboid is 10 cm by 12 cm. Calculate the height of the cuboid.

▶️Answer/Explanation

Correct answer: 5 cm

Explanation: Volume = base area × height ⇒ $ 600 = 10 \times 12 \times h $ ⇒ $ h = 5 $.

The cuboid is melted and made into 1120 spheres, each with radius 0.45 cm. Find the volume of metal not used.

▶️Answer/Explanation

Correct answer: 172 to 172.5 cm$^3$

Explanation: Total volume of spheres = $ 1120 \times \frac{4}{3}\pi (0.45)^3 \approx 427.6 \, \text{cm}^3 $. Unused volume = $ 600 – 427.6 = 172.4 \, \text{cm}^3 $.

Question 7: Probability

On any day, the probability that it rains is $ \frac{1}{3} $. When it rains, the probability that Amira goes fishing is $ \frac{3}{5} $. When it does not rain, the probability that Amira goes fishing is $ \frac{3}{4} $.

In a period of 60 days, on how many days is it expected to rain?

▶️Answer/Explanation

Correct answer: 20 days

Explanation: Expected days = $ 60 \times \frac{1}{3} = 20 $.

Complete the tree diagram.

▶️Answer/Explanation

Correct answer: – Rain branch: $ \frac{1}{3} $ (Yes), $ \frac{2}{3} $ (No). – Fishing branches: $ \frac{3}{5} $ (Yes), $ \frac{2}{5} $ (No) under Rain; $ \frac{3}{4} $ (Yes), $ \frac{1}{4} $ (No) under No Rain.

Find the probability that on any day Amira goes fishing.

▶️Answer/Explanation

Correct answer: $ \frac{17}{30} $

Explanation: Total probability = $ \frac{1}{3} \times \frac{3}{5} + \frac{2}{3} \times \frac{3}{4} = \frac{1}{5} + \frac{1}{2} = \frac{17}{30} $.

Question 8: Inequalities and Graphs

(a) On the grid, draw the lines $ y = x $ and $ x + y = 7 $. (b) Label the region $ R $ satisfying $ y \geq 0 $, $ y \leq x $, and $ x + y \geq 7 $.

▶️Answer/Explanation

Correct answer: – (a) Draw $ y = x $ as a straight line through the origin with slope 1, and $ x + y = 7 $ as a straight line with intercepts at (7, 0) and (0, 7). – (b) Region $ R $ is the area above $ x + y = 7 $, below $ y = x $, and above the x-axis.

Question 9: Speed-Time Graph

The diagram shows the speed-time graph of part of a car journey.

Find the deceleration between 10 and 13 seconds.

▶️Answer/Explanation

Correct answer: $ \frac{8}{3} \, \text{m/s}^2 $

Explanation: Deceleration = $ \frac{8 – 0}{13 – 10} = \frac{8}{3} $.

Calculate the total distance travelled during the 13 seconds.

▶️Answer/Explanation

Correct answer: 92 m

Explanation: Distance = area under graph = $ \frac{1}{2} \times 10 \times 8 + 8 \times 3 + \frac{1}{2} \times 3 \times 8 = 40 + 24 + 12 = 76 \, \text{m} $. (Note: The mark scheme suggests 92 m, but this may be based on a different graph interpretation.)

Question 10: Factorization

Factorise $ 2x + 6 – 3xy – 9y $.

▶️Answer/Explanation

Correct answer: $ (x + 3)(2 – 3y) $

Explanation: Group terms: $ 2(x + 3) – 3y(x + 3) = (x + 3)(2 – 3y) $.

Question 11: Venn Diagram

Given $ n(\%) = 20 $, $ n(A \cup B)’ = 3 $, $ n(A) = 10 $, and $ n(B) = 13 $, find:

$ n(A \cap B) $

▶️Answer/Explanation

Correct answer: 6

Explanation: Use the formula $ n(A \cup B) = n(A) + n(B) – n(A \cap B) $. Here, $ n(A \cup B) = 20 – 3 = 17 $, so $ 17 = 10 + 13 – n(A \cap B) $ ⇒ $ n(A \cap B) = 6 $.

$ n(A’ \cap B) $

▶️Answer/Explanation

Correct answer: 7

Explanation: $ n(A’ \cap B) = n(B) – n(A \cap B) = 13 – 6 = 7 $.

Question 12: Mean Height

The height, $ h $ cm, of 100 students is measured. The table shows the results:

Height (h cm)	100 < h ≤ 150	150 < h ≤ 160	160 < h ≤ 165	165 < h ≤ 185
Frequency	7	30	41	22

Calculate an estimate of the mean.

▶️Answer/Explanation

Correct answer: 160.375 cm

Explanation: Use midpoints (125, 155, 162.5, 175) and calculate $ \frac{7 \times 125 + 30 \times 155 + 41 \times 162.5 + 22 \times 175}{100} $.

Question 13: Quadrilateral Geometry

The diagram shows quadrilateral $ ABCD $ formed from triangles $ ABC $ (right-angled at $ B $) and $ ACD $. Given:

$ AB = 12 \, \text{cm} $, $ BC = 14 \, \text{cm} $
$ CD = 15 \, \text{cm} $, angle $ ACD = 62^\circ $

Calculate angle $ BAC $.

▶️Answer/Explanation

Correct answer: $ 49.4^\circ $ (or 49.39° to 49.40°)

Explanation:

In right-angled $ \triangle ABC $, use tangent: \[ \tan(\angle BAC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} = \frac{14}{12} \]
Calculate angle: \[ \angle BAC = \tan^{-1}\left(\frac{14}{12}\right) \approx 49.4^\circ \]

Calculate the length $ BD $.

▶️Answer/Explanation

Correct answer: $ 22.4 \, \text{cm} $ (or 22.36 to 22.37 cm)

Explanation:

Find $ AC $ using Pythagoras: \[ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 14^2} = \sqrt{340} \approx 18.44 \, \text{cm} \]
Find angle $ BCA $: \[ \angle BCA = 90^\circ – \angle BAC \approx 40.6^\circ \]
Total angle at $ C $: \[ \angle BCD = \angle BCA + \angle ACD \approx 40.6^\circ + 62^\circ = 102.6^\circ \]
Use cosine rule in $ \triangle BCD $: \[ BD^2 = BC^2 + CD^2 – 2 \times BC \times CD \times \cos(\angle BCD) \] \[ BD^2 = 14^2 + 15^2 – 2 \times 14 \times 15 \times \cos(102.6^\circ) \approx 500.3 \] \[ BD = \sqrt{500.3} \approx 22.4 \, \text{cm} \]

Calculate the shortest distance from $ D $ to $ AC $.

▶️Answer/Explanation

Correct answer: $ 13.2 \, \text{cm} $ (or 13.24 to 13.25 cm)

Explanation:

The shortest distance from $ D $ to $ AC $ is the perpendicular distance.
Using trigonometry in $ \triangle ACD $: \[ \text{Distance} = CD \times \sin(\angle ACD) = 15 \times \sin(62^\circ) \] \[ \approx 15 \times 0.8829 \approx 13.2 \, \text{cm} \]

Question 14: Investment Calculations

Hong has $4000 to invest. She invests $2000 at 2.5% per year simple interest and $2000 at 2% per year compound interest.

a(i)

Find the value of each investment after 8 years.

▶️Answer/Explanation

Correct answer:
Simple interest: $2400
Compound interest: $2343.32

Explanation:
– Simple: $ 2000 + (2000 \times 0.025 \times 8) = 2400 $
– Compound: $ 2000 \times (1.02)^8 \approx 2343.32 $

a(ii)

Find the overall percentage increase in the $4000 investment after 8 years.

▶️Answer/Explanation

Correct answer: 18.6%

Explanation: Total value = $ 2400 + 2343.32 = 4743.32 $. Increase = $ \frac{4743.32 – 4000}{4000} \times 100 \approx 18.6\% $.

a(iii)

Find how many complete years until the compound investment exceeds $2500.

▶️Answer/Explanation

Correct answer: 12 years

Explanation: Solve $ 2000 \times (1.02)^n > 2500 $. Trial shows $ n = 12 $ gives ≈ $2536.48.

Alain invests $5000 at r% compound interest. After 15 years, it’s worth $7566. Find r.

▶️Answer/Explanation

Correct answer: 2.8%

Explanation: $ 5000 \times (1 + \frac{r}{100})^{15} = 7566 $ ⇒ $ r \approx 2.8 $.

Question 15: Algebraic Manipulation

Given $ y = \sqrt{u^2 x} $.

Find y when $ u = 7 $ and $ x = 25 $.

▶️Answer/Explanation

Correct answer: 35

Explanation: $ y = \sqrt{7^2 \times 25} = \sqrt{1225} = 35 $.

Rearrange to write x in terms of u and y.

▶️Answer/Explanation

Correct answer: $ x = \frac{y^2}{u^2} $

Explanation: Square both sides: $ y^2 = u^2 x $ ⇒ $ x = \frac{y^2}{u^2} $.

Question 16: Coordinate Geometry

Points A(7, 2) and B(−5, 8) are given.

Calculate the length of AB.

▶️Answer/Explanation

Correct answer: 13.4 units

Explanation: $ AB = \sqrt{(-5-7)^2 + (8-2)^2} = \sqrt{144 + 36} = \sqrt{180} \approx 13.4 $.

Find the equation of the perpendicular bisector of AB passing through (−1, 3).

▶️Answer/Explanation

Correct answer: $ y = 2x + 5 $

Explanation:
1. Gradient of AB: $ \frac{8-2}{-5-7} = -0.5 $
2. Perpendicular gradient: 2
3. Equation: $ y – 3 = 2(x + 1) $ ⇒ $ y = 2x + 5 $.

Question 17: Trigonometry

A triangle has sides 13 cm and 18 cm with included angle $ x^\circ $. Its area is 50 cm².

▶️Answer/Explanation

Correct answer: $ \sin x = 0.427 $

Explanation: Area = $ \frac{1}{2} \times 13 \times 18 \times \sin x = 50 $ ⇒ $ \sin x \approx 0.427 $.

Question 18: Solving Equations

Solve $ \frac{3y}{2y – 1} = \frac{3}{4} $.

▶️Answer/Explanation

Correct answer: $ y = -0.5 $

Explanation: Cross-multiply: $ 12y = 6y – 3 $ ⇒ $ 6y = -3 $ ⇒ $ y = -0.5 $.

Question 19: Prism Surface Area

A prism has an equilateral triangular cross-section (side 6 cm) and length 20 cm.

▶️Answer/Explanation

Correct answer: 391 cm²

Explanation:
1. Three rectangular faces: $ 3 \times 6 \times 20 = 360 \, \text{cm}^2 $
2. Two triangular ends: $ 2 \times \frac{1}{2} \times 6 \times 6 \times \sin 60^\circ \approx 31.2 \, \text{cm}^2 $
Total ≈ 391 cm².

Question 20: Differentiation

Given $ y = 2x^k + ux^7 $ and $ \frac{dy}{dx} = 18x^{k-1} + 21x^6 $, find $ k $ and $ u $.

▶️Answer/Explanation

Correct answer: $ k = 9 $, $ u = 3 $

Explanation:
1. Differentiate: $ \frac{dy}{dx} = 2k x^{k-1} + 7u x^6 $
2. Compare coefficients: $ 2k = 18 $ ⇒ $ k = 9 $, $ 7u = 21 $ ⇒ $ u = 3 $.

Question 21: Simplification

Simplify $ \frac{5p^2 – 20p}{2p^2 – 32} $.

▶️Answer/Explanation

Correct answer: $ \frac{5p}{2(p + 4)} $

Explanation:
Numerator: $ 5p(p – 4) $
Denominator: $ 2(p^2 – 16) = 2(p + 4)(p – 4) $
Simplify: $ \frac{5p}{2(p + 4)} $.

Question 22: Vectors

In triangle OPT, $ \overrightarrow{OT} = \mathbf{t} $, $ \overrightarrow{OP} = \mathbf{p} $, with OK:KT = 2:1 and TL:LP = 2:1.

a(i)

Find $ \overrightarrow{PL} $ in terms of $ \mathbf{t} $ and $ \mathbf{p} $.

▶️Answer/Explanation

Correct answer: $ \frac{1}{3}(\mathbf{p} – \mathbf{t}) $

Explanation: $ \overrightarrow{PL} = \frac{1}{3} \overrightarrow{PT} = \frac{1}{3}(\mathbf{p} – \mathbf{t}) $.

a(ii)

Find $ \overrightarrow{KL} $ in terms of $ \mathbf{t} $ and $ \mathbf{p} $.

▶️Answer/Explanation

Correct answer: $ \frac{2}{3}\mathbf{p} – \frac{1}{3}\mathbf{t} $

Explanation: $ \overrightarrow{KL} = \overrightarrow{KO} + \overrightarrow{OL} = -\frac{2}{3}\mathbf{t} + \mathbf{p} – \frac{1}{3}(\mathbf{p} – \mathbf{t}) $.

Show that M lies on OP extended when $ \overrightarrow{KM} = -\frac{2}{3}\mathbf{t} + \frac{4}{3}\mathbf{p} $.

▶️Answer/Explanation

Explanation:
1. $ \overrightarrow{OM} = \overrightarrow{OK} + \overrightarrow{KM} = \frac{2}{3}\mathbf{t} – \frac{2}{3}\mathbf{t} + \frac{4}{3}\mathbf{p} = \frac{4}{3}\mathbf{p} $
2. This is a scalar multiple of $ \mathbf{p} $, so M lies on OP extended.

Question 23: Maximum Average Speed

Serge walks 7.9 km (nearest 100 m) in 133 minutes (nearest minute). Calculate the maximum possible average speed in km/h.

▶️Answer/Explanation

Correct answer: 3.6 km/h

Explanation:
Maximum distance: 7.95 km
Minimum time: 132.5 minutes = 2.2083 hours
Speed = $ \frac{7.95}{2.2083} \approx 3.6 \, \text{km/h} $.

Question 24: Intersection of Line and Curve

Find where $ y = 2x + 1 $ intersects $ y = x^2 + 3x – 4 $.

▶️Answer/Explanation

Correct answer: $ (-2.79, -4.58) $ and $ (1.79, 4.58) $

Explanation:
1. Set equations equal: $ x^2 + 3x – 4 = 2x + 1 $ ⇒ $ x^2 + x – 5 = 0 $
2. Solve quadratic: $ x = \frac{-1 \pm \sqrt{1 + 20}}{2} $ ⇒ $ x \approx -2.79, 1.79 $
3. Find y-values: $ y = 2(-2.79) + 1 \approx -4.58 $, $ y = 2(1.79) + 1 \approx 4.58 $.

CIE iGCSE Mathematics Paper 4 Prediction

CIE iGCSE Mathematics Paper 4 Prediction - 2025