🏆 Grade 8 Practice Paper 3

Cambridge Lower Secondary Stage 8 — Challenge Paper (Calculator Permitted)

⏱ 60 Minutes
📊 60 Marks
✅ Calculator Permitted
🏆 Multi-step problems
⏱ 60:00 — Click to start
Section A — Number & Proportion (15 marks)
1 [3]

A laptop costs $800. It depreciates by 25% in year 1 and 20% in year 2. Find the value after 2 years. Is the total percentage decrease more or less than 45%? Enter value after 2 years.

Year 1: 800 × 0.75 = 600. Year 2: 600 × 0.80 = 480. Less than 45% decrease (it's 40%).
2 [3]

Calculate (3.2 × 105) ÷ (8 × 10−2). Give your answer in standard form.

3.2/8 = 0.4, 10^5/10^{-2} = 10^7 → 0.4 × 10^7 = 4 × 10^6
3 [3]

Three friends share a bill. Ratio of contributions is 2 : 3 : 5. The middle person pays $42. What is the total bill, and what does the largest contributor pay?

Total:   Largest:
3 parts = $42, 1 part = $14, total = 10 × 14 = $140, largest = 5 × 14 = $70
4 [3]

A train travels at 90 km/h. Convert this to m/s (1 d.p.). Then find how far it travels in 25 seconds.

Speed m/s:   Distance: m
90 ÷ 3.6 = 25 m/s. Distance = 25 × 25 = 625 m.
5 [3]

y is inversely proportional to x². When x = 3, y = 8. Find y when x = 6.

k = 8 × 9 = 72. y = 72/36 = 2
Section B — Algebra (18 marks)
6 [3]

Expand and simplify: (2x + 3)(x − 4)

2x² − 8x + 3x − 12 = 2x² − 5x − 12
7 [3]

Solve the inequality: 2(3x + 1) > 5x + 8. List the first three integer solutions greater than the boundary.

Boundary x >   First 3 integers:
6x + 2 > 5x + 8 → x > 6. First 3: 7, 8, 9
8 [3]

f(x) = 2x − 1 and g(x) = x² + 3. Find: (a) gf(3)  (b) fg(2)  (c) f⁻¹(9)

gf(3) =   fg(2) =   f⁻¹(9) =
gf(3)=g(5)=28, fg(2)=f(7)=13, f⁻¹(9)=(9+1)/2=5
9 [3]

The nth term of a sequence is 4n − 3. Find the first term, the 10th term, and which term equals 97.

T₁ =   T₁₀ =   n =
T₁=1, T₁₀=37, 4n−3=97 → n=25
10 [3]

Two lines: L₁: y = 3x − 2 and L₂ passes through (0, 5) and is perpendicular to L₁. Find the equation of L₂ and their intersection point.

Perpendicular gradient = −1/m. Substitute to find intersection.
L₂: y =   Intersection:
L₂: y=−(1/3)x+5. Set equal: 3x−2=−x/3+5 → 10x/3=7 → x=2.1, y=4.3. Exact: x=21/10, y=43/10
11 [3]

A rectangle has length (3x + 2) cm and width (x + 1) cm. Its area is 52 cm². Form a quadratic equation and solve for x (take positive solution only, 1 d.p.).

3x²+5x+2=52 → 3x²+5x−50=0. x=(−5+√(25+600))/6=(−5+25)/6=20/6≈3.3
Section C — Geometry & Measures (15 marks)
12 [3]

A hemisphere has diameter 12 cm. Find: (a) its total surface area (= 2πr² + πr² = 3πr²)  (b) its volume. Give both answers to 1 d.p.

SA:   Vol:
r=6. SA=3π×36=339.3 cm². Vol=(2/3)π×216=452.4 cm².
13 [3]

Two similar solids have volumes in ratio 27 : 125. Find the linear scale factor and the ratio of their surface areas.

Scale factor:   SA ratio (smaller:larger):
k³=27/125 → k=3/5=0.6. SA ratio=k²=9:25.
14 [3]

A 3D shape has 6 faces, 12 edges. Using Euler's formula, find the number of vertices. Name a common solid with these properties.

V =   Name:
F+V−E=2 → 6+V−12=2 → V=8. A cuboid (rectangular prism).
15 [3]

A shape is enlarged by scale factor 2 from centre (0, 0). The original triangle has vertices A(1, 2), B(3, 2), C(2, 4). What are the coordinates of B'?

B' =
B' = 2×(3,2) = (6,4)
16 [3]

A composite shape is a rectangle (8 cm × 5 cm) with a semicircle of diameter 5 cm removed from one end. Find the area. (2 d.p.)

Rectangle = 40. Semicircle = π×(2.5)²/2 = 9.82. Area = 40 − 9.82 = 30.18 cm²
Section D — Statistics & Probability (12 marks)
17 [3]

The back-to-back stem-and-leaf diagram shows test scores for two classes:

Class A (leaves)StemClass B (leaves)
8 6 432 5 7
9 7 5 240 3 6 8
8 6 152 5 9
461

Find the median of Class A.

 

Which class has the higher median?

Class A sorted: 34,36,38,42,45,47,49,51,56,58,64. n=11, median=47 (6th value). Class B sorted: 32,35,37,40,43,46,48,52,55,59,61. n=11, median=46. Class A has higher median.
18 [3]

A bag has 4 red and 6 blue balls. Two balls are drawn WITHOUT replacement. Find P(both red).

P(RR) = (4/10) × (3/9) = 12/90 = 2/15
19 [3]

A grouped frequency table:

Time (min)Frequency
0 – 103
10 – 207
20 – 3015
30 – 4010
40 – 505

Find: (a) estimated mean (1 d.p.)  (b) frequency density for the 20–30 class

Mean:   Freq density:
Mean=(3×5+7×15+15×25+10×35+5×45)/40=(15+105+375+350+225)/40=1070/40=26.75≈26.8. Freq density=15/10=1.5
20 [3]

P(A) = 0.5, P(B) = 0.4. A and B are independent. Find: (a) P(A and B)  (b) P(A or B)  (c) P(neither)

P(A∩B) =   P(A∪B) =   P(neither) =
P(A∩B)=0.2. P(A∪B)=0.5+0.4−0.2=0.7. P(neither)=1−0.7=0.3.

Your Result