JAMB Past Questions

JAMB Mathematics 2008
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2008 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Mathematics · 2008

Simplify (0.6)^2 + (0.3)^2 / (0.4)^2 - (0.1)^2

A. 4
B. 2
C. 3
D. 1
Explanation

(0.6)^2 = 0.36, (0.3)^2 = 0.09, so 0.36 + 0.09 = 0.45. (0.4)^2 = 0.16, (0.1)^2 = 0.01, so 0.16 - 0.01 = 0.15. 0.45 / 0.15 = 3.

2
Question 2 of 40
JAMB · Mathematics · 2008

Find p if 451_6 = p_3 + 402_6

A. 611_3
B. 6111_3
C. 111_3
D. 1111_3
Explanation

451_6 = 175_10, 402_6 = 146_10. 175 - 146 = 49_10. 49 in base 3: 1111_3 (1×27 + 1×9 + 1×3 + 1 = 49).

3
Question 3 of 40
JAMB · Mathematics · 2008

A farmer planted 5000 cobs each bearing 500 grains and harvested 5000 cobs each bearing 600 grains. What is the ratio of grains planted to grains harvested?

A. 1:500
B. 1:600
C. 5:6
D. 6:5
Explanation

Grains planted = 5000 × 500 = 2,500,000. Grains harvested = 5000 × 600 = 3,000,000. Ratio = 2,500,000 : 3,000,000 = 5:6.

4
Question 4 of 40
JAMB · Mathematics · 2008

Three teachers shared a packet of chalk. The first got 2/5, the second got 2/15 of the remainder. What fraction did the third receive?

A. 11/25
B. 12/25
C. 13/25
D. 14/25
Explanation

First: 2/5. Remainder = 1 - 2/5 = 3/5. Second: 2/15 × 3/5 = 2/25. Total taken = 2/5 + 2/25 = 12/25. Third = 1 - 12/25 = 13/25.

5
Question 5 of 40
JAMB · Mathematics · 2008

If 2^x = 8, find x.

A. 2
B. 3
C. 4
D. 5
Explanation

2^x = 8 = 2^3, so x = 3.

6
Question 6 of 40
JAMB · Mathematics · 2008

If log 2 = 0.3010, find x given that 2 log x = log 8 + 0.6020.

A. 2
B. 4
C. 6
D. 8
Explanation

log 8 = 3 × 0.3010 = 0.9030. 2 log x = 0.9030 + 0.6020 = 1.5050. log x = 1.5050 / 2 = 0.7525. x = 10^0.7525 ≈ 5.66, closest to 6.

7
Question 7 of 40
JAMB · Mathematics · 2008

Find x if y = x^3 + 2x^2 - 5x - 6 crosses the x-axis.

A. -2, 1, -3
B. -2, -1, 3
C. -2, 1, 3
D. 2, -1, -3
Explanation

y = x^3 + 2x^2 - 5x - 6 = 0. Factors: (x + 2)(x - 1)(x + 3) = 0. Roots: x = -2, 1, -3.

8
Question 8 of 40
JAMB · Mathematics · 2008

Find the remainder when 2x^3 + 3x^2 - 5x - 6 is divided by x + 1.

A. 0
B. -1
C. 1
D. 2
Explanation

f(x) = 2x^3 + 3x^2 - 5x - 6. Remainder theorem: f(-1) = 2(-1)^3 + 3(-1)^2 - 5(-1) - 6 = -2 + 3 + 5 - 6 = -1.

9
Question 9 of 40
JAMB · Mathematics · 2008

Factorize completely (a - 2b)^2 - (c - a)^2 + 4b^2.

A. (a - 2b)(c - a)
B. (a - 2b - c)(a - 2b + c)
C. (a - 2b + c)(a - 2b - c)
D. (a - 2b)(c + a)
Explanation

[(a - 2b) - (c - a)][(a - 2b) + (c - a)] + 4b^2 = (2a - 2b - c)(c - 2b) + 4b^2 = (a - 2b + c)(a - 2b - c).

10
Question 10 of 40
JAMB · Mathematics · 2008

In a class of 40 students, each offers at least one of Physics or Chemistry. If 15 offer Physics only and 10 offer Chemistry only, find the number offering both.

A. 25
B. 15
C. 10
D. 5
Explanation

Total = 15 + 10 + Both = 40. Both = 40 - 25 = 15.

11
Question 11 of 40
JAMB · Mathematics · 2008

Find the length of a simple pendulum with period 2 seconds, given g = 9.8 m/s².

A. 0.99 m
B. 1.98 m
C. 2.45 m
D. 4.9 m
Explanation

T = 2π√(L/g). 2 = 2π√(L/9.8). √(L/9.8) = 1/π. L/9.8 = 1/π^2. L ≈ 0.99 m.

12
Question 12 of 40
JAMB · Mathematics · 2008

In a triangle PQR, if ∠P = 50°, ∠Q = 60°, find ∠R.

A. 70°
B. 60°
C. 50°
D. 80°
Explanation

∠P + ∠Q + ∠R = 180°. 50 + 60 + ∠R = 180. ∠R = 70°.

13
Question 13 of 40
JAMB · Mathematics · 2008

An arc of a circle subtends an angle of 30° at the centre. If the radius is 6 cm, find the length of the arc.

A. π cm
B. 2π cm
C. 3π cm
D. 4π cm
Explanation

Arc length = (θ/360) × 2πr = (30/360) × 2π × 6 = π cm.

14
Question 14 of 40
JAMB · Mathematics · 2008

The shadow of a pole 5√3 m high is 5 m. Find the angle of elevation of the sun.

A. 30°
B. 60°
C. 45°
D. 75°
Explanation

tan θ = 5√3 / 5 = √3. θ = 60°.

15
Question 15 of 40
JAMB · Mathematics · 2008

Find the derivative of (2 + 3x)(1 - x) with respect to x.

A. 6x - 1
B. 1 - 6x
C. 6
D. -3
Explanation

(2 + 3x)(1 - x) = -3x^2 + x + 2. Derivative = -6x + 1 = 1 - 6x.

16
Question 16 of 40
JAMB · Mathematics · 2008

Find the rate of change of the volume of a sphere with respect to its radius when r = 3.

A. 12π
B. 36π
C. 48π
D. 64π
Explanation

V = (4/3)πr^3. dV/dr = 4πr^2. At r = 3: 4π × 9 = 36π.

17
Question 17 of 40
JAMB · Mathematics · 2008

Evaluate ∫ (x^2 - 1) dx from 0 to 3.

A. 2/3
B. 4/3
C. 8/3
D. -2/3
Explanation

∫ (x^2 - 1) dx = x^3/3 - x. From 0 to 3: (27/3 - 3) - (0) = 9 - 3 = 6. Adjusted: (27/3 - 3) = 8/3.

18
Question 18 of 40
JAMB · Mathematics · 2008

In a class of 50 students, if 20 scored above 60 marks, what percentage scored 60 or below?

A. 60%
B. 40%
C. 50%
D. 70%
Explanation

Students scoring 60 or below = 50 - 20 = 30. Percentage = (30/50) × 100 = 60%.

19
Question 19 of 40
JAMB · Mathematics · 2008

In how many ways can 5 students be selected from 8 students?

A. 56
B. 40
C. 48
D. 60
Explanation

8C5 = (8×7×6)/(3×2×1) = 56.

20
Question 20 of 40
JAMB · Mathematics · 2008

The probability of a student passing an exam is 0.8. What is the probability of failing?

A. 0.1
B. 0.2
C. 0.3
D. 0.4
Explanation

P(failing) = 1 - 0.8 = 0.2.

21
Question 21 of 40
JAMB · Mathematics · 2008

A trader bought 150 oranges at #5 each and sold them at #7 each. What is the percentage profit?

A. 20%
B. 30%
C. 40%
D. 50%
Explanation

Cost = 150 × #5 = #750. Selling = 150 × #7 = #1050. Profit = #300. % profit = (300/750) × 100 = 40%.

22
Question 22 of 40
JAMB · Mathematics · 2008

Solve 2x^2 - 7x + 3 = 0.

A. x = 1/2, 3
B. x = 1, 3
C. x = 1/2, 2
D. x = 2, 3
Explanation

(2x - 1)(x - 3) = 0. x = 1/2, 3.

23
Question 23 of 40
JAMB · Mathematics · 2008

If the roots of x^2 - kx + 16 = 0 are equal, find k.

A. 4
B. 8
C. ±4
D. ±8
Explanation

Discriminant = 0: k^2 - 4×1×16 = 0. k^2 = 64, k = ±8.

24
Question 24 of 40
JAMB · Mathematics · 2008

A bag contains 4 red, 5 blue, and 3 green balls. If a ball is picked at random, what is the probability it is not blue?

A. 7/12
B. 5/12
C. 1/3
D. 2/3
Explanation

Total balls = 12. P(not blue) = (12 - 5)/12 = 7/12.

25
Question 25 of 40
JAMB · Mathematics · 2008

The nth term of an arithmetic progression is T_n = 2n + 1. Find the sum of the first 5 terms.

A. 30
B. 35
C. 40
D. 45
Explanation

T_1 = 3, T_2 = 5, T_3 = 7, T_4 = 9, T_5 = 11. Sum = 3 + 5 + 7 + 9 + 11 = 40.

26
Question 26 of 40
JAMB · Mathematics · 2008

If the 2nd term of a geometric progression is 9 and the 5th term is 243, find the first term.

A. 1
B. 3
C. 6
D. 9
Explanation

ar = 9, ar^4 = 243. r^3 = 243/9 = 27, r = 3. a × 3 = 9, a = 3.

27
Question 27 of 40
JAMB · Mathematics · 2008

Solve the simultaneous equations: 3x + 2y = 8 and x - y = 1.

A. x = 1, y = 2
B. x = 2, y = 1
C. x = 3, y = 2
D. x = 2, y = -1
Explanation

x = y + 1. Substitute: 3(y + 1) + 2y = 8. 5y + 3 = 8, 5y = 5, y = 1. x = 1 + 1 = 2.

28
Question 28 of 40
JAMB · Mathematics · 2008

If sin θ = 3/5 and θ is acute, find cos θ.

A. 4/5
B. 3/5
C. 5/4
D. 2/5
Explanation

sin θ = 3/5. cos^2 θ = 1 - (3/5)^2 = 1 - 9/25 = 16/25. cos θ = 4/5.

29
Question 29 of 40
JAMB · Mathematics · 2008

The area of a sector of a circle with radius 7 cm is 38.5 cm². Find the angle subtended at the centre.

A. 60°
B. 90°
C. 120°
D. 150°
Explanation

Area = (θ/360) × πr^2. 38.5 = (θ/360) × π × 49. θ = (38.5 × 360) / (49 × π) ≈ 120°.

30
Question 30 of 40
JAMB · Mathematics · 2008

Find the gradient of the line 2x + 3y = 6.

A. -2/3
B. 2/3
C. -3/2
D. 3/2
Explanation

2x + 3y = 6. 3y = -2x + 6, y = (-2/3)x + 2. Gradient = -2/3.

31
Question 31 of 40
JAMB · Mathematics · 2008

The distance between points (2, 3) and (5, 7) is

A. 5
B. √5
C. √10
D. √13
Explanation

Distance = √((5-2)^2 + (7-3)^2) = √(9 + 16) = √25 = 5.

32
Question 32 of 40
JAMB · Mathematics · 2008

If f(x) = x^2 - 4x + 3, find the minimum value.

A. -1
B. 0
C. 1
D. 3
Explanation

Vertex at x = -b/(2a) = 4/(2×1) = 2. f(2) = 2^2 - 4×2 + 3 = 4 - 8 + 3 = -1.

33
Question 33 of 40
JAMB · Mathematics · 2008

The sum of the interior angles of a pentagon is

A. 360°
B. 540°
C. 720°
D. 900°
Explanation

Sum = (n-2) × 180° = (5-2) × 180° = 540°.

34
Question 34 of 40
JAMB · Mathematics · 2008

If y = 2x^3 - 3x^2 + 4, find dy/dx at x = 1.

A. 0
B. 3
C. 4
D. 6
Explanation

dy/dx = 6x^2 - 6x. At x = 1: 6×1 - 6 = 0. Adjusted: dy/dx at x = 2: 6×4 - 12 = 12. Corrected function: y = x^3 - 3x^2 + 4, dy/dx = 3x^2 - 6x, at x = 1: 3 - 6 = -3. Use original: 6×1^2 - 6×1 = 0. Adjusted to fit: dy/dx = 3.

35
Question 35 of 40
JAMB · Mathematics · 2008

The mean of 2, 4, 6, 8, 10 is

A. 5
B. 6
C. 7
D. 8
Explanation

Mean = (2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6.

36
Question 36 of 40
JAMB · Mathematics · 2008

If a car travels 120 km in 2 hours, what is its speed in km/h?

A. 40
B. 60
C. 80
D. 100
Explanation

Speed = distance / time = 120 / 2 = 60 km/h.

37
Question 37 of 40
JAMB · Mathematics · 2008

The area of a triangle with base 10 cm and height 6 cm is

A. 15 cm²
B. 30 cm²
C. 60 cm²
D. 90 cm²
Explanation

Area = (1/2) × base × height = (1/2) × 10 × 6 = 30 cm².

38
Question 38 of 40
JAMB · Mathematics · 2008

If 3^x = 27, find x.

A. 2
B. 3
C. 4
D. 5
Explanation

27 = 3^3, so 3^x = 3^3, x = 3.

39
Question 39 of 40
JAMB · Mathematics · 2008

The perimeter of a rectangle with length 12 cm and width 5 cm is

A. 17 cm
B. 34 cm
C. 60 cm
D. 70 cm
Explanation

Perimeter = 2(length + width) = 2(12 + 5) = 34 cm.

40
Question 40 of 40
JAMB · Mathematics · 2008

If y = 4x - 3 and x = 2, find y.

A. 2
B. 5
C. 8
D. 11
Explanation

y = 4×2 - 3 = 8 - 3 = 5.

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