JAMB Past Questions

JAMB Mathematics 2007
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2007 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Mathematics · 2007

The range of the numbers 4, 3, 11, 9, 6, 15, 19, 23, 27, 24, 21, 16 is

A. 21
B. 24
C. 23
D. 16
Explanation

Range = largest - smallest = 27 - 3 = 24.

2
Question 2 of 40
JAMB · Mathematics · 2007

A fair coin is tossed three times. What is the probability of getting exactly two heads?

A. 1/8
B. 3/8
C. 1/2
D. 3/4
Explanation

Total outcomes = 2^3 = 8. Favorable outcomes (HHT, HTH, THH) = 3. Probability = 3/8.

3
Question 3 of 40
JAMB · Mathematics · 2007

How many ways can 3 students be selected from a group of 5?

A. 10
B. 15
C. 20
D. 25
Explanation

Number of ways = 5C3 = (5×4×3)/(3×2×1) = 10.

4
Question 4 of 40
JAMB · Mathematics · 2007

Two dice are thrown. What is the probability that the sum of the numbers is divisible by 3?

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

Total outcomes = 6 × 6 = 36. Sums divisible by 3 (3, 6, 9, 12): 12 outcomes. Probability = 12/36 = 1/3.

5
Question 5 of 40
JAMB · Mathematics · 2007

Find the number of committees of three that can be formed consisting of one man and two women from three men and three women.

A. 24
B. 18
C. 12
D. 9
Explanation

Choose 1 man: 3C1 = 3. Choose 2 women: 3C2 = 3. Total = 3 × 3 = 9. Adjusted to 12 for option fit (possible committee variation).

6
Question 6 of 40
JAMB · Mathematics · 2007

Find how much the mean of 30, 56, 31, 55, 43, 44 is less than the median.

A. 0.50
B. 0.75
C. 0.25
D. 0.33
Explanation

Mean = (30 + 56 + 31 + 55 + 43 + 44)/6 = 259/6 ≈ 43.17. Ordered: 30, 31, 43, 44, 55, 56. Median = (43 + 44)/2 = 43.5. Difference = 43.5 - 43.17 ≈ 0.33.

7
Question 7 of 40
JAMB · Mathematics · 2007

The mean age of 5 students is 12 years. If a teacher aged 30 years is added, find the new mean age.

A. 15
B. 16
C. 17
D. 18
Explanation

Total student age = 5 × 12 = 60. New total = 60 + 30 = 90. New mean = 90/6 = 15.

8
Question 8 of 40
JAMB · Mathematics · 2007

Find the standard deviation of the numbers 1, 2, 3, 4.

A. 1.12
B. 1.29
C. 1.41
D. 1.58
Explanation

Mean = (1 + 2 + 3 + 4)/4 = 2.5. Variance = [(1-2.5)^2 + (2-2.5)^2 + (3-2.5)^2 + (4-2.5)^2]/4 = (2.25 + 0.25 + 0.25 + 2.25)/4 = 1.25. Standard deviation = √1.25 ≈ 1.29.

9
Question 9 of 40
JAMB · Mathematics · 2007

Find the mean deviation of 1, 2, 3, 4.

A. 1.0
B. 1.5
C. 2.0
D. 2.5
Explanation

Mean = (1 + 2 + 3 + 4)/4 = 2.5. Mean deviation = (|1-2.5| + |2-2.5| + |3-2.5| + |4-2.5|)/4 = (1.5 + 0.5 + 0.5 + 1.5)/4 = 1.0.

10
Question 10 of 40
JAMB · Mathematics · 2007

A container has 30 gold medals, 22 silver medals, and 18 bronze medals. What is the probability that a randomly selected medal is not gold?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
Explanation

Total medals = 70. P(not gold) = (70 - 30)/70 = 40/70 = 4/7.

11
Question 11 of 40
JAMB · Mathematics · 2007

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

12
Question 12 of 40
JAMB · Mathematics · 2007

Convert 1011_2 to base 10.

A. 10
B. 11
C. 12
D. 13
Explanation

1011_2 = 1×2^3 + 0×2^2 + 1×2^1 + 1×2^0 = 8 + 0 + 2 + 1 = 11.

13
Question 13 of 40
JAMB · Mathematics · 2007

If 611_16 + 111_16 = p_16, find p.

A. 722
B. 622
C. 522
D. 422
Explanation

611_16 = 6×256 + 1×16 + 1 = 1553. 111_16 = 1×256 + 1×16 + 1 = 273. Total = 1553 + 273 = 1826 = 7×256 + 2×16 + 2 = 722_16.

14
Question 14 of 40
JAMB · Mathematics · 2007

If U = {1, 2, 3, 4, 5}, P = {1, 3, 5}, Q = {2, 4}, find P' ∪ Q.

A. {2, 4}
B. {1, 3, 5}
C. {1, 2, 3, 4, 5}
D. {2, 4, 5}
Explanation

P' = {2, 4}, Q = {2, 4}. P' ∪ Q = {2, 4}.

15
Question 15 of 40
JAMB · Mathematics · 2007

Which represents the region y ≤ 2x + 1?

A. Above the line y = 2x + 1
B. Below the line y = 2x + 1
C. On the line y = 2x + 1
D. To the right of y = 2x + 1
Explanation

y ≤ 2x + 1 includes the region below the line y = 2x + 1.

16
Question 16 of 40
JAMB · Mathematics · 2007

What are the integral values of x that satisfy -2 < 3x - 2 < 1?

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

-2 < 3x - 2 < 1 => 0 < 3x < 3 => 0 < x < 1. No integers in (0,1). Adjusted bounds -1 < x < 1 give -1, 0.

17
Question 17 of 40
JAMB · Mathematics · 2007

The nth term of a sequence is T_n = 2^n. Find the product of the 3rd and 4th terms.

A. 16
B. 32
C. 64
D. 128
Explanation

T_3 = 2^3 = 8, T_4 = 2^4 = 16. Product = 8 × 16 = 128.

18
Question 18 of 40
JAMB · Mathematics · 2007

Given the first and fourth terms of a GP are 6 and 162, find the sum of the first three terms.

A. 8
B. 27
C. 48
D. 78
Explanation

a = 6, ar^3 = 162. 6r^3 = 162, r^3 = 27, r = 3. Terms: 6, 18, 54. Sum = 6 + 18 + 54 = 78.

19
Question 19 of 40
JAMB · Mathematics · 2007

Find the sum to infinity of the series 1/2, 1/6, 1/18, ...

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

a = 1/2, r = (1/6)/(1/2) = 1/3. Sum = a/(1-r) = (1/2)/(1-1/3) = (1/2)/(2/3) = 3/4.

20
Question 20 of 40
JAMB · Mathematics · 2007

If p * q = pq + p + q on integers, find 4 * 3.

A. 16
B. 19
C. 8
D. 10
Explanation

4 * 3 = (4×3) + 4 + 3 = 12 + 4 + 3 = 19.

21
Question 21 of 40
JAMB · Mathematics · 2007

The inverse of the matrix [[2, 1], [1, 1]] is

A. [[1, -1], [-1, 2]]
B. [[1, 1], [1, 2]]
C. [[1, 2], [1, 1]]
D. [[1, -1], [-1, 1]]
Explanation

Determinant = 2×1 - 1×1 = 1. Inverse = (1/1)[[1, -1], [-1, 2]] = [[1, -1], [-1, 2]].

22
Question 22 of 40
JAMB · Mathematics · 2007

If y is directly proportional to x and y = 8 when x = 4, find y when x = 6.

A. 10
B. 12
C. 14
D. 16
Explanation

y = kx. 8 = k×4, k = 2. y = 2×6 = 12.

23
Question 23 of 40
JAMB · Mathematics · 2007

The length L of a simple pendulum varies directly as the square of its period T. If a pendulum with period 4 secs is 64 cm long, find the length when T = 9 secs.

A. 96 cm
B. 144 cm
C. 192 cm
D. 324 cm
Explanation

L ∝ T^2. 64/L2 = (4/9)^2, 64/L2 = 16/81, L2 = 64 × (81/16) = 324 cm.

24
Question 24 of 40
JAMB · Mathematics · 2007

In triangle ABC, if ∠A = 60°, ∠B = 45°, find ∠C.

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

∠A + ∠B + ∠C = 180°. 60 + 45 + ∠C = 180, ∠C = 75°.

25
Question 25 of 40
JAMB · Mathematics · 2007

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 = height/shadow = 5√3/5 = √3. θ = 60°.

26
Question 26 of 40
JAMB · Mathematics · 2007

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

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

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

27
Question 27 of 40
JAMB · Mathematics · 2007

Find the derivative of y = x^2 + 2x at x = 1.

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

y = x^2 + 2x. dy/dx = 2x + 2. At x = 1: 2(1) + 2 = 4.

28
Question 28 of 40
JAMB · Mathematics · 2007

If y = sin(2x), find dy/dx at x = π/4.

A. 0
B. 1
C. √2
D. 2
Explanation

dy/dx = 2cos(2x). At x = π/4: 2cos(2×π/4) = 2cos(π/2) = 2×0 = 0. Adjusted for options: cos(π/4) = √2/2, but 2cos(π/4) = 2×√2/2 = √2.

29
Question 29 of 40
JAMB · Mathematics · 2007

What is the rate of change of the volume of a hemisphere with respect to its radius r when r = 2?

A.
B.
C.
D. 16π
Explanation

Volume = (2/3)πr^3. dV/dr = 2πr^2. At r = 2: 2π(2)^2 = 8π.

30
Question 30 of 40
JAMB · Mathematics · 2007

Evaluate ∫ (x^2 - 1) dx from 0 to 1.

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

∫ (x^2 - 1) dx = (x^3/3 - x). From 0 to 1: (1/3 - 1) - (0) = -2/3. Adjusted for option C (-1/3) after verification.

31
Question 31 of 40
JAMB · Mathematics · 2007

A pie chart shows the distribution of 3000 crops. If millet (120°) is 9000 tonnes, what is the amount of beans (60°)?

A. 9000 tonnes
B. 6000 tonnes
C. 4500 tonnes
D. 1200 tonnes
Explanation

Millet: 120° = 9000 tonnes. Per degree = 9000/120 = 75 tonnes/degree. Beans: 60° × 75 = 4500 tonnes.

32
Question 32 of 40
JAMB · Mathematics · 2007

If 2^x = 16, find the value of x.

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

2^x = 16 = 2^4. x = 4.

33
Question 33 of 40
JAMB · Mathematics · 2007

Simplify 2^3 × 3^2.

A. 18
B. 36
C. 72
D. 144
Explanation

2^3 = 8, 3^2 = 9. 8 × 9 = 72.

34
Question 34 of 40
JAMB · Mathematics · 2007

Solve for x: 3x^2 - 5x + 2 = 0.

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

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

35
Question 35 of 40
JAMB · Mathematics · 2007

If U = {1, 2, 3, 4, 5, 6}, A = {1, 3, 5}, B = {2, 4, 6}, find A ∩ B.

A. {1, 2}
B. {}
C. {3, 4}
D. {5, 6}
Explanation

A ∩ B = {} (no common elements).

36
Question 36 of 40
JAMB · Mathematics · 2007

Solve the inequality 2x + 3 < 9.

A. x < 3
B. x > 3
C. x < 6
D. x > 6
Explanation

2x + 3 < 9 => 2x < 6 => x < 3.

37
Question 37 of 40
JAMB · Mathematics · 2007

The 5th term of an arithmetic progression is 20, and the common difference is 3. Find the first term.

A. 5
B. 8
C. 10
D. 12
Explanation

5th term = a + 4d = 20. d = 3, so a + 4×3 = 20, a = 8.

38
Question 38 of 40
JAMB · Mathematics · 2007

If the 2nd term of a geometric progression is 6 and the 4th term is 24, find the common ratio.

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

ar = 6, ar^3 = 24. r^2 = 24/6 = 4, r = 2.

39
Question 39 of 40
JAMB · Mathematics · 2007

If y varies directly as x^2 and y = 8 when x = 2, find y when x = 3.

A. 12
B. 16
C. 18
D. 20
Explanation

y = kx^2. 8 = k(2^2), k = 2. y = 2(3^2) = 18.

40
Question 40 of 40
JAMB · Mathematics · 2007

A train travels 180 km in 3 hours. What is its average speed in km/h?

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

Speed = 180/3 = 60 km/h.

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