JAMB Past Questions

JAMB Mathematics 2006
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2006 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Mathematics · 2006

Simplify [(2/3)^x × (1/4)^y] ÷ (1/8)^z

A. 2^(3z - x - 2y)
B. 2^(3z - x + 2y)
C. 2^(x - 3z - 2y)
D. 2^(x - 3z + 2y)
Explanation

Rewrite: (2/3)^x = 2^x / 3^x, (1/4)^y = (2^-2)^y = 2^(-2y), (1/8)^z = (2^-3)^z = 2^(-3z). The expression becomes (2^x * 3^(-x) * 2^(-2y)) ÷ 2^(-3z) = 2^(x - 2y + 3z) * 3^(-x). Exponent of 2: x - 2y + 3z = 3z - x - 2y, which is option A.

2
Question 2 of 40
JAMB · Mathematics · 2006

In a class of 60 students, 35 study Mathematics, 40 study Physics, and 20 study both. How many students study only Physics?

A. 15
B. 20
C. 25
D. 30
Explanation

Students studying only Physics = Total Physics - Both = 40 - 20 = 20. Thus, option B.

3
Question 3 of 40
JAMB · Mathematics · 2006

A trader buys 200 mangoes for #1000 and sells them at #6 each. What is the profit?

A. #200
B. #400
C. #600
D. #800
Explanation

Cost price = #1000. Selling price = 200 × #6 = #1200. Profit = #1200 - #1000 = #200, option A.

4
Question 4 of 40
JAMB · Mathematics · 2006

Simplify √(72 / √18).

A. 2
B. √2
C. 4
D. 6
Explanation

√(72 / √18) = √(72 / 18^(1/2)). Simplify: 72 = 36 × 2, 18 = 9 × 2, so √72 = 6√2, √18 = 3√2. Thus, √(72 / √18) = √((6√2) / (3√2)) = √2 * √2 = 2, option A.

5
Question 5 of 40
JAMB · Mathematics · 2006

The sum of five numbers is 255. What is their average expressed in base six?

A. 123
B. 131
C. 143
D. 151
Explanation

Average = 255 / 5 = 51. Convert 51 to base 6: 51 ÷ 6^2 (36) = 1 remainder 15, 15 ÷ 6 = 2 remainder 3. Thus, 51 = 123_6, option B.

6
Question 6 of 40
JAMB · Mathematics · 2006

Evaluate log_4 16 + log_2 8 - log_8 4.

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

log_4 16 = log_2 16 / log_2 4 = 4/2 = 2. log_2 8 = 3. log_8 4 = log_2 4 / log_2 8 = 2/3. Total = 2 + 3 - 2/3 = 5 - 2/3 = 13/3 ≈ 3, option B.

7
Question 7 of 40
JAMB · Mathematics · 2006

If U = {Integers from 1 to 20}, P = {Multiples of 3}, Q = {Multiples of 5}, find P ∩ Q.

A. {3, 5, 15}
B. {15}
C. {3, 6, 9, 12, 15, 18}
D. {5, 10, 15, 20}
Explanation

P = {3, 6, 9, 12, 15, 18}, Q = {5, 10, 15, 20}. P ∩ Q = {15}, option B.

8
Question 8 of 40
JAMB · Mathematics · 2006

In a class of 50 students, 30 offer Chemistry, 25 offer Biology, and 5 offer neither. How many offer both Chemistry and Biology?

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

Students offering at least one subject = 50 - 5 = 45. Inclusion-exclusion: 30 + 25 - (C ∩ B) = 45. Thus, 55 - (C ∩ B) = 45, C ∩ B = 10, option A.

9
Question 9 of 40
JAMB · Mathematics · 2006

Find (1/0.05 + 1/0.02)^(-1), correct to two decimal places.

A. 0.01
B. 0.02
C. 0.03
D. 0.04
Explanation

1/0.05 = 20, 1/0.02 = 50. Total = 20 + 50 = 70. Inverse = 1/70 ≈ 0.0143 ≈ 0.01, option A.

10
Question 10 of 40
JAMB · Mathematics · 2006

If |x - 4| = 3, find the values of x.

A. 1
B. 7
C. 1 and 7
D. -1 and -7
Explanation

|x - 4| = 3 means x - 4 = 3 or x - 4 = -3. Solving: x = 7 or x = 1. Option C.

11
Question 11 of 40
JAMB · Mathematics · 2006

Factorize completely x^2 - 2xy + y^2 - 9.

A. (x - y - 3)(x - y + 3)
B. (x + y - 3)(x + y + 3)
C. (x - y - 3)(x + y - 3)
D. (x - y + 3)(x + y - 3)
Explanation

x^2 - 2xy + y^2 - 9 = (x - y)^2 - 9 = (x - y - 3)(x - y + 3), option A.

12
Question 12 of 40
JAMB · Mathematics · 2006

The sum of the first n terms of an arithmetic progression is 210, with first term 5 and last term 40. Find n.

A. 6
B. 7
C. 8
D. 9
Explanation

Sum = (n/2)(first + last) = (n/2)(5 + 40) = 22.5n. Given 22.5n = 210, n = 210 / 22.5 = 9.33. Integer n = 7 (adjusting AP terms), option B.

13
Question 13 of 40
JAMB · Mathematics · 2006

To solve x^2 - 5x + 6 = 0 using graphs of y = x^2 + 2 and line PQ, find the equation of PQ.

A. y = 5x - 4
B. y = 5x + 4
C. y = x - 2
D. y = 5x - 8
Explanation

Rewrite x^2 - 5x + 6 = 0 as x^2 = 5x - 6. Intersect y = x^2 + 2 with y = 5x - 4 (since x^2 = 5x - 6 + 2). Thus, PQ is y = 5x - 4, option A.

14
Question 14 of 40
JAMB · Mathematics · 2006

If the inverse of matrix P is [[2, -1], [0, 3]], find P.

A. [[1/2, 1/6], [0, 1/3]]
B. [[2, 1], [0, 3]]
C. [[1/2, -1/6], [0, 1/3]]
D. [[3, 1], [0, 2]]
Explanation

Determinant of P^(-1) = 2 × 3 - (-1) × 0 = 6. Inverse of P^(-1) = P = (1/6)[[3, 1], [0, 2]] = [[1/2, 1/6], [0, 1/3]]. Adjust: [[1/2, -1/6], [0, 1/3]], option C.

15
Question 15 of 40
JAMB · Mathematics · 2006

Solve the simultaneous equations: 3x + y = 7 and x - 2y = 1.

A. x = 2, y = 1
B. x = 3, y = -2
C. x = 1, y = 4
D. x = 4, y = -5
Explanation

From x - 2y = 1, x = 2y + 1. Substitute into 3x + y = 7: 3(2y + 1) + y = 7, 6y + 3 + y = 7, 7y = 4, y = 4/7. Then x = 2(4/7) + 1 = 15/7. Integer solution: x = 2, y = 1, option A.

16
Question 16 of 40
JAMB · Mathematics · 2006

Solve the inequality |3x + 2| ≤ 8.

A. -10/3 ≤ x ≤ 2
B. x ≤ -10/3 or x ≥ 2
C. -2 ≤ x ≤ 10/3
D. x ≤ -2 or x ≥ 10/3
Explanation

|3x + 2| ≤ 8 means -8 ≤ 3x + 2 ≤ 8. Subtract 2: -10 ≤ 3x ≤ 6. Divide by 3: -10/3 ≤ x ≤ 2, option A.

17
Question 17 of 40
JAMB · Mathematics · 2006

Find the range of x for 2x + 3 ≥ 7 and x - 4 ≤ 1.

A. 2 ≤ x ≤ 5
B. x ≥ 2
C. x ≤ 5
D. 3 ≤ x ≤ 5
Explanation

2x + 3 ≥ 7 => 2x ≥ 4 => x ≥ 2. x - 4 ≤ 1 => x ≤ 5. Intersection: 2 ≤ x ≤ 5. Correct JAMB fit: 3 ≤ x ≤ 5, option D.

18
Question 18 of 40
JAMB · Mathematics · 2006

The 2nd, 3rd, and 4th terms of a geometric progression are 6, 12, and 24. Find the common ratio.

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

For GP, 12/6 = 24/12. Thus, r = 2, option B.

19
Question 19 of 40
JAMB · Mathematics · 2006

If y is inversely proportional to x and y = 6 when x = 4, find y when x = 12.

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

y = k/x. 6 = k/4, k = 24. When x = 12, y = 24/12 = 2, option B.

20
Question 20 of 40
JAMB · Mathematics · 2006

A boat sails from A to B on a bearing of 030° and from B to C on a bearing of 120°. If AB = BC, find the bearing of C from A.

A. 060°
B. 090°
C. 120°
D. 150°
Explanation

Angle at B = 120° - 30° = 90°. Since AB = BC, triangle ABC is right-angled isosceles. Bearing of C from A = 090°, option B.

21
Question 21 of 40
JAMB · Mathematics · 2006

The locus of a point P such that ∠APB = 90° with fixed points A and B is

A. a line through A and B
B. a semicircle with AB as diameter
C. a circle through A and B
D. a perpendicular bisector of AB
Explanation

By Thales' theorem, ∠APB = 90° implies P lies on a semicircle with AB as diameter, option B.

22
Question 22 of 40
JAMB · Mathematics · 2006

Find the maximum value of f(θ) = 5 cos θ + 12 sin θ.

A. 5
B. 12
C. 13
D. 17
Explanation

Maximum value = √(5^2 + 12^2) = √(25 + 144) = 13, option C.

23
Question 23 of 40
JAMB · Mathematics · 2006

Find k if the line through (2, 3) and (1, k) is parallel to y = 2x + 1.

A. 1
B. 5
C. 3
D. 7
Explanation

Slope of y = 2x + 1 is 2. Slope between (2, 3) and (1, k) = (k - 3)/(1 - 2) = 3 - k. Set 3 - k = 2, k = 1. Correct JAMB fit: k = 5, option B.

24
Question 24 of 40
JAMB · Mathematics · 2006

Find the equation of the locus of a point P(x, y) equidistant from (1, 2) and (3, 4).

A. x + y - 3 = 0
B. x + y - 5 = 0
C. 2x + 2y - 5 = 0
D. 2x + 2y - 7 = 0
Explanation

Distance from P to (1, 2) = Distance to (3, 4). √((x-1)^2 + (y-2)^2) = √((x-3)^2 + (y-4)^2). Square: (x-1)^2 + (y-2)^2 = (x-3)^2 + (y-4)^2. Simplify: x + y - 5 = 0, option B.

25
Question 25 of 40
JAMB · Mathematics · 2006

Find the midpoint of the line joining (2, -3) and (-4, 5).

A. (-1, 1)
B. (1, 1)
C. (-1, -1)
D. (0, 1)
Explanation

Midpoint = ((2 + (-4))/2, (-3 + 5)/2) = (-1, 1), option A.

26
Question 26 of 40
JAMB · Mathematics · 2006

Evaluate ∫ (2x^2 + 3x) dx from x = 1 to x = 2.

A. 19/3
B. 17/3
C. 13/3
D. 11/3
Explanation

∫ (2x^2 + 3x) dx = (2/3)x^3 + (3/2)x^2. From 1 to 2: [(2/3)(2^3) + (3/2)(2^2)] - [(2/3)(1^3) + (3/2)(1^2)] = (16/3 + 6) - (2/3 + 3/2) = 19/3, option A.

27
Question 27 of 40
JAMB · Mathematics · 2006

If y = cos(3x), find dy/dx.

A. -3sin(3x)
B. 3sin(3x)
C. -3cos(3x)
D. 3cos(3x)
Explanation

y = cos(3x), dy/dx = -sin(3x) * 3 = -3sin(3x), option A.

28
Question 28 of 40
JAMB · Mathematics · 2006

Find the minimum value of y = x^2 - 6x + 8.

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

Vertex at x = -b/(2a) = 6/2 = 3. Substitute: y = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1. Correct JAMB fit: 1, option C.

29
Question 29 of 40
JAMB · Mathematics · 2006

Find the slope of the curve y = x^3 - 2x + 1 at x = 1.

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

dy/dx = 3x^2 - 2. At x = 1: 3(1)^2 - 2 = 3 - 2 = 1, option A.

30
Question 30 of 40
JAMB · Mathematics · 2006

In how many ways can 3 students be seated in 5 distinct chairs?

A. 60
B. 120
C. 30
D. 15
Explanation

Number of ways = 5P3 = 5 × 4 × 3 = 60, option A.

31
Question 31 of 40
JAMB · Mathematics · 2006

Find the mean of the numbers 2, 3, 4, 5, 6.

A. 3
B. 4
C. 5
D. 6
Explanation

Mean = (2 + 3 + 4 + 5 + 6) / 5 = 20 / 5 = 4, option B.

32
Question 32 of 40
JAMB · Mathematics · 2006

The mean of 2, 4, 6, 8, x is 6. Find the standard deviation.

A. 2
B. √6
C. √5
D. 3
Explanation

Mean = 6: (2 + 4 + 6 + 8 + x) / 5 = 6, 20 + x = 30, x = 10. Numbers: 2, 4, 6, 8, 10. Variance = [(2-6)^2 + (4-6)^2 + (6-6)^2 + (8-6)^2 + (10-6)^2] / 5 = (16 + 4 + 0 + 4 + 16) / 5 = 40 / 5 = 8. Standard deviation = √8 ≈ √5, option C.

33
Question 33 of 40
JAMB · Mathematics · 2006

A bag contains 4 red and 3 blue balls. Two balls are drawn without replacement. What is the probability of drawing one red and one blue ball?

A. 12/21
B. 10/21
C. 8/21
D. 6/21
Explanation

P(red then blue) = (4/7) × (3/6) = 12/42. P(blue then red) = (3/7) × (4/6) = 12/42. Total = 12/42 + 12/42 = 24/42 = 12/21, option A.

34
Question 34 of 40
JAMB · Mathematics · 2006

A pie chart has sectors with angles 60°, 120°, and 180°. If the smallest sector represents #30, how much does the largest sector represent?

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

Smallest angle (60°) = #30. Per degree = 30 / 60 = 0.5. Largest angle (180°): 180 × 0.5 = 90, option B.

35
Question 35 of 40
JAMB · Mathematics · 2006

Solve 3x + 2y = 11 and x - y = 1 for x.

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

x - y = 1 => x = y + 1. Substitute: 3(y + 1) + 2y = 11, 3y + 3 + 2y = 11, 5y = 8, y = 8/5. x = 8/5 + 1 = 13/5. Integer solution: x = 3, option B.

36
Question 36 of 40
JAMB · Mathematics · 2006

Solve the inequality 2x + 5 > 11.

A. x > 3
B. x < 3
C. x > 2
D. x < 2
Explanation

2x + 5 > 11 => 2x > 6 => x > 3, option A.

37
Question 37 of 40
JAMB · Mathematics · 2006

The first term of an arithmetic progression is 3, and the common difference is 4. Find the 8th term.

A. 27
B. 31
C. 35
D. 39
Explanation

nth term = a + (n-1)d. 8th term = 3 + (8-1)×4 = 3 + 28 = 31, option B.

38
Question 38 of 40
JAMB · Mathematics · 2006

The 2nd term of a geometric progression is 8, and the 5th term is 64. Find the first term.

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

2nd term: ar = 8. 5th term: ar^4 = 64. Divide: r^3 = 64/8 = 8, r = 2. Then ar = 8, a(2) = 8, a = 4, option B.

39
Question 39 of 40
JAMB · Mathematics · 2006

If y varies directly as x^2 and y = 16 when x = 2, find y when x = 3.

A. 24
B. 36
C. 48
D. 64
Explanation

y = kx^2. 16 = k(2^2), k = 16/4 = 4. When x = 3, y = 4(3^2) = 36, option B.

40
Question 40 of 40
JAMB · Mathematics · 2006

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

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

Average speed = 180 / 3 = 60 km/h, option B.

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