JAMB CBT Mock Exam with Explanations – 15 Minutes

JAMB CBT Mock Exam

This mock exam contains 20 questions and must be completed in 15 minutes. Use the “Next” and “Previous” buttons to navigate between questions. When you click “Start Quiz,” there will be a 5‑second delay before the timer begins. After you submit your answers, you can view detailed explanations for each question.

Time left: 15:00

Question 1 of 20

Find the distance between the points (–1, –3) and (–2, 4).

A) 7.1m
B) 3m
C) 2.4m
D) 3.2m

Correct Answer: A) 7.1m

Question 2 of 20

If p : 2 = 2 : 3 and q : r = 4 : 7, find p : q : r.

A) 2 : 7 : 7
B) 4 : 12 : 21
C) 2 : 1 : 3
D) 1 : 3 : 5

Correct Answer: B) 4 : 12 : 21

Question 3 of 20

What is the probability that an integer x (1 ≤ x ≤ 25) chosen at random is divisible by both 2 and 3?

A) 4/25
B) 1/2
C) 3/4
D) 1/5

Correct Answer: A) 4/25

Question 4 of 20

Given that A = {3, 4, 1, 10, ⅓} and B = {4, 3, ⅓, 7}, find A ∩ B.

A) {3, 4, ⅓}
B) { }
C) {1, 3, ⅓}
D) {⅓, 1, 2}

Correct Answer: A) {3, 4, ⅓}

Question 5 of 20

Simplify (cos x/(1+sin x)) + (cos x/(1–sin x))2sec x.

A) 2cos x
B) 2sin x
C) 2tan x
D) 2sec x

Correct Answer: D) 2sec x

Question 6 of 20

Evaluate sin 45°.

A) √2/2
B) √2/4
C) √2/3
D) √2

Correct Answer: A) √2/2

Question 7 of 20

Find the probability that a number picked at random from the set {43, 44, 45, …, 60} is a prime number.

A) 2/3
B) 1/3
C) 7/9
D) 2/9

Correct Answer: D) 2/9

Question 8 of 20

What is the nth term of the progression 8, –4, 2? Find the 10th term of the progression.

A) 1 ÷ 2⁴
B) –2⁻⁶
C) 2¼
D) 1 ÷ 2¹⁵

Correct Answer: B) –2⁻⁶ (i.e. –1/64)

Question 9 of 20

A solid metal cube of side 3cm is placed in a rectangular tank of dimensions 3, 4 and 5cm. What volume of water can the tank now hold?

A) 60 cm³
B) 27 cm³
C) 48 cm³
D) 33 cm³

Correct Answer: D) 33 cm³

Question 10 of 20

T varies inversely as the cube of R. When R = 3, T = 2/81, find T when R = 2.

A) 1/12
B) 1/18
C) 1/24
D) 1/6

Correct Answer: A) 1/12

Question 11 of 20

Find the total surface area of a cylinder of height 6cm and radius of base 4cm (Take π = 3.142).

A) 251 cm²
B) 157 cm²
C) 152 cm²
D) 315 cm²

Correct Answer: A) 251 cm²

Question 12 of 20

In a town of 6250 inhabitants, there were 62 births during 1984. Find the percentage birth rate.

A) 1.00%
B) 3.00%
C) 2.50%
D) 5.40%

Correct Answer: A) 1.00%

Question 13 of 20

The nth term of a sequence is n² – 6n – 4. Find the sum of the 3rd and 4th terms.

A) 23
B) –24
C) –25
D) 24

Correct Answer: C) –25

Question 14 of 20

Rationalize 5 ÷ (2 – √3).

A) 3(2 + √5)
B) 5(2 + √3)
C) 3√2 + 1
D) 2(3 + √5)

Correct Answer: B) 5(2 + √3)

Question 15 of 20

Simplify 4√[(390695T – 8)½].

(For this mock exam, assume the simplified form is 1/T.)

A) 1/T
B) 2T
C) T/5
D) 5/T

Correct Answer: A) 1/T

Question 16 of 20

An arc subtends an angle of 50° at the centre of a circle of radius 6 cm. Calculate the area of the sector formed.

A) 110/7 cm²
B) 80/7 cm²
C) 90/7 cm²
D) 100/7 cm²

Correct Answer: A) 110/7 cm²

Question 17 of 20

In a right-angled triangle, if tan θ = ¾, what is cos θ – sin θ?

A) 3/5
B) 4/5
C) 2/5
D) 1/5

Correct Answer: D) 1/5

Question 18 of 20

Simplify 1/(x + 1) + 1/(x – 1).

A) 2x/(x + 1)²
B) 2x(x + 1)²
C) 2x/((x + 1)(x – 3))
D) 2x/((x + 1)(x – 1))

Correct Answer: D) 2x/((x + 1)(x – 1))

Question 19 of 20

Evaluate ∫ 2(2x – 3)^(2/3) dx.

A) (3/5)(2x – 3)^(5/3) + k
B) 2x – 3 + k
C) (3/4)(2x – 3)^(5/3) + k
D) 2(2x – 3) + k

Correct Answer: A) (3/5)(2x – 3)^(5/3) + k

Question 20 of 20

Simplify √30 × √40.

A) 10√3
B) 20√3
C) 5√3
D) 15√3

Correct Answer: B) 20√3

Explanations

Question 1:

Using the distance formula, d = √[(–2 – (–1))² + (4 – (–3))²] = √[(-1)² + (7)²] = √(1+49) = √50 ≈ 7.1 m.

Question 2:

From p:2 = 2:3, we get p/2 = 2/3 so p = 4/3. To express in whole numbers, multiply by 3 to get p = 4. For q:r = 4:7, let q = 4k and r = 7k. Adjusting the ratios so that the second term (from p:2) is 6, we find q = 12 and r = 21. Thus, p : q : r = 4 : 12 : 21.

Question 3:

A number divisible by both 2 and 3 must be divisible by 6. Between 1 and 25, the multiples of 6 are 6, 12, 18, and 24. Hence, the probability is 4/25.

Question 4:

The intersection of sets A = {3, 4, 1, 10, ⅓} and B = {4, 3, ⅓, 7} consists of the common elements: 3, 4, and ⅓.

Question 5:

Combine the fractions: (cos x/(1+sin x)) + (cos x/(1–sin x)) = 2cos x/(1–sin² x). Since 1 – sin² x = cos² x, the expression becomes 2cos x/cos² x = 2sec x.

Question 6:

sin 45° is a standard value equal to √2/2.

Question 7:

The set {43, 44, …, 60} contains 18 numbers. The prime numbers in this range are 43, 47, 53, and 59 (4 primes). Hence, the probability = 4/18 = 2/9.

Question 8:

The progression is geometric with first term 8 and common ratio r = –½. The nth term is 8 × (–½)^(n–1). For n = 10, the term is 8 × (–½)^9 = 8 × (–1/512) = –8/512 = –1/64.

Question 9:

The tank’s volume is 3 × 4 × 5 = 60 cm³. The cube’s volume is 3³ = 27 cm³. The remaining volume for water = 60 – 27 = 33 cm³.

Question 10:

Since T varies inversely as R³, we have T = k/R³. Given that when R = 3, T = 2/81, we get k = (2/81) × 27 = 2/3. Then, when R = 2, T = (2/3) / (2³) = (2/3)/8 = 1/12.

Question 11:

The total surface area of a cylinder is 2πr(h + r). Here, r = 4 and h = 6, so the area is 2 × 3.142 × 4 × (6 + 4) ≈ 251 cm².

Question 12:

The percentage birth rate = (number of births/total population) × 100 = (62/6250) × 100 ≈ 1.00%.

Question 13:

For n = 3, term = 3² – 6×3 – 4 = 9 – 18 – 4 = –13; for n = 4, term = 4² – 6×4 – 4 = 16 – 24 – 4 = –12. Their sum = –13 + (–12) = –25.

Question 14:

To rationalize 5 ÷ (2 – √3), multiply numerator and denominator by (2 + √3): 5(2 + √3)/[(2)² – (√3)²] = 5(2 + √3)/(4 – 3) = 5(2 + √3).

Question 15:

As stated in the question, assume that the given expression simplifies to 1/T.

Question 16:

The area of a sector = (θ/360) × πr² = (50/360) × 3.142 × 36 ≈ 15.71 cm², which is approximately 110/7 cm².

Question 17:

In a right-angled triangle with tan θ = ¾, one can take sin θ = 3/5 and cos θ = 4/5. Therefore, cos θ – sin θ = 4/5 – 3/5 = 1/5.

Question 18:

To add 1/(x + 1) + 1/(x – 1), use a common denominator: (x – 1) + (x + 1) = 2x, so the sum is 2x/[(x + 1)(x – 1)].

Question 19:

Let u = 2x – 3 so that du = 2 dx. Then the integral becomes ∫ u^(2/3) du = (3/5)u^(5/3) + k = (3/5)(2x – 3)^(5/3) + k.

Question 20:

√30 × √40 = √(30 × 40) = √1200 = √(400 × 3) = 20√3.

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