JAMB Mathematics 2018 past questions

Checked by the Skolra question team · Last updated 23 October 2026

Ten questions from the 2018 JAMB Mathematics paper, with answers and explanations. We hold 59 questions from this paper, and the Skolra app has all of them.

JAMB Mathematics 2018 questions and answers

1.
If P = [ $\frac{Q(R - T)}{15}$ ] $^{\frac{1}{3}}$ make T the subject of the relation
1. A
  T = R − 15P³/Q
  Correct
2. B
  T = R − P³/15Q
3. C
  T = 15P³/Q − R
4. D
  T = R − 15P^(1/3)/Q
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Answer: A
We start with P = [Q(R − T)/15]^1/3.
Step 1: Cube both sides to remove the cube root: P³ = Q(R − T)/15.
Step 2: Multiply both sides by 15: 15P³ = Q(R − T).
Step 3: Divide both sides by Q: 15P³/Q = R − T.
Step 4: Make T the subject by rearranging: T = R − 15P³/Q.
2.
Find the average of the first four prime numbers greater than 10
1. A
  20
2. B
  15
  Correct
3. C
  18
4. D
  14
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Answer: B
Step 1: List the prime numbers greater than 10.
A prime number has only two factors: 1 and itself. Starting just after 10, the primes are 11, 13, 17, 19, 23, 29, ... (note that 12, 14, 15, 16, 18, 20 are not prime).
Step 2: Pick the first four.
These are 11, 13, 17, 19.
Step 3: Find their average.
Sum = 11 + 13 + 17 + 19 = 60
Average = 60 ÷ 4 = 15
3.
How many times, correct to the nearest whole number, will a man run round a circular track of diameter 100m to cover a distance of 1000m?
1. A
  3 times
  Correct
2. B
  4 times
3. C
  10 times
4. D
  6 times
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Answer: A
The track is circular with diameter = 100 m. One complete round of the track equals its circumference.
Circumference = π × diameter = π × 100 ≈ 314.16 m (using π ≈ 22/7 gives 314.29 m — essentially the same).
To cover 1000 m, the number of rounds needed is:
1000 ÷ 314.16 ≈ 3.18
Correct to the nearest whole number, 3.18 rounds down to 3 times, which is the correct answer.
4.
An arc of a circle of radius 14cm subtends angle 300° at the centre. Find the perimeter of the sector formed by the arc (take π = $\frac{22}{7}$ )
1. A
  101.3cm
  Correct
2. B
  50.7cm
3. C
  42.7cm
4. D
  73.3cm
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Answer: A
The sector's perimeter is made up of the arc plus the two straight radii that join the arc's ends to the centre — not the arc alone.
First find the full circumference: C = 2πr = 2 × 22/7 × 14 = 88cm.
The arc subtends 300° out of the full 360°, so:
Arc length = 300/360 × 88 = 5/6 × 88 = 73.33cm.
Perimeter of sector = arc length + 2r = 73.33 + (2 × 14) = 73.33 + 28 = 101.3cm.
So the correct perimeter of the sector is 101.3cm.
5.
Convert 0.04945 to two significant figures
1. A
  0.0495
2. B
  0.049
  Correct
3. C
  0.0494
4. D
  0.05
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Answer: B
In 0.04945, the leading zeros after the decimal point are not significant — they only fix the position of the decimal point. The significant figures start from the first non-zero digit, so they are 4, 9, 4, 5.
To round to two significant figures, keep the first two of these digits, 4 and 9, giving 0.04|945. Then look at the next digit, which is 4. Since 4 is less than 5, we round down and simply drop the remaining digits.
This gives 0.049 as the value correct to two significant figures.
6.
if y = 23 $_{five}$ + 101 $_{three}$ find y leaving your answer in base two
1. A
  10011₂
2. B
  11001₂
3. C
  10111₂
  Correct
4. D
  10001₂
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Answer: C
First convert each number to base ten before adding, since you cannot add numbers directly while they are in different bases.
23 in base five: (2 × 5) + (3 × 1) = 10 + 3 = 13.
101 in base three: (1 × 9) + (0 × 3) + (1 × 1) = 9 + 0 + 1 = 10.
Adding these in base ten: y = 13 + 10 = 23.
Now convert 23 to base two by repeated division by 2 or by powers of 2: 23 = 16 + 4 + 2 + 1 = 2⁴ + 2² + 2¹ + 2⁰, which gives 10111 in base two. This is the correct answer.
7.
Given that sin (5 $x$ − 28) $^o$ = cos(3 $x$ − 50) $^o$ , O $x$ < 90 $^o$
Find the value of $x$
1. A
  x = 34
2. B
  x = 21
  Correct
3. C
  x = 11
4. D
  x = 24
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Answer: B
Given: sin(5x − 28)° = cos(3x − 50)°, with 0° < x < 90°.
The trick is to turn the sine into a cosine using the co-function rule: sinθ = cos(90° − θ). Applying this to the left side:
sin(5x − 28)° = cos[90° − (5x − 28)]° = cos(118° − 5x)°
Now both sides are cosines, so their angles must be equal:
118° − 5x = 3x − 50°
118° + 50° = 3x + 5x
168° = 8x
x = 21
Check: 5(21) − 28 = 77°, and 3(21) − 50 = 13°. sin77° = cos13° = 0.9744, which checks out, so 21 is correct.
8.
If $\text{N} = \begin{pmatrix} 3 & 5 & -4 \\ 6 & -3 & -5 \\ -2 & 2 & 1 \\ \end{pmatrix}$ , find |N|.
1. A
  -17
2. B
  7
3. C
  -23
4. D
  17
  Correct
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Answer: D
To find the determinant of a 3×3 matrix N, expand along the first row using the standard rule (alternating signs +, −, +).
N = 3 5 −4
6 −3 −5
−2 2 1
|N| = 3[(−3)(1) − (−5)(2)] − 5[(6)(1) − (−5)(−2)] + (−4)[(6)(2) − (−3)(−2)]
= 3[−3 + 10] − 5[6 − 10] + (−4)[12 − 6]
= 3(7) − 5(−4) + (−4)(6)
= 21 + 20 − 24
= 17
So |N| = 17 is correct.
A candidate who forgets to alternate the cofactor signs and adds every term instead of subtracting the middle one lands on −23. Someone who gets the arithmetic right but flips the overall sign of the answer at the end gets −17. And a student who stops after evaluating only the first 2×2 minor (3 × 7) without completing the other two terms wrongly reports 7.
9.
Express 495g as a percentage of 16.5kg
1. A
  30%
2. B
  6%
3. C
  3%
  Correct
4. D
  0.3%
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Answer: C
Before comparing two quantities as a percentage, they must be in the same unit.
Convert 16.5kg to grams: 16.5 x 1000 = 16500g.
Now express 495g as a percentage of 16500g:
(495 / 16500) x 100% = 3%.
The answer 30% comes from a common slip: dividing 495 by 16.5 directly without converting kilograms to grams first (495/16.5 = 30), which ignores the unit mismatch entirely.
0.3% comes from misplacing the decimal point after the division, while 6% comes from doubling the correct value by mistake. Always convert to matching units first, then apply the percentage formula: (part/whole) x 100%.
10.
Tossing a coin and rolling a die are two separate events. What is the probability of obtaining a tail on the coin and an even number on the die?
1. A
  1/3
2. B
  1/4
  Correct
3. C
  1/2
4. D
  1/6
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Answer: B
The coin toss and the die roll are independent events, so we multiply their individual probabilities.
For the coin: there are 2 equally likely outcomes (head, tail), so P(tail) = 1/2.
For the die: the even numbers are 2, 4 and 6, giving 3 favourable outcomes out of 6, so P(even) = 3/6 = 1/2.
P(tail and even) = 1/2 × 1/2 = 1/4.
A candidate who picks 1/2 has stopped after only one of the two events. A candidate who picks 1/3 has wrongly counted the even numbers on the die as 2 out of 6. A candidate who picks 1/6 has mistakenly added the two probabilities as 1/2 + 1/2 and then divided incorrectly, or confused the problem with a single die-only question of getting a specific number.

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JAMB Mathematics 2018 past questions

JAMB Mathematics 2018 questions and answers

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