JAMB Mathematics 2017 past questions

Checked by the Skolra question team · Last updated 15 October 2026

Ten questions from the 2017 JAMB Mathematics paper, with answers and explanations. We hold 41 questions from this paper, and the Skolra app has all of them.

JAMB Mathematics 2017 questions and answers

1.What is the next number in the series 2, 1, $\frac{1}{2}$ , $\frac{1}{4}$ ...
1. A
  1/16
2. B
  1/6
3. C
  1/3
4. D
  1/8
  Correct
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Answer: D
The numbers 2, 1, 1/2, 1/4, ... form a geometric progression in which each term is gotten by dividing the one before it by 2 (equivalently, multiplying by 1/2).
Check the pattern: 2 ÷ 2 = 1, then 1 ÷ 2 = 1/2, then 1/2 ÷ 2 = 1/4. Following the same rule, the next term is 1/4 ÷ 2 = 1/8.
A candidate who divides by 4 instead of 2 lands on 1/16, and one who mistakenly thinks the pattern subtracts a fixed amount might guess 1/6 or 1/3 — neither fits a consistent ratio between consecutive terms.
2.
Given T = { even numbers from 1 to 12 }
N = {common factors of 6, 8 and 12}
Find T ∩ N
1. A
  {2}
  Correct
2. B
  { }
3. C
  {2, 6}
4. D
  {1, 2}
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Answer: A
Step 1: Find T. T = even numbers from 1 to 12 = {2, 4, 6, 8, 10, 12}.
Step 2: Find N. N = common factors of 6, 8 and 12.
Factors of 6 = {1, 2, 3, 6}
Factors of 8 = {1, 2, 4, 8}
Factors of 12 = {1, 2, 3, 4, 6, 12}
The numbers common to all three lists are 1 and 2, so N = {1, 2}.
Step 3: Find T ∩ N. This means the elements that appear in BOTH T and N. Checking N's members against T: 2 is even, so it is in T; 1 is odd, so it is not in T. Therefore T ∩ N = {2}, which is correct.
3.
Simplify $(\sqrt[3]{64}a^3)^{−1}$ .
1. A
  a³/4
2. B
  1/(64a³)
3. C
  4a³
4. D
  1/(4a³)
  Correct
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Answer: D
Step 1: Evaluate the cube root first: $\sqrt[3]{64} = 4$ , because $4 \times 4 \times 4 = 64$ .
Step 2: This leaves $4a^3$ inside the bracket, raised to the power $-1$ . A negative power means "take the reciprocal": $(4a^3)^{-1} = \dfrac{1}{4a^3}$ .
So the correct simplification is 1/(4a³).
A candidate who forgets to apply the inverse and just simplifies the cube root ends up with 4a³. Applying the negative sign only to the variable and not the whole bracket gives a³/4. Forgetting to take the cube root of 64 before inverting (treating 64 itself as needing inversion) gives 1/(64a³).
4.Find the range of the following set of numbers 0.4, −0.4, 0.3, 0.47, −0.53, 0.2 and −0.2
1. A
  0.87
2. B
  0.73
3. C
  0.90
4. D
  1.00
  Correct
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Answer: D
The range of a set of numbers is the highest value minus the lowest value.
Looking at the numbers 0.4, −0.4, 0.3, 0.47, −0.53, 0.2 and −0.2:
The highest value is 0.47 and the lowest value is −0.53.
Range = 0.47 − (−0.53) = 0.47 + 0.53 = 1.00
A candidate who mistakenly picks −0.4 as the lowest value (instead of the more negative −0.53) would wrongly get 0.47 − (−0.4) = 0.87. Another common slip is subtracting the smallest positive number, 0.2, from the highest, giving 0.27, or simple rounding mistakes that give 0.90. The correct approach must use the true minimum, −0.53, and the true maximum, 0.47.
5.Find the sum to infinity of the series
$\frac{1}{4}$ , $\frac{1}{8}$ , $\frac{1}{16}$ ,..........
1. A
  1/4
2. B
  1/6
3. C
  1
4. D
  1/2
  Correct
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Answer: D
This is a geometric series with first term a = 1/4 and common ratio r = 1/2 (since dividing any term by the one before it gives 1/8 ÷ 1/4 = 1/2).
Because the common ratio is between -1 and 1, the series converges, so we use the sum to infinity formula:
S∞ = a / (1 − r) = (1/4) / (1 − 1/2) = (1/4) / (1/2) = 1/2
So the correct sum to infinity is 1/2.
A candidate who mistakes the common ratio itself for the answer would wrongly pick 1/4. Using the wrong denominator (1 + r instead of 1 − r) gives the mismatched result 1/6. Doubling the first term incorrectly, or overestimating the total, leads some candidates to the too-large value of 1.
6.What is the product of 2x² − x + 1 and 3 − 2x
1. A
  -4x³ - 8x² - 5x + 3
2. B
  -4x³ + 8x² + 5x + 3
3. C
  -4x³ + 8x² - 5x + 3
  Correct
4. D
  4x³ + 8x² - 5x + 3
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Answer: C
To multiply 2x² − x + 1 by 3 − 2x, share each term of the second bracket across the first bracket.
Multiply by 3: 3(2x² − x + 1) = 6x² − 3x + 3
Multiply by −2x: −2x(2x² − x + 1) = −4x³ + 2x² − 2x
Now add the two results together:
−4x³ + (6x² + 2x²) + (−3x − 2x) + 3 = −4x³ + 8x² − 5x + 3
7.Simplify 3 $^{n − 1}$ × 27 $\frac{n + 1}{81^n}$
1. A
  9
  Correct
2. B
  3
3. C
  27
4. D
  81
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Answer: A
Rewrite every term with base 3: 27 = 3³ and 81 = 3⁴.
So the expression becomes:
3^{n − 1} × 3^{3(n + 1)} ÷ 3⁴ⁿ
Combine the exponents in the numerator: (n − 1) + 3(n + 1) = n − 1 + 3n + 3 = 4n + 2.
Now divide by 3⁴ⁿ: subtract the exponents: (4n + 2) − 4n = 2.
So the whole expression simplifies to 3² = 9. Notice the n terms cancel out completely — the answer is a plain constant, not an expression in n.
8.Evaluate $1 - (\frac{1}{5} \times \frac{2}{3}) + (5 + \frac{2}{3})$
1. A
  6 7/15
2. B
  6 8/15
  Correct
3. C
  5 13/15
4. D
  6 2/15
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Answer: B
First find the product inside the bracket: 1/5 × 2/3 = 2/15.
Subtract from 1: 1 - 2/15 = 13/15.
Next evaluate 5 + 2/3, which is a mixed number: 5 + 2/3 = 17/3.
Convert 17/3 to fifteenths: 17/3 = 85/15.
Add the two results: 13/15 + 85/15 = 98/15 = 6 8/15, which is the correct value.
A candidate who subtracts instead of adding the fractions, or who mishandles the multiplication step, easily lands on 6 2/15. Forgetting to add the 1 correctly, or errors converting 5 + 2/3, produce 5 13/15 or 6 7/15 as tempting but wrong results.
9.Given $M = N\sqrt{\frac{SL}{T}}$ , make $T$ the subject of the formula
1. A
  T = NSL/M²
2. B
  T = M²/(N²SL)
3. C
  T = SLN²/M
4. D
  T = N²SL/M²
  Correct
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Answer: D
Start with M = N√(SL/T).
Divide both sides by N: M/N = √(SL/T).
Square both sides to remove the square root: M²/N² = SL/T.
Now T is in the denominator, so cross-multiply: T × M² = SL × N².
Finally divide both sides by M² to isolate T: T = N²SL/M², which is the correct subject of the formula.
The option T = M²/(N²SL) comes from inverting the fraction instead of cross-multiplying properly. The option T = NSL/M² comes from forgetting to square N when squaring both sides. The option T = SLN²/M comes from forgetting to square M, leaving it to the first power instead of squared.
10.A cylindrical tank has a capacity of 3080m³. What is the depth of the tank if the diameter of its base is 14m? Take pi = 22/7.
1. A
  5m
2. B
  20m
  Correct
3. C
  140m
4. D
  70m
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Answer: B
The tank's depth is 20m.
The volume of a cylinder is V = πr²h. The diameter of the base is 14m, so the radius r = 7m (half the diameter — always halve the diameter before squaring it).
Using π = 22/7:
V = (22/7) × 7 × 7 × h = 3080
(22/7) × 49 = 154
154h = 3080
h = 3080 ÷ 154 = 20m
A candidate who mistakenly uses the diameter itself (14m) as the radius gets an area of (22/7) × 14 × 14 = 616m², giving h = 3080 ÷ 616 = 5m — the wrong-side error of not halving the diameter.
A candidate who forgets to square the radius at all gets (22/7) × 7 = 22, giving h = 3080 ÷ 22 = 140m — far too large for a real tank.

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JAMB Mathematics 2017 past questions

JAMB Mathematics 2017 questions and answers

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