JAMB Mathematics 2015 past questions

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

Ten questions from the 2015 JAMB Mathematics paper, with answers and explanations. We hold 68 questions from this paper, and the Skolra app has all of them.

JAMB Mathematics 2015 questions and answers

1.Make x the subject of the equation
s = 2 + $\frac{t}{5}$ (x + ⅗y)
1. A
  x = 5(s − 2)/t + ⅗y
2. B
  x = t(s − 2)/5 − ⅗y
3. C
  x = 5(s − 2)/t − 3y/5
4. D
  x = 5(s − 2)/t − ⅗y
  Correct
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Answer: D
Start with s = 2 + (t/5)(x + ⅗y). Subtract 2 from both sides: s − 2 = (t/5)(x + ⅗y).
Multiply both sides by 5/t to clear the fraction: 5(s − 2)/t = x + ⅗y.
Subtract ⅗y to isolate x: x = 5(s − 2)/t − ⅗y.
Getting the sign wrong when moving ⅗y across the equals sign gives x = 5(s − 2)/t + ⅗y instead. Mis-simplifying the fraction ⅗, writing it as 3/5 with a changed sign, gives x = 5(s − 2)/t − 3y/5, which does not match the original ⅗y term correctly through the algebra. Leaving t and 5 un-swapped when clearing the fraction (multiplying by t/5 instead of 5/t) gives the wrong x = t(s − 2)/5 − ⅗y.
2.
How many sides has a regular polygon whose interior angles are 120° each?
1. A
  6
  Correct
2. B
  5
3. C
  8
4. D
  4
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Answer: A
Every interior angle of a regular polygon has a matching exterior angle, and interior + exterior = 180°.
Exterior angle = 180° − 120° = 60°.
The exterior angles of any polygon add up to 360°, so the number of sides = 360° ÷ 60° = 6.
Dividing 360 by the interior angle itself (360 ÷ 120 = 3, adjusted wrongly) or misreading the exterior angle as 90° (giving 4 sides) are common slips. Getting 5 or 8 sides usually comes from an arithmetic error in the 180 − 120 step or the final division.
3.
The extension of a stretched string is directly proportional to its tension. If the extension produced by a tension of 8 Newtons is 2cm, find the extension produced by a tension of 12 Newtons.
1. A
  4.5cm
2. B
  3cm
  Correct
3. C
  1.33cm
4. D
  2cm
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Answer: B
Since extension is directly proportional to tension, extension ÷ tension is a constant.
2cm/8N = x/12N → x = 2 × 12/8 = 3cm.
Simply repeating the original 2cm extension ignores that the tension has changed. Multiplying instead of using the correct ratio (2 × 12/8 done incorrectly) can give 4.5cm. Inverting the ratio (using 8/12 instead of 12/8) gives the wrong 1.33cm.
4.Simplify √30 × √40
1. A
  10√3
2. B
  20√3
  Correct
3. C
  20√6
4. D
  40√3
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Answer: B
√30 × √40 = √(30 × 40) = √1200.
1200 = 400 × 3, and 400 is a perfect square (20²), so √1200 = √400 × √3 = 20√3.
Taking out only √100 instead of the full √400 leaves an extra factor unsimplified, giving 10√3. Multiplying 30 and 40 correctly but factoring out the wrong perfect square (mistaking 1200 for 6 × 400 rather than 3 × 400) gives 20√6. Doubling the coefficient by mistake gives 40√3.
5.Simplify 1¼ ÷ (2 ÷ ¼ of 28)
1. A
  4 3/8
  Correct
2. B
  1/8
3. C
  3 4/5
4. D
  8
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Answer: A
Work the bracket first: ¼ of 28 = 28 ÷ 4 = 7.
Then 2 ÷ 7 = 2/7.
Now the whole expression: 1¼ ÷ (2/7) = 5/4 ÷ 2/7 = 5/4 × 7/2 = 35/8 = 4 3/8.
Working the operations in the wrong order, dividing before finding ¼ of 28, can lead to the unrelated 3 4/5. Forgetting to flip the second fraction when dividing (multiplying by 2/7 instead of its reciprocal 7/2) gives the much smaller 1/8. Simply reporting the value of ¼ of 28 ÷ something too roughly gives 8.
6.
Calculate the value of x and y if 27^x ÷ 81^x+2y = 9, x + 4y = 0
1. A
  x = 2, y = −1/2
  Correct
2. B
  x = −2, y = 1/2
3. C
  x = 1/2, y = −2
4. D
  x = 2, y = 1/2
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Answer: A
Rewrite with base 3: 27 = 3³, 81 = 3⁴, 9 = 3².
27^x ÷ 81^(x+2y) = 9 becomes 3^(3x) ÷ 3^(4(x+2y)) = 3², so 3x − 4x − 8y = 2, i.e. −x − 8y = 2.
From x + 4y = 0, x = −4y. Substituting: 4y − 8y = 2 → −4y = 2 → y = −1/2, and then x = −4(−1/2) = 2.
Check: 27² ÷ 81^(2 − 1) = 729 ÷ 81 = 9 ✓.
Flipping both signs gives x = −2, y = 1/2, which fails the check. Dropping the negative sign on y only (x = 2, y = 1/2) also fails when substituted back. Swapping which variable takes which value (x = 1/2, y = −2) does not satisfy x + 4y = 0 at all.
7.Solve the equation $3x^2 − 4x − 5 = 0$
1. A
  x = (4 ± √19)/3
2. B
  x = (2 ± √76)/3
3. C
  x = (−2 ± √19)/3
4. D
  x = (2 ± √19)/3
  Correct
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Answer: D
Use the quadratic formula on 3x² − 4x − 5 = 0, with a=3, b=−4, c=−5: x = [−b ± √(b²−4ac)]/2a.
Discriminant = (−4)² − 4(3)(−5) = 16 + 60 = 76. √76 = 2√19.
x = [4 ± 2√19]/6 = (2 ± √19)/3 after dividing every term by 2.
Getting the sign of −b wrong (using −4 instead of +4 at the start) gives the wrong (−2 ± √19)/3. Miscalculating −b as 4 but forgetting to halve it correctly when simplifying the fraction gives (4 ± √19)/3. Forgetting to simplify √76 down to 2√19 at all, and leaving it unreduced inside the fraction with the wrong denominator, gives (2 ± √76)/3.
8.
Evaluate log₅( $y^2x^5 ÷ 125b^{\frac{1}{2}})$
1. A
  5log₅x + 2log₅y + 3 − ½log₅b
2. B
  log₅x + log₅y − 3 − ½log₅b
3. C
  5log₅x + 2log₅y − 3 − ½log₅b
  Correct
4. D
  5log₅x + 2log₅y − 3 + ½log₅b
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Answer: C
Expand log₅(y²x⁵ ÷ 125b^½) using the log laws: log of a product/quotient becomes addition/subtraction of logs, and a power inside the log becomes a multiplying coefficient.
log₅(x⁵) + log₅(y²) − log₅(125) − log₅(b^½) = 5log₅x + 2log₅y − log₅(5³) − ½log₅b.
Since log₅(5³) = 3, this simplifies to 5log₅x + 2log₅y − 3 − ½log₅b.
Treating 125 as if it added rather than subtracted (since it's in the denominator) gives the wrong +3. Getting the sign of the b½ term wrong the same way gives +½log₅b instead of −½log₅b. Forgetting to bring the powers 5 and 2 down as coefficients on x and y leaves the incomplete log₅x + log₅y − 3 − ½log₅b.
9.A man with an annual salary of ₦2000, has allowances of ₦600. If Income Tax is 5%, how much tax does he pay each year?
1. A
  ₦100
2. B
  ₦130
3. C
  ₦70
  Correct
4. D
  ₦30
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Answer: C
Allowances are deducted from the salary before tax is charged, so the taxable income = 2000 − 600 = ₦1400.
Tax = 5% of 1400 = (5/100) × 1400 = ₦70.
Taxing the full salary of ₦2000 without first removing the allowance gives the inflated wrong figure of ₦100. Taxing only the allowance amount itself (5% of 600) gives ₦30. Miscalculating the taxable base or the percentage step can also give a stray figure like ₦130.
10.
The area of an ellipse is 132cm². The length of its major axis is 14cm. Find the length of it minor axis.
1. A
  12cm
  Correct
2. B
  6cm
3. C
  9.43cm
4. D
  18.86cm
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Answer: A
The area of an ellipse is A = πab, where a and b are the semi-major and semi-minor axes (half of each full axis length).
Major axis = 14cm, so semi-major axis a = 7cm.
132 = π × 7 × b. Using π = 22/7: 132 = (22/7) × 7 × b = 22b, so b = 132 ÷ 22 = 6cm (the semi-minor axis).
The full minor axis is twice this: 2 × 6 = 12cm.
Reporting the semi-minor axis (6cm) instead of doubling it to get the full minor axis is a common slip. Using the full major axis (14) instead of the semi-major axis (7) in the formula throws the answer off to 18.86cm or, after halving incorrectly, 9.43cm.

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JAMB Mathematics 2015 past questions

JAMB Mathematics 2015 questions and answers

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