NECO Mathematics past questions and answers

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

NECO Mathematics: 38 objective past questions from the 2008 paper, each with the answer and an explanation. Newer NECO papers are not in our bank yet. Here are ten.

1. 1.The ages of 10 students in a class are; 15, 16, 15.5, 17, 14.9, 14.5, 14.1, 15.1, 14.8. find the range of their ages.
   1. A
      2.5
   2. B
      1.9
   3. C
      3.1
   4. D
      2.9
      Correct
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   Answer: D

   The range of a set of data is found by subtracting the lowest value from the highest value.

   Looking through the ages: 15, 16, 15.5, 17, 14.9, 14.5, 14.1, 15.1, 14.8 — the highest age is 17 and the lowest age is 14.1.

   Range = 17 − 14.1 = 2.9

   A student who mistakenly takes 16 as the highest value (missing 17) would get 16 − 14.1 = 1.9, which is wrong.

   A student who mistakenly picks 14.5 as the lowest value (missing 14.1) would get 17 − 14.5 = 2.5, which is also wrong.

   A student who subtracts carelessly, 17 − 14.1, and slips in the decimal borrowing may arrive at 3.1 — this is a simple subtraction error, not the correct range.

2. 2.

   Which of the following properties is/are common to parallelograms and rectangles?

   I. all angles are right angles

   II. the diagonals are equals.

   III. the opposite sides are equal and parallel.

   1. A
      I, II and III
   2. B
      III only
      Correct
   3. C
      I and II only
   4. D
      II only
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   Answer: B

   A rectangle is a special type of parallelogram, so it keeps every parallelogram property, but a general parallelogram does not gain the extra properties that make a rectangle special.

   Statement I (all angles are right angles): this is true only for the rectangle. A general parallelogram can have slanted (acute/obtuse) angles, so this is not shared by both shapes.

   So the correct answer is that only statement III is common to parallelograms and rectangles. A student who forgets that a rectangle is itself a parallelogram may wrongly pick the right-angle or equal-diagonal option, thinking those special rectangle features must also belong to the general parallelogram.

3. 3.Given that x² + y² + k = 60 and x = $\sqrt{10}$ when y = 4, find the value of k.
   1. A
      26
   2. B
      34
      Correct
   3. C
      44
   4. D
      50
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   Answer: B

   Given: x² + y² + k = 60, with x = √10 and y = 4.

   Square each value: x² = (√10)² = 10, and y² = 4² = 16.

   Substitute into the equation: 10 + 16 + k = 60, so 26 + k = 60.

   Therefore k = 60 − 26 = 34.

   A candidate who stops after adding x² and y² and forgets to subtract from 60 wrongly reports 26 as the answer. Leaving out x² and subtracting only y² from 60 gives 44, while leaving out y² and subtracting only x² gives 50 — both come from forgetting one of the two squared terms before solving for k.

4. 4.A pair of shoes was sold for N2,250.00 at a loss of 10%. What was the cost price?
   1. A
      N2,500.00
      Correct
   2. B
      N2,475.00
   3. C
      N2,025.00
   4. D
      N2,050.00
   Show answerHide answer

   Answer: A

   A loss of 10% means the pair of shoes was sold for only 90% of what it cost.

   Selling Price = 90% × Cost Price

   2,250.00 = 0.9 × Cost Price

   Cost Price = 2,250.00 ÷ 0.9 = N2,500.00

   Check: 10% of N2,500.00 is N250.00, and N2,500.00 − N250.00 = N2,250.00, which matches the selling price given.

5. 5.P = {x:(x - 2)(x + 3) (x - 4)(x + 5) = 0}. The elements in P are
   1. A
      {-2, -3, -4, -5}
   2. B
      {-2, 3, -4, 5}
   3. C
      {2, -3, 4, -5}
      Correct
   4. D
      {2, 3, 4, 5}
   Show answerHide answer

   Answer: C

   P = {2, -3, 4, -5} is correct.

   The equation (x - 2)(x + 3)(x - 4)(x + 5) = 0 is a product of four factors set equal to zero. A product is zero only when at least one of its factors is zero, so we set each factor to zero and solve:

   x - 2 = 0 → x = 2
   x + 3 = 0 → x = -3
   x - 4 = 0 → x = 4
   x + 5 = 0 → x = -5

   These four values are the elements of P.

   The wrong sign on every root, giving {-2, 3, -4, 5}, comes from mixing up which side of the bracket gives a positive or negative solution — for a factor like (x + 3), the root is -3, not +3, and for (x - 2), the root is +2, not -2.

6. 6.T varies inversely as the square root of F when T = 7, F = 2$\frac{1}{4}$. Find T when F = $\frac{27}{9}$
   1. A
      T = 21√3⁄4
   2. B
      T = 7⁄2
   3. C
      T = 7√3⁄2
      Correct
   4. D
      T = 21√3⁄2
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   Answer: C

   Since T varies inversely as the square root of F, we write T = k/√F, where k is the constant of variation.

   Step 1: Find k.
   When T = 7, F = 2¼ = 9/4, so √F = √(9/4) = 3/2.
   k = T × √F = 7 × 3/2 = 21/2.

   Step 2: Find T when F = 27/9.
   27/9 simplifies to 3, so √F = √3.
   T = k/√F = (21/2) ÷ √3 = 21/(2√3).

   Rationalising the denominator: 21/(2√3) × (√3/√3) = 21√3/6 = 7√3/2, which is correct.

7. 7.Find the value of log$_{\sqrt{3}}$ 81
   1. A
      16
   2. B
      4
   3. C
      2
   4. D
      8
      Correct
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   Answer: D

   We want log base √3 of 81.

   Write both numbers as powers of 3: 81 = 3⁴, and √3 = 3^1/2.

   So log_√3 81 = log_3^1/2 3⁴. Using the rule that log_a^k aⁿ = n ÷ k, we get 4 ÷ (1/2) = 4 × 2 = 8.

   A candidate who finds log₃81 = 4 but forgets to adjust for the fractional base index gets 4 instead. Dividing 4 by 2 instead of multiplying gives 2. Squaring 4 by mistake, thinking the root doubles the exponent the wrong way, gives 16. The correct working is to divide the exponent 4 by the base's own exponent 1/2, which is the same as multiplying by 2, giving 8.

8. 8.Construct a quadratic equation whose roots are -$\frac{3}{2}$ and 7
   1. A
      2x² - 11x - 21 = 0
      Correct
   2. B
      2x² + 11x + 21 = 0
   3. C
      2x² - 11x + 21 = 0
   4. D
      2x² + 11x - 21 = 0
   Show answerHide answer

   Answer: A

   The roots are -3/2 and 7, so:

   Sum of roots = -3/2 + 7 = 11/2
   Product of roots = -3/2 × 7 = -21/2

   Using the rule x² - (sum of roots)x + (product of roots) = 0, we get:

   x² - (11/2)x - 21/2 = 0

   Multiplying through by 2 to clear the fractions gives 2x² - 11x - 21 = 0, which is the required equation.

   You can confirm this by factoring: (2x + 3)(x - 7) = 2x² - 14x + 3x - 21 = 2x² - 11x - 21, and setting each factor to zero gives x = -3/2 and x = 7, matching the given roots.

9. 9.If the 3rd and the 5th terms of an A.P are 6 and 10 respectively, find the 1st term and the common difference respectively.
   1. A
      2 and 4
   2. B
      6 and 2
   3. C
      4 and 2
   4. D
      2 and 2
      Correct
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   Answer: D

   In an A.P, the nth term is given by a + (n-1)d, where a is the first term and d is the common difference.

   From the 3rd term: a + 2d = 6

   From the 5th term: a + 4d = 10

   Subtracting the first equation from the second: 2d = 4, so d = 2.

   Putting d = 2 back into a + 2d = 6 gives a + 4 = 6, so a = 2.

   The correct pair is first term 2 and common difference 2.

10. 10.Which of the following angular inequalities defines an obtuse angle?
    1. A
       90° < x < 180°
       Correct
    2. B
       x = 90°
    3. C
       180° < x < 360°
    4. D
       0° < x < 90°
    Show answerHide answer

    Answer: A

    An obtuse angle is any angle that is bigger than a right angle but smaller than a straight line, that is, it lies strictly between 90° and 180°.

    So the inequality 90° < x < 180° is the correct definition of an obtuse angle.

    The inequality 0° < x < 90° describes an acute angle, not an obtuse one — this is a common mix-up. The statement x = 90° describes a right angle, a single fixed value rather than a range. The inequality 180° < x < 360° describes a reflex angle, which is larger than a straight line, not an obtuse angle.

    Remember the order as angle size grows: acute, right, obtuse, straight, reflex.

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

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