WAEC Mathematics 2013 past questions

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

Ten questions from the 2013 WAEC Mathematics paper, with answers and explanations. We hold 33 questions from this paper, and the Skolra app has all of them.

WAEC Mathematics 2013 questions and answers

1. 1.

   If y varies directly as the square root of (x + 1) and y = 6 when x = 3, find x when y = 9

   1. A
      8
      Correct
   2. B
      15
   3. C
      35
   4. D
      24
   Show answerHide answer

   Answer: A

   Write the direct variation equation and find the constant k:

   1. y = k√(x + 1)
   2. When x = 3, y = 6:  6 = k√(3 + 1) = k × 2, so k = 3
   3. When y = 9:  9 = 3√(x + 1)
   4. 3 = √(x + 1)
   5. 9 = x + 1
   6. x = 8
2. 2.If $\frac{1}{2}$x + 2y = 3 and $\frac{3}{2}$x and $\frac{3}{2}$x - 2y = 1, find (x + y)
   1. A
      5
   2. B
      2
   3. C
      3
      Correct
   4. D
      4
   Show answerHide answer

   Answer: C

   Treat the two equations as: (i) ½x + 2y = 3, and (ii) (3/2)x − 2y = 1, then solve simultaneously.

   1. Add (i) and (ii): ½x + (3/2)x = 4 → 2x = 4 → x = 2.
   2. Substitute x = 2 into (i): 1 + 2y = 3 → y = 1.
   3. x + y = 2 + 1 = 3.
3. 3.

   If $\sqrt{50} - K\sqrt{8} = \frac{2}{\sqrt{2}}$, find $K$

   1. A
      K = 3
   2. B
      K = 4
   3. C
      K = 1
   4. D
      K = 2
      Correct
   Show answerHide answer

   Answer: D

   Simplify each surd in terms of √2:

   1. √50 = √(25 × 2) = 5√2
   2. K√8 = K√(4 × 2) = 2K√2
   3. 2/√2 = 2√2/2 = √2
   4. Equation becomes: 5√2 − 2K√2 = √2
   5. Divide by √2: 5 − 2K = 1
   6. 2K = 4  ⇒  K = 2
4. 4.If 9^{(2 - x)} = 3, find x
   1. A
      2/3
   2. B
      1/2
   3. C
      3/2
      Correct
   4. D
      5/2
   Show answerHide answer

   Answer: C

   Express both sides as powers of the same base (3).

   1. 9 = 3², so 9^(2−x) = 3^2(2−x).
   2. Equation becomes 3^2(2−x) = 3¹.
   3. Equate exponents: 2(2 − x) = 1 → 4 − 2x = 1 → 2x = 3 → x = 3/2.
5. 5.

   The area of a circle is $154cm^{2}$. It is divided into three sectors such that two of the sectors are equal in size and the third sector is three times the size of the other two put together. Calculate the perimeter of the third sector. [Take $\pi = \frac{22}{7}$].

   1. A
      44 cm
   2. B
      33 cm
   3. C
      38.5 cm
   4. D
      47 cm
      Correct
   Show answerHide answer

   Answer: D

   Find the radius, the angle of the third sector, then compute its perimeter:

   1. Area = πr² = 154 ⇒ r² = 154 × 7/22 = 49 ⇒ r = 7 cm
   2. Let each equal sector subtend angle α. Third sector = 3 × 2α = 6α.
   3. Total: 2α + 6α = 360° ⇒ α = 45°, so third sector = 270°.
   4. Arc length = (270/360) × 2πr = (3/4) × 2 × (22/7) × 7 = (3/4) × 44 = 33 cm
   5. Perimeter = 2r + arc = 2(7) + 33 = 14 + 33 = 47 cm
6. 6.

   When a number is subtracted from 2, the result equals 4 less than one-fifth of the number. Find the number

   1. A
      4
   2. B
      5
      Correct
   3. C
      10
   4. D
      15
   Show answerHide answer

   Answer: B

   Let the unknown number be n and translate the words into an equation.

   1. "A number subtracted from 2" = 2 − n.
   2. "4 less than one-fifth of the number" = (1/5)n − 4.
   3. Equation: 2 − n = (1/5)n − 4.
   4. 2 + 4 = (1/5)n + n = (6/5)n → n = 6 × (5/6) = 5.
7. 7.

   The bearing of Y from X is 060° and the bearing of Z from Y = 060°. Find the bearing of X from Z

   1. A
      060°
   2. B
      300°
   3. C
      120°
   4. D
      240°
      Correct
   Show answerHide answer

   Answer: D

   Since both successive bearings are 060°, the three points are collinear in the 060° direction. Use the back-bearing rule:

   1. Bearing of Y from X = 060°, so X, Y, Z lie along the line at 060° from X.
   2. Z is further along this same line from X.
   3. Bearing of X from Z = back-bearing = 060° + 180° = 240°.
8. 8.

   Given that (x + 2)(x² - 3x + 2) + 2(x + 2)(x - 1) = (x + 2) M, find M

   1. A
      x² − 2x
   2. B
      x² + x
   3. C
      x² − 3x + 2
   4. D
      x² − x
      Correct
   Show answerHide answer

   Answer: D

   Factor the left side by recognising (x + 2) and a common factor in the remaining expression.

   1. Factorise x² − 3x + 2 = (x − 1)(x − 2).
   2. LHS = (x + 2)(x − 1)(x − 2) + 2(x + 2)(x − 1).
   3. Factor out (x + 2)(x − 1): (x + 2)(x − 1)[(x − 2) + 2] = (x + 2)(x − 1)(x).
   4. Divide both sides by (x + 2): M = x(x − 1) = x² − x.
9. 9.

   The graph of the relation y = x² + 2x + k passes through the point (2, 0). Find the values of k

   1. A
      4
   2. B
      8
   3. C
      -8
      Correct
   4. D
      -4
   Show answerHide answer

   Answer: C

   Substitute the given point (2, 0) into the equation to find k:

   1. y = x² + 2x + k, with x = 2 and y = 0
   2. 0 = (2)² + 2(2) + k
   3. 0 = 4 + 4 + k
   4. 0 = 8 + k
   5. k = −8
10. 10.

    The distance between two towns is 50km. It is represented on a map by 5cm. Find the scale used

    1. A
       1 : 500,000
    2. B
       1 : 10,000,000
    3. C
       1 : 1,000,000
       Correct
    4. D
       1 : 100,000
    Show answerHide answer

    Answer: C

    Convert the actual distance to centimetres, then write the ratio in simplest form.

    1. Actual distance = 50 km = 50 × 1000 m × 100 cm/m = 5,000,000 cm.
    2. Map distance = 5 cm.
    3. Scale = 5 : 5,000,000 = 1 : 1,000,000.

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

Spotted a wrong answer? Tell us from the app or on the contact page and we check it.

Practise all 33 questions we hold from the 2013 paper, on a timed CBT screen

Every question has the answer and an explanation, in the Skolra app.

Download for WindowsDownload for Mac

About the app · Skolra for Windows · Skolra for Mac