WAEC Mathematics 2019 past questions

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

Ten questions from the 2019 WAEC Mathematics paper, with answers and explanations. We hold 40 questions from this paper, and the Skolra app has all of them.

WAEC Mathematics 2019 questions and answers

1. 1.

   If 3p = 4q and 9p = 8q - 12, find the value of pq.

   1. A
      12
      Correct
   2. B
      24
   3. C
      −12
   4. D
      6
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   Answer: A

   Step 1: Express p in terms of q from the first equation.
   3p = 4q → p = 4q/3

   Step 2: Substitute into the second equation.
   9 × (4q/3) = 8q − 12
   12q = 8q − 12
   4q = −12
   q = −3

   Step 3: Find p.
   p = 4(−3)/3 = −4

   Step 4: Calculate pq.
   pq = (−4)(−3) = 12

2. 2.

   Evaluate: 2$\sqrt{28} - 3\sqrt{50} + \sqrt{72}$

   1. A
      4√7 + 9√2
   2. B
      4√7 − 9√2
      Correct
   3. C
      4√7 − 3√2
   4. D
      2√7 − 9√2
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   Answer: B

   Simplify each surd separately:

   • 2√28 = 2√(4 × 7) = 2 × 2√7 = 4√7
   • 3√50 = 3√(25 × 2) = 3 × 5√2 = 15√2
   • √72 = √(36 × 2) = 6√2

   Combining:
   4√7 − 15√2 + 6√2 = 4√7 − 9√2

3. 3.

   If log$_x$ 2 = 0.3, evaluate log$_x$ 8.

   1. A
      2.4
   2. B
      2.1
   3. C
      0.6
   4. D
      0.9
      Correct
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   Answer: D

   Step 1: Express 8 as a power of 2.
   8 = 2³

   Step 2: Apply the logarithm power rule.
   log_x 8 = log_x 2³ = 3 × log_x 2

   Step 3: Substitute the given value.
   log_x 8 = 3 × 0.3 = 0.9

4. 4.

   An arc subtends an angle of 72$^o$ at the centre of a circle. Find the length of the arc if the radius of the circle is 3.5 cm. [Take $\pi = \frac{22}{7}$]

   1. A
      4.4 cm
      Correct
   2. B
      5.2 cm
   3. C
      3.4 cm
   4. D
      3.8 cm
   Show answerHide answer

   Answer: A

   Step 1: Use the arc length formula.
   Arc length = (θ/360°) × 2πr

   Step 2: Substitute values.
   Arc length = (72/360) × 2 × (22/7) × 3.5
   = (1/5) × 2 × (22/7) × 3.5
   = (1/5) × 2 × 11
   = (1/5) × 22 = 4.4 cm

5. 5.

   If 2$^{a}$ = $\sqrt{64}$ and $\frac{b}{a}$ = 3, evaluate a$^2+ b^{2}$

   1. A
      100
   2. B
      45
   3. C
      90
      Correct
   4. D
      117
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   Answer: C

   Step 1: Solve for a.
   √64 = 8 = 2³
   So 2^a = 2³, giving a = 3

   Step 2: Solve for b.
   b/a = 3 → b/3 = 3 → b = 9

   Step 3: Evaluate a² + b².
   a² + b² = 3² + 9² = 9 + 81 = 90

6. 6.

   Factorize completely; (2x + 2y)(x - y) + (2x - 2y)(x + y)

   1. A
      4(x + y)²
   2. B
      4(x² + y²)
   3. C
      4(x − y)²
   4. D
      4(x² − y²)
      Correct
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   Answer: D

   Step 1: Factor each term.
   (2x + 2y)(x − y) = 2(x + y)(x − y)
   (2x − 2y)(x + y) = 2(x − y)(x + y)

   Step 2: Combine like terms.
   2(x + y)(x − y) + 2(x − y)(x + y) = 4(x + y)(x − y)

   Step 3: Apply the difference of squares identity.
   4(x + y)(x − y) = 4(x² − y²)

7. 7.

   The fourth term of an Arithmetic Progression (A.P) is 37 and the first term is -20. Find the common difference.

   1. A
      21
   2. B
      19
      Correct
   3. C
      23
   4. D
      17
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   Answer: B

   Step 1: Write the formula for the nth term of an AP.
   T_n = a + (n − 1)d

   Step 2: Substitute the known values.
   T₄ = −20 + (4 − 1)d = 37
   −20 + 3d = 37

   Step 3: Solve for d.
   3d = 37 + 20 = 57
   d = 57 ÷ 3 = 19

8. 8.

   Find the angle at which an arc of length 22 cm subtends at the centre of a circle of radius 15cm. [Take $\pi = \frac{22}{7}$]

   1. A
      90°
   2. B
      84°
      Correct
   3. C
      72°
   4. D
      96°
   Show answerHide answer

   Answer: B

   Step 1: Use the arc length formula.
   Arc length = (θ/360°) × 2πr

   Step 2: Substitute known values.
   22 = (θ/360) × 2 × (22/7) × 15
   22 = (θ/360) × (660/7)

   Step 3: Solve for θ.
   θ = 22 × 360 × 7 / 660
   θ = 55,440 / 660 = 84°

   Verification: (84/360) × (44/7) × 15 = (7/30) × (44/7) × 15 = 44/2 = 22 ✓

9. 9.

   Simplify: $\frac{x^2 - 5x - 14}{x^2 - 9x + 14}$

   1. A
      (x + 7)/(x − 7)
   2. B
      (x − 2)/(x + 2)
   3. C
      (x − 7)/(x + 7)
   4. D
      (x + 2)/(x − 2)
      Correct
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   Answer: D

   Step 1: Factorise the numerator.
   x² − 5x − 14 = (x − 7)(x + 2)

   Step 2: Factorise the denominator.
   x² − 9x + 14 = (x − 7)(x − 2)

   Step 3: Cancel the common factor.
   (x − 7)(x + 2) / [(x − 7)(x − 2)] = (x + 2)/(x − 2)

10. 10.

    If T = {prime numbers} and M = {odd numbers} are subsets of $\mu$ = {x : 0 < x ≤10}and x is an integer, find (T$^{\prime}$ nM$^{\prime}$).

    1. A
       {1, 4, 6, 9, 10}
    2. B
       {4, 6, 8, 10}
       Correct
    3. C
       {1, 2, 4, 6}
    4. D
       {2, 4, 6, 8, 10}
    Show answerHide answer

    Answer: B

    Step 1: List the universal set:
    μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

    Step 2: Identify T and M:
    T = {2, 3, 5, 7} (prime numbers ≤ 10)
    M = {1, 3, 5, 7, 9} (odd numbers ≤ 10)

    Step 3: Find complements:
    T′ = {1, 4, 6, 8, 9, 10}
    M′ = {2, 4, 6, 8, 10}

    Step 4: T′ ∩ M′ = elements in both T′ and M′ = {4, 6, 8, 10}

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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