WAEC Mathematics 2012 past questions

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

Ten questions from the 2012 WAEC Mathematics paper, with answers and explanations. We hold 30 questions from this paper, and the Skolra app has all of them.

WAEC Mathematics 2012 questions and answers

1. 1.Which of these statements about y = 8$\sqrt{m}$ is correct?
   1. A
      m varies inversely as y
   2. B
      y varies inversely as √m
   3. C
      y varies directly as m
   4. D
      y² varies directly as m
      Correct
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   Answer: D

   Square both sides to find the true variation relationship.

   1. Given: y = 8√m.
   2. Square both sides: y² = 64m.
   3. Since y² = 64m, y² is directly proportional to m (with constant of proportionality 64).
   4. Option C is wrong — y is proportional to √m, not m.
   5. Therefore the correct statement is y² varies directly as m.
2. 2.In $\bigtriangleup$ XYZ, /XY/ = 8cm, /YZ/ = 10cm and /XZ/ = 6cm. Which of these relation is true?
   1. A
      /XY/² = /YZ/² + /XZ/²
   2. B
      /XY/² = /XZ/² − /YZ/²
   3. C
      /XZ/² = /XY/² + /YZ/²
   4. D
      /YZ/² = /XY/² + /XZ/²
      Correct
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   Answer: D

   Check which Pythagorean relationship holds among the three sides.

   1. XY = 8, YZ = 10, XZ = 6.
   2. Test option D: /YZ/² = /XY/² + /XZ/² → 10² = 8² + 6² → 100 = 64 + 36 = 100. ✓
   3. The triangle is right-angled at X (the angle opposite YZ).
   4. The true relation is /YZ/² = /XY/² + /XZ/².
3. 3.

   The volume of a cuboid is 54 cm³. If the length, width and height of the cuboid are in the ratio 2:1:2 respectively, find its total surface area

   1. A
      144 cm²
   2. B
      36∛2 cm²
   3. C
      72∛2 cm²
      Correct
   4. D
      90 cm²
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   Answer: C

   Let the common factor be k, so l = 2k, w = k, h = 2k:

   1. Volume: l × w × h = 2k × k × 2k = 4k³ = 54, therefore k³ = 27/2
   2. k = (27/2)^1/3 = 3/∛2, so k² = 9/∛4 = 9∛2/2
   3. Total surface area = 2(lw + wh + lh) = 2(2k·k + k·2k + 2k·2k) = 2(2k² + 2k² + 4k²) = 16k²
   4. TSA = 16 × 9∛2/2 = 72∛2 cm² (≈ 90.7 cm²)

   Option (d) 90 cm² results from incorrectly using ratio 2:1:1 instead of 2:1:2. Option (a) 144 cm² results from erroneously setting k = 3.

4. 4.

   Factorise completely: 32x²y - x³y³

   1. A
      xy(32x − x²y²)
   2. B
      x²y(32 − xy²)
      Correct
   3. C
      x²(32y − xy³)
   4. D
      2xy(16x − xy²)
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   Answer: B

   Find the highest common factor (HCF) of both terms:

   1. Terms: 32x²y and x³y³
   2. HCF of coefficients: GCF(32, 1) = 1
   3. Lowest power of x: x² (since min(2, 3) = 2)
   4. Lowest power of y: y¹ (since min(1, 3) = 1)
   5. HCF = x²y
   6. Divide each term: 32x²y ÷ x²y = 32; x³y³ ÷ x²y = xy²
   7. 32x²y − x³y³ = x²y(32 − xy²)

   Options (a) and (c) are partially factored only — both can be factored further to reach x²y(32 − xy²).

5. 5.If x + y = 2y - x + 1 = 5, find the value of x
   1. A
      4
   2. B
      1
   3. C
      2
      Correct
   4. D
      3
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   Answer: C

   Use the two equations implied by the triple equality.

   1. From x + y = 5: y = 5 − x.
   2. From 2y − x + 1 = 5: 2y − x = 4.
   3. Substitute y = 5 − x: 2(5 − x) − x = 4 → 10 − 2x − x = 4 → 10 − 3x = 4.
   4. 3x = 6 → x = 2.
6. 6.In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y
   1. A
      14 : 17
   2. B
      15 : 17
      Correct
   3. C
      3 : 7
   4. D
      15 : 19
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   Answer: B

   Simplify the ratio 1050 : 1190 by dividing both terms by their GCD:

   1. Find GCD(1050, 1190): 1190 = 1 × 1050 + 140; 1050 = 7 × 140 + 70; 140 = 2 × 70, so GCD = 70
   2. 1050 ÷ 70 = 15
   3. 1190 ÷ 70 = 17
   4. Ratio X : Y = 15 : 17
7. 7.The n^th term of a sequence is T_n = 5 + (n - 1)². Evaluate T₄ - T₆
   1. A
      16
   2. B
      −20
   3. C
      −16
      Correct
   4. D
      20
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   Answer: C

   Substitute n = 4 and n = 6 into the formula, then subtract.

   1. T₄ = 5 + (4 − 1)² = 5 + 9 = 14.
   2. T₆ = 5 + (6 − 1)² = 5 + 25 = 30.
   3. T₄ − T₆ = 14 − 30 = −16.
8. 8.Simplify: $\frac{54k^2 - 6}{3k + 1}$
   1. A
      6(3k + 1)
   2. B
      18k − 6
      Correct
   3. C
      18k + 6
   4. D
      6k − 6
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   Answer: B

   Factor the numerator then cancel.

   1. 54k² − 6 = 6(9k² − 1) = 6(3k + 1)(3k − 1).
   2. (54k² − 6)/(3k + 1) = 6(3k + 1)(3k − 1)/(3k + 1) = 6(3k − 1).
   3. 6(3k − 1) = 18k − 6.
9. 9.The curved surface area of a cylindrical tin is 704cm². If the radius of its base is 8cm, find the height. [Take $\pi = \frac{22}{7}$]
   1. A
      16 cm
   2. B
      12 cm
   3. C
      14 cm
      Correct
   4. D
      10 cm
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   Answer: C

   Curved surface area of a cylinder = 2πrh.

   1. 2 × (22/7) × 8 × h = 704.
   2. (44/7) × 8 × h = 704 → 352h/7 = 704.
   3. h = 704 × 7 / 352 = 4928 / 352 = 14 cm.
10. 10.Express $3 - \left(\frac{x - y}{y}\right)$ as a single fraction
    1. A
       $\frac{4y - x}{y}$
       Correct
    2. B
       $\frac{y - x}{y}$
    3. C
       $\frac{2y - x}{y}$
    4. D
       $\frac{3y - x}{y}$
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    Answer: A

    Write 3 with denominator y, then combine.

    1. 3 = 3y/y.
    2. 3 − (x − y)/y = 3y/y − (x − y)/y = (3y − x + y)/y.
    3. Simplify numerator: 3y + y − x = 4y − x.
    4. Result: (4y − x)/y.

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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