WAEC MathematicsPast Questions & Explanations

Practice 59+ WAEC Mathematics questions from the SuperPrep question bank. Every question comes with a full explanation so you learn the method, not just the answer.

1. 1.

   $23_{x} = 32_{5}$. Find the value of $x$.

   1. A.7
   2. B.6
   3. C.5
   4. D.4

   Show explanation

   $23_x = 32_5$ Converting both numbers to base $10$: $2x + 3 = 3(5) + 2$ $2x + 3 = 17$ $2x = 14$ $\therefore x = 7$ Hence, the value of $x$ is $7$.
2. 2.

   In the diagram below, TPS is a straight line, PQRS is a parallelogram with base QR and height 8cm. |QR| = 6cm and the area of triangle QST is 52cm2. Find the area of ΔQPT

   1. A.39 cm²
   2. B.28 cm²
   3. C.24 cm²
   4. D.21cm²

   Show explanation

   Area of ΔQST = 52cm2; height = 8cm b=base length; but A = 1/2 b x h =52=1/2 x b x 8cm = 104 =8b b=104/8 = 13cm; ST= 13cm; PS = 6cm TP = 13 - 6 = 7cm; area of ΔQST = 1/2 b x h = 1/2 x 7 x 8/1 = 28cm2

3. 3.

   In the diagram O is the center of the circle, if ∠QRS = $62^{0}$, find the value of ∠SQR.

   1. A.14°
   2. B.28°
   3. C.31°
   4. D.45°

   Show explanation

   < QSR = 90° (angle in a semi-circle)

   ∴ < SQR = 180° - (90° + 62°)

   = 28°

4. 4.

   Convert $1101_2$ to base ten.

   1. A.12
   2. B.13
   3. C.14
   4. D.15

   Show explanation

   $1101_2 = (1\times2^3)+(1\times2^2)+(0\times2^1)+(1\times2^0) = 8+4+0+1 = 13$.
5. 5.

   Convert $25_{ten}$ to base two.

   1. A.11001
   2. B.10011
   3. C.11100
   4. D.10101

   Show explanation

   $25 = 16+8+1 = (1\times2^4)+(1\times2^3)+(0\times2^2)+(0\times2^1)+(1\times2^0) = 11001_2$.
6. 6.

   Convert $342_5$ to base ten.

   1. A.87
   2. B.92
   3. C.97
   4. D.102

   Show explanation

   $342_5 = (3\times5^2)+(4\times5^1)+(2\times5^0) = 75+20+2 = 97$.
7. 7.

   Convert $58_{ten}$ to base five.

   1. A.$203_5$
   2. B.$213_5$
   3. C.$223_5$
   4. D.$123_5$

   Show explanation

   Dividing repeatedly by 5: $58 \div 5 = 11$ remainder $3$; $11 \div 5 = 2$ remainder $1$; $2 \div 5 = 0$ remainder $2$. Reading the remainders from last to first gives $213_5$. Check: $(2\times25)+(1\times5)+3 = 50+5+3 = 58$.
8. 8.

   Evaluate $101_2 + 111_2$, leaving your answer in base two.

   1. A.1011
   2. B.1100
   3. C.1101
   4. D.1110

   Show explanation

   $101_2 = 5$ and $111_2 = 7$. $5+7 = 12$. Converting $12$ back to base two gives $1100_2$.
9. 9.

   Evaluate $1011_2 - 110_2$, leaving your answer in base two.

   1. A.100
   2. B.101
   3. C.110
   4. D.111

   Show explanation

   $1011_2 = 11$ and $110_2 = 6$. $11-6 = 5$. Converting $5$ back to base two gives $101_2$.
10. 10.

    Find $11_2 \times 101_2$, leaving your answer in base two.

    1. A.1101
    2. B.1110
    3. C.1111
    4. D.10000

    Show explanation

    $11_2 = 3$ and $101_2 = 5$. $3\times5 = 15$. Converting $15$ back to base two gives $1111_2$.

    Want thousands more with instant AI explanations after every answer?

    Practice free on SuperPrep
11. 11.

    If $24_x = 20_{ten}$, find the value of the base $x$.

    1. A.6
    2. B.7
    3. C.8
    4. D.9

    Show explanation

    $24_x$ means $(2\times x)+4$. Setting this equal to $20$: $2x+4=20$, so $2x=16$ and $x=8$.
12. 12.

    Convert $123_4$ to base ten.

    1. A.27
    2. B.21
    3. C.24
    4. D.31

    Show explanation

    $123_4 = (1\times4^2)+(2\times4^1)+(3\times4^0) = 16+8+3 = 27$.
13. 13.

    Evaluate $1100_2 \div 10_2$, leaving your answer in base two.

    1. A.101
    2. B.110
    3. C.111
    4. D.100

    Show explanation

    $1100_2 = 12$ and $10_2 = 2$. $12 \div 2 = 6$. Converting $6$ back to base two gives $110_2$.
14. 14.

    If $23_r = 19_{ten}$, find the value of the base $r$.

    1. A.5
    2. B.6
    3. C.7
    4. D.8

    Show explanation

    $23_r$ means $(2\times r)+3$. Setting this equal to $19$: $2r+3=19$, so $2r=16$ and $r=8$.
15. 15.

    Evaluate $11011_2 - 1101_2$, leaving your answer in base two.

    1. A.1101
    2. B.1110
    3. C.1010
    4. D.1111

    Show explanation

    $11011_2 = 27$ and $1101_2 = 13$. $27-13 = 14$. Converting $14$ back to base two gives $1110_2$.
16. 16.

    Evaluate $\sqrt{25}$.

    1. A.3
    2. B.4
    3. C.5
    4. D.6

    Show explanation

    $\sqrt{25} = 5$ because $5^2 = 25$.
17. 17.

    Simplify $\sqrt{36} + \sqrt{49}$.

    1. A.11
    2. B.12
    3. C.13
    4. D.14

    Show explanation

    $\sqrt{36} = 6$ and $\sqrt{49} = 7$. Therefore, $6 + 7 = 13$.
18. 18.

    Find the value of $\sqrt{81} - \sqrt{16}$.

    1. A.3
    2. B.4
    3. C.5
    4. D.6

    Show explanation

    $\sqrt{81} = 9$ and $\sqrt{16} = 4$. Thus, $9 - 4 = 5$.
19. 19.

    Express $\sqrt{50}$ in its simplest form.

    1. A.$5\sqrt{2}$
    2. B.$2\sqrt{5}$
    3. C.$10\sqrt{2}$
    4. D.$25\sqrt{2}$

    Show explanation

    $\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$.
20. 20.

    Simplify $3\sqrt{5} + 2\sqrt{5}$.

    1. A.$5\sqrt{5}$
    2. B.$6\sqrt{5}$
    3. C.$5$
    4. D.$\sqrt{25}$

    Show explanation

    Since the surds are alike, add the coefficients: $3 + 2 = 5$. Therefore, $3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}$.