JAMB MathematicsPast Questions & Explanations

Practice 97+ JAMB 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.

   Solve the simultaneous equations $\frac{x}{2} - \frac{y}{5} = 1$ and $y - \frac{x}{3} = 8$

   1. A.$x = 10, y = -6$
   2. B.$x = 10, y = 6$
   3. C.$x = 6, y = 10$
   4. D.$x = 5, y = -7$

   Show explanation

   Solve the simultaneous equations: $\frac{x}{2} - \frac{y}{5} = 1$, \quad $y - \frac{x}{3} = 8$ Step 1: Eliminate denominators in the first equation Multiply through by $10$: $10\left(\frac{x}{2} - \frac{y}{5}\right) = 10$ $5x - 2y = 10$ Step 2: Express $y$ from the second equation $y = 8 + \frac{x}{3}$ Step 3: Substitute into the first equation $5x - 2\left(8 + \frac{x}{3}\right) = 10$ Step 4: Expand $5x - 16 - \frac{2x}{3} = 10$ Step 5: Multiply through by $3$ $15x - 48 - 2x = 30$ $13x - 48 = 30$ Step 6: Solve for $x$ $13x = 78$ $x = 6$ Step 7: Substitute back to find $y$ $y = 8 + \frac{6}{3} = 10$ Final Answer: $x = 6,\; y = 10$
2. 2.

   Calculate the sum of all elements in the inverse of the matrix $\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$.

   1. A.-1
   2. B.1
   3. C.-2
   4. D.2

   Show explanation

   Let $A=\begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$. The determinant is $\det(A)=(2)(3)-(1)(5)=1$. Therefore, $A^{-1}=\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$. The sum of all elements is $3+(-1)+(-5)+2=-1$. Hence, the answer is $\boxed{-1}$.
3. 3.

   Given that $\cos 30° = \dfrac{\sqrt{3}}{2}$, find the exact value of $\sin 60° + \cos 60°$.

   1. A.A. $\dfrac{\sqrt{3} + 1}{2}$
   2. B.B. $\dfrac{\sqrt{3} - 1}{2}$
   3. C.C. $\sqrt{3}$
   4. D.D. $1$

   Show explanation

   Using standard trigonometric values: $$\sin 60° = \frac{\sqrt{3}}{2}, \quad \cos 60° = \frac{1}{2}$$ Therefore: $$\sin 60° + \cos 60° = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2}$$ The correct answer is A.
4. 4.

   If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find the value of $\dfrac{\tan\theta - \cos\theta}{\sin\theta + \cos\theta}$.

   1. A.A. $\dfrac{1}{20}$
   2. B.B. $\dfrac{-1}{20}$
   3. C.C. $\dfrac{1}{28}$
   4. D.D. $\dfrac{-1}{28}$

   Show explanation

   Given $\sin\theta = \dfrac{3}{5}$, and $\theta$ is acute (i.e. $0° < \theta < 90°$, so all trig ratios are positive). Step 1 — Find $\cos\theta$ using the Pythagorean identity: $$\sin^2\theta + \cos^2\theta = 1$$ $$\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1 \implies \cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$ Since $\theta$ is acute, $\cos\theta > 0$, so $\cos\theta = \dfrac{4}{5}$. Step 2 — Find $\tan\theta$: $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4}$$ Step 3 — Substitute into the expression: $$\frac{\tan\theta - \cos\theta}{\sin\theta + \cos\theta} = \frac{\dfrac{3}{4} - \dfrac{4}{5}}{\dfrac{3}{5} + \dfrac{4}{5}}$$ Step 4 — Simplify the numerator (common denominator 20): $$\frac{3}{4} - \frac{4}{5} = \frac{15}{20} - \frac{16}{20} = -\frac{1}{20}$$ Step 5 — Simplify the denominator: $$\frac{3}{5} + \frac{4}{5} = \frac{7}{5}$$ Step 6 — Divide (multiply by the reciprocal): $$-\frac{1}{20} \div \frac{7}{5} = -\frac{1}{20} \times \frac{5}{7} = -\frac{5}{140} = -\frac{1}{28}$$
5. 5.

   Convert $1101_2$ to base ten.

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

   Show explanation

   $1101_2 = (1\times2^3)+(1\times2^2)+(0\times2^1)+(1\times2^0) = 8+4+0+1 = 13$.
6. 6.

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

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

   Show explanation

   $25 = 16+8+1 = (1\times2^4)+(1\times2^3)+(0\times2^2)+(0\times2^1)+(1\times2^0) = 11001_2$.
7. 7.

   Convert $342_5$ to base ten.

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

   Show explanation

   $342_5 = (3\times5^2)+(4\times5^1)+(2\times5^0) = 75+20+2 = 97$.
8. 8.

   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: $(2\times25)+(1\times5)+3 = 50+5+3 = 58$.
9. 9.

   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 $1100_2$.
10. 10.

    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 $101_2$.

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11. 11.

    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 $1111_2$.
12. 12.

    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 $x=8$.
13. 13.

    Convert $123_4$ to base ten.

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

    Show explanation

    $123_4 = (1\times4^2)+(2\times4^1)+(3\times4^0) = 16+8+3 = 27$.
14. 14.

    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 $110_2$.
15. 15.

    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 $r=8$.
16. 16.

    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 $1110_2$.
17. 17.

    Evaluate $\sqrt{25}$.

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

    Show explanation

    $\sqrt{25} = 5$ because $5^2 = 25$.
18. 18.

    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, $6 + 7 = 13$.
19. 19.

    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, $9 - 4 = 5$.
20. 20.

    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

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