JAMB Mathematics Practice Questions & Explanations

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: $x = 6,\; y = 10$

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 $\boxed{-1}$.

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.

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): $$-\frac{1}{20} \div \frac{7}{5} = -\frac{1}{20} \times \frac{5}{7} = -\frac{5}{140} = -\frac{1}{28}$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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