JAMB Mathematics Past Questions 2025

Answer: C. $x=6,y=10$

Multiply the first equation by 10: $5x-2y=10$. From the second, $x=3y-24$. Substituting: $5(3y-24)-2y=10$, so $15y-120-2y=10$, $13y=130$, $y=10$, then $x=3(10)-24=6$.

Answer: B. $-1$

The determinant is $(1)(5)-(2)(3)=5-6=-1$. The inverse is $\frac{1}{-1}\left(\begin{array}{cc}5& -2\\ -3& 1\end{array}\right)=\left(\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right)$. Sum of entries $=-5+2+3-1=-1$.

Answer: A. $68$

Let $P=a+bQ$. Then $32=a+16b$ and $20=a+12b$. Subtracting: $12=4b$, so $b=3$, and $a=32-16(3)=-16$. When $Q=28$: $P=-16+3(28)=-16+84=68$.

Answer: B. $10$

Factor the numerator: $x^3-2x^2+6x-12=x^2(x-2)+6(x-2)=(x-2)(x^2+6)$. So $y=x^2+6$ for $x\ne 2$. As $x$ approaches $2$, $y$ approaches $4+6=10$.

Answer: D. $(1,5)$

Midpoint $=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-3+5}{2},\frac{4+6}{2}\right)=(1,5)$.

Answer: C. $130{\text{ cm}}^2$

Area $=\frac{1}{2}\times \text{base}\times \text{height}$, using the full base $QR=QM+MR=11+9=20\text{ cm}$ and height $PM=13\text{ cm}$ (since $PM$ is perpendicular to $QR$). Area $=\frac{1}{2}\times 20\times 13=130{\text{ cm}}^2$.

Answer: C. $\theta =\frac{360A}{\pi r^2}$

Multiply both sides by 360: $360A=\theta \pi r^2$. Divide both sides by $\pi r^2$: $\theta =\frac{360A}{\pi r^2}$.

Answer: A. $1022_5$

$137÷5=27$ remainder $2$. $27÷5=5$ remainder $2$. $5÷5=1$ remainder $0$. $1÷5=0$ remainder $1$. Reading remainders bottom to top: $1022_5$.

Answer: C. $\frac{1}{2}$

$\frac{T_5}{T_2}=r^3=\frac{1/8}{1}=\frac{1}{8}$, so $r=\frac{1}{2}$.

Answer: C. $0$

Sum $=0+(x+2)+(3x+6)+(4x+8)=8x+16$. Mean $=\frac{8x+16}{4}=4$, so $8x+16=16$, giving $x=0$.