Mathematics 2025 Jamb Examination
Mathematics 2025 Jamb Examination
Question 21
Find the value of t for which (\(\displaystyle \frac{1}{2}\))\(^{t - 1}\) = 64
Express both sides with the same base (base 2):\[ (2^{-1})^{t-1} = 2^{6} \]\[ 2^{-t+1} = 2^{6} \]Equating the powers:\[ -t + 1 = 6 \]\[ -t = 6 - 1 = 5 \]\[ t = -5 \]
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Question 22
If 54\(_{ten}\) = X\(_{four}\), find the value of X
Divide 54 by 4 repeatedly and record the remainders:\[ \begin{array}{r|l}
\text{Divide} & \text{Remainder} \\ \hline
54 \div 4 = 13 & 2 \\
13 \div 4 = 3 & 1 \\
3 \div 4 = 0 & 3 \\
\end{array}
\quad \uparrow\] \[ \text{Reading the remainders from bottom to top: } 312_4\]
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Question 23
Integrate the function y = 3x\(^{2}\) + 2x - 5 with respect to x.
Using \(\displaystyle \int x^{n} dx = \displaystyle \frac{x^{n+1}}{n+1}\):\[ \displaystyle \int (3x^{2} + 2x - 5) dx = \displaystyle \frac{3x^{3}}{3} + \displaystyle \frac{2x^{2}}{2} - 5x + C \]\[ = x^{3} + x^{2} - 5x + C \]
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Question 24
Express \(\sqrt[4]{0.16}\) in standard form
First, express 0.16 as a fraction or power: \(0.16 = 16 \times 10^{-2}\).\[ \sqrt[4]{16 \times 10^{-2}} = (16)^{1/4} \times (10^{-2})^{1/4} \]\[ = 2 \times 10^{-2/4} = 2 \times 10^{-1/2} \]
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Question 25
Solve x\(^{2}\) + 3x - 4 ≤ 0
Factorize the quadratic:\[ (x + 4)(x - 1) \le 0 \]The critical values are \(x = -4\) and \(x = 1\). For the expression to be less than or equal to zero, \(x\) must lie between the roots.\[ -4 \le x \le 1 \]
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Question 26
Solve for y in \(\sqrt{75} - \sqrt{12} + \sqrt{27} = y\sqrt{3}\)
Simplify each surd:
\(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
\(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\)
\[ 5\sqrt{3} - 2\sqrt{3} + 3\sqrt{3} = 6\sqrt{3} \]Comparing to \(y\sqrt{3}\), \(y = 6\).
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Question 27
If cos \(\theta\) = \(\displaystyle \frac{x}{y}\), find tan \(\theta\) in terms of x and y
Given \(\cos \theta = \text{Adj}/\text{Hyp} = x/y\).
By Pythagoras, \(\text{Opposite} = \sqrt{\text{Hyp}^{2} - \text{Adj}^{2}} = \sqrt{y^{2} - x^{2}}\).
\(\tan \theta = \text{Opposite}/\text{Adjacent} = \displaystyle \frac{\sqrt{y^{2} - x^{2}}}{x}\).
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Question 28
Given that \(P = \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix}\) and Q = \(\begin{pmatrix} 3 & -7 \\ 1 & 2 \end{pmatrix}\) . Find P + 2Q
\[ 2Q = \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} \]\[ P + 2Q = \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix} + \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 1+6 & 3-14 \\ 2+2 & -5+4 \end{pmatrix} = \begin{pmatrix} 7 & -11 \\ 4 & -1 \end{pmatrix} \]
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Question 29

The figure above is a pie chart. Use it to find in degrees those who are doctors
The pie chart represents a full circle of \(360^\circ\).
The total number of people is 16 (Doctors) + 8 (Engineers) + 24 (Lawyers) + 21Â (Information Technologists) + 27 (Teachers) = 96
The fraction of the circle for doctors is \( \dfrac{16}{96}\)
Hence, the angle representing doctors in the pie chart is\[ \text{Angle for doctors} = \frac{16}{96} \times 360^\circ\]\[ = 60^\circ\]
The total number of people is 16 (Doctors) + 8 (Engineers) + 24 (Lawyers) + 21Â (Information Technologists) + 27 (Teachers) = 96
The fraction of the circle for doctors is \( \dfrac{16}{96}\)
Hence, the angle representing doctors in the pie chart is\[ \text{Angle for doctors} = \frac{16}{96} \times 360^\circ\]\[ = 60^\circ\]
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Question 30

In the Venn diagram above, the shaded region is
The Venn diagram has three overlapping circles labeled P, Q, and R.
The shaded region has cross-hatching and covers only the part that belongs to Q but not to P or RÂ (the exclusive part of Q outside the overlaps with P and R).
In set notation, the shaded region is \( Q \cap (P \cup R)^c \)Â
