Mathematics 2025 Jamb Examination

First, find the total number of fruits: \(6 + 11 + 13 = 30\).
Since the events "choosing a grape" and "choosing a banana" are mutually exclusive, the probability of choosing either is the sum of their individual probabilities:\[ P(\text{Grape or Banana}) = \frac{6}{30} + \frac{11}{30} = \displaystyle \frac{17}{30} \]

The formula for the sum of interior angles of an \(n\)-sided polygon is \((n - 2) \times 180^{\circ}\).
For a pentagon (\(n = 5\)):\[ \text{Sum} = (5 - 2) \times 180^{\circ} = 3 \times 180^{\circ} = 540^{\circ} \]For a regular polygon, each interior angle is equal:\[ \text{Interior Angle} = \displaystyle \frac{540^{\circ}}{5} = 108^{\circ} \]

Let the total salary be \(1\). Fraction spent on shirts: \[ \frac{1}{5} \] Remainder after buying shirts: \[ 1 - \frac{1}{5} = \frac{4}{5} \] Fraction spent on transport: \[ \frac{1}{3} \text{ of } \frac{4}{5} = \frac{4}{15} \] Total fraction spent: \[ \frac{1}{5} + \frac{4}{15} = \frac{3 + 4}{15} = \frac{7}{15} \] Fraction of salary left: \[ 1 - \frac{7}{15} = \frac{8}{15} \]

The sample space for a six-sided die is \(S = \{1, 2, 3, 4, 5, 6\}\), so total outcomes = 6.
The even numbers are \(E = \{2, 4, 6\}\), so successful outcomes = 3.\[ P(\text{Even}) = \displaystyle \frac{3}{6} = \displaystyle \frac{1}{2} \]

Use the power rule \(\displaystyle \int x^{n} dx = \displaystyle \frac{x^{n+1}}{n+1}\) and trig integral \(\displaystyle \int \cos x dx = \sin x\):\[ \displaystyle \int (4x^{3} + 2x + \cos x) dx = \displaystyle \frac{4x^{4}}{4} + \displaystyle \frac{2x^{2}}{2} + \sin x + C \]\[ = x^{4} + x^{2} + \sin x + C \]

First, express 0.00001 as a power of 10:
\(0.00001 = 10^{-5}\).
Now substitute into the expression:\[ - \log_{10}(10^{-5}) \]Using the power rule \(\log a^{b} = b \log a\):\[ -(-5 \log_{10} 10) = -(-5 \times 1) = 5 \]

A perpendicular line from the centre to a chord bisects the chord. This forms a right-angled triangle where:
Hypotenuse = radius = 17 cm.
Base = half of the chord = \(\displaystyle \frac{30}{2} = 15\) cm.
Let \(d\) be the distance from the centre. By Pythagoras:\[ d^{2} + 15^{2} = 17^{2} \]\[ d^{2} + 225 = 289 \]\[ d^{2} = 64 \implies d = 8 \text{ cm} \]

Factor the numerator by grouping:\[ x^{2}(x - 2) + 6(x - 2) = (x^{2} + 6)(x - 2) \]The expression becomes:\[ \lim_{x \to 2} \displaystyle \frac{(x^{2} + 6)(x - 2)}{x - 2} = \lim_{x \to 2} (x^{2} + 6) \]Substitute \(x = 2\):\[ 2^{2} + 6 = 4 + 6 = 10 \]

In a GP, \(T_{n} = ar^{n-1}\).
\(T_{2} = ar = 1\) (Eq. 1)
\(T_{5} = ar^{4} = \displaystyle \frac{1}{8}\) (Eq. 2)
Divide Eq. 2 by Eq. 1:\[ \displaystyle \frac{ar^{4}}{ar} = \displaystyle \frac{1/8}{1} \]\[ r^{3} = \displaystyle \frac{1}{8} \implies r = \sqrt[3]{\displaystyle \frac{1}{8}} = \displaystyle \frac{1}{2} \]

The Total Amount (A) is given by \(A = P + I\), where \(I = \displaystyle \frac{PRT}{100}\).\[ A = P\left(1 + \displaystyle \frac{RT}{100}\right) \]Given \(A = 600,000\), \(R = 5\), and \(T = 4\):\[ 600,000 = y\left(1 + \displaystyle \frac{5 \times 4}{100}\right) \]\[ 600,000 = y(1 + 0.20) \]\[ 600,000 = 1.2y \]\[ y = \displaystyle \frac{600,000}{1.2} = 500,000 \]