Mathematics 2025 Jamb Examination

The area of triangle \(PQR\) is given by the formula
\[ \text{Area of } \triangle PQR = \frac{1}{2} \times \text{base} \times \text{height}\]
Here, \(QR\) is the base and \(P_c\) is the corresponding height. Therefore,
\[ \text{Area of } \triangle PQR = \frac{1}{2} \times 20 \times 13\] \[ = 10 \times 13\] \[ = 130\text{ cm}^2\]

Two sets are equal if they contain exactly the same elements, regardless of order. They are equivalent if they have the same number of elements (cardinality).

The set {1, 2, 3, 4, 5} has 5 elements.
Option A has 4 elements.
Option B has 5 unique elements (repeated elements in set notation are counted once).
Option C has 4 elements.
Option D {4, 3, 1, 5, 2} has 5 elements and they are identical to the original set.

Since Option D contains the exact same five elements in a different order, it is both equal and equivalent to the original set.

Equation 1: \(\frac{x}{2} - \frac{y}{5} = 1\). Multiply by 10: \(5x - 2y = 10\).
Equation 2: \(y - \frac{x}{3} = 8\). Multiply by 3: \(3y - x = 24 \implies x = 3y - 24\).

Substitute \(x\) into the modified Equation 1:\[ 5(3y - 24) - 2y = 10 \]\[ 15y - 120 - 2y = 10 \]\[ 13y = 130 \]\[ y = 10 \]Now find \(x\):\[ x = 3(10) - 24 = 30 - 24 = 6 \]The solution is \(x = 6, y = 10\).

Collect like terms:\[ 2x - 5x > 8 - 3 \]\[ -3x > 5 \]When dividing by a negative number, the inequality sign must be reversed:\[ x < -\frac{5}{3} \]\[ x < -1\frac{2}{3} \]

Substitute \(a = 3\) and \(b = 2\) into the given operation:\[ 3 * 2 = \frac{3 + 2}{3 \times 2} + (3 - 2) \]\[ 3 * 2 = \frac{5}{6} + 1 \]\[ 3 * 2 = 1\frac{5}{6} \text{ or } \frac{11}{6} \]

\(125^{-1/3} = (5^{3})^{-1/3} = 5^{-1} = 1/5\)
\(49^{-1/2} = (7^{2})^{-1/2} = 7^{-1} = 1/7\)
\(10^{0} = 1\)

Multiply the results:\[ \frac{1}{5} \times \frac{1}{7} \times 1 = \frac{1}{35} \]

The midpoint formula is \(M = \left( \dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2} \right)\).
Substitute the values for P(-3, 4) and Q(5, 6):\[ M = \left( \frac{-3 + 5}{2}, \frac{4 + 6}{2} \right) \]\[ M = \left( \frac{2}{2}, \frac{10}{2} \right) = (1, 5) \]

Since \(\cos A = 12/13\) (Adjacent/Hypotenuse), let the adjacent side be 12 and the hypotenuse be 13.
Find the opposite side (opp) using Pythagoras:\[ \text{opp} = \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = \sqrt{25} = 5 \]Now, calculate \(\tan A = \text{Opposite}/\text{Adjacent} = 5/12\).

Angles that are multiples of 15° or 7.5° (derived from bisecting 60°, 90°, etc.) can be constructed using only a compass and ruler.
- 60° is constructed using an equilateral triangle.
- 90° is a perpendicular bisector.
- 135° is 90° + 45° (bisected 90).
145° is not a multiple of 7.5° or 15° and cannot be constructed using a compass and straightedge (though the prompt mentions a protractor, in the context of examination theory, it refers to standard Euclidean construction limits).

The word SCHOOL has 6 letters in total (n=6).
The letter 'O' is repeated 2 times.
The formula for permutations with repetitions is:\[ P = \frac{n!}{p!q!...} \]\[ P = \frac{6!}{2!} = \frac{720}{2} = 360 \]