Mathematics 2025 Jamb Examination

To find the modal weight, we first determine the value of \( x \) by summing the frequencies to 12: \[ x + 1 + 2x + 2 + 3 = 12 \] \[ 3x + 6 = 12 \] \[ 3x = 6 \implies x = 2 \] Now, substitute \( x \) back into the table to find the frequencies:
Weight 54: \( x = 2 \)
Weight 58: \( 2x = 4 \)
Since 4 is the highest frequency, the modal weight is \( 58 \text{ kg} \).

Substitute the values \( a = 1 \) and \( b = 3 \) into the defined rule: \[ 1 * 3 = (1 \times 3) + 1 + 3 \] \[ 1 * 3 = 3 + 1 + 3 = 7 \]

First, find the value of \( x \) by summing the frequencies to 12: \[ x + 1 + 2x + 2 + 3 = 12 \] \[ 3x + 6 = 12 \] \[ 3x = 6 \implies x = 2 \] Now, substitute \( x \) back into the frequencies to find the highest occurrence:
Weight 54: \( x = 2 \)
Weight 58: \( 2x = 4 \)
Weight 60: 2
Weight 62: 3
Since 4 is the highest frequency, the modal weight is \( 58 \text{ kg} \).

The modal class is the range with the highest frequency, which is \( 21 - 30 \) (frequency 10).
The upper-class boundary is the midpoint between the current upper limit (30) and the next lower limit (31): \[ \text{Upper Boundary} = \displaystyle \frac{30 + 31}{2} = 30.5 \]

In \( \triangle PQR \), the interior angle at \( Q \) is \( 60^{\circ} + 30^{\circ} = 90^{\circ} \). Since R is due East of P, the angle at \( P \) is \( 90^{\circ} - 60^{\circ} = 30^{\circ} \).
We need the length \( QR \) (the side opposite the \( 30^{\circ} \) angle), and we have \( PQ = 200 \text{ km} \) (the adjacent side): \[ \tan(30^{\circ}) = \displaystyle \frac{QR}{PQ} \] \[ QR = 200 \times \tan(30^{\circ}) = 200 \times \displaystyle \frac{1}{\sqrt{3}} \] If we look at the triangle from the perspective where \( PQ \) is opposite the \( 30^{\circ} \) angle, then \( QR = 200 \times \tan(60^{\circ}) = 200\sqrt{3} \).

First, solve for the length \( XY \) (the side opposite \( \theta \)): \[ XY = \sqrt{13^{2} - 11^{2}} = \sqrt{169 - 121} = \sqrt{48} \] Next, calculate \( \cot \theta \), which is the ratio of the Adjacent side to the Opposite side: \[ \cot \theta = \displaystyle \frac{\text{Adjacent}}{\text{Opposite}} = \frac{11}{\sqrt{48}} \] Therefore, the values are \( \displaystyle \frac{11}{\sqrt{48}} \) and \( \sqrt{48} \).