Mathematics 2025 Jamb Examination

The equation for partial variation is \(y = a + bx\), where \(a\) and \(b\) are the constants.
Substituting the first set of values (x=3, y=7):\[ 7 = a + 3b \quad \text{(Eq. 1)} \]Substituting the second set of values (x=5, y=11):\[ 11 = a + 5b \quad \text{(Eq. 2)} \]Subtract Eq. 1 from Eq. 2:\[ (11 - 7) = (a - a) + (5b - 3b) \]\[ 4 = 2b \implies b = 2 \]Substitute \(b = 2\) into Eq. 1:\[ 7 = a + 3(2) \]\[ 7 = a + 6 \implies a = 1 \]The constants are 1 and 2.

1. Total weight of 15 bars = \(15 \times 1000 = 15,000\text{ kg}\).
2. When the heaviest is removed, 14 bars remain.
3. New average = \(1000 - 5 = 995\text{ kg}\).
4. Total weight of 14 bars = \(14 \times 995 = 13,930\text{ kg}\).
5. Weight of heaviest bar = \(15,000 - 13,930 = 1,070\text{ kg}\).

For a matrix \(M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
The inverse is \(\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\).
Determinant = \((1 \times 5) - (2 \times 3) = 5 - 6 = -1\).
Inverse = \(\frac{1}{-1} \begin{bmatrix} 5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix}\).
Sum of entries = \(-5 + 2 + 3 + (-1) = -1\).

The relation is \(A = kb^{2}\).
Find \(k\): \(4 = k(1)^{2} \implies k = 4\).
Now find \(A\) when \(b = 2\):\[ A = 4(2)^{2} = 4(4) = 16 \]

A \(2 \times 2\) identity matrix is \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\).
Determinant = \((1 \times 1) - (0 \times 0) = 1 - 0 = 1\).

Expand along the first row:\[ -2\begin{vmatrix} 2 & 1 \\ 4 & 2 \end{vmatrix} - 3\begin{vmatrix} p & 1 \\ 1 & 2 \end{vmatrix} + 1\begin{vmatrix} p & 2 \\ 1 & 4 \end{vmatrix} = -5 \]\[ -2(4 - 4) - 3(2p - 1) + (4p - 2) = -5 \]\[ 0 - 6p + 3 + 4p - 2 = -5 \]\[ -2p + 1 = -5 \]\[ -2p = -6 \implies p = 3 \]

\[ \frac{0 + (x + 2) + (3x + 6) + (4x + 8)}{4} = 4 \]\[ \frac{8x + 16}{4} = 4 \]\[ 2x + 4 = 4 \]\[ 2x = 0 \implies x = 0 \]

Slope \(m = 16/5\). Use the formula \(y - y_{1} = m(x - x_{1})\):\[ y - 15 = \frac{16}{5}(x - 3) \]\[ 5(y - 15) = 16(x - 3) \]\[ 5y - 75 = 16x - 48 \]\[ 5y - 16x - 75 + 48 = 0 \]\[ 5y - 16x - 27 = 0 \]

Total Cost Price (CP) = \(270,000 + 70,000 = 340,000\).
Selling Price (SP) = 490,000.
Profit = \(490,000 - 340,000 = 150,000\).
Percentage Profit = \(\dfrac{150,000}{340,000} \times 100 \approx 44.11\%\).
Rounding to the nearest option provided in JAMB context, 45% is the most appropriate choice.

1. Brackets first: \(3\dfrac{2}{3} - \dfrac{15}{6} = \dfrac{11}{3} - \dfrac{5}{2} = \dfrac{22-15}{6} = \dfrac{7}{6}\).
2. Division in bracket: \(\displaystyle \frac{3}{4} \div \frac{9}{4} = \frac{3}{4} \times \frac{4}{9} = \frac{1}{3}\).
3. 'Of' (multiplication): \(\displaystyle\frac{1}{3} \times \frac{18}{11} = \frac{6}{11}\).
4. Final multiplication with first bracket result: \(\displaystyle\frac{6}{11} \times \frac{7}{6} = \frac{7}{11}\).