Mathematics 2018 Jamb Examination

First, find the number of students who offer at least one of the two subjects: \[ n(M \cup P) = \text{Total} - \text{Neither} = 40 - 4 = 36 \] Using the set formula \( n(M \cup P) = n(M) + n(P) - n(M \cap P) \): \[ 36 = 32 + 24 - n(M \cap P) \] \[ 36 = 56 - n(M \cap P) \implies n(M \cap P) = 56 - 36 = 20 \]

Multiply the entire inequality by the LCM of 4 and 3, which is 12, to clear the fractions: \[ 3(x + 2) - 4(2x - 3) < 48 \] \[ 3x + 6 - 8x + 12 < 48 \] \[ -5x + 18 < 48 \] \[ -5x < 30 \] When dividing by a negative number, the inequality sign reverses: \[ x > \frac{30}{-5} \implies x > -6 \]

First, find the total degrees remaining for Housing and School Fees: \[ 360^\circ - (90^\circ + 120^\circ) = 360^\circ - 210^\circ = 150^\circ \] Let the angle for School Fees be \( x \) and Housing be \( 2x \): \[ x + 2x = 150^\circ \implies 3x = 150^\circ \implies x = 50^\circ \] So, the angle for Housing is \( 2(50^\circ) = 100^\circ \). \[ \text{Expenditure} = \frac{100}{360} \times 7200 = 100 \times 20 = 2000 \]

To find the maximum profit, differentiate the profit function \( P = 10x - x^2 \) and set it to zero: \[ \frac{dP}{dx} = 10 - 2x \] \[ 10 - 2x = 0 \implies 2x = 10 \implies x = 5 \] Alternatively, for a quadratic \( ax^2 + bx + c \), maximum occurs at \( x = -b/2a \): \[ x = \frac{-10}{2(-1)} = 5 \]

\[ 23_{5} = (2 \times 5^{1}) + (3 \times 5^{0}) = 10 + 3 = 13 \] \[ 101_{3} = (1 \times 3^{2}) + (0 \times 3^{1}) + (1 \times 3^{0}) = 9 + 0 + 1 = 10 \] \[ y = 13 + 10 = 23_{10} \] Now convert \(23_{10}\) to base two. \[ \begin{array}{c|c} \text{Division} & \text{Remainder} \\ \hline 23 \div 2 = 11 & 1 \\ 11 \div 2 = 5 & 1 \\ 5 \div 2 = 2 & 1 \\ 2 \div 2 = 1 & 0 \\ 1 \div 2 = 0 & 1 \end{array} \uparrow \] \[ 23_{10} = 10111_{2} \]

The sum of angles at a point is \(360^\circ\). \[ (x - 10) + (4x - 50) + 3x + (2x + 20) = 360 \] \[ x - 10 + 4x - 50 + 3x + 2x + 20 = 360 \] \[ 10x - 40 = 360 \] \[ 10x = 400 \] \[ x = 40 \]

Multiply the entire equation by the LCM of 4 and 3, which is 12: \[ 12 \left( \frac{3}{4}t \right) + 12 \left( \frac{1}{3}(21 - t) \right) = 12(11) \] \[ 9t + 4(21 - t) = 132 \] \[ 9t + 84 - 4t = 132 \] \[ 5t = 132 - 84 \] \[ 5t = 48 \implies t = \frac{48}{5} = 9\frac{3}{5} \]

First, find the total fraction spent: \[ \frac{1}{4} + \frac{1}{3} = \frac{3 + 4}{12} = \frac{7}{12} \] To find the remaining fraction, subtract the amount spent from the whole (1): \[ 1 - \frac{7}{12} = \frac{12 - 7}{12} = \frac{5}{12} \]

Isolate the term containing \( y \): \[ \frac{y}{b} = 1 - \frac{x}{a + 1} \] Combine the right side into a single fraction: \[ \frac{y}{b} = \frac{(a + 1) - x}{a + 1} \] Multiply both sides by \( b \): \[ y = \frac{b(a + 1 - x)}{a + 1} \]

 The total surface area (TSA) of a cuboid is given by \( 2(lw + lh + wh) \): \[ TSA = 2(12 \times 10 + 12 \times 8 + 10 \times 8) \] \[ TSA = 2(120 + 96 + 80) \] \[ TSA = 2(296) = 592\text{cm}^2 \]