Mathematics 2018 Jamb Examination

This forms a right-angled triangle with PQ (base) = 5km and QR (height) = 12km. Using Pythagoras' theorem: \[ \displaystyle PR^2 = 5^2 + 12^2 = 25 + 144 = 169 \] \[ \displaystyle PR = \sqrt{169} = 13\text{km} \]

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

Total balls = 6. Probability of Red (R) = \( \displaystyle \frac{2}{6} = \frac{1}{3} \). Since the ball is replaced, the probability remains the same for the second draw. \[ \displaystyle P(R, R) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \]

A1 = Elements in the universal set but not in A = {s, w, x, y, z}

  B = {r, s. t, u}

  C = {t, u, v, w, x}

  A1 n B n C = elements common to the three sets = none = empty set = Φ

The area of an equilateral triangle is given by \( \displaystyle \frac{\sqrt{3}}{4}s^2 \), where \( \displaystyle s \) is the side length. \[ \displaystyle \text{Area} = \frac{\sqrt{3}}{4} \times 8^2 \] \[ \displaystyle \text{Area} = \frac{\sqrt{3}}{4} \times 64 = 16\sqrt{3} \]

To round to 1 decimal place, look at the second decimal digit (8). Since it is 5 or greater, round up the first decimal digit (9). Adding 1 to 9 makes it 10, which carries over to the units place. \[ \displaystyle 0.9875 \approx 1.0 \]

Evaluate each part using index laws: 1. \( \displaystyle 25^{\frac{1}{2}} = \sqrt{25} = 5 \) 2. \( \displaystyle 8^{-\frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}} = \frac{1}{(\sqrt[3]{8})^2} = \frac{1}{2^2} = \frac{1}{4} \) Multiply them together: \[ \displaystyle 5 \times \frac{1}{4} = \frac{5}{4} = 1\frac{1}{4} \]

Point R is 300m North of P, and Q is East of R. In triangle PRQ, PR = 300m (adjacent) and PQ = 600m (hypotenuse). \[ \displaystyle \cos \angle RPQ = \frac{300}{600} = 0.5 \implies \angle RPQ = 60^\circ \] Since Q is East of R, \( \displaystyle \angle PQR = 180 - (90 + 60) = 30^\circ \). The bearing of P from Q (measured clockwise from North at Q) is: \[ \displaystyle 180 + (90 - 30) = 180 + 60 = 240^\circ \]

Add the digits in base 8: - Units: \( \displaystyle 4 + 7 = 11 \). In base 8, \( \displaystyle 11 = 1 \text{ remainder } 3 \). Write 3, carry 1. - Eights: \( \displaystyle 5 + 6 + 1 (\text{carry}) = 12 \). In base 8, \( \displaystyle 12 = 1 \text{ remainder } 4 \). Write 4, carry 1. - Result: 143.

First, find the total number of possible outcomes: \( \displaystyle 4 \times 4 = 16 \). Now, list the pairs whose product is \( \displaystyle \le 6 \): (1,1), (1,2), (1,3), (1,4) (2,1), (2,2), (2,3) (3,1), (3,2) (4,1) Total favorable outcomes = 10. \[ \displaystyle \text{Probability} = \frac{10}{16} = \frac{5}{8} \]