Mathematics 2023 Jamb Examination

First, calculate the constant speed of the car using the formula \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\):\[\text{Speed} = \frac{225}{4.5} = 50 \text{ km/h}\]Now, find the time required to cover 150 km at this speed:\[\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{150}{50} = 3 \text{ hrs}\]

Convert the binary numbers to base ten to simplify the division:\[1101001_{two} = (1 \times 2^6) + (1 \times 2^5) + (0 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)\]\[= 64 + 32 + 0 + 8 + 0 + 0 + 1 = 105_{ten}\]\[101_{two} = (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 4 + 0 + 1 = 5_{ten}\]Divide the base ten numbers: \(105 \div 5 = 21_{ten}\).
Convert 21 back to binary:
21 ÷ 2 = 10 R 1
10 ÷ 2 = 5 R 0
5 ÷ 2 = 2 R 1
2 ÷ 2 = 1 R 0
1 ÷ 2 = 0 R 1
Reading remainders upwards: \(10101_{two}\).

Rationalize the denominator by multiplying the numerator and denominator by the conjugate \(2 - \sqrt{3}\):\[\frac{3 - \sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{6 - 3\sqrt{3} - 2\sqrt{3} + 3}{2^2 - (\sqrt{3})^2}\]\[= \frac{9 - 5\sqrt{3}}{4 - 3} = 9 - 5\sqrt{3}\]Comparing \(9 - 5\sqrt{3}\) with \(a + b\sqrt{3}\), we get \(a = 9\) and \(b = -5\).

Use the distance formula \(d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\):\[9^2 = (t - (-3))^2 + (-5 - (-14))^2\]\[81 = (t + 3)^2 + (9)^2\]\[81 = (t + 3)^2 + 81\]\[(t + 3)^2 = 0 \implies t + 3 = 0 \implies t = -3\]

Let the Principal be \(P\). Amount after 10 years = \(3P\). Since \(A = P + I\), interest \(I = 2P\) in 10 years.
Using \(I = PRT/100\): \(2P = P \times R \times 10 / 100 \implies 2 = R/10 \implies R = 20\%\).
We want the amount to be \(5P\), which means interest \(I = 4P\):\[4P = P \times 20 \times T / 100 \implies 4 = 20T/100 \implies 400 = 20T \implies T = 20 \text{ years}\]

Direct substitution gives \(\dfrac{0}{0}\).
Factorize the numerator and denominator:\[x^2 + 4x - 12 = (x + 6)(x - 2)\]\[x^2 - 2x = x(x - 2)\]The expression becomes: \(\dfrac{(x + 6)(x - 2)}{x(x - 2)}\). Cancel \((x - 2)\):\[\lim_{x\to2} \frac{x + 6}{x} = \frac{2 + 6}{2} = \frac{8}{2} = 4\]

First, find the mid-points (x) of each class interval:
5-6: \(5.5\); 7-8: \(7.5\); 9-10: \(9.5\)
Calculate \(\sum fx\):\[(29 \times 5.5) + (40 \times 7.5) + (38 \times 9.5) = 159.5 + 300 + 361 = 820.5\]Calculate total frequency \(\sum f = 29 + 40 + 38 = 107\):\[\text{Mean} = \frac{820.5}{107} \approx 7.668...\]Rounding to one decimal place gives 7.7.

A majority of women in a committee of 5 means having 3, 4, or 5 women. Since there are only 4 women available, the possible cases are:
Case 1: 3 women and 2 men → \(^4C_3 \times ^6C_2 = 4 \times 15 = 60\)
Case 2: 4 women and 1 man → \(^4C_4 \times ^6C_1 = 1 \times 6 = 6\)
Total committees = 60 + 6 = 66.

Expand along the first row:\[|D| = 2\begin{vmatrix} 1 & 2 \\ -3 & 1 \end{vmatrix} - (-1)\begin{vmatrix} 4 & 2 \\ 1 & 1 \end{vmatrix} + 3\begin{vmatrix} 4 & 1 \\ 1 & -3 \end{vmatrix}\]\[= 2(1 - (-6)) + 1(4 - 2) + 3(-12 - 1)\]\[= 2(7) + 1(2) + 3(-13) = 14 + 2 - 39 = -23\]

The movement forms a right-angled triangle PQR. Let the angle at P inside the triangle be \(\theta\).\[\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{6}{8} = 0.75\]\[\theta = \tan^{-1}(0.75) \approx 36.87^o\]Since the boat moved North and then West, the bearing is measured clockwise from North. The angle \(\theta\) is West of North.\[\text{Bearing} = 360^o - 36.87^o = 323.13^o\]To the nearest degree, the bearing is 323°.