Explanation

P : 5r, 3b, 2w

Q : 3r, 2b, 4w

For Mr P: \(P(r) = \frac{1}{2}; P(b) = \frac{3}{10}; P(w) = \frac{1}{5}\)

For Mr Q: \(P(r) = \frac{1}{3}; P(b) = \frac{2}{9}; P(w) = \frac{4}{9}\)

\(P(\text{ bus of the same color}) = \frac{1}{2} \times \frac{1}{3} + \frac{3}{10} \times \frac{2}{9} + \frac{1}{5} \times \frac{4}{9}\)

= \(\frac{29}{90}\)

\(P(\text{not of the same color}) = 1 - \frac{29}{90} = \frac{61}{90}\)

(b) 6 airlines, 4 shipping lines

If 1 airline is used from Nigeria to Ghana, there will be 5 out of 6 for the return journey. This can be done in \(^{6}C_{5}\) ways. Similarly, if 1 shipping line is used from Ghana to Liberia, there will be 3 shipping lines left for his return journey. This can be done in \(^{4}C_{3}\) ways.

\(\therefore \text{The total number of ways =} ^{6}C_{5} + ^{4}C_{3}\)

= \(6 + 4 = 10\) ways.

Correct Answer: B. 75

Explanation

No. of ways = 3 x 5 x 5 = 75

Correct Answer: B. 6.5

Explanation

\(f(x) = (x^{2} + 3)^{2}\)

Using the chain rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{\mathrm d y}{\mathrm d u} \times \frac{\mathrm d u}{\mathrm d x}\)

Let \(u = x^{2} + 3\) so that \(y = u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 2u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2x\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = 2u(2x) = 4xu\)

But \(u = x^{2} + 3\),

\(\frac{\mathrm d y}{\mathrm d x} = 4x(x^{2} + 3)\)

At \(x = \frac{1}{2}, \frac{\mathrm d y}{\mathrm d x} = 4(\frac{1}{2})((\frac{1}{2})^{2} + 3)\)

= \(2 \times \frac{13}{4} = \frac{13}{2} = 6.5\)

Correct Answer: A. \(\frac{3}{2}\)

Explanation

\(\frac{log √27 - log √8}{log 3 - log 2}\)

= \(\frac{log √3^3 - log √2^3}{log 3 - log 2}\)

= \(\frac{log3^{3/2} - log2^{3/2}}{log3}\)

=\(\frac{^3/_2(log 3 - log 2)}{log 3 - log 2}\)

\(\therefore\frac{3}{2}\)

Explanation

(a)(i) \(F_{1} = \begin{pmatrix} 3 \cos 210° \\ 3 \sin 210° \end{pmatrix} ; F_{2} = \begin{pmatrix} 4 \cos 120° \\ 4 \sin 120° \end{pmatrix}\)

\(R = F_{1} + F_{2}\)

= \(\begin{pmatrix} 3 \cos 210° \\ 3 \sin 210° \end{pmatrix} + \begin{pmatrix} 4 \cos 120° \\ 4 \sin 120° \end{pmatrix}\)

= \(\begin{pmatrix} -3 \cos 30° \\ -3 \sin 210° \end{pmatrix} + \begin{pmatrix} -4 \cos 60° \\ 4 \sin 60° \end{pmatrix}\)

= \(\begin{pmatrix} -3 \times 0.866 \\ -3 \times 0.500 \end{pmatrix} + \begin{pmatrix} -4 \times 0.500 \\ 4 \times 0.866 \end{pmatrix}\)

= \(\begin{pmatroix} -2.598 \\ -1.500 \end{pmatrix} + \begin{pmatrix} -2.00 \\ 3.464 \end{pmatrix}\)

= \(\begin{pmatrix} -4.598 \\ 1.964 \end{pmatrix}\)

\(|R| = \sqrt{(-4.598)^{2} + (1.964)^{2}}\)

= \(\sqrt{21.14 +3.86} = \sqrt{25}\)

= 5N

Acceleration = \(\frac{force}{mass}\)

= \(\frac{5}{7} = 0.71 ms^{-2}\)

(ii) \(Velocity = acceleration \times time\)

= \(0.71 \times 3 = 2.13 ms^{-1}\).

(b) Let fourth force be \(F_{4} = \begin{pmatrix} a \\ b \end{pmatrix}\).

\(F_{1} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}; F_{2} = \begin{pmatrix} 0 \\ -5 \end{pmatrix} ; F_{3} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}\)

Particle is in equilibrium when the sum of all forces acting on the particle = 0

\(\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 0 \\ -5 \end{pmatrix} + \begin{pmatrix} 6 \\ -4 \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(\implies 8 + a = 0 ; a = -8\)

\(-6 + b = 0 \implies b = 6\)

\(F_{4} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}\)

\(|F_{4}| = \sqrt{(-8)^{2} + 6^{2}} = 10N\)

Let the angle be \(\theta\).

\(\tan \theta = \frac{6}{-8} = -0.75\)

\(\theta = -36.87°\)

= \(180° - 36.87° = 143.13°\)

The equilibriant is 10N and acts in the direction of N53.13°W.

Correct Answer: C. 355,000

Explanation

Angle of Men = 360\(^o\) - (120 + 169)

= 360\(^o\) - 289\(^o\) = 71\(^o\)

No. of men \(\frac{71}{360}\) x 1800,000

= 355,000

Correct Answer: B. 2

Explanation

\(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)

= \(\log_{10} (\frac{25 \times 32}{8})\)

= \(\log_{10} 100 \)

= 2

Explanation

\(y = \frac{x - 1}{2x + 1}, x \neq -\frac{1}{2}\)

Using the quotient rule, we have

\(\frac{\mathrm d y}{\mathrm d x} = \frac{(2x + 1). 1 - (x - 1). 2}{(2x - 1)^{2}}\)

= \(\frac{2x + 1 - 2x + 2}{(2x + 1)^{2}}\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{3}{(2x + 1)^{2}}\)

At (1, 0), \(\frac{\mathrm d y}{\mathrm d x} = \frac{3}{(2(1) + 1)^{2}} = \frac{3}{9} = \frac{1}{3}\)

Equation : \(\frac{y - 0}{x - 1} = \frac{1}{3}\)

\(y = \frac{1}{3}(x - 1)\)

Correct Answer: A. 1

Explanation

\(\frac{2x + 1}{6} - \frac{3x - 1}{4}\) = 0

\(\frac{4(2n + 1) - 6(3x - 1)}{24}\) = 0

-10x + 10 = 0

-10x = -10

x = \(\frac{-10}{-10}\)

x = 1

Correct Answer: C. 1.701 x 10³

Explanation

2.7 x 10^-4 x 6.3 x 10⁶

= 2.7 x 6.3 x 10^-4 x 10⁶

= 17.01 x 10^{-4 + 6}

= 17.01 x 10²

= 1.701 x 10¹ x 10²

= 1.701 x 10^{1 + 2}

= 1.701 x 10³