Correct Answer: B. \(\frac{19 + 4\sqrt{15}}{11}\)

Correct Answer: D. VIP = 80, Regular = 120

Explanation

Let \(x\) = number of VIP tickets sold and

\(y\) = number of regular tickets sold

Total number of tickets sold = 200

⇒ \(x\) + \(y\) = 200 ---- (i)

If it costs N1,200 for a VIP ticket, then it costs N1200x for \(x\) number of VIP tickets sold and

If it costs N700 for a regular ticket, then it costs N700\(y\) for \(y\) number of VIP tickets sold

The total amount realised from the sale of tickets = N180,000

⇒ 1200\(x\) + 700\(y\) = 180000 ----- (ii)

From equation (i)

\(x\) = 200 - \(y\) ----- (iii)

Substitute (200 - \(y\)) for \(x\) in equation (ii)

⇒ 1200(200 - \(y\)) + 700\(y\) = 180000

⇒ 240000 - 1200\(y\) + 700\(y\) = 180000

⇒ 240000 - 500\(y\) = 180000

Collect like terms

⇒ 240000 - 180000 = 500\(y\)

⇒ 60000 = 500\(y\)

⇒ \(y = \frac{60000}{500} = 120\)

Substitute 120 for \(y\) in equation (iii)

⇒ \(x = 200 - 120\)

⇒ \(x = 80\)

∴ The total number of VIP tickets sold is 80 and regular is 120

Explanation

(a) Formula : Volume = \(\frac{1}{3} \pi r^{2} h\)

\(\pi = 3.142; r = 4cm; h = 6cm\)

\(Volume = \frac{1}{3} \times 3.142 \times 4^{2} \times 6 = 100.544 cm^{3}\)

(b) Volume of hemisphere = \(\frac{1}{2} \times \frac{4}{3} \pi r^{3}\)

= \(\frac{2}{3} \times \frac{22}{7} \times 5^{3}\)

= \(\frac{5500}{21} cm^{3} = 261.905 cm^{3}\)

Volume of water discharged per second = \(\pi r^{2} h\)

= \(\frac{22}{7} \times 0.2m \times 0.2m \times 2m = \frac{1.76}{7}\)

= \(0.2514 m^{3} / sec\)

\(\therefore \text{The time to fill the tank = } \frac{261.905}{0.2514 \times 60}\)

= \(\frac{261.905}{15.084} = 17.363 mins\)

\(\approxeq 17.4 mins\)

Explanation

(a)

(b)(i) From the Venn diagram,

x + x + 7 + 8 = n(B)

2x + 15 = 47

2x = 47 - 15 = 32

x = 16

(ii) Number of people who travelled by at least 2 means

= 16 + 7 + 8 + 3 = 34

Explanation

\(R_{1} = \begin{pmatrix} -20 \cos 60° \\ -20 \sin 60° \end{pmatrix} ; R_{2} = \begin{pmatrix} -10 \cos 60° \\ 10 \sin 60° \end{pmatrix} \)

\(R_{3} = \begin{pmatrix} 8\sqrt{2} \cos 45° \\ 8\sqrt{2} \sin 45° \end{pmatrix}\)

\(R = R_{1} + R_{2} + R_{3}\)

\(R = \begin{pmatrix} -20 \cos 60° \\ -20 \sin 60° \end{pmatrix} + \begin{pmatrix} -10 \cos 60° \\ 10 \sin 60° \end{pmatrix} + \begin{pmatrix} 8\sqrt{2} \cos 45° \\ 8\sqrt{2} \sin 45° \end{pmatrix}\)

\(R = \begin{pmatrix} -20 \times \frac{1}{2} \\ -20 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} -10 \times \frac{1}{2} \\ 10 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} 8\sqrt{2} \times \frac{1}{\sqrt{2}} \\ 8\sqrt{2} \times \frac{1}{\sqrt{2}} \end{pmatrix}\)

= \(\begin{pmatrix} - 10 - 5 + 8 \\ -10\sqrt{3} + 5\sqrt{3} + 8 \end{pmatrix}\)

= \(\begin{pmatrix} -7 \\ 8 - 5\sqrt{3} \end{pmatrix} = \begin{pmatrix} -7 \\ -0.66 \end{pmatrix}\)

Let \(\theta\) be the direction from the horizontal.

\(\tan \theta = \frac{-0.66}{-7}\)

\(\tan \theta = 0.0943\)

\(\theta = \tan^{-1} (0.0943) = 5.39°\)

Correct Answer: A. 5

Correct Answer: B. 132°

Explanation

In the diagram; a = b = 48^o (alternate< S)

x = 180^o - b (angles on a str. line)

x = 180^o - 48^o

= 132^o

Correct Answer: D. 2.5m

Correct Answer: A. 2

Correct Answer: D. 2x+3y=4

Explanation

\(2x+3y = 6 \implies 3y = 6-2x\)

\(y = \frac{6}{3} - \frac{2x}{3}\)

Parallel lines have the same gradient

\(\therefore\) Gradient of the line = \(\frac{-2}{3}\)

Line passes through (-1,2)

Equation: \(\frac{y-2}{x-(-1)} = \frac{y-2}{x+1} = \frac{-2}{3}\)

\(3y-6 = -2x-2 \implies 3y+2x = -2+6 =4\)