Explanation

(a) Maximum height reached

Using the equation, \(v^{2} = u^{2} - 2gs\) (Upward movement against gravity)

velocity at maximum height = 0 m/s

\(0 = 80^{2} - 2(10)s\)

\(0 = 6400 - 20s \implies 6400 = 20s \)

\(s = 320 m\)

(b) Time to reach the maximum height

\(v = u + at = u - gt\)

\(0 = 80 - 10t \implies 80 = 10t\)

\(t = 8s\)

Correct Answer: C. \(\frac{3}{5}\)

Explanation

Cos (x + y)

= cos x cos y - sin x sin y

cos (x + y) = \(\frac{4}{5} \times \frac{24}{25} - \frac{3}{5} \times \frac{7}{25}\)

= \(\frac{96}{125} - \frac{21}{125} = \frac{96 - 21}{125}\)

= \(\frac{75}{125}\)

= \(\frac{3}{5}\)

Correct Answer: C. six

Explanation

\(32_4 = 22_x\)

\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)

12 + 2 x 1 = 2x + 2 x 1

14 = 2x + 2

14 - 2 = 2x

12 = 2x

x = \(\frac{12}{2}\)

x = 6

Explanation

(a)

From \(\Delta\) AMC, \(\sin 60° = \frac{\sqrt{3}}{2}\)

\(\cos 30° = \frac{\sqrt{3}}{2}\)

\(\tan 60° = \frac{\sqrt{3}}{1} = \sqrt{3}\)

Hence, \(\frac{2 \tan 60° + \cos 30°}{\sin 60} = \frac{2 \times \sqrt{3} + \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\)

= \(\frac{4\sqrt{3} + \sqrt{3}}{2} \div \frac{\sqrt{3}}{2}\)

= \(\frac{5\sqrt{3}}{2} \times \frac{2}{\sqrt{3}} = 5\)

(b)

In \(\Delta\) POT, \(\tan 36° = \frac{|TO|}{1050}\)

\(|OT| = 1050 \tan 36° = 1050 \times 0.7265\)

= 762.87m

In \(\Delta\) POB, \(\tan 41° = \frac{|OB|}{1050}\)

\(|OB| = 1050 \tan 41° = 1050 \times 0.8693\)

= 912.751m

\(|TB| = 912.751m - 762.87m = 149.88m\)

Hence, the height of the control tower = 150 m (to the nearest meter).

(ii) In \(\Delta\) POB, \(\cos 41° = \frac{1050}{|PB|}\)

\(|PB| = \frac{1050}{\cos 41°} = \frac{1050}{0.7547}\)

= 1,391.28m

The shortest distance between the aeroplane and the base of the control tower = 1,391m (to the nearest metre).

Correct Answer: C. 3/4

Correct Answer: D. 27

Explanation

(a)

(b)(i) Upper quartile \(Q_{3}\) = 57 marks

(ii) Pass mark if 60% of the class passed = 42 marks

Correct Answer: B. 12cm²

Explanation

\(tan \hspace{1mm}y = \frac{opp}{adj}=\frac{2}{3}\Rightarrow \frac{2}{3}=\frac{

LN}{MN}=\frac{4}{MN}\\

? MN = \frac{12}{2}=6\\

Area \hspace{1mm}of \hspace{1mm}\Delta LMN = \frac{1}{2}\times MN\times LN=\frac{1}{2}\times 6\times 4=12cm²\)

Correct Answer: C. 7.15m

Explanation

Tan 50^o = \(\frac{h}{6}\)

h = 6 tan 50

= 6 x 1.1917

= 7.1505

= 7.15

Correct Answer: B. 21

Explanation

\(Mean = \frac{sum of items}{total number of items}\)

\(20 = \frac{x}{8} \implies x = 160\)

The sum of the 7 nos = 160 - 17 = 143

Correct mean = \(\frac{143 + 25}{8} = \frac{168}{8} = 21\)