Correct Answer: (c) 140

Explanation

(a) Let B represent those with degree.

Let D represent those with diploma.

Let x represent those who have both degree and diploma.

(17 - x) + x + (15 - x) + 4 = 30

36 - x = 30

x = 36 - 30 = 6.

(b)

\(< PNM = 75°\) (corresponding angle)

\(\therefore NQR = 125° - 75° \)

= \(50°\)

\(< NPM = \frac{180° - 125°}{2}\)

= \(\frac{55°}{2}\)

= \(27.5°\). (Base angles of an isosceles triangle).

Correct Answer: B. 54

Explanation

\(T_{n} = a + (n - 1)d\)

\(380 = 9 + (n - 1)7\)

\(380 = 9 + 7n - 7 \implies 380 = 2 + 7n\)

\(7n = 380 - 2 = 378 \therefore n = \frac{378}{7} = 54\)

Correct Answer: C. 72 N

Explanation

\(F = ma \)

\(a = \frac{v - u}{t}\)

\(\therefore F = m(\frac{v - u}{t})\)

\(F = 12(\frac{35 - 5}{5}) = 12 \times 6 = 72 N\)

Correct Answer: D. 15

Explanation

First four prime numbers greater than 10 are 11, 13, 17, 19

The average \(= \frac{11+13+17+19}{4}=15\)

Correct Answer: A. y = x² - x - 2

Correct Answer: B. 0.0084N

Explanation

\(F = mass \times acceleration\) but \(accl = \frac{v - u}{t}\)

\(\therefore F = m(\frac{v - u}{t})\)

\(Mass = 28g = 0.028kg\)

\(v = 5.4 ms^{-1}; u = 0; t = 18secs\)

\(\therefore F = 0.028(\frac{5.4 - 0}{18}) = 0.028 \times 0.3 = 0.0084N\)

Explanation

\(\tan 2A = \frac{2 \tan A}{1 - \tan^{2} A}\)

\(\tan 30 = \tan 2(15°)\)

\(\frac{\sqrt{3}}{3} = \frac{2 \tan 15}{1 - \tan^{2} 15°}\)

\(\frac{\sqrt{3}}{6} = \frac{\tan 15}{1 - tan^{2} 15°}\)

Let \(\tan 15° = x\)

\(\frac{\sqrt{3}}{6} = \frac{x}{1 - x^{2}}\)

\(\sqrt{3} (1 - x^{2}) = 6x \implies \sqrt{3} - \sqrt{3} x^{2} = 6x \)

\(\sqrt{3} x^{2} + 6x - \sqrt{3} = 0\)

Divide through by the coefficient of x\(^{2}\).

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

\(\implies x^{2} + 2\sqrt{3} x - 1 = 0\)

\(x = \frac{-(2\sqrt{3}) \pm \sqrt{(2\sqrt{3})^{2} - 4(1)(-1)}}{2(1)}\)

\(x = \frac{-2\sqrt{3} \pm \sqrt{16}}{2} = \frac{-2\sqrt{3} \pm 4}{2}\)

\(x = -\sqrt{3} + 2 ; x = -\sqrt{3} - 2\)

Since 15° is in the first quadrant, \(\tan 15° > 0\)

\(\therefore \tan 15° = -\sqrt{3} + 2 \equiv 2 - \sqrt{3}\).

Correct Answer: C. 0

Explanation

\(2\log_{4} 2 = x + 1\)

\(\log_{4} 2^{2} = \log_{4} 4 = 1\)

\(x + 1 = 1 \implies x = 0\)

Correct Answer: A. 33.3\(ms^{-1}\)

Explanation

\(120 kmh^{-1} = \frac{120 \times 1000}{3600} = \frac{100}{3} = 33.3ms^{-1}\)