Explanation

(a) p(man wins) = p = 0.8 ; p(man loses) = q = 0.2.

4 different races : \((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2} q^{2} + 4pq^{3} + q^{4}\)

(i) p(man wins all) = \(p^{4} = (0.8)^{4} = 0.4096\)

(ii) p(man wins no race) = \(q^{4} = (0.2)^{4} = 0.0016\)

(iii) p(man wins at most 3) = 1 - p(man wins all)

= 1 - 0.4096 = 0.5904.

(b)(i) 5 girls, 10 boys ; Total = 15.

Ways of selecting 3 boys = \(^{10}C_{3} \times ^{5}C_{2}\)

= \(120 times 10\)

= 1200

Without restriction, selection = \(^{15}C_{5}\)

= \(\frac{15!}{10! 5!}\)

= 3003

p(selecting 3 boys) = \(\frac{1200}{3003} = 0.3996\)

(ii) Ways of selecting no girls = \(^{10}C_{5} \times ^{5}C_{0}\)

= 252

p(at least one girl) = 1 - p(no girl)

= \(1 - \frac{252}{3003}\)

= \(2751}{3003} = 0.9161\)

Correct Answer: C. \(\frac{1}{5}\)

Explanation

Probability of obtaining a 5 = \(\frac{\text{frequency of 5}}{\text{total frequency}}\)

\(12+18+y+30+2y+45 = 150 \implies 105+3y = 150\)

\(3y = 45; y = 15\)

Probability of 5 = \(\frac{2\times 15}{150} = \frac{1}{5}\)

Correct Answer: A. m > 2

Explanation

3m + 3 > 9

3m > 9 - 3

3m > 6

m > 2

Correct Answer: A. 2(3n- 1)

Explanation

\(T_{1} = 4; T_2 = 10; T_3 = 16\)

\(T_{2} - T_1 = T_3 - T_1 = 6\)

\(T_n = a + (n - 1) d\)

= \(4 + (n - 1) \times 6\)

= \(4 + 6n - 6\)

= \(6n - 2\)

= 2(3n - 1)

Correct Answer: A. 16/3cm

Explanation

L = θ/360 x 2πR = 240/360 x 2 x 22/7 x 8/1 ... (1)

This must be equal to the circumference of the circle which is 2πr = 44R/7 .... (2)

equate (1) and (2)

r = 16/3 = 51/3

Correct Answer: B. 10% loss

Explanation

First S.P = N230.00

% profit = 15%

% profit = \(\frac{S.P - C.P}{C.P}\) x 100%

⇒ 15% = \(\frac{230 - C.P}{C.P}\) x 100%

⇒ \(\frac{15}{100}= \frac{230 - C.P}{C.P}\)

⇒15C.P = 100 (230 - C.P)

⇒15C.P = 23000 - 100C.P

⇒15C.P + 100C.P = 23000

⇒115C.P = 23000

⇒ C.P = \(\frac{23000}{115}\) = N200.00

Second S.P = ?180.00

Since C.P is greater than S.P, therefore it's a loss

% loss = \(\frac{C.P - S.P}{C.P}\) x 100%

⇒ \(\frac{200 - 180}{200}\) x 100%

⇒ \(\frac{20}{200}\) x 100%

⇒ \(\frac{1}{10}\) x 100% = 10%

∴ It's a loss of 10%

Explanation

< RPQ = 50° + 75° = 125°

By cosine rule,

\(|QR|^{2} = |RP|^{2} + |PQ|^{2} - 2(|PR|)(|PQ|) \cos 125°\)

= \(90^{2} + 50^{2} + 2(90)(50) \cos 55°\)

= \(8100 + 2500 + 9000 \times 0.5736\)

= \(10600 + 5162.4\)

\(|QR|^{2} = 15762.4\)

\(|QR| = \sqrt{15762.4} = 125.548km\)

(b) By Sine rule,

\(\frac{RQ}{\sin 125°} = \frac{PR}{\sin <PQR}\)

\(\frac{125.548}{\sin 125°} = \frac{90}{\sin < PQR}\)

\(\sin < PQR = \frac{90 \times 0.8192}{125.548}\)

\(\sin < PQR = 0.5782 \implies < PQR = \sin^{-1} (0.5872)\)

= \(35.96°\)

Bearing of R from Q = 180° + 75° + 35.96° = 290.96°\)

\(\approxeq 291°\)

Correct Answer: C. 28

Correct Answer: D. \(x - \frac{5}{x} - \frac{1}{2x^{2}} + c\)

Explanation

\(\frac{x^{3} + 5x + 1}{x^{3}} \equiv 1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}\)

\(\equiv \int (1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}) \mathrm {d} x = \int (1 + 5x^{-2} + x^{-3}) \mathrm {d} x\)

= \((x - 5x^{-1} - \frac{1}{2}x^{-2} + c)\)

= \(x - \frac{5}{x} - \frac{1}{2x^{2}} + c\).

Explanation

\(y = ax^{2} + bx + c\)

Gradient = \(\frac{\mathrm d y}{\mathrm d x} = 2ax + b = 8x + 4\)

Equating, we have

\(2a = 8 \implies a = 4\)

\(b = 4\)

For minimum value, gradient = 0

\(8x + 4 = 0 \implies x = -\frac{1}{2}\)

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

\(1 = 4(-\frac{1}{2})^{2} + 4(-\frac{1}{2}) + c\)

\(1 = 1 - 2 + c\)

\(1 = c - 1 \implies c = 2\)

\(a, b, c = 4, 4, 2\).