Correct Answer: C. 37°

Correct Answer: D. \(\frac{3a+11}{(a-3)(a+2)}\)

Explanation

\(\frac{4}{a-3}-\frac{1}{a+3}\)

= \(\frac{4(a+2)-1(a-3)}{(a-3)(a+2)}\)

\(\frac{4a+8 -a +3}{(a-3)(a+2)}\)

= \(\frac{3a+11}{(a-3)(a+2)}\)

Correct Answer: A. D8,000.00

Explanation

A = P + 1

I = A - P

= 10,400 - P

Now using I = \(\frac{P \times T \times R}{100}\)

i.e. 10,400 - P = \(\frac{P \times 5 \times 6}{100}\)

= 100(10,400 - P) = 30P

10(10,400 - P) = 3P

104,000 - 10P = 3P

104,000 - 10P = 3P

104,000 = 3P + 10P

= 104,000 = 13P

P = \(\frac{104,000}{100}\)

P = D8,000

Explanation

\(p(Ali) = p(A) = \frac{2}{3} ; p(Baba) = p(B) = \frac{3}{4} ; p(Katty) = p(K) = \frac{4}{5}\)

(a) \(p(\text{only Katty and Baba}) = p(K) \times p(B) \times p(A')\)

= \(\frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}\)

= \(\frac{1}{5}\)

(b) \(p(\text{none gains admission}) = p(A') \times p(K') \times p(B')\)

= \(\frac{1}{3} \times \frac{1}{4} \times \frac{1}{5}\)

= \(\frac{1}{60}\).

(c) \(p(\text{at most two will gain admission}) = p(\text{none gains}) + p(\text{one gains}) + p(\text{two gains})\)

\(p(\text{one gains}) = p(A) \times p(B') \times p(K') + p(B) \times p(A') \times p(K') + p(K) \times p(A') \times p(B')\)

= \(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5} + \frac{3}{4} \times \frac{1}{3} \times \frac{1}{5} + \frac{4}{5} \times \frac{1}{3} \times \frac{1}{4}\)

= \(\frac{2}{60} + \frac{3}{60} + \frac{4}{60}\)

= \(\frac{9}{60}\)

\(p(\text{two gains}) = p(A) \times p(B) \times p(K') + p(A) \times p(K) \times p(B') + p(B) \times p(K) \times p(A')\)

= \(\frac{2}{3} \times \frac{3}{4} \times \frac{1}{5} + \frac{2}{3} \times \frac{4}{5} \times \frac{1}{4} + \frac{3}{4} \times \frac{4}{5} \times \frac{1}{3}\)

= \(\frac{6}{60} + \frac{8}{60} + \frac{12}{60}\)

= \(\frac{26}{60}\)

\(p(\text{at most two gains admission}) = \frac{1}{60} + \frac{9}{60} + \frac{26}{60}\)

= \(\frac{36}{60}\)

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

Correct Answer: B. 73.40°

Explanation

\(F_{1} = 7i + 0j\)

\(F_{2} = (4\cos\theta)i + (4\sin\theta)j\)

\(9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}\)

\(9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta\)

\(81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)\)

Recall, \(\cos^{2}\theta + \sin^{2}\theta = 1\)

\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)

\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)

\(\theta = \cos^{-1} 0.2857 = 73.40°\)

Explanation

(a) Number of triangles that can be formed = \(^{6}C_{3}\)

= \(\frac{6!}{(6 - 3)! 3!} = \frac{6 \times 5 \times 4}{3 \times 2}\)

= 20 ways.

(b) Tie the father and mother together so that there are 5 persons to arrange in 5! ways.

Changing the positions of the father and mother, this can be done in 2! ways.

\(\therefore \text{Ways of arranging the family = } 2!5!\)

Without restriction, they can be arranged in 6! ways.

\(\therefore \text{Father and mother not together =} 6! - 2!5! = 720 - 240 = 480 \)

Correct Answer: B. 1.25

Correct Answer: D. x + 2y = 8

Correct Answer: D. 7

Correct Answer: C. 25km

Explanation

x² = 24² + 7² = 576 + 49

x² = 625

x = √625 = 25