Correct Answer: A. x = \(\frac{6y}{z}\)

Explanation

\(x\) x \(\frac{y}{z}\)

x = \(\frac{ky}{z}\)

15 = \(\frac{10k}{4}\)

\(\frac{60}{10}\) = k = 6

Therefore; x = \(\frac{6y}{z}\)

Explanation

(a) \(3(3n - 1) + 5(2n + 1) = 135\)

\(9n - 3 + 10n + 5 = 135\)

\(19n + 2 = 135\)

\(19n = 135 - 2 = 133\)

\(n = \frac{133}{19} = 7\)

(b) Radius of a cylinder = 14 cm

side of a cube = 22 cm

\(\therefore \frac{22}{7} \times 14 \times 14 \times h = 22 \times 22 \times 22\)

\(44 \times 14 \times h = 22 \times 22 \times 22\)

\(h = \frac{22 \times 22 \times 22}{44 \times 14}\)

= \(\frac{121}{7} cm\)

TSA of a cylinder = \(2\pi r (r + h) = 2 \times \frac{22}{7} \times 14 (14 + \frac{121}{7})\)

= \(88(\frac{98 + 121}{7}) = \frac{88 \times 219}{7}\)

= \(\frac{19272}{7} cm^{2}\)

TSA of a cube = \(6s^{2}\)

= \(6 \times 22 \times 22\)

= \(2904 cm^{2}\)

Ratio = \(\frac{19272}{7} : 2904\)

= \(73 : 77\)

Correct Answer: B. 1.54m

Explanation

for 20 students, mean = 1.67

∑fx = μf

∑fx = 20 × 1.67 = 33.4

for group 2

∑fx = 16 × 1.50 = 24

for group 3

∑fx = 14 × 1.40 = 19.6

33.4 + 24 + 19.6 = 77

∑f = 20+16+14 = 50

Correct Answer: A. 32

Explanation

Reacting force - Weight = Net force

R = W + ma = mg + ma

480 = 10p + 5p

480 = 15p

p = 32 kg

Correct Answer: B. 200

Correct Answer: C. the point of intersection of the perpendicular bisectors of PQ and PR

Explanation

\(Mean, (\bar {x}) = A + \frac{\sum fd}{\sum f}\)

= \(24.5 + \frac{-300}{120}\)

= \(24.5 - 2.5\)

= 22.0

Explanation

(a) \(y = (2x + 3)^{7} + \frac{x - 1}{2x - 1}\)

\(\frac{\mathrm d y}{\mathrm d x} = 7(2x + 3)^{6}. 2 + \frac{(2x - 1).1 - (x - 1).2}{(2x - 1)^{2}}\)

= \(14(2x + 3)^{6} + \frac{2x - 1 - 2x + 2}{(2x - 1)^{2}}\)

= \(14(2x + 3)^{6} + \frac{1}{(2x - 1)^{2}}\)

At x = -1, \(\frac{\mathrm d y}{\mathrm d x} = 14(-1)^{6} + \frac{1}{(2(-1) - 1)^{2}}\)

= \(\frac{\mathrm d y}{\mathrm d x} = 14 + \frac{1}{9} = 14\frac{1}{9}\)

(b) \(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x\)

\( u = x + 2 \implies \mathrm d u = \mathrm d x\)

If x = 1, u = 1 + 2 = 3; x = 2, u = 2 + 2 = 4.

\(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x = \int_{3} ^{4} \frac{u - 3}{u^{4}}\)

= \(\int_{3} ^{4} (\frac{u}{u^{4}} - \frac{3}{u^{4}}) \mathrm d u\)

= \(\int_{3} ^{4} (u^{-3} - 3u^{-4}) \mathrm d u\)

= \([\frac{u^{-2}}{-2} + u^{-3}]_{3} ^{4}\)

= \([\frac{1}{u^{3}} - \frac{1}{2u^{2}}]_{3} ^{4}\)

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

= \(\frac{1}{64} - \frac{1}{32} - \frac{1}{27} + \frac{1}{18}\)

= \(\frac{5}{1728}\).

Correct Answer: A. 5n - 2

Explanation

Given the sequence 3, 8, 13, 18, ... which is an arithmetic sequence

a = 3

d = T\(_2\) - T\(_1\) = 8 - 3 = 5

The general term of an A.P is:

T\(_n\) = a + (n - 1)d

⇒ T\(_n\) = 3 + (n - 1)5

= T\(_n\) = 3 + 5n - 5

∴ T\(_n\) = 5n - 2

Correct Answer: C. 0.85

Explanation

Note that the group that reads both papers is being counted twice.

(percentages add up beyond 100%)

Let M = those reading the Morning Times,

E = those reading the Evening Dispatch

P( M or E) = P(M) + P(E) – P(M and E)

= 0.45 + 0.60 – 0.20

= 0.85