Explanation

(b) Two possible ways

(c) The distances of the locations from town X = 27.6 km and 12.4 km.

(d) 12.4 km

Correct Answer: C. 7/25

Correct Answer: A. -4

Explanation

Substituting for x in the equation

\(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4\)

Correct Answer: C. 32

Explanation

\(\sqrt{128} +\sqrt{18}-\sqrt{k}=7\sqrt{2}\\

\sqrt{2\times 64}+\sqrt{9\times 2} - \sqrt{k} = 7\sqrt{2}\\

8\sqrt{2} + 3\sqrt{2} - \sqrt{k} = 7\sqrt{2}; 11\sqrt{2} - \sqrt{k} = 7\sqrt{2}\\

-\sqrt{k}=7\sqrt{2}-11\sqrt{2}; -\sqrt{k} = -4\sqrt{2}; \sqrt{k}=4\sqrt{2}\\

=\sqrt{4^2\times 2} = \sqrt{16\times 2}; \sqrt{k}=\sqrt{32}; k = 32\)

Correct Answer: D. 0.299 cms\(^{-1}\)

Explanation

Area of a circle (A) = \(\pi r^2\)

Given

\(\frac{dA}{dt} = 1.5cm^2s^{-1}\)

\(\frac{dr}{dt}\) = ?

A = 2cm\(^2\)

Now

2 = \(\pi r^2\)

= r\(^2 = \frac {2}{\pi}\)

r = \(\sqrt \frac {2}{\pi}\) cm = 0.798cm

\(\frac {dr}{dt} = \frac {dA}{dt} \times \frac {dr}{dt}\)

\(\frac {dA}{dr} = 2\pi r\) (differentiating A = \(\pi r^2)\)

\(\frac {dr}{dA} = \frac {1}{2\pi r}\)

\(\frac {dr}{dt} = 1.5 \times \frac {1}{2\times \pi \times 0.798} = 1.5 \times 0.199\)

\(\frac {dr}{dt} = 0.299cms^{-1}\) (to 3 s.f)

Correct Answer: C. 0.00247

Explanation

(a) \(d = 15 cm \implies r = \frac{15}{2} cm\)

\(V = \frac{4\pi r^{3}}{3}\)

\(V = \frac{4}{3} \times \frac{22}{7} \times (\frac{15}{2})^{3}\)

\(V = \frac{1125 \times 22}{7 \times 2}\)

\(V = 1,767.857 cm^{3}\)

(b)(i) Longitude difference : 105° + 35° = 140°

Distance along the parallel of latitude = \(\frac{140}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 28\)

= \(\frac{140 \times 44 \times 6400 \times 0.8829}{360 \times 7}\)

= \(13812.48 km\)

\(\approxeq 13800 km\)

(ii) Distance from the equator : \(\frac{28}{360} \times 2 \times \frac{22}{7} \times 6400\)

= \(3128.89 km\)

\(\approxeq 3130 km\)

Explanation

p(defective) = \(\frac{8}{100} = 0.08\)

p(non- defective) = \(1 - 0.08 = 0.92\)

p(at least 3 defective) = \(1 - [p(0) + p(1) + p(2)]\)

\(p(x) = ^{n}C_{x} p^{x} q^{n - x}\)

\(p(0) = ^{10}C_{0} (0.08)^{0} (0.92)^{10} = 0.434\)

\(p(1) = ^{10}C_{1} (0.08)^{1} (0.92)^{9} = 0.378\)

\(p(2) = ^{10}C_{2} (0.08)^{2} (0.92)^{8} = 0.148\)

p(at least 3 defective) = 1 - [0.434 + 0.378 + 0.148]

= 1 - 0.960 = 0.04

Correct Answer: D. 4

Explanation

\(\log_{3}(x - y) = 1 \implies x - y = 3^{1} = 3 .... (1)\)

\(\log_{3}(2x + y) = 2 \implies 2x + y = 3^{2} = 9 ..... (2)\)

From (1), y = x - 3

From (2), y = 9 - 2x

\(\implies 9 - 2x = x - 3\)

\(9 + 3 = x + 2x = 3x\)

\(x = 4\)

Explanation

(a)(i) Volume of outer pipe = \(\pi r^{2} h\)

= \(\frac{22}{7} \times (5.0)^{2} \times 28 \times 100\)

Volume of inner pipe = \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)

Volume of metal used = Volume of outer pipe - Volume of inner pipe

= \(22 \times 400 \times (5.0^{2} - 3.5^{2})\)

= \(22 \times 400 \times (5.0 - 3.5) \times (5.0 + 3.5)\)

= \(112,200 cm^{3}\)

(ii) Vol of water the pipe can hold = vol of inner pipe

= \(\frac{22}{7} \times (3.5)^{2} \times 28 \times 100\)

= \(107,800 cm^{3}\)

= 107.8 litres.

(b) < PMN = < LNM (alternate angles)

< PMN = < LMN (angles in alternate segment)

\(\therefore\) < LNM = < NLM, then \(\Delta\) LMN is isosceles.