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From \(\Delta\) BCD, \(\sin \theta = \frac{50}{56}\) \(\theta = \sin^{-1} (0.8929)\) \( < BCD = 63.23^\circ\) \(\therefore < ACB = 63.23^\circ \times 2 = 126.46^\circ\) \(< ACB (reflex) = 360^\circ - 126.46^\circ\) = \(233.54^\circ\) Perimeter of the arc = \(\frac{233.54}{360} \times 2 \times \frac{22}{7} \times 56\) = \(228.35 m\) Perimeter of the railway tunnel = Perimeter of the arc + |AB| = 228.35 m + 100 m = 328.35 m.

(a) \(P(Good) = \frac{18}{24} = \frac{3}{4}\) \(P(Bad) = \frac{6}{24} = \frac{1}{4}\) P(2 are good and 1 bad) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}\) = \(\frac{9}{64}\) (b) P(the 3 are bad) = \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\) = \(\frac{1}{64}\) (c) P(the 3 are good) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}\) = \(\frac{27}{64}\)

(a) \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\) \(3\frac{1}{12} + \frac{7}{8} = \frac{37}{12} + \frac{7}{8}\) = \(\frac{74 + 21}{24}\) = \(\frac{95}{24}\) \(2\frac{1}{4} - \frac{1}{6} = \frac{9}{4} - \frac{1}{6}\) = \(\frac{27 - 2}{12}\) = \(\frac{25}{12}\) \(\therefore \frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}} = \frac{95}{24} \div \frac{25}{12}\) \(\frac{95}{24} \times \frac{12}{25} = \frac{19}{10}\) = \(1.9\) (b)(i) \(p = \frac{m}{2} - \frac{n^{2}}{5m}\) \(\frac{n^{2}}{5m} = \frac{m}{2} - p\) \(n^{2} = 5m(\frac{m}{2} - p)\) \(n = \pm \sqrt{5m(\frac{m}{2} - p)}\) (ii) When p = 14 and m = -8, \(n = \sqrt{5(-8)(\frac{-8}{2} - 14)}\) \(n = \sqrt{-40(- 4 - 14)}\) \(n = \sqrt{720}\) \(n = 12\sqrt{5}\) = \(\pm 26.83\)

(a) No Log 0.004592 \(\bar{3}.6620\) \((0.004592)^{\frac{1}{3}}\) \(\frac{\bar{3}.6620}{3} = \bar{1}.2207\) Antilog of \(\bar{1}.2207 = 0.1663\) (b) \(\log_{10} y + 3 \log_{10} x = 2\) \(\log_{10} y + \log_{10} x^{3} = 2\) \(\log_{10} (yx^{3}) = 2\) \(yx^{3} = 10^{2}\) \(yx^{3} = 100\) \(y = \frac{100}{x^{3}}\) (c) \(3x - 2y = 21 .... (1)\) \(4x + 5y = 5 ..... (2)\) Multiply (1) by 4 and (2) by 3, \(12x - 8y = 84 .... (1a)\) \(12x + 15y = 15 ... (2a)\) (1a) - (2a), \(- 8y - 15y = 84 - 15\) \(- 23y = 69 \implies y = -3\) \(3x - 2y = 21 \implies 3x - 2(-3) = 21\) \(3x + 6 = 21 \implies 3x = 15\) \(x = 5\)

The diagram above is the solution

(a) \(\frac{x^{2} - y^{2}}{3x + 3y}\) \(\frac{(x + y)(x - y)}{3(x + y)}\) (Using difference of two squares) = \(\frac{x - y}{3}\) (b)(i) \(\sin 37 = \frac{/PS/}{15}\) \(/PS/ = 15 \times 0.6018\) = \(9.03 cm\) (ii) \(\cos 37 = \frac{/SK/}{15}\) \(/SK/ = 15 \times 0.7986\) = \(11.98 cm \) \(\approxeq 12.0 cm\) (iii) Area of the shaded portion = Area of rectangle PQRS - Area of triangle PSK. /SR/ = 2(/SK/) = 2(11.98) = 23.96 cm Area of rectangle PQRS = \(23.96 \times 9.03 \) = \(216.3588 cm^{2}\) Area of triangle PKS = \(\frac{1}{2} \times 11.98 \times 9.03\) = \(54.088 cm^{2}\) Area of shaded portion : \((216.3588 - 54.088)cm^{2}\) \(162.2708 cm^{2}\) \(\approxeq 162 cm^{2}\)

(a) \(3(x + y) + 45 = 180^\circ\) ( sum of angles on a straight line) \(3x + 3y + 45 = 180 \implies 3x + 3y = 135\) \(3x = 135 - 3y \implies x = 45 - y\) Also, \(5x + y + y = 180\) \(5x + 2y = 180..... (*)\) Put x = 45 - y into (*) above, \(5(45 - y) + 2y = 180 \implies 225 - 5y + 2y = 180\) \(225 - 3y = 180 \implies 3y = 45\) \(y = \frac{45}{3} = 15^\circ\) \(x = 45 - y\) \(x = 45 - 15 = 30^\circ\) (b) Let the distance BD = x and CD = y. In \(\Delta ABD\), \(\tan 53 = \frac{33}{x}\) \(x = \frac{33}{\tan 53}\) = \( 24.868 cm\) In \(\Delta ACD\), \(\tan 67 = \frac{33}{y}\) \(y = \frac{33}{\tan 67}\) \(y = 14.01 cm\) \(\therefore BC = x + y \) \(24.868 + 14.01 = 38.878 cm\) \(\approxeq 38.9 cm\)

(a) \(T_{n} = a + (n - 1)d\) (terms of an A.P) 3rd term = -9 ; 8th term = 26 \(\implies T_{3} = a + 2d = -9 ... (1)\) \(\implies T_{8} = a + 7d = 26 ..... (2)\) (i) Solving (2) - (1), we have \(7d - 2d = 26 - (-9)\) \(5d = 35 \implies d = 7\) (ii) Putting d = 7 in (1) above, we have \(a + 2(7) = -9\) \(a + 14 = -9\) \(a = - 9 - 14 = -23\) (b) \(\frac{XP}{XY} = \frac{2}{5}\) (similar triangles) \(\frac{24}{\text{Area of \Delta XYZ} = \frac{2^{2}}{5^{2}}\) \(\text{Area of } \Delta XYZ = \frac{24 \times 25}{4} = 150 cm^{2}\) \(\therefore \text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\)

\(x\) mid-value \(f\) \(fx\) \((x - \bar{x})\) \((x - \bar{x})^{2}\) \(f(x - \bar{x})^{2}\) 0 - 4 2 1 2 -12 144 144 5 - 9 7 5 35 -7 49 245 10 - 14 12 10 120 -2 4 40 15 - 19 17 9 153 3 9 81 20 - 24 22 5 110 8 64 320 \(\sum \) 30 420 830 (a) Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\) = \(\frac{420}{30} = 14\) (b) Standard deviation = \(\sqrt{\frac{\sum f(x - \bar{x})^{2}}{\sum f}\) = \(\sqrt{\frac{830}{30}}\) = \(\sqrt{27.66}\) = 5.26.

Deprecated; to express disapproval of something Condemn is to express complete disapproval of something.

“Reversed” means the decision must be withdrawn or cancelled officially. "Resciended" means officially cancelled or withdrawn, which is the closest in meaning to “reversed” in this context.

Depraved implies morally corrupt; wicked

'Made in ignorance' implies made without information or knowledge. The most appropriate answer is option A as other options can be considered as antonyms

Fend off means to avoid, refrain or stay away from; prevent. abstain; defend.

Recrimination is an accusation in response to one from someone else.