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Class Interval Freq (f) Mid-point (x) \(fx\) \(x^{2}\) \(fx^{2}\) 60 - 64 2 62 124 3844 7688 65 - 69 3 67 201 4489 13487 70 - 74 6 72 432 5184 31104 75 - 79 11 77 847 5929 65219 80 - 84 8 82 656 6724 53792 85 - 89 7 87 609 7569 52983 90 - 94 2 92 184 8464 16928 95 - 99 1 97 97 9409 9409 \(\sum\) 40 3150 250590 (a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3150}{40}\) = 78.75 (b) \(SD = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\) = \(\sqrt{\frac{250590}{40} - (\frac{3150}{40})^{2}}\) = \(\sqrt{6264.75 - (78.75)^{2}}\) = \(\sqrt{6264.75 - 6201.5625}\) = \(\sqrt{63.1875}\) = 7.95.

(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 = \(\frac{19272}{7} : 2904\) = \(73 : 77\)

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ 240^\circ 270^\circ 300^\circ y -3.0 -2.5 -1.0 1.0 3.0 4.5 5.0 4.5 3.0 1.0 -1.0 (b) (c)(i) x = 75^\circ or x = 285^\circ (ii) When x = 105^\circ, y = 2. (iii) When y = 1.5, x = 96^\circ or x = 264^\circ.

(a) \(\cos R = \frac{5}{13} = 0.3846\) \(R = \cos^{-1} (0.3846) = 67.38^\circ\) \(/PS/^{2} = 5^{2} + 19^{2} - 2(5)(19)\cos 67.38\) = \(25 + 361 - 190(0.3872)\) = \(386 - 73.568\) \(/PS/^{2} = 312.432\) \(/PS/ = \sqrt{312.432}\) = \(17.68 \approxeq 17.7 cm\) (b) Slant height of a cone = \(l = \sqrt{7^{2} + 24^{2}}\) \(\sqrt{49 + 576} = \sqrt{625}\) = \(25 cm\) C.S.A of a cone = \(\pi rl\) = \(\frac{22}{7} \times 7 \times 25\) = \(550 cm^{2}\) Area of hemisphere = \(\frac{1}{2} \times 4\pi r^{2} = 2 \pi r^{2}\) \(2 \times \frac{22}{7} \times 7 \times 7 = 308 cm^{2}\) TSA of the structure = \(550 cm^{2} + 308 cm^{2} = 858 cm^{2}\)

Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{150}{360} \times \pi \times 12^{2}\) = 60\(\pi\) cm\(^{2}\)

\(\tan 16 = \frac{x}{150}\) \(x = 150 \tan 16\) = \(43 cm\)

If \(y = \cos x\) \(\frac{\mathrm d y}{\mathrm d x}\) =—\(\sin x\)

mn - nq - n\(^2\) + mq mn - n\(^2\) - nq + mq (mn - n\(^2\)) (mq - nq) n(m - n) + q(m - n) = (n +q) (m - n)

(a)(i) < PTS = 38^\circ (alternate angle) 42^\circ + x + 38^\circ = 180^\circ (angles on a straight line) 80^\circ + x = 180^\circ x = 180^\circ - 80^\circ = 100^\circ (ii) < SQT = 42^\circ (alternate angle) < SQR = 60^\circ (sum of angles in a triangle) y + 42^\circ + 60^\circ = 180^\circ (angle on a straight line) y + 102^\circ = 180^\circ y = 180^\circ - 102^\circ = 78^\circ (b) \(< QSP = \frac{150}{2}\) (angle at the centre is twicw angle at the circumference) \(< QSP = 75^\circ\) \(\hat{R} + \hat{P} = 180^\circ\) (opposite angles in a cyclic quad are supplementary) \(\hat{P} + 75^\circ + 45^\circ = 180^\circ\) (sum of angle in a triangle) \(\hat{P} = 180^\circ - 120^\circ\) \(\hat{P} = 60^\circ\) \(\hat{R} + \hat{P} = 180^\circ\) \(\hat{R} = 180^\circ - 60^\circ\) \(\hat{R} = 120^\circ\) \(< RQS + < QRS + < QSR = 180^\circ\) (sum of angle in a triangle) \(< RQS = 180^\circ - (40^\circ + 120^\circ)\) = \(180^\circ - 160^\circ\) \(< RQS = 20^\circ\)

x2 + 2x > 15 x2 + 2x - 15 > 0 (x2 + 5x) - (3x - 15) > 0 x(x + 5) - 3(x + 5) >0 (x - 3)(x + 5) > 0 therefore, x = 3 or -5 i.e. x< 3 or x > -5

volume of rectangular container = L x B x H 1 litre = 1000cm3 96 litres = \(\frac{1000cm^3}{1 litres}\) x 96 litres volume = 96000cm3 96000 = 220 x 40 x 11 96--- = 8800H H = \(\frac{96000}{8800}\) = 10.97cm = 11cm

2q35 = 778 2 x 52 + q x 51 + 3 x 50 = 7 x 81 + 7 x 80 2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1 50 + 5q + 3 = 56 + 7 5q = 63 - 53 q = \(\frac{10}{5}\) q = 2

\(\frac{3\frac{2}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\) \(\frac{\frac{11}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\) \(\frac{110}{54} \div \frac{66}{1620}\) 50

S.I. = \(\frac{P \times R \times T}{100}\) If T = 9 months, it is equivalent to \(\frac{9}{12}\) years S.I. = \(\frac{5000 \times 4 \times 9}{100 \times 12}\) S.I. = N150

M:N:Q == 5:4:3 i.e M = 5, N = 4, Q = 3 Substituting values into equation, we have... \(\frac{2N - Q}{M}\) = \(\frac{2(4) - 3}{5}\) = \(\frac{8 - 3}{5}\) = \(\frac{5}{5}\) = 1