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\(x \propto \frac{1}{y}\) \(x = \frac{k}{y}\) \(y \propto t^{2}\) \(y = ct^{2}\) k and c are constants. \(x = \frac{k}{ct^{2}}\) Let \(\frac{k}{c} = d\) (a constant) \(x = \frac{d}{t^{2}}\) \(1 = \frac{d}{3^{2}} \implies d = 9\) \(\therefore x = \frac{9}{t^{2}}\) \(x = 9 \div (\frac{1}{3})^{2} \) = \( 9 \div \frac{1}{9} = 9 \times 9 = 81\)

\(\sin x = \cos x\) \(\implies x = 45^\circ\) In radians, \(x = \frac{\pi}{4}\).

Ratio of bread to sugar = r:t 25% increase in bread = \(\frac{125r}{100}\) 10% increase in sugar = \(\frac{100t}{100}\) New ratio = \(\frac{125r}{100}\):\(\frac{110t}{100}\) = 25r:22t = 50r:44t

Let the tens digits of the number be x and the unit digit be y 3x = 2y - 2 3x - 2y = -2.......(i) If the digits are interchanged, the tens digit becomes y, the unit digit becomes x. Hence 2(10x + y) = 10y + x + 20 (20x + 2y) - (10y + x) = 20 19x - 8y = 20.....(ii) Multiply eqn.(i) by 8 and eqn.(ii) by 2 24x - 16y = -16......(iii) 38x - 16y = 40........(iv) eqn(iv) - eqn(iii) 14x = 56 x = 4 Sub. for x = 4 in eqn(i) 3(4) - 2y = -2 14 = 2y y = 7 So the original number is 10(4) + 7 i.e. 10x + y = 47

\(\frac{x}{2} - \frac{1}{3} < \frac{2x}{5} + \frac{1}{6}\) \(\frac{x}{2} - \frac{2x}{5} < \frac{1}{6} + \frac{1}{3}\) \(\frac{x}{10} < \frac{1}{2}\) \(2x < 10 \implies x < 5\)

Arranging the scores in ascending order: 12, 17, 25, 25, 25, 25, 26, 26, 26, 29, 29, 29, 35, 35. The median is the average of the 7th and 8th marks. = \(\frac{26 + 26}{2} = 26\)

All others are greater than 0.333 when converted to their fractions except \(\frac{15}{46}\)

The formula for calculating the diagonal of a cuboid is given by : \(\sqrt{l^2 + w^2 + h^2}\) where w = width, l = length and h = height 9 = \(\sqrt{l^2 + w^2 + h^2}\) 81 = 4\(^2\) + 4\(^2\) + h\(^2\) 81 - 16 - 16 = h\(^2\) h\(^2\) = 49 h = \(\sqrt{49}\) = 7cm NOTE:[l = w = 4 (square base)]

827.51 x 0.015 By normal multiplication or use of four figure table, 827.51 x 0.015 = 12.4127 (to 4 decimal places).

Area of circle 1 with diameter 26cm: \(\pi r^{2} = \pi \times (\frac{26}{2})^{2} \) = \(169 \pi cm^{2}\) Area of circle 2 with diameter 20 cm: \(\pi R^{2} = \pi \times (\frac{20}{2})^{2}\) = \(100 \pi cm^{2}\) Area between the two circles = \((169 - 100) \pi cm^{2}\) = \(69 \pi cm^{2}\)