Question Library

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 = \((169 - 100) \pi cm^{2}\) = \(69 \pi cm^{2}\)

Add all the No.s together to give 100 \(\sum\)xf = 100 N = 20 x = \(\frac{\sum xf}{N}\) = \(\frac{100}{20}\) = 5

x 5 6 7 8 9 f 6 4 8 7 5 Mode is the No having the highest frequency = 7

\(\sin x = \frac{a}{b}\) \(\sin^{2} x + \cos^{2} x = 1\) \(\sin^{2} x = \frac{a^{2}}{b^{2}}\) \(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\) \(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\) \(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\) = \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\) = \(\frac{\sqrt{b^{2} - a^{2}}}{b}\)

Total age of the 7 pupils = 7 x 12 = 84 Total age of the 25 pupils = 25 x 14 = 350 Total age of the 6 pupils = 6 x 11 = 66 7 pupils leaving a class of 25 pupils given 18 pupils Total age of the 18 pupils = 350 - 84 = 266 6 pupils joining 18 pupils = 24 Total age of 24 pupils = 266 + 66 = 332 Average age of 24 pupils now in class \(\frac{332}{24}\) = 13yrs 10 months

\(A = P(1 + \frac{R}{100})^{T}\) \(285.20 = P(1 + \frac{5}{100})^{3}\) \(285.20 = P(1.05)^{3} \implies 1.16P = 285.20\) \(P = \frac{285.20}{1.16} = $245.86\) Given Amount = $434.00 Interest = Amount - Principal Interest = $434.00 - $245.86 = $188.14 \(T = \frac{100I}{PR}\) \(T = \frac{100 \times 188.14}{245.86 \times 7.5}\) \(T = \frac{18814}{1843.95} = 10.2\) Approximately 10 years.

When C.P = N45.00 at 8% profit C.P = \(\frac{100}{108}\) x 45 at profit of 32%, S.P = \(\frac{132}{100}\) x c.p = \(\frac{132}{100}\) x \(\frac{100 \times 45}{108}\) = \(\frac{5940}{108}\) = N55.00

In \(\Delta DAC, \stackrel\frown{DAC} = \theta\) \(\sin \theta = \frac{5}{13}\) \(\theta = 22.6^\circ\) \(< DOC = 22.6^\circ \times 2 = 45.2^\circ\) \(\sin 45.2 = \frac{5}{r} \implies r = \frac{5}{\sin 45.2}\) \(r = 7.046cm\) = \(7\frac{1}{24} cm\)