Question Library

Let the sum of the 12 numbers be x and the 13th number be y. \(\frac{x}{12} = 3 \implies x = 36\) \(\frac{36 + y}{13} = 5 \implies 36 + y = 65\) \(y = 65 - 36 = 29\)

\(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} -1 + \frac{3}{4})]\) \(\frac{1}{3} \div [\frac{5}{7}(\frac{9}{10} - \frac{10}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-1}{10} + \frac{3}{4})]\) = \(\frac{1}{3} \div [\frac{5}{7}(\frac{-2 + 15}{20})]\) = \(\frac{1}{3} \div [\frac{5}{7} \times \frac{13}{20}]\) \(\frac{1}{3} + [\frac{13}{28}]\) = \(\frac{1}{3} \times \frac{28}{13}\) = \(\frac{28}{39}\)

\(\frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4}\) = \(\frac{36}{420} \times \frac{54}{90} \times \frac{63}{240}\) = \(\frac{6}{70} \times \frac{18}{30} \times \frac{21}{80}\) = \(\frac{27}{2000}\) = 0.0135 \(\approx\) = 0.014

\(\frac{log_5 0.04}{log_3 18 - log_3 2}\) = \(\frac{log_5 0.04}{log_3(\frac{18}{2})}\) = \(\frac{log_5 0.04}{log_3 9}\) = \(\frac{-2}{2}\) = -1 Let log5 0.04 = x 5x = 0.04 x = \(\frac{4}{100}\) = 5\(^{-2}\) Let log\(_3\) 9 = z 3\(^2\) = 3\(^z\) z = 3

8x-2 = \(\frac{2}{25}\) = 200x-2 = 2 = 100x-2 = 1 x-2 = \(\frac{1}{100}\) x-2 = 10-2 x = 10

\(\sqrt{48}\) - \(\frac{9}{\sqrt{3}}\) + \(\sqrt{75}\) Rearrange = \(\sqrt{48}\) + \(\sqrt{75}\) - \(\frac{9}{\sqrt{3}}\) = (\(\sqrt{16}\) x \(\sqrt{3}\)) + (\(\sqrt{25}\) x \(\sqrt{3}\)) - \(\frac{9}{\sqrt{3}}\) = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) - \(\frac{9}{\sqrt{3}}\) Rationalize \(\frac{9}{\sqrt{3}}\) \(\to\) = \(\frac{9}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = 3\(\sqrt{3}\) = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) - 3\(\sqrt{3}\) = 6\(\sqrt{3}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\) x \(\frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{\sqrt{2}}{2}\) = \(\frac{1.414}{2}\) = 0.707

A \(\subset\) B means A is contained in B i.e. A is a subset of B(A \(\cap\) B)' = A' A(A \(\cap\) B)' = A \(\cap\) A' The intersection of complement of a set P and P' has no element i.e. n(A \(\cap\) A') = \(\phi\)

\(\frac{(2m - u)^2 - (m - 2u)^2}{5m^2 - 5u^2}\) = \(\frac{2m - u + m - 2u)(2m - u - m + 2u)}{5(m + u)(m - u)}\) = \(\frac{3(m - u)(m + u)}{5(m + u)(m - u)}\) = \(\frac{3}{5}\)

Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero \(\therefore (x - 1) = 0\) x = 1 Substitute in the polynomial the value x = 1 = \(p(1)^3 + q(1)^2 + 11(1) - 6 = 0\) p + q + 5 = 0 .....(i) Also since x - 3 is a factor, \(\therefore\) x - 3 = 0 x = 3 Substitute \(p(3)^3 + q(3)^2 + 11(3) - 6 = 0\) 27p + 9q = -27 ......(2) Combine eqns. (i) and (ii) Multiply equation (i) by 9 to eliminate q 9p + 9q = -45 Subtract (ii) from (i), \(18p = 18\) \(\therefore\) p = 1 Put p = 1 in (i), \(1 + q = -5 \implies q = -6\) \((p, q) = (1, -6)\)

\(\frac{a}{a - x}\) = \(\frac{b}{x - b}\) \(\frac{1}{1 - x}\) = \(\frac{3}{x - 3}\) ∴ 3(1 - x) = x - 3 3 - 3x = x - 3 Rearrange 6 = 4x; x = \(\frac {6}{4}\) = \(\frac{3}{2}\)

\(\frac{1}{r - 1}\) + \(\frac{2}{r + 1}\) = \(\frac{3}{r}\) Multiply through by r(r -1) which is the LCM = (r)(r + 1) + 2(r)(r - 1) = 3(r - 1)(r + 1) = r2 + r + 2r2 - 2r 3r2 - 3 = 3r2 r = 3r2 - 3 -r = -3 ∴ r = 3

\(\frac{x - 3}{(1 - x)(x + 2)}\) = \(\frac{p}{1 - x}\) + \(\frac{Q}{x + 2}\) Multiply both sides by LCM i.e. (1 - x(x + 2)) ∴ x - 3 = p(x + 2) + Q(1 - x) When x = +1 (+1) - 3 = p(+1 + 2) + Q(1 - 1) -2 = 3p + 0(Q) 3p = -2 ∴ p = \(\frac{-2}{3}\)

\(\frac{1}{x}\) > 2 = \(\frac{x}{x^2}\) > 2 x > 2x2 = 2x2 2 - x 0, 2x - 1 0, x 0 x < 0, x , \(\frac{1}{2}\) = 0

In an AP, Tn = a + (n - 1)d T6 = a + 5d = 11 The first term = a = 1 ∴ T6 = 1 + 5d = 11 5d = 11 - 1 5d = 10 ∴ d = 2