Question Library

\(\frac{2 - \sqrt5}{3 - \sqrt5}\) x \(\frac{3 + \sqrt5}{3 + \sqrt5}\) \(\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}\) = \(\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}\) = \(\frac{6 - \sqrt5 - 5}{9 - 5}\) = \(\frac{1 - \sqrt5}{4}\)

\(\frac{\log \sqrt{27}}{\log \sqrt{81}}\) = \(\frac{\log 27\frac{1}{2}}{81\frac{1}{2}}\) = \(\frac{\log 3\frac{1}{2}}{\log 3^2}\) \(\frac{\frac{3}{2} \log 3}{2 \log 3} = \frac{3}{2} \div \frac{2}{1}\) = \(\frac{3}{2} \times \frac{1}{2}\) = \(\frac{3}{4}\)

\(\sqrt{72} + \sqrt{32} - 3\sqrt{18} = x\sqrt{8}\) = \(\sqrt{36 \times 2} + \sqrt{16 \times 2} - 3\sqrt{2 \times 9}\) = x\(\sqrt{2 \times 4}\) = 6\(\sqrt{2} + 4\sqrt{2} - 9\sqrt{2} = 2 \times \sqrt{2}\) \(\sqrt{2} (6 + 4 - 9) = 2x\sqrt{2}\) \(\sqrt{2} = 2x \sqrt{2}\) divide both sides by \(\sqrt{2}\) \(\frac{\sqrt{2}}{\sqrt{2}} = \frac{ 2x\sqrt{2}}{\sqrt{2}}\) 1 = 2x 2x = 1 x = \(\frac{1}{2}\)

let the number of question = n n(n + 50) = 600 n2 + 50n - 600 - 0 by quadratic formular x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2}\) n = \(\frac{-50 \pm \sqrt{50^2 - 4(1) (-600)}}{2(1)}\) = \(\frac{-50 \pm \sqrt{4900}}{2}\) = \(\frac{-50 + 70}{2}\) = \(\frac{20}{2}\) = 10

(a) \(\frac{\frac{1}{2} of \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}}\) Numerator - \(\frac{1}{2} \times \frac{1}{4} \div \frac{1}{3} = (\frac{1}{2} \times \frac{1}{4}) \div \frac{1}{3}\) \(\frac{1}{8} \times 3 = \frac{3}{8}\) Denominator - \(\frac{1}{6} - \frac{3}{4} + \frac{1}{2} = \frac{1}{6} - (\frac{3}{4} - \frac{1}{2})\) \(\frac{1}{6} - (\frac{3 - 2}{4}) = \frac{1}{6} - \frac{1}{4}\) \(\frac{2 - 3}{12} = \frac{-1}{12}\) \(\therefore \frac{\frac{1}{2} \times \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}} = \frac{3}{8} \div \frac{-1}{12}\) \(\frac{3}{8} \times -12 = \frac{-9}{2}\) = \(-4.5\) (b) \(\sqrt{x} = 10^{\bar{1}.6741}}\) Squaring both sides, we have \(x = (10^{\bar{1}.6741}})^{2}\) \(x = 10^{\bar{1}.3482}\) (checking antilogarithm) \(x = 0.2229\)

(a) \(t = \sqrt{\frac{pq}{r} - r^{2}q}\) Squaring both sides, \(t^{2} = \frac{pq}{r} - r^{2}q\) \(t^{2} = \frac{pq - r^{3}q}{r}\) \(t^{2}r = pq - r^{3}q \implies t^{2}r = q(p - r^{3})\) \(q = \frac{t^{2}r}{p - r^{3}}\). (b) \(9^{(1 - x)} = 27^{y}\) \(\implies 3^{2(1 - x)} = 3^{3y}\) \(2 - 2x = 3y ..... (1)\) \(x - y = -1\frac{1}{2} ....... (2)\) From (2), \(x = -1\frac{1}{2} + y\) \(\therefore 2 - 2(-1\frac{1}{2} + y) = 3y\) \(2 + 3 - 2y = 3y \implies 5 = 3y + 2y = 5y\) \(\implies y = 1\) \(x = -1\frac{1}{2} + 1 = -\frac{1}{2}\) \(\therefore x + y = -\frac{1}{2} + 1 = \frac{1}{2}\)

cot\(\theta\) = \(\frac{1}{\cos \theta}\) = \(\frac{8}{15}\)(given) tan\(\theta\) = \(\frac{15}{8}\) By Pythagoras' theorem, x\(^2\) = 15\(^2\) + 8\(^2\) x\(^2\) = 225 + 64 = 289 x = \(\sqrt{289}\) = 17 Hence sin\(\theta\) = \(\frac{15}{x}\) = \(\frac{15}{17}\)

\(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1}\) = \(\frac{1}{\left(\frac{81}{16}\right)^{\frac{1}{4}}}\times \frac{1}{2}\) = \(\left(\frac{16}{81}\right)^{\frac{1}{4}}\times \frac{1}{2}\) = \(\left(\frac{2}{3}\right)^{4\times \frac{1}{4}}\) =\(\frac{2}{3}\times \frac{1}{2}\) = \(\frac{1}{3}\)

\(\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}}\) = \((\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}})(\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}})\) = \(\frac{2\sqrt{15} + 6 + 5 + \sqrt{15}}{5 - 3}\) = \(\frac{3\sqrt{15} + 11}{2}\)

Area of a triangle = \(\frac{1}{2}\) ab Sin\theta 12\(\sqrt{3}\) = \(\frac{1}{2}\) x 8 x q sin 60 12\(\sqrt{3}\) = 4 x q x \(\frac{\sqrt{3}}{2}\) 12\(\sqrt{3}\) = 2q\(\sqrt{3}\) q = \(\frac{12\sqrt{3}}{2\sqrt{3}}\) q = 6cm.

(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\) Applying the rule of BODMAS, we have: (\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\) (\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\) \(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\) = \(\frac{1}{36}\)

\(\frac{5^{n +3}}{25^{2n -3}} = 5^0\) \(\frac{5^{n + 3}}{5^{2(2n - 3)}} = 5^0\) n + 3 - 4n + 6 = 0 -3n + 9 = 0 -3n = -9 n = \(\frac{-9}{-3}\) n = 3

2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\) = 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\) = 2\(\sqrt{3} - \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{3\sqrt{3}}\) = 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\) = 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\) = \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)

Given tan x = 1 x = tan-1(1) x = 45o Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\) = \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\) = \(\frac{1}{2} + \frac{1}{\sqrt{2}}\) = \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\) = \(\frac{\sqrt{2}}{2}\)

Log 2 = 0.3010, Log 7 = 0.8451 ∴Log 280 = Log 28 x 10 = Log 7x4x10 =Log7 + Log4 + Log10 = Log7 + Log2\(^2\) + Log10 = Log7 + 2Log2 + Log10 = 0.8451 + 2(0.3010) + 1 = 0.8451 + 0.6020 + 1 = 2.4471