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\(\frac{\sqrt{98} - \sqrt{50}}{\sqrt{32}}\) = \(\frac{\sqrt{49 \times 2} - \sqrt{25 \times 2}}{\sqrt{16 \times 2}}\) => \(\frac{7\sqrt{2} - 5\sqrt{2}}{4\sqrt{2}}\) = \(\frac{2\sqrt{2}}{4\sqrt{2}}\) = \(\frac{1}{2}\)

\(\frac{9^{2x - 1}}{27^{x + 1}} = 1\) \(\implies 9^{2x - 1} = 27^{x + 1}\) \((3^{2})^{2x - 1} = (3^{3})^{x + 1}\) \(2(2x - 1) = 3(x + 1) \implies 4x - 2 = 3x + 3\) \(4x - 3x = 3 + 2 \implies x = 5\)

\(\frac{4\sqrt{18}}{\sqrt{8}}\\ = \frac{4\sqrt{9\times 2}}{\sqrt{4 \times 2}} = \frac{12\sqrt{2}}{2\sqrt{2}}=\frac{6\sqrt{2}}{\sqrt{2}}\\ Rationalising \frac{6\sqrt{2}}{\sqrt{2}}\\ =\frac{6\sqrt{2}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 6\)

\(\frac{log8}{log\frac{1}{4}}=\frac{log2^3}{log2^{-2}}=\frac{3log2}{-2log2}=-\frac{3}{2}\)

\(4P4_5 = 119_{10}\) \(4 \times 5^2 + P \times 5^1 + 4 \times 5^0 = 119\) \(100 + 5P + 4 = 119\) \(5P = 119 - 104 = 15\) \(P = 3\)

\(\frac{3^{(1-n)}}{9^{-2n}}\) =\(\frac{1}{9}\) 3\(^{1-n}\times 3^{-2(-2n)} = 3^{-2}\) 1 - n - 2(-2n) = -2 1 - n + 4n = -2 n = -1

(a) \(\sqrt{50} - 3\sqrt{2}(2\sqrt{2} - 5) - 5\sqrt{32}\) \(\sqrt{25 \times 2} - (3 \times 2)\sqrt{2 \times 2} + (3 \times 5)\sqrt{2} - 5\sqrt{16 \times 2}\) = \(5\sqrt{2} - 6(2) + 15\sqrt{2} - 5(4\sqrt{2})\) = \(5\sqrt{2} - 12 + 15\sqrt{2} - 20\sqrt{2}\) = \((5 + 15 - 20)\sqrt{2} - 12\) = -12. (b) \(\frac{1}{2} \log_{10} (\frac{25}{4}) - 2 \log_{10} (\frac{4}{5}) + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{25}{4})^{\frac{1}{2}} - \log_{10} (\frac{4}{5})^{2} + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{5}{2}) - \log_{10} (\frac{16}{25}) + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{\frac{5}{2} \times \frac{320}{125}}{\frac{16}{25}}\) = \(\log_{10} (\frac{\frac{160}{25}}{\frac{16}{25}}\) = \(\log_{10} 10 = 1\)

\((343)^{\frac{1}{3}} \times (0.14)^{-1} \times (25)^{-\frac{1}{2}}\) = \(\sqrt[3]{343} \times (\frac{14}{100})^{-1} \times (\sqrt{25})^{-1}\) = \(7 \times \frac{100}{14} \times \frac{1}{5}\) = 10.

\((\sqrt{0.7} + \sqrt{70})^{2}\) = \((\sqrt{0.7} + \sqrt{70})(\sqrt{0.7} + \sqrt{70})\) = \(0.7 + 2\sqrt{0.7 \times 70} + 70\) = \(0.7 + 14 + 70 \) = 84.7

\(\frac{0.21 \times 0.072 \times 0.00054}{0.006 \times 1.68 \times 0.063}\) = \(\frac{9 \times 10^{-10}}{7 \times 10^{-8}}\) = \(1.2857 \times 10^{-2}\) = \(0.012857 \approxeq 0.01286\) (to 4 s.f)

\(x \propto \sqrt{n}\) \(x = k \sqrt{n}\) \(9 = k \sqrt{9} \implies 9 = 3k\) \(k = 3\) \(x = 3 \sqrt{n}\) When n = 17/9, \(x = 3 \times \sqrt{\frac{17}{9}} = \sqrt{17}\)

Diameter = 4\(\sqrt{3}\) cm radius = 2\(\sqrt{3}\) cm Area of major sector = \(\frac{\theta}{360} \times \pi r^{2}\) \(\theta = 360 - 120 = 240^\circ\) = \(\frac{240}{360} \times \pi \times 12\) = \(8\pi cm^{2}\)

Volume of a frustrum with top of radius R and bottom r and height h = \(\frac{1}{3} \pi (R^{2} + Rr + r^{2})\) V = \(\frac{1}{3} \pi \times 4 \times (6^2 + 6 \times 4 + 4^2)\) = \(\frac{304}{3} \pi cm^{3}\)

Since radius = \(\frac{15}{2}\) diameter = 2 * (\(\frac{15}{2}\)) = 15cm XYZ is a right-angled triangle (since

Source: Myschool Past Questions Archive (Record #7728)