(a) Without using mathematical table or calculator, evaluate : \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\).
(b) Simplify : \(\frac{8 – 4\sqrt{18}}{\sqrt{50}}\).
(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.
Explanation
(a) \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\)
\(\frac{0.18 \times 12.5}{0.05 \times 0.2} = \frac{18 \times 10^{-2} \times 125 \times 10^{-1}}{5 \times 10^{-2} \times 2 \times 10^{-1}}\)
= \(9 \times 25 \times 10^{-3 - (-3)}\)
= \(9 \times 25\)
= \(225\)
\(\therefore \sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}} = \sqrt{225}\)
= \(15\).
(b) \(\frac{8 - 4\sqrt{18}}{\sqrt{50}}\)
\(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\)
\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\)
\(\frac{8 - 4(3\sqrt{2})}{5\sqrt{2}} = \frac{8 - 12\sqrt{2}}{5\sqrt{2}}\)
Rationalising, we have
= \(\frac{8 - 12\sqrt{2}}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)
= \(\frac{8\sqrt{2} - 24}{10}\)
= \(0.8\sqrt{2} - 2.4\)
= \(-2.4 + 0.8\sqrt{2}\)
(c) \(x \propto \frac{y^{3}}{z^{2}}\)
\(\implies x = \frac{ky^{3}}{z^{2}}\)
\(108 = \frac{k \times 3^{3}}{4^{2}}\)
\(k = \frac{108 \times 16}{27}\)
\(k = 64\)
\(\therefore x = \frac{64y^{3}}{z^{2}}\)
\(\therefore 4000 = \frac{64 \times 10^{3}}{z^{2}}\)
\(4 = \frac{64}{z^{2}}\)
\(z^{2} = \frac{64}{4} = 16\)
\(z = \sqrt{16} = \pm 4\)