(a) If \(2^{x + y} = 16\) and \(4^{x - y} = \frac{1}{32}\), find the value of x and y.
(b) P, Q and R are related in such a way that \(P \propto \frac{Q^{2}}{R}\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.
(b) P, Q and R are related in such a way that \(P \propto \frac{Q^{2}}{R}\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.
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Correct Answer: Option n
Explanation:
(a) \(2^{x + y} = 16 \implies 2^{x + y} = 2^{4}\)
\(\implies x + y = 4 ..... (1)\)
\(4^{x - y} = \frac{1}{32} \implies (2^{2})^{(x - y)} = 2^{-5}\)
\(\implies 2x - 2y = -5 .... (2)\)
From (1), x = 4 - y.
\(\therefore 2(4 - y) - 2y = -5\)
\(8 - 2y - 2y = -5 \implies -4y = - 5 - 8 = -13\)
\(y = \frac{13}{4}\)
\(x = 4 - \frac{13}{4} \)
= \(\frac{3}{4}\)
(b) \(P \propto \frac{Q^{2}}{R}\)
\(\implies P = \frac{kQ^{2}}{R}\)
\(k = \frac{PR}{Q^{2}}\)
\(k = \frac{36 \times 4}{9} = 16\)
\(\implies P = \frac{16Q^{2}}{R}\)
\(Q^{2} = \frac{PR}{k}\)
\(Q^{2} = \frac{200 \times 2}{16}\)
\(Q^{2} = 25\)
\(Q = \pm 5\).
(a) \(2^{x + y} = 16 \implies 2^{x + y} = 2^{4}\)
\(\implies x + y = 4 ..... (1)\)
\(4^{x - y} = \frac{1}{32} \implies (2^{2})^{(x - y)} = 2^{-5}\)
\(\implies 2x - 2y = -5 .... (2)\)
From (1), x = 4 - y.
\(\therefore 2(4 - y) - 2y = -5\)
\(8 - 2y - 2y = -5 \implies -4y = - 5 - 8 = -13\)
\(y = \frac{13}{4}\)
\(x = 4 - \frac{13}{4} \)
= \(\frac{3}{4}\)
(b) \(P \propto \frac{Q^{2}}{R}\)
\(\implies P = \frac{kQ^{2}}{R}\)
\(k = \frac{PR}{Q^{2}}\)
\(k = \frac{36 \times 4}{9} = 16\)
\(\implies P = \frac{16Q^{2}}{R}\)
\(Q^{2} = \frac{PR}{k}\)
\(Q^{2} = \frac{200 \times 2}{16}\)
\(Q^{2} = 25\)
\(Q = \pm 5\).