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CAT 2019 Shift 1 Paper

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Question : 77 of 100

Marks: +1, -0

Let x and y be positive real numbers such that

\log_5 (x + y) + \log_5 (x - y) = 3

\log_2 y - \log_2 x = 1 - \log_2 3

. Then xy equals

250
100
25
150

Solution:

\log_{5}(x+y)+\log_{5}(x-y)=3

\Rightarrow \log_{5}[(x+y)(x-y)]=3

\Rightarrow (x+y)(x-y)=5^{3}=125

\Rightarrow x^{2}-y^{2}=125 \ldots (1)

And

\log_{2} y - \log_{2} x = 1 - \log_{2} 3

\Rightarrow \log_{2}\left( \; \frac{y}{x} \right) = \log_{2} 2 - \log_{2} 3

\Rightarrow \log_{2}\left( \; \frac{y}{x} \right) = \log_{2}\left( \; \frac{2}{3} \right)

\Rightarrow \; \frac{y}{x} = \; \frac{2}{3}

Let

x=3 k

and

y=2 k

. Putting the values in (1)

(3 k)^{2}-(2 k)^{2}=125

\Rightarrow 5 k^{2}=125

\Rightarrow k=5

Hence

x \times y=3 k \times 2 k=6 \times 25=150

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