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

, and

\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

⇒ \log _{5}[(x+y)(x-y)]=3

⇒(x+y)(x-y)=5^{3}=125

⇒ x^{2}-y^{2}=125 . . .(1)

And

\log _{2} y-\log _{2} x=1-\log _{2} 3

⇒ \log _{2}(\;{y}/{x})=\log _{2} 2-\log _{2} 3

⇒ \log _{2}(\;{y}/{x})=\log _{2}(\;{2}/{3})

⇒ \;{y}/{x}=\;{2}/{3}

Let

x=3 k

and

y=2 k

. Putting the values in (1)

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

⇒ 5 k^{2}=125

⇒ k=5

Hence

x × y=3 k × 2 k=6 × 25=150

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