CAT 2018 Quant Questions

Question:
If ${\log _2}\left( {5 + {{\log }_3}a} \right) = 3$ and ${\log _5}\left( {4a + 12 + {{\log }_2}b} \right) = 3$, then a + b is equal to

 67 40 32 59

$\log_{2}({5+\log_{3}{a}})=3$

=>$5 + \log_{3}{a}$ = 8

=>$\log_{3}{a}$ = 3

or $a$ = 27

$\log_{5}({4a+12+\log_{2}{b}})=3$

=>$4a+12+\log_{2}{b}$ = 125

Putting $a$ = 27, we get

$\log_{2}{b}$ = 5

or, $b$ = 32

So, $a + b$ = 27 + 32 = 59

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CAT 2018 Quant Questions
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