Three numbers are in the ratio $3 : 4 : 5$ . If the sum of the largest and the smallest numbers exceeds the third number by $64$ , find the numbers.

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Solution

The ratio of the no. are given as, $3 : 4 : 5$

So. let the no. as, $3 x, 4 x, 5 x$

according to question,

$3 x + 5 x = 4 x + 64$

$⟹ 4 x = 64$

$⟹ x = 16$

Hence the numbers are,

$3 x = 3 (16) = 48$
$4 x = 4 (16) = 64$
$5 x = 5 (16) = 80$

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