What quantity must be added to each term of the ratio $6 : 11$ so that it becomes equal to $2 : 3$ ?

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Solution

The correct option is C: $4$
Let $^{'} x^{'}$ be added to each term in the ratio $6 : 11.$
The resultant ratio is equal to $2 : 3.$
So we have,

$\frac{6 + x}{11 + x} = \frac{2}{3}$
$\Rightarrow 3 (6 + x) = 2 (11 + x)$

$\Rightarrow 18 + 3 x = 22 + 2 x$
$\Rightarrow 3 x - 2 x = 22 - 18$
$\Rightarrow x = 4$

So, the quantity that should be added to each term of the initial ratio is $4.$

Verification: LHS $= \frac{6 + 4}{11 + 4} = \frac{10}{15} = \frac{2}{3} =$ RHS

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