Question

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

Answer 1

Let 'x' be added to each term in the ratio 6 : 11.
The resultant ratio is equal to 2 : 3.
So we have,

$\frac{6+x}{11+x}=\frac{2}{3}$
i.e. 3(6+x) = 2(11+x)

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