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Word Problems

A can contain...

Question

A can contains liquids $L_{1}$ and $L_{2}$ in the ratio $11 : 13$ . Draw $24 L$ of mixture from this Can and refill it by liquid $L_{2}$ . If the can has now $L_{1}$ and $L_{2}$ in the ratio $2 : 3$ , how much quantity of $L_{1}$ was in the can initially?

14 \frac{2}{7} L

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86 \frac{3}{7} L

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187 \frac{1}{7} L

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55 \frac{5}{7} L

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Solution

The correct option is B $86 \frac{3}{7} L$
$G i v e n r a t i o o f l i q u i d s i n t h e c a n L_{1} a n d L_{2} = 11 : 13$
$l e t t h e c o m m o n r a t i o b e x$
$H e n c e q u a n t i t y o f L_{1} = 11 x$
$q u a n t i t y o f L_{2} = 13 x$
$T o t a l a m o u n t o f m i x t u r e = 24 x$
$n o w 24 L o f {mixture is drawn from the mixture and 24L of liquid L}_{2} i s added to mixture$
$Amount of liquid left in can=24x-24$
$A mount of liquid L_{1} i n the can=11x- (24 / 24 x) \times 11$
$A m {ount of liquid L}_{2} in the can =13x- (24 / 24 x) \times 13 x + 24$
$N e w r a t i o o f l i q u i d s 2 : 3$
$\frac{L_{1}}{L_{2}} = \frac{11 x - 11}{13 x - 13 + 24}$
$\Rightarrow \frac{2}{3} = \frac{11 x - 11}{13 x + 17}$
$\Rightarrow 26 x + 34 = 33 x - 33$
$\Rightarrow 55 = 7 x$
$x = \frac{55}{7}$
$A {mount of L}_{1} = 11 \times \frac{55}{7}$
$= 86 \frac{3}{7} l i t e r s$

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