Question

A tin contains a mixture of two liquids $A$ and $B$ in the ratio $7:5.$ When $9$ litres of mixture are drawn off and the tin is filled with $B,$ the ratio of $A$ and $B$ becomes $7:9.$ How many litres of liquid $A$ was contained by the tin initially?

• A. $10$
• B. $20$
• C. $21$
• D. $25$

Answer 1

The correct option is C $21$

Suppose the tin initially contains $7x$ litres and $5x$ litres of mixtures $A$ and $B$ respectively.
Quantity of $A$ in mixture left
$=(7x-\frac{7}{12}\times 9)$ litres $=(7x-\frac{21}{4})$ litres
Quantity of $B$ in mixture left
$=(5x-\frac{5}{12}\times 9)$ litres $=(5x-\frac{15}{4})$ litres

According to question,
$\frac{7x-\frac{21}{4}}{5x-\frac{15}{4}+9}=\frac{7}{9}$
$\Rightarrow \frac{28x-21}{20x+21}=\frac{7}{9}$
$\Rightarrow 252x-189=140x+147$
$\Rightarrow 112x=336$
$\Rightarrow x=3$
So, the tin contained $21$ litres of liquid $A.$