A container contains a mixture of two liquids ‘P’ and ‘Q’ in the ratio of 7 : 5. When 9 litres of mixture is drawn off and then the container is filled with the same amount of 'Q', then the ratio of ‘P’ and ‘Q’ becomes 7:9. How many litres of liquid ‘P’ was contained by the container initially?
Solution
Correct Answer: Option B
Suppose the container initially contains 7x and 5x litres of ‘P’ and ‘Q’ respectively. Then,
According to the question,
Quantity of ‘P’ in the mixture left
\(= \left( {7x - \frac{7}{{12}} \times 9} \right) = \left( {7x - \frac{{21}}{4}} \right)\ litres\)
Quantity of ‘Q’ in the mixture left:
\(= \left( {5x + 9 - \frac{5}{{12}} \times 9} \right) litres = \left( {5 x + \frac{{21}}{4}} \right)\ litres\)
According to the question:
\(\begin{array}{l} \Rightarrow \frac{{\left( {7x - \frac{{21}}{4}} \right)}}{{5x + \frac{{21}}{4}}} = \frac{7}{9}\\ \Rightarrow \frac{{28x - 21}}{{20x + 21}} = \frac{7}{9} \end{array}\)
⇒ 252x – 189 = 140x + 147
⇒ 112x = 336
⇒ x = 3
∴ Litres of liquid ‘P’ that was contained by the container initially = 7 × 3 = 21
Hence, the required answer is 21 litres
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