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12. A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially? | |

A. 21 | B. 19 |

C. 17 | D. 23 |

Answer: Option A

Explanation:

Let initial quantity of P in the container be $7x$

and initial quantity of Q in the container be $5x$

Now 9 litres of mixture is drawn off from the container.

Quantity of P in 9 litres of the mixture drawn off

$=9× \dfrac{7}{12} = \dfrac{63}{12}=\dfrac{21}{4}$

Quantity of Q in 9 litres of the mixture drawn off

$=9× \dfrac{5}{12} = \dfrac{45}{12}=\dfrac{15}{4}$

Hence,

Quantity of P remaining in the mixture after 9 litres is drawn off

$=7x-\dfrac{21}{4}$

Quantity of Q remaining in the mixture after 9 litres is drawn off

$=5x-\dfrac{15}{4}$

Since the container is filled with Q after 9 litres of mixture is drawn off, quantity of Q in the mixture

$= 5x-\dfrac{15}{4}+9=5x+\dfrac{21}{4}$

Given that the ratio of P and Q becomes 7 : 9

$\Rightarrow \left(7x-\dfrac{21}{4}\right) : \left(5x+\dfrac{21}{4}\right) = 7 : 9\\\Rightarrow 9\left(7x-\dfrac{21}{4}\right)=7\left(5x+\dfrac{21}{4}\right)\\\Rightarrow 63x-\left(\dfrac{9× 21}{4}\right)=35x+\left(\dfrac{7× 21}{4}\right)\\\Rightarrow 28x=\left(\dfrac{16× 21}{4}\right)\\\Rightarrow x=\left(\dfrac{16× 21}{4× 28}\right)$

Litres of P contained in the container initially

$=7x = \left(\dfrac{7 ×16× 21}{4× 28}\right)\\= \dfrac{16× 21}{4× 4}= 21$

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