Question

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?

Answer 1

Container contains mixtures of two liquids p and q in ratio of 7:5.
Let us assume p = 7x  litres and q = 5x litres
9 litres of mixture is removed.
Hence, quantity of p after removal
= 7x - ($\frac{7}{12}$×9) = 7x - $\frac{21}{4}$ litres
quantity of q after removal
= 5x - ($\frac{5}{12}$×9) = 5x - $\frac{15}{4}$ litres
New ratio is 7:9 after removal of 9 litres of mixture and adding 9 litres of q.
Hence, 
$\frac{7x - \frac{21}{4} }{5x - \frac{15}{4} +9 } = \frac{7}{9}$
=> $\frac{28x-21}{20x+21} = \frac{7}{9}$
=> 252x -189 = 140x +147
=> 112x = 147+ 186
=>x = $\frac{336}{112}$ = 3

Quantity of p initially= 7x = 7*3 = 21 litres

Answer 2

(x-9)7/12=(x)7/16

(x-9)*4=x*3

4x-36=3x

X=36

Where 7p:5p=36

P=3

7p=21