A beaker contains a mixture of two liquids X and Y in the ratio 5 : 3. When 6 litres of the mixture is drawn off and then replaced with Y, the ratio of X and Y becomes 5 : 7. How many litres of liquid X was contained in the beaker initially?

GIVEN:

Two liquids X and Y in the ratio 5 : 3

6 l of the mixture  is drawn off and replaced with Y

The final ratio of X and Y = 5 : 7

CALCULATION:

Let the beaker initially contains 5z and 3zof mixtures  of X and Y respectively,

⇒Quantity of X in mixture left =(5z - \(5\over8\) × 6) = (5z - \(15\over4\) ) L

⇒ Quantity of Y in mixture left = ( 3z - \(3\over8\) × 6) = ( 3z - \(9\over4\) )L

⇒ \((5z -\frac{15}{4})\over(3z -\frac{9}{4}) +6\) = \(\frac{5}{7}\) 

⇒  \((20z -15)\over(12z +15)\) = \(5\over7\)  

⇒ (140z - 60z) = (105 +75) 

⇒ 80z = 180 ⇔ z = 180/80 

⇒ Liters of liquid initially in X = 5z = 5× 180/80  = 11.25

Hence, initially the liquid in X was 11.25.

Shortcut Trick 

Quantity of X in step 1 and Final will be same.

So, for Y the quantity is increased by 4 unit.

⇒ 4 unit = 6 lt

So 1 unit = (6/4) lt

Then the quantity of X in Step 1 = (6/4) × 5 = 30/4 lt

So, the initial quantity of X = (30/4) + (5/8) of initial mixture drawn

⇒ X = (30/4) + (5/8) × 6

⇒ X = (30/4) + (30/8)

⇒ X = 11.25

Hence, initially the liquid X was 11.25.