Question

A solution contains alcohol and water in the ratio 3 : 1. 16 litres of the solution is drawn and 12 litres of water is added. Now 11 litres of the mixture is replaced by 11 litres of the water. If the final ratio of alcohol to water in the solution is 9 : 13, find the initial quantity of the solution.

Answer 1

Answer:

54

Step-by-step explanation:

Let the initial solution be = 4xL.

Then the quantity of alcohol =3xL  

And the quantity of water =xL.

Out of the 16L of mixture is drawn, alcohol =12L and water =4L. And again 12L of water is added.

Thus, quantity of alcohol =(3x−12)L  

Quantity of water =(x−4+12)L.

Again, 11L of mixture is replaced by 11L of water.

Quantity of alcohol =(3x−12−334)L  

Quantity of water =(x−4+12−114+11)L

Therefore,

(3x−12−33/4)/x−4+12−11/4+11 = 9/13

Thus,

9:13=36:52

So,  (66-x)/(x-36)=12/11

On Solving

X=50.347

50.347+4=54.347

Answer 2

............total..48

3:1 i.e............................. 36:12

Withdraw 16 liter 24:8

12 liter water added 24:20

Now 11 liter withdraw 18:15

11 liter water added 18::26 or 9:13

So , initial quantity =48