Question

A flask of 500 ml is filled with dilute HCL consisting of 3 parts HCL and one part water. Now, 20% of this liquid is removed and replaced with water. 10% of this mixture is again removed and replaced with water. Calculate quantity of water is present in the flask. 
1. 220 ml
2. 230 ml
3. 240 ml
4. 250 ml
5. None of the above

Answer 1

Correct Answer - Option 2 : 230 ml

GIVEN :

Capacity of flask = 500 ml

Mixture 1 = 3 parts HCL + 1 part water

Process: 20% liquid replaced with water + 10% of that mixture replaced with water

 

FORMULA USED :

Quantity to be found = Initial quantity × (1 – (a/Q)ⁿ

Where, a = quantity taken out, Q = Capacity of flask, n = number of process

 

CALCULATION :

Lets first find quantity of HCL and water in original mixture

⇒ Mixture 1 = 3 parts HCL + 1 part water

⇒ Mixture 1 consist of 3 parts HCL = (3/4) × 500

⇒ Mixture 1 consist of 375 ml HCL and 125 ml water

Now, calculating liquid replaced

⇒ 20% of 500 ml = 100 ml

⇒ 10% of 500 ml = 50 ml

⇒ 500 × (1 – (100/500) × (1 – (50/500)

⇒ 500 × (4/5) × (9/10)

⇒ 500 × 36/50

⇒ 360 ml of mixture

This 360 ml of mixture contains 3 parts HCL and one part water

⇒ Water in final mixture = (360 × (1/4)) + (500 – 360)

⇒ water in final mixture = 90 + 140

∴ water in final mixture is 230 ml