Question

Two vessels A and B contains spirit and water in the ratio 5 : 2 and 8 : 3. Their contents are mixed in the ratio 7 : 11. In 360 ml of this solution, how much water should be added so that the mixture contains spirit and water in the ratio 5 : 3?
1. 58 ml
2. 54 ml
3. 60 ml
4. 56 ml

Answer 1

Correct Answer - Option 4 : 56 ml

Given- 

Ratio of Spirit and Water in vessel A = 5 : 2

Ratio of Spirit and Water in vessel B = 8 : 3

Concept Used-

If A and B are mixed in ratio of x : y, then part of A in Mixture = x/(x + y) 

Calculation-

Part of Spirit and Water in vessel A = 5/7 and 2/7

Part of Spirit and Water in vessel B = 8/11 and 3/11

If Vessel A and Vessel B are mixed in ratio of 7 : 11 then,

Part of Spirit and Water in vessel A = 7 × 5/7 and 7 × 2/7

⇒ 5 and 2

Part of Spirit and Water in vessel B = 11 × 8/11 and 11 × 3/11

⇒ 8 and 3

Total Spirit = 8 + 5 = 13

Total Water = 2 + 3 = 5

Ratio of spirit and water in the new mixture = 13 : 5 

Now, 

18 units = 360

1 units = 360/18 = 20 

13 units = 20 × 13 = 260

And 5 units = 20 × 5 = 100

Let x ml water should be added.

According to the question:

260/(100 + x) = 5/3

⇒ 100 + x = 156

⇒ x = 56 ml

∴ 56 ml water should be added so that the mixture contains spirit and water in the ratio 5 : 3.