Question

A container has acid solution with acid and water in the ratio 8:5 .How much water should be added to this solution such that resultant solution contains acid and water in the ratio 4:3 and has quantity 70 L

Answer 1

Answer:

5 liters

Step-by-step explanation:

8:5 -> 4:3

4x + 3x = 7x

7x = 70

x = 10

4x = 8y = 40

y = 5

5y = 5×5 = 25

3x = 3×10 = 30

30-25 = 5

Answer 2

Step-by-step explanation:

Final solution quantity = 70 L

devided into 4 : 3

acid = 40 and water = 30.

water should be added that means acid remain same in previous mixture.

so, Initial mixture acid = 40

ratio 8 : 5

8 = = 40

13 = = 65 ( 8 + 5 = 13)

Total initial quantity = 65

final quantity = 70

increase = 70 - 65 = 5

so, 5 L water should be added.