Question

By adding 10 liters of water in a mixture of two liquids A and B, the percentage of A in total solution is reduced by 30% and B is 10% of the solution. In this solution, 2 liters of A is added. What is the ratio of water, A and B in the solution respectively? 

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Answer 1

Answer:

Let the quantity of A and B in the solution be X and Y.

⇒ After adding 10ltr of water, total solution = X + Y + 10

B is only 10% of the total solution

⇒ Y/ (X + Y + 10) = 0.1         ----(1)

Initial percentage of A in the solution = (X × 100)/ (X + Y)

New percentage of A in solution = (X × 100)/(X + Y + 10)

This new percentage is 30% less than before

[(X × 100)/(X+Y) - (X × 100)/ (X + Y + 10)]/ [(X × 100)/(X+Y)] = 0.3

10 /(X+Y+10) = 0.3

X + Y + 10 = 100/3

Putting this in (1) we get

Y = 0.1 × 100/3 = 10/3

X = 70/3 - 10/3 = 20

If 2lts of A is added the amount of A becomes 2 + 20 = 22

Ratio of water ? A ? B is = 10 ? 22 ? 10/3 = 15 ? 33 ? 5