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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? 1). 13 ∶ 15 ∶ 52). 15 ∶ 33 ∶ 53). 12 ∶ 11 ∶ 154). 13 ∶ 33 ∶ 8 |
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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 |
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