A mixture of lime and water contains 3a liters lime and (a + 40) liters water. Upon adding 20 liters more water, the ratio of quantity of water to that of lime in the mixture becomes 2:3. How much water should be added in the resultant mixture to further decrease the % of quantity of lime in the mixture to 50%?
ATQ, we can say that {(3a)/(a + 40 + 20)} = (3/2) Or, 6a = 3a + 180 Or, 3a = 180 So, a = 60 So, quantity of milk present in the mixture = (60 × 3) = 180 liters And quantity of water present in the mixture = 60 + 40 + 20 = 120 liters So, quantity of mixture after more water is added = (180/0.5) = 360 liters So, quantity of water to be added = 360 - 300 = 60 liters
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