In a mixture of alcohol and water contains ‘3a’ liters alcohol and (a + 40) liters water. On adding 20 liters more water, the ratio of quantity of water to that of alcohol in the mixture becomes 2:3. How much water should be added in the resultant mixture to further decrease the % of quantity of alcohol in the mixture to 50%?
ATQ; we can say that [(3a)/(a + 40 + 20)] = (3/2) Or, 6a = 3x + 180 Or, 3a = 180 So, a = 60 So, quantity of alcohol 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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