In a vessel the ratio of alcohol to water is 6:1. If 21 liter of mixture is taken out from the vessel and 1 liter water is added, then the ratio of alcohol to water in the final mixture becomes 21:4. Then, what will be the initial quantity (in liters) of alcohol in the vessel?
ATQ, we can say that Let the initial alcohol & water in vessel be 6x & x liters respectively. So, {(6x − 21 × 6/7)}/{(x − 21 × 1/7 +1)} = 21/4 (6x−18)/(x−3+1) = 21/4 24x − 72 = 21x − 42 3x = 30 x = 10 liters Initial the quantity of alcohol in vessel = 6x = 60 liters
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