A cistern contains 400 liters of water. 120 litres of water is taken out of it and replaced by alcohol. The process is repeated two times again. Find the quantity of alcohol in the resulting mixture.

Explanation

In such type of question , just see that how much % solution is drawn . Here we can say that 120L is drawn out of 400L i,e. 30% is drawn. So first time , water left = 400L – 30% of 400L = 400L -120L = 280L Now this process is repeated again Now Water left = 280L – 30% of 280L = 196L Now this process is repeated again Now Water left = 196L – 30% of 196L = 137.2L Hence water is replaced with alcohol So rest quantity of alochol = 400L – 137.2L = 262.8L Alternate Method: Let x be the orginal pure alcohol & y be the alcohol withdrawn & replaced with water n times So after n times, net alcohol left = x(1 -y/x) n = 400(1-120/400) 3 = 400(3/10) 3 = 137.2L. Hence water is replaced with alcohol So rest quantity of alcohol = 400L – 137.2L = 262.8L