A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?
Let x be the quantity of solution one in mixture. so solution 2 would be (400-x) L
according to ques. 80% of x + 30% of (400-x) = 62% of 400 i.e. 80*x/100 + 30*(400-x)/100= 62*400/100 => 80x - 30x = 62*400 - 30*400 => 50x = 32*400 x=256 L therefore 256L of solution one is required
