Quandaries and Queries Having a problem with word problems. example A chemist mixes a 10% acid solution with a 50% acid solution to form 400mL of a 40% solution. How much of each solution should be used. Need to identify the variables. Thank you Hi, You have two solutions, a weak solution (10%) and a strong solution (50%). You need to mix some of the weak solution and some of the strong solution and end up with 400 ml of a mild solution. Suppose that you take x ml of the weak solution then you know that you will need (400 - x) ml of the strong solution since the the total amount must be 400 ml. You also know that the final solution is 40% acid, so in the final solution the amount of acid is 40% of 400 ml, that is 40% of 400 = 40/100  400 = 160 ml of acid. 10% of the weak solution is acid, that is 10% of x ml is acid, so the amount of acid that comes from the weak solution is 10% of x = 10/100  x = x/10 ml of acid. 50% of the weak solution is acid, that is 50% of (400 - x) ml is acid, so the amount of acid that comes from the weak solution is 50% of (400 - x) = 50/100  (400 - x) = (400 - x)/2 ml of acid. Since the total amount of acid is 160 ml you have x/10 + (400 - x)/2 = 160. Solve for x. Cheers, Penny Go to Math Central