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