Question

a chemist has one solution that is 25% acid and a second that is 50% acid. How many liters of each chould be mixed of each to get a 10L of a solution that is 40% acid? WHAT ARE THE STEPS TO BREAKING THIS DOWN

Answer 1

well, assume he has x L of the first solution, so 25% of x L is acid therefore he has 0.25x L of acid. assume he has y L of the second solution, so he has 50% of y L of acid therefore he has 0.50y L of acid.

he wants to mix the two amounts to get x + y L of solution, but he wants that solution to be 40% acid, so he wants to get 0.40(x + y) L of acid

0.25x + 0.50y = 0.40(x + y)

hence 0.25x + 0.50y = 0.40x + 0.40y

hence 0.15x - 0.10y = 0

therefore 15x - 10y = 0

therefore 3x - 2y = 0

But we also know that x + y = 10

multiply equation 2 by 3: 3x + 3y = 30

now subtract the first from the second: 3x + 3y - 3x + 2y = 30

hence 5y = 30

hence y = 6

thus x = 4

so he should mix 4 L of the 25% solution and 6 L of the 50% solution