A chemist needs to mix an 18% acid solution with a 45% acid solution to obtain a 12 liter mixture consisting of 36% acid. How many liters of each of the solutions must be used?

Question

anonymous · Answer

a = liters of first solution
b = liters of second solution

a+b=12
and
0.18a+0.45b=0.36*12
0.18a+0.45b=4.32

Let's solve for a in the first equation.
a+b=12
a=12-b
Substitute it into the second.

0.18(12-b)+0.45b=4.32
2.16-0.18b+0.45b=4.32
0.27b=2.16
b=8

Because a+b=12,
a+8=12, and
a=4.

You need 4L of 18% and 8L of 45%.