BTGmoderatorDC wrote: ↑

Wed Dec 30, 2020 4:36 pm

Solution A is 40% chlorine by volume, and Solution B is 60% chlorine by volume. If a tank currently holds 40 gallons of Solution A, how many gallons of Solution B must be added so that the liquid in the tank is 50% chlorine?

A. 40 gallons
B. 50 gallons
C. 60 gallons
D. 80 gallons
E. 100 gallons



OA A

Solution:

Currently there are 40 x 0.4 = 16 gallons of chlorine in solution A. We can let the amount of solution B = n and create the equation:

(16 + 0.6n)/(40 + n) = 1/2

2(16 + 0.6n) = 40 + n

32 + 1.2n = 40 + n

0.2n = 8

n = 40

Alternate Solution:

Notice that 50 is the average of 40 and 60. Notice that when an equal amount of 40% and 60% solutions are mixed, the resulting mixture will have a concentration of 50%. Thus, we should add as much 60% solution as the amount of 40% solution already in the tank (which is 40 gallons) to obtain a 50% chlorine solution.

Answer: A