Practice Algebra Question

Question

How many gallons of 40 percent alcohol solution must be mixed with 70 percent solution to obtain 30 gallons of a 52 percent alcohol solution?

Answer 1

40x + 70(30-x) = 52(30)

-30x = 1560-2100

30x = 540

x= 540/30 = 18 gallons of 40%

30-x = 12 gallons of 70%

to get a mixture of 30 gallons of 50%

.4(18) + .7(12) = .52(30)

7.2 + 8.4 = 15.6

.4 is within .12 of .52

.7 is within .18 of .52

.18 is 3/2 times .12 so you need 3/2 as many gallons of 40% as of 70%

12 times 3/2 = 18