Math textbook question

Question

a lab technician needs to mix a 5% fungicide solution with a 10% fungicide solution to obtain a 50-liter mixture consisting of 8% fungicide. how many liters of each of the fungicide solutions must be used? begin by filling in the missing entries in the following table. then use the fact that the amount of fungicide in the 5% solution plus the amount of fungicide in the 10% solution must equal the amount of fungicide in the 8% mixture.

Answer 1

mix 5% and 10% to get 50 Liters of 8%

10-8=2

8-5 = 3

the tech needs 3/2 = 1.5 times as much 10% as 5%

3/2 =x/y where x = amount of 10%, y=amount of 5%

3/2 = x/(50-x)

cross multiply 2x =150-3x

5x = 150

x =150/5 =30 Liters of 10%

y =20 Liters of 5%

check the answers

.10(x) +.05(y) = .08(50)

.10(30) + .05(20) = .08(50)

3 +1 = 4

Usual method though is start with

.10(x) +.05(y) = .08(50)

plug in 50-x for y, since x+y=50

.10(x) +.05(50-x) = .08(50)

expand, collect/combine like terms and solve for x

.1x -05x = (.08-.05)(50)

.05x = .03(50) = 1.5

x = 1.5/.05= 30 Liters of 10%

y=50-x=50-30 = 20 Liters of 5%