A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains

80

%

pure antifreeze. The company wants to obtain

320

gallons of a mixture that contains

55

%

pure antifreeze. How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?

Question

Jovanni

Mathematics

12 August, 18:13

A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains

80

%

pure antifreeze. The company wants to obtain

320

gallons of a mixture that contains

55

%

pure antifreeze. How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?

+4

Answers (1)

Know the Answer?

Trevin Austin · Answer 1

let w = amt of water required

:

An equation based on the the percent water, an 80% solution is 20% water.

.20 (320-w) + w =.50 (320)

64 -.2w + w = 160

.8w = 160 - 64

w = 96%2F. 8

w = 120 gal of water to produce 320 gal of 50% antifreeze

:

320 - 120 = 200 gal of 80% solution

Water: _120___ gallons

Premium Antifreeze: __200___ gallons