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?

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

Step-by-step explanation:

72 is the answer to this problem

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.