Question

A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solutioncontains 80% pure antifreeze. The company wants to obtain 320 gallons of a mixture that contains 15% pure antifreeze. How many gallons of water and howmany gallons of the premium antifreeze solution must be mixed?

Answer 1

Amount of water = 260 gallons

Amount of antifreeze = 60 gallons

Step-by-step explanation:

let the amount of water = w

amount of water + amount of antifreeze will give 320 gallons of mixture

w + amount of antifreeze = 320

amount of antifreeze = 320 - w ...(1)

The concentration for the antifreeze = 80% = 0.8

The concentration of the mixture = 15% = 0.15

From our question, we are mixing antifreeze with pure water (two different things). In our calculation, we can write the concentration in terms of water or in terms of antifreeze.

Writing the concentration in terms of amount of antifreeze

0.8(amount of antifreeze) + percent of antifreeze (amount of water) = concentraton of the mixture

The mixture contains 15% antifreeze

water contains no antifreeze, percent = 0

0.8(320 - w) + 0(w) = 0.15(amount of mixture)

0.8(320 - w) + 0 = 0.15(320)

256 - 0.8w + 0 = 48

256 - 0.8w = 48

collect like terms:

256 - 48 = 0.8w

208 = 0.8w

divide both sides by 0.8:

208/0.8 = w

w = 260

substitute for w in equation (1)

amount of antifreeze = 320 - 260 = 60

Amount of water = 260 gallons

Amount of antifreeze = 60 gallons