Question

A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. Thepremium antifreeze solution contains 95% pure antifreeze. The company wants to obtain 285 gallons of a mixture that contains20% pure antifreeze. How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?Water:gallonsх5?Premium antifreeze: I gallons

Answer 1

Let the number of antifreeze gallon is x

Since the solution contains 95% pure antifreeze, then multiply x by 95% after changing it to decimal

`x*(95)/(100)=0.95x`

Since they need 285 gallons containing 20% pure antifreeze, then multiply 285 by 20%

`285*(20)/(100)=57`

Equate 0.95x by 57

`0.95x=57`

Divide both sides by 0.95

`\begin{gathered} (0.95x)/(0.95)=(57)/(0.95) \\ x=60 \end{gathered}`

Subtract 60 from 285

`285-60=225`

They need 225 gallons of water and 60 gallons of premium antifreeze to mixe them

Water: 225 gallons

Premium antifreeze: 60 gallons