Question

A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. the premium antifreeze solution contains 90% pure antifreeze. the company want to obtain 180 gallons of a muxture that contains 45% pure antifreeze how many and how many gallons of the premium antifreeze solution must be mixed

Answer 1

Both should be 90 gallons

Step-by-step explanation:

Let the gallons of pure water used = x gallons

Since the company want to obtain 180 gallons of a mixture, the gallons of 90% pure antifreeze needed = (180-x) gallons

We therefore have that:

90% of (180-x) gallons = 45% of 180 gallons

`\begin{gathered} 0.9(180-x)=0.45*180 \\ 162-0.9x=81 \\ 0.9x=162-81 \\ 0.9x=81 \\ x=(81)/(0.9) \\ x=90 \end{gathered}`

• The number of gallons of pure water used = 90 gallons

• The number of gallons of premium antifreeze solution

= 180-90

= 90 gallons.