Question

A chemical company makes 2 brands of antifreeze. the first brand is 35% pure antifreeze and the second is 85% pure aantifreeze. in order to obtain 70 gallons of a mixture that contains 80% pure antifreeze how many gallons of each brand must be used?

Answer 1

Let A be the amount of the brand of 35% pure antifreeze, and B the amount of the brand of 85% pure antifreeze.

Since a total of 70 gallons is to be obtained, then:

$A+B=70$

The total amount of pure antifreeze at the end should be 80% of 70, which is:

$\frac{80}{100}\cdot70=56$

On the other hand, the total amount of antifreeze in A gallons of the first brand and B gallons of the second brand is:

$\frac{35}{100}A+\frac{85}{100}B$

Therefore:

$0.35A+0.85B=56$

Isolate A from the first equation and substitute in the second one to find B:

$A=70-B$$\begin{gathered} \Rightarrow0.35(70-B)+0.85B=56 \\ \Rightarrow24.5-0.35B+0.85B=56 \\ \Rightarrow0.5B=56-24.5=31.5 \\ \Rightarrow B=\frac{31.5}{0.5} \\ \Rightarrow B=63 \end{gathered}$

Substitute B=63 into the expression for A:

$\begin{gathered} A=70-63 \\ =7 \end{gathered}$

Therefore, to make a mixture of 70 gallons of 80% pure antifreeze, we need 7 gallons of the 35% pure brand and 63 gallons of the 85% pure brand.