Question

A chemical company makes two brands of antifreeze the first brand 65% pure antifreeze the second brand is 95% pour antifreeze in order to obtain 30 gallons of mix fixture that contains 85% pure antifreeze how many gallons of each brand of antifreeze must be used

Answer 1

Given:

Let x be the amount of 65% pure antifreeze.

Let (30-x) be the amount of 95% pure antifreeze.

We need to obtain 30 gallons of mix fixture that contains 85% pure antifreeze.

To find the number of gallons in each brand:

According to the question,

Let us frame the equation,

$65\text{ \% (x)+9}5\text{ \% (30-x)}=85\text{ \% (30)}$

On simplification we get,

$\begin{gathered} \frac{65}{100}\times x+\frac{95}{100}\times(30-x)=\frac{85}{100}\times(30) \\ \frac{65x+95(30-x)}{100}=\frac{85(30)}{100} \\ 65x+95(30-x)=85(30) \\ 65x+2850-95x=2550 \\ -30x=2550-2850 \\ -30x=-300 \\ x=10 \end{gathered}$

Therefore,

The amount of 65% pure antifreeze needed is 10 gallons.

The amount of 95% pure antifreeze needed is 20 gallons.