Question

A chemical company makes two brands of antifreeze. The first brand is 60% pure antifreeze, and the second brand is 90% pure antifreeze. In order to obtain 180 gallons of a mixture that contains 85% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Answer:

30 gallons of the first brand and 150 gallons of the second one.

Explanation:

We have two unknown quantities, $V_1$ and $V_2$, so it is necessary to write two equations that relate that variables. First, we are going to consider that the volume is an additive property, which means that if you mix $V_1$ and $V_2$ gallons the final volume V will be:

$V=V_1+V_2$

The other equation we will write is based on a mass balance of the antifreeze. We know that the content of antifreeze in a given volume is the multiplication of that volume by the concentration, so the mass balance would be:

$V_M*x_M=V_1*x_1+V_2*x_2$

It means that the total of the antifreeze in the final mix (which is $V_M*x_M$) is equal to the sum of the antifreeze that was on the first brand volume (V_1*x_1) and the antifreeze that was on the second brand volume (V_2*x_2).

Now we have a two equation two unknown system, so let's solve it:

$180=V_1+V_2$

$0.85*180=0.60*{V_1}+0.90*{V_2}\\153=0.60*{V_1}+0.90*{V_2}$

From first equation:

$V_1=180-{V_2}$,

Replace $V_1$ on the second one.

$153=0.60*(180-{V_2})+0.90*{V_2}\\153=0.60*180-0.60*{V_2}+0.90*{V_2}\\153=108+0.30*{V_2}\\{153-108}=0.30*{V_2}\\\\0.30*{V_2}=45\\{V_2}={\frac{45}{0.30} }= 150$

Finally, from $V_1=180-{V_2}$, $V_1=30$