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

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

Step-by-step 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}={(45)/(0.30) }= 150`

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