Question

A chemical company makes two brands of antifreeze. The first brand is 30% pure antifreeze, and the second brand is 70% pure antifreeze. In order to obtain 60 gallons of a mixture that contains 40% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Given:

The first brand is 30% anti-freeze

The second brand is 70% anti-freeze.

We want to make a 60 gallons mixture that has 40% anti-freeze.

Solution

Let the gallons of the first brand be x

Hence the gallons of the second brand would be:

`=60\text{ - x}`

The gallons of the first brand that would be pure anti-freeze:

`\begin{gathered} =\text{ }(30)/(100)\text{ }*\text{ x} \\ =\text{ 0.3x} \end{gathered}`

The gallons of the second brand that would be pure anti-freeze:

`\begin{gathered} =\text{ }(70)/(100)\text{ }*\text{ (60-x)} \\ =\text{ 0.7(60-x)} \end{gathered}`

The gallons of the mixture that would be pure anti-freeze:

`\begin{gathered} =\text{ }(40)/(100)\text{ }*\text{ 60} \\ =\text{ 0.4 }*\text{ 60} \\ =\text{ 24} \end{gathered}`

On mixing, we have the equation:

`0.3x\text{ + 0.7(60-x) = 24}`

When we solve for x, we have:

`\begin{gathered} 0.3x\text{ + 42 -0.7x = 24} \\ \text{Collect like terms} \\ 0.3x\text{ - 0.7x = 24 -42} \\ -0.4x\text{ = -18} \\ \text{Divide both sides by -0.4} \\ (-0.4x)/(-0.4)\text{ =}(-18)/(-0.4) \\ x\text{ = 45} \end{gathered}`

Hence, the gallons of the first brand required to obtain a mixture of 40% anti-freeze is 45 gallons.

The gallons of the second brand required would be:

`\begin{gathered} =60\text{ - 45} \\ =\text{ 15 gallons} \end{gathered}`

Answer:

First brand = 45 gallons

Second brand = 15 gallons