Question

A 5-gallon radiator is full and contains a 40% solution of antifreeze. how much needs to be drained out and replaced with pure antifreeze to obtain a 70% solution?

Answer 1

Let x be the number of antifreeze solution. Thus;

`(0.4)x+1(5-x)=0.7x`

We would solve for x;

`\begin{gathered} 0.4x+5-x=0.7x \\ \\ 5=x-0.4x+0.7x \\ \\ 5=1.3x \\ \\ x=(5)/(1.3) \\ \\ x=3.85 \end{gathered}`

Hence, the amount needs to be drained out is;

`\begin{gathered} 5-x=5-3.85 \\ \\ =1.15 \end{gathered}`

Answer: 1.15 gallon