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The antifreeze in my car is a mixture of pure antifreeze and water; the concentration of antifreeze is $50 \%$. Today, I added some pure antifreeze (the volume of liquid I added is about $\dfrac 14$ the volume of the liquid that was already present). What is the new concentration of the mix?

This is how I solved the probem, I would just like to double check.

I defined the unit $U$ to be the ammount of antifreeze present originally. So I had $\dfrac {1 U \text{ Antifreeze}}{2 U \text{ Mix}}$. Then I added $\dfrac 14 U$ antifreeze with concentration of $100 \%$. So now I have $\dfrac {(1 + \dfrac 14)U \text{ Antifreeze}}{(2 + \dfrac 14) U \text{ Mix}} = \dfrac {\dfrac 54 U \text{ Antifreeze}}{\dfrac 94 U \text{ Mix}} = \dfrac 59 \cdot \dfrac { U \text{ Antifreeze}}{U \text{ Mix}}$

So my new concentration is $55.56 \%$, right?

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    You have edited the exercise in such a way that it became a different exercise. Why you have done that ? – 2017-01-16
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    @callculus I edited it to make it a little bit more clear, I don't think I changed anything. What do you think I changed about the exercise? – 2017-01-16

2 Answers 2

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When in doubt, draw a table!

$$\begin{array}&& \text{By Volume} \\ \hline \bf\text{Antifreeze}&\bf\text{Water}&\bf\text{Total}\\ \hline \frac 12 &\frac 12 &1\\ \frac 14 &0 &\frac 14\\ \hline \frac 34 &\frac 12&\frac 54\\ \hline \end{array}$$

Final concentration of Antifreeze is $\frac 34\big/\frac 54=\color{red}{\frac 35}$.

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Consider that you car contains one liter of mixture (which then contain $50$ units of antifreeze).

Now, you add $\frac 14$ liter of pure antifreeze.

This makes that the resulting volume is $1+\frac 14=\frac 54$ and it contains $50+\frac 14 \times 100=75$ of antifreeze.

Then, the concentration is $\frac {75}{\frac 54}=\frac {300} 5=60$.