# Thinking about mixtures and percentages

The question in the book was "What percentage of a $35\%$ solution of alcohol in water should be replaced by pure alcohol to give a solution containing 75% alcohol? Assume that the original solution contains 100 units."

Here's how I answered it: $35\%(100) + 100\%(x) = 75\%(100 + x)$

after solving, I got $x = 160$ and then I tried to express 100 as a percentage of 160 which resulted to $61 + \frac{7}{13}\%$ which is the correct answer. But the answer was a complete guess so my question is, how do I go think about or rather, how do I visualize "what percentage should be replaced" so that I could answer similar problems in the future.

Thanks!

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It's a bit simpler, in my opinion, to solve the problem more directly:

Let $x$ be the amount of the original solution replaced.

The new amount of alcohol is $$\underbrace{\vphantom{(}.35}_{35\% \text{ of}}\cdot\underbrace{ {(100-x) }}_{\text{remaining amount}}+\underbrace{\vphantom{(} x}_{\text{amount}\atop\text{ added}};$$ or, upon simplification: $$35+.65x.$$

The new total amount of solution is still 100 and you know it is 75 percent alcohol, so $${35+.65x \over 100 }= .75$$ Solving this equation for $x$ gives $x={800/ 13}$. This is also the percentage replacement, since the new total amount is still 100 (you replaced ${800/ 13}$ units of the original solution with pure alcohol).

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That was great! Thanks a lot for the help! –  Juan de la John Dec 21 '11 at 14:53
So maybe I should think about it this way: I have a jar containing 35% alcohol and I know that x amount of pure alcohol added would cause the jar to spill x amount of the original solution. Did I get it right? –  Juan de la John Dec 21 '11 at 22:17

When you remove $x$ units of liquid, the number of alcohol units drops by $\frac{35x}{100}$,
and when you replace it with $x$ units of pure alcohol, the number of alcohol units increases by $x$, such that the alcohol content is now 75%.
We can write all this as $$35 - \frac{35x}{100} + x = 75$$ and solve to give $x = 61\frac{7}{13}$.