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! 
 A: You start with 35 units of alcohol.
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 $x = 61\frac{7}{13}$.
A: 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).
