Starting with some solution of volume <em>v</em> and an alcohol concentration of 100<em>c </em>% , if you add 180 mL of 45% isopropyl to it, you get a mixture with a total volume of 180 mL + <em>v</em>.
Each mL of the starting solution contains <em>c</em> mL of alcohol. For example, if the concentration of the starting solution is 80% (so <em>c</em> = 0.8), then each mL contains 80% = 0.8 of one mL of alcohol.
Similarly, each mL of the added solution with 45% concentration contains 0.45 mL of alcohol.
You want the new solution to have a concentration of 70%, so that the ratio of the amount of alcohol in it to the total volume is 70%, meaning
<em>cv</em> + 0.45 (180 mL) = 0.7 (180 mL + <em>v</em>)
and you want to solve for <em>v</em> :
<em>cv</em> + 81 mL = 126 mL + 0.7<em>v</em>
<em>cv</em> - 0.7<em>v</em> = 126 mL - 81 mL
(<em>c</em> - 0.7) <em>v</em> = 45 mL
<em>v</em> = (45 mL) / (<em>c</em> - 0.7)
Judging by context clues, <em>c</em> lies somewhere between 0.7 and 1. (It can't be less than 0.7 because mixing this solution with any other solution of smaller concentration will never yield a solution of higher concentration.)
If, for example, <em>c</em> = 0.8, so that your starting solution is at 80% concentration, then you would need
<em>v</em> = (45 mL) / (0.8 - 0.7) = 450 mL
to be mixed with 180 mL of 45% solution to end up with 180 mL + 450 mL = 630 mL of 70% solution.
As another example, if you're mixing pure alcohol, so that <em>c</em> = 1, you would need
<em>v</em> = (45 mL) / (1 - 0.7) = 150 mL
to make a 180 mL + 150 mL = 330 mL batch of 70% solution. The fact that you would need less of a higher concentration solution is not surprising.
