Hey rtaha2412:
I like this question and a pretty healthy takeaway that it brings up: When you're dealing with three-part problems (ratios, inequalities, etc.) it's often best to break them into two, two-part problems so that you can perform the algebra more easily.
In this problem, the ratio of:
Soap: Alcohol: Water is
2: 50: 100
Which means that you have multiple two-part ratios:
S:A = 2:50
S:W = 2:100
A:W = 50:100
When they say to double the ratio of soap to alcohol, it's probably best to see that ratio as a fraction (2/50) which makes it easier to "double" (2/50 * 2 = 4/50) to 4/50.
Then they say to halve (divide by 2) the ratio of soap to water: 1/2 * 2/100 = 1/100.
So now you have two ratios:
Soap:Alcohol = 4:50, or you could say 2:25
Soap:Water = 1:100
Now, when dealing with multi-part ratios/inequalities/etc., the key becomes putting it all back together around the common variable. Here, that common term is soap, which we express as 2 (in the 2:25 ratio) and as 1 (in the 1:100 ratio). In order to have a common ratio, we need to have soap in the same terms, and 2 is the Least Common Multiple. To get common terms, we then express the ratios with a base of 2 for soap:
Soap:Alcohol = 2:25
Soap: Water = 2:200
Soap:Alcohol:Water = 2:25:200
Because we now then have 100 cubic centimeters of alcohol, we need to multiply everything by 4 to keep it relative (because our base for alcohol is 25), so we would then have:
8
800
And therefore there are 800 ccs of water, and the correct answer is E.
As a big-picture takeaway, when you have a 3-part ratio or inequality, it's often best to break it into two 2-part problems, and then you have to synchronize your results around that common term to finish if off.
800
