sw = 2:50:100 ----> 1
s:a is doubled = 2*2/50 = 4:50 ----> 2
s:w is halved = 2/100 * 1/2 = 2:200 ----> 3
Express soap value of 3 in terms of 2
s:w = 4: 400

Now 1 becomes 4:50:400

The question asks for 100 cc of alcohol so everything doubles

8800

So E.

The basic concept is similar to addition of fractions where we bring all values to common denominator the similar method should be implemented here keeping the same ratio.

S|A|W = 2|50|100

Now altered;

S|A = 4|50 = 8|100
S|W = 1|100 = 8|800

Ratio becomes 8|100|800
E

GmatKiss wrote:The ratio, by volume, of soap to alcohol to water in a certain solution is 2:50:100. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved.
If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain?

A. 50
B. 200
C. 400
D. 625
E. 800

Need a generic method (faster) to solve similar problems.

TIA,
GK

OK, so they want to double 2:50 to make the ratio of soap to alcohol in the altered solution to become 4:50 now; and they further want to make half of 2:100 to make the ratio of soap to water in the altered solution to become 1:100 now.

In the altered solution we symbolically have
S:A = 4:50 and S:W = 1:100

And the question cleverly asks us to find A:W first and then answer to the question. See the shortest we can do from here is to divide the two ratios so that the unwanted S cancels out and we have

(S/W)/ (S/A) = A/W

Or A:W in the altered solution = (1/100)/ (4/50) = 1/100 × 50/4 = 1/8 or 1:8.

In other words, in the altered solution, for every 1 cubic centimeters of alcohol, there are 8 cubic centimeters of water contents; likewise for every 100 cubic centimeters of alcohol, there are [spoiler]800 cubic centimeters[/spoiler] of water contents.

[spoiler]E[/spoiler]

A line-by-line solution:

The ratio, by volume, of soap to alcohol to water in a certain solution is 2:50:100.
S:A:W = 2:50:100

The solution will be altered so that the ratio of soap to alcohol is doubled
S:A:W= 4:50:100

while the ratio of soap to water is halved.
S:A:W= 4:50:400

If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain?

A = 100
W = ?
a/w = 1/8 => W = 800 Option E

A. 50
B. 200
C. 400
D. 625
E. 800

Need a generic method (faster) to solve similar problems.

TIA,
GK[/quote]

The only 2 generic ways I know to do this are to set up a table with the 3 parts as the 3 columns. Or to look at the two key ratios that are changing. I'll do it the 2nd way.

Our two key ratios here are:

Soap to alcohol: 2:50 or 1:25
Soap to water: 2:100 or 1:50

We don't need the third ratio of alcohol to water, because it is built into these two relationships.

We want to double the ratio of soap to alchohol, so the new ratio is 2:25 and we want to half the ratio of soap to water so the new ratio is 1:100 or 2:200 (since our first term had 2 parts of soap).

So our new soap:alcohol:water ratio is 2:25:200

So if it contains 100 cm^3 of alcohol, it will contain 200*4 or 800 cm^3 of water.

E

S:A:W = 2:50:100

AFTER ALTERED

S/A = 2 (2/50) = 2/25 S/W = 0.5 (2/100) = 1/100

Now, we know altered solution contains 100cc of alcohol

S/A = 2/25
S/100 = 2/25
S= 8

Now solve for W

S/W = 1/100
8/W=1/100
W=800