The ratio, by volume, of soap to alcohol to water in a

intial soap to alcohol = 2/50
final soap to alcohol = 4/50 = 2/25

initial soap to water = 2/100
final soap to water = 2/200

in fianl solution soap = 2x, alcohol = 25x, and water = 200x

as per question 25x=100, x = 4

water = 200 * 4 = 800 option E

INITIAL-
S:A:W
2:50:100

DOUBLED AND HALVED
S:A:W
4:50:-
1:-:100 OR 4:-:400

so A/W=50/400
100/x=50/400
x=800

How you get 1/100=4/400?? Why you multiply by 4??

We have that s/a=4/50 (s corresponds to 4) and s/w=1/100 (s corresponds to 1). To get s/a/w ratio we need s to correspond to the same number in both ratio, so we multiply 1/100 by 4 to get s/w=4/100.

The first thing we want to do is set up the ratio of the original solution using variable multipliers.

soap : alcohol : water = 2x : 50x : 100x

We are given that the ratio of soap to alcohol is doubled. Our original ratio of soap to alcohol is:

soap/alcohol = 2x/50x = x/25x

(Notice that we only canceled the 2 and 50; we didn’t cancel the x because it’s our ratio multiplier.)

If we double the ratio we multiply the entire ratio by 2, so we have:

2(x/25x) = 2x/25x

So now our new ratio is:

soap/alcohol = 2x/25x

Next we are given that the ratio of soap to water is halved. Our original ratio of soap to water is:

soap/water = 2x/100x = x/50x

If we halve the ratio, we multiply the entire ratio by 1/2, so we have:

(1/2)(x/50x) = x/100x

So now our new ratio is:

soap/water = x/100x

Next we must notice that the amount of soap in the two new ratios is not the same value. In the first new ratio, soap = 2x and in the second new ratio, soap = x. Thus, we need to make these values equal before continuing to the answer. To do this, we multiply the second ratio by 2/2, so we have:

soap/water = (2/2)(x/100x) = 2x/200x

Now we can set up the ratio of the altered solution.

soap : alcohol : water = 2x : 25x : 200x

Lastly, we are given that the altered solution will contain 100 cubic centimeters of alcohol.

With this we can set up the following equation and determine x:

25x = 100

x = 4 (This means that the ratio multiplier is 4.)

Thus, the altered solution will contain (200)(4) = 800 cubic centimeters of water.

Answer is E.

I think this problem can be solved with accuracy by setting up a table. Please see the attached table. The table is self explanatory. Let know if an explanation needed.

Answer = 800

option E

S : A : W = 2 : 50 : 100

The solution will be altered so that the ratio of soap to alcohol is doubled
The ORIGINAL ratio is S : A = 2 : 50.
NEW ratio is S : A = 4 : 50

the ratio of soap to water is halved
The ORIGINAL ratio is S : W = 2 : 100
NEW ratio is S : W = 1 : 100


If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain?
We need to COMBINE our two ratios.
Take S : W = 1 : 100 and create the EQUIVALENT ratio S : W = 4 : 400

We can now combine both ratios to get: S : A : W = 4 : 50 : 400
We want water = 100
So, take S : A : W = 4 : 50 : 400 and rewrite as S : A : W = 8 : 100 : 800

Answer: E

Cheers,
Brent

My method is as follows:-
Step 1 : Present ratio S:A:W = 2 : 50 : 100
Step 2 : Altered Ratio S:A = 4:50 and S:W= 1:100
Step 3 : To make S common multiply S:W x 4 = 4:400
Step 4: The new ratio S:A:W = 4 : 50 : 400
Step 5: Let amount of water be x. Therefore, 50/400=100/x ...which on solving gives x=800

If soap and alcohol is doubled, why it is not 4:100 instead of 4:50

Posted from my mobile device

Initially i also use to get confused with halving or doubling of ratio. Best way to understand is that instead of considering ratio in form S:A use form S/A and than perform the intended operation. In case doubled, S/A*2 =4:100 ....Similarly when halving the ratio S/A*1/2

I hope it is clear now

­Fun ratio problem where you have to adjust the ratios step by step and preserve some relationships as you go along:

Interesting question, it was not clear in my mind how you can double a ratio? So I ended up with final ratio as 4:100:200 and marked 200cc as water content. Wrong answer.

S:A:W
2:5:10

S:A needs to be doubled we get S:A 4:50
S:W needs to be halved we get S:W 1:100

Alcohol is at 100 -> S:A = 8:100

current level of S:W is 1:100 S is at 8 from previous step. Thus, 100*8 = 800.

I think its inappropriate to use X till the end since the solution will be altered.
This worked here since we can assume X as constant even if the solution will be altered.

We are accounting for the alteration in our operations; "x" is not the volume, the volumes are 2x, 50x, and 100x for soap, alcohol, and water respectively. "x" is just an unknown multiplier, there is no issue in keeping "x" till the end as long as we are performing the given operations on volumes.

Before:

S/A=2/50 and A/W=50/100 ---> Multiplying both ratios: S/W= 2/100


After

S/A= (2/50)*2= 4/50
S/W= (2/100)*1/2= 2/200---> W/S=200/2
Multiplying W/S and S/A--> W/A= (4/50)*(200/2)--->W/A=8/1--->A/W=1/8---->W=8A--->8*100
Answer choice E