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by rishijhawar » Wed Jul 06, 2011 1:31 pm
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
OA E. Though it is an old post, does someone has some ready material on how to deal with three or more ratios quickly. Thanks
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by winniethepooh » Wed Jul 06, 2011 2:54 pm
800
Let soap be s , alcohol be a and water be w.
s:a:w = 2:50:100 (given)
Altered solution:
s:a = 2:50 = 2/50 x 2 = 4 :50
s:w = 2:100 = 2/100 x 1/2= 1:100
So the new ratio is :
4:50:400(As 1 : 100 needs to be equated with 4 the new ratio of s:w will be multiplied by 4 on both sides = 4:400)
If a = 100 , then w = 800
Hence, E.
Last edited by winniethepooh on Thu Jul 07, 2011 12:27 pm, edited 2 times in total.
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by krishnasty » Wed Jul 06, 2011 11:47 pm
winniethepooh wrote:800
Let soap be s , alcohol be a and water be w.
s:a:w = 2:50:100 (given)
Altered solution:
s:a = 2:50 = 2/50 x 2 = 4 :50
s:w = 2:100 = 2/100 x 1/2= 1:100
So the new ration is :
4:50:100
If a = 100 , then w = 800
Hence, E.
The new ratio which you formed does not depic the ratio of s:w being half from the original ratio solution. Can you please explain how did you arrive at this solution?
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by krishnasty » Wed Jul 06, 2011 11:55 pm
The ans is definately 800, but my new ratio is coming as:
s:a:w :: 4:50:400

it can be verified that s:a ratio has been doubled from 2:50 and s:w ratio is half of 2:100.

solving by this
50x = 100
x = 2
400x =800
IMO, E
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by rishijhawar » Thu Jul 07, 2011 11:23 am
krishnasty, could you please help me how to deal with ratio problems involving three or more variables in the test day quickly.
Fellas, if some one has any link/materials/flashcard, plz share.
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by winniethepooh » Thu Jul 07, 2011 12:27 pm
@krishnasty: You may check the edited solution.
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by MBA.Aspirant » Fri Jul 08, 2011 4:15 am
rishijhawar 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
OA E. Though it is an old post, does someone has some ready material on how to deal with three or more ratios quickly. Thanks
S:A:W 2:50:100

S:A 2/50 doubled = 4:50

S:W 2:100 halved = 2:200

common ratio

A:S:W
100:8: 800

so 800
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by Tani » Fri Jul 08, 2011 1:54 pm
Original ratios:

___S_______A_________W____

____50______2______________

____________2_________100____

New ratios:


___S_______A_________W____

____50______4______________

____________1_________100____


Combining new ratios:

___S_______A_________W____

____50______4______________

____________1_________100____

___400______4_______400
Tani Wolff
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by Tani » Fri Jul 08, 2011 1:56 pm
Correcting the typo on the last step! :-(


Original ratios:

___S_______A_________W____

____50______2______________

____________2_________100____

New ratios:


___S_______A_________W____

____50______4______________

____________1_________100____


Combining new ratios:

___S_______A_________W____

____50______4______________

____________1_________100____

___50______4_______400


Needing Soap to be 100, we double everything giving 100::8:800
Tani Wolff
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