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A solution of X pounds of water and sugar is boiled until 20% of the

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A solution of X pounds of water and sugar is boiled until 20% of the  [#permalink]

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New post Updated on: 26 Jul 2016, 23:57
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66% (02:52) correct 34% (03:13) wrong based on 72 sessions

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A solution of X pounds of water and sugar is boiled until 20% of the water content evaporates. Originally y% of sugar was present in the solution. After evaporation, what percentage of the solution is sugar?

A. 100y/(1-y)
B. 80y/(1-y)
C. 75y/(1-y)
D. 100y/(80-0.2y)
E. 100y/(80+0.2y)

Originally posted by rahulprasad90 on 26 Jul 2016, 19:14.
Last edited by Bunuel on 26 Jul 2016, 23:57, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: A solution of X pounds of water and sugar is boiled until 20% of the  [#permalink]

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New post 26 Jul 2016, 20:18
Hi rahulprasad90,

You would likely get more of a response if you posted your PS question in the PS Forum here:

gmat-problem-solving-ps-140/

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Re: A solution of X pounds of water and sugar is boiled until 20% of the  [#permalink]

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New post 26 Jul 2016, 22:02
rahulprasad90 wrote:
Hello Guys,

Please suggest me an alternative approach to solve this problem.

Q. A solution of X pounds of water and sugar is boiled until 20% of the water content evaporates. Originally y% of sugar was present in the solution. After evaporation, what percentage of the solution is sugar?

a. 100y/(1-y)
b. 80y/(1-y)
c. 75y/(1-y)
d. 100y/80-0.2y
e 100y/80+0.2y

The answer in the guide is (d.)


Assume values when the options have a variable.

Say, x is 100 pounds. Say y% is 50% i.e. the solution had 50 pounds of sugar and 50 pounds of water.
20% water evaporates so 20/100 * 50 = 10 pounds of water evaporates leaving 40 pounds of water.

%age of sugar solution = 50/(50+40) * 100 = 55%

Put y = 50 in the options. Options (A), (B) and (C) make no sense since they will have a negative denominator. Option (E) gives 55% when you put y = 50.

Answer (E)

And yes, please post Problem Solving questions in the the PS forum.
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Re: A solution of X pounds of water and sugar is boiled until 20% of the  [#permalink]

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New post 26 Jul 2016, 23:01
EMPOWERgmatRichC wrote:
Hi rahulprasad90,

You would likely get more of a response if you posted your PS question in the PS Forum here:

gmat-problem-solving-ps-140/

GMAT assassins aren't born, they're made,
Rich



Thanks Rich, I have posted the question.
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Re: Please suggest me an alternative approach to solve this question.  [#permalink]

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New post 26 Jul 2016, 23:52
rahulprasad90 wrote:
Hello Guys,

Please suggest me an alternative approach to solve this problem.

Q. A solution of X pounds of water and sugar is boiled until 20% of the water content evaporates. Originally y% of sugar was present in the solution. After evaporation, what percentage of the solution is sugar?

a. 100y/(1-y)
b. 80y/(1-y)
c. 75y/(1-y)
d. 100y/80-0.2y
e 100y/80+0.2y

The answer in the guide is (d.)




We'll go for the ALTERNATIVE approach since there are variables in all the answers (which means that any number could work). Since the question deals with percents, the easiest number to use is X = 100. Now let's say that y = 10. Before the evaporation we had 10 pound of sugar and 90 pounds of water. Since only water evaporated, after the evaporation the 10 pounds of sugar remained the same, but the water reduced by 20% of 90 (18 pounds), so we have only 72 pounds of water. 10 out of 82 is the fraction of sugar, so if we multiply it by 100 we get the percents. The correct answer is E: 100y/(80+0.2y) >>> 100x10 / 82
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Re: A solution of X pounds of water and sugar is boiled until 20% of the  [#permalink]

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New post 27 Jul 2016, 14:07
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rahulprasad90 wrote:
A solution of X pounds of water and sugar is boiled until 20% of the water content evaporates. Originally y% of sugar was present in the solution. After evaporation, what percentage of the solution is sugar?

A. 100y/(1-y)
B. 80y/(1-y)
C. 75y/(1-y)
D. 100y/(80-0.2y)
E. 100y/(80+0.2y)


% of sugar in the new solution = 100 *sugar/ total solution

y% sugar was present means (100-y)% water was present.

In new solution 20% water is evaporated and 80% is remaining. .80(100-y)= 80-.8y

Sugar in the new solution= y
Total solution = y+ 80+.8y= 80+.2y

% of sugar= 100y/(80+.2y)

E is the answer
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Re: A solution of X pounds of water and sugar is boiled until 20% of the  [#permalink]

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Re: A solution of X pounds of water and sugar is boiled until 20% of the &nbs [#permalink] 03 Feb 2018, 01:34
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