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01 Sep 2010, 16:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (02:14) correct
53%
(03:17)
wrong
based on 39
sessions
History
Date
Time
Result
Not Attempted Yet
Each of four girls, A,B,C and D, had a few chocolates with her. A first gave 1/3rd of the chocolates with her to B, B gave 1/4th of what she then had to C and C gave 1/5th of what she then had to D. Finally, all the four girls had an equal number of chocolates. If A had 80 more chocolates then B initially, find the difference between the number of chocolates that C and D initially had
A) 20
B) 30
C) 15
D) 25
E) 50
A) 20
B) 30
C) 15
D) 25
E) 50
whiplash2411
Joined: 09 Jun 2010
Last visit: 02 Mar 2015
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Concentration: General Management, Nonprofit
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01 Sep 2010, 16:34
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The easiest way to solve this would be by setting up a table of sorts.
Let us assume that initially A had a chocolates, B had b chocolates, C had c chocolates and D had d chocolates.
Given: \(a = b+80\)
We need to find: \(c-d\)
Here's a table I created with the information we've been given.
Table .jpg [ 33.25 KiB | Viewed 19623 times ]
So now let us look at the other things we've been given. Finally, each girl has the same number of chocolates. So that means:
\(\frac{2a}{3}= \frac{3}{4} (b + \frac{a}{3}) = \frac{4}{5} (c + \frac{b}{4} + \frac{a}{12}) = d + \frac{1}{5} ( c + \frac{b}{4} + {a}{12})\)
So now take this one at a time.
\(\frac{2a}{3}= \frac{3}{4} (b + \frac{a}{3})\)
But you know that a = b + 80. So substitute that into this equation.
\(\frac{2b + 160}{3}= \frac{3}{4} (b + \frac{b + 80}{3})\)
Simplifying this we get:
\(\frac {3b+a}{4} = \frac{4c}{5} + \frac{b}{5} + \frac{a}{15}\)
Substituting a = 80 and b = 100 in this, we get c = 110
Now solve for the last part of the equation:
\(\frac{4}{5} (c + \frac{b}{4} + \frac{a}{12}) = d + \frac{1}{5} ( c + \frac{b}{4} + \frac{a}{12})\)
Taking everything except d to the left hand side we get:
\(\frac{3}{5} (c + \frac{b}{4} + \frac{a}{12} = d\)
Now substituting values of a, b and c into the equation as 180, 100 and 110 respectively we get the value of d as:
\(d = 90\)
Thus \(c - d = 110 - 90 = 20.\)
Hence the answer is A.
P.S - This problem seems way too lengthy.
Let us assume that initially A had a chocolates, B had b chocolates, C had c chocolates and D had d chocolates.
Given: \(a = b+80\)
We need to find: \(c-d\)
Here's a table I created with the information we've been given.
Attachment:
Table .jpg [ 33.25 KiB | Viewed 19623 times ]
So now let us look at the other things we've been given. Finally, each girl has the same number of chocolates. So that means:
\(\frac{2a}{3}= \frac{3}{4} (b + \frac{a}{3}) = \frac{4}{5} (c + \frac{b}{4} + \frac{a}{12}) = d + \frac{1}{5} ( c + \frac{b}{4} + {a}{12})\)
So now take this one at a time.
\(\frac{2a}{3}= \frac{3}{4} (b + \frac{a}{3})\)
But you know that a = b + 80. So substitute that into this equation.
\(\frac{2b + 160}{3}= \frac{3}{4} (b + \frac{b + 80}{3})\)
Simplifying this we get:
\(\frac {3b+a}{4} = \frac{4c}{5} + \frac{b}{5} + \frac{a}{15}\)
Substituting a = 80 and b = 100 in this, we get c = 110
Now solve for the last part of the equation:
\(\frac{4}{5} (c + \frac{b}{4} + \frac{a}{12}) = d + \frac{1}{5} ( c + \frac{b}{4} + \frac{a}{12})\)
Taking everything except d to the left hand side we get:
\(\frac{3}{5} (c + \frac{b}{4} + \frac{a}{12} = d\)
Now substituting values of a, b and c into the equation as 180, 100 and 110 respectively we get the value of d as:
\(d = 90\)
Thus \(c - d = 110 - 90 = 20.\)
Hence the answer is A.
P.S - This problem seems way too lengthy.
gurpreetsingh
Joined: 12 Oct 2009
Last visit: 15 Jun 2019
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GMAT 1: 670 Q49 V31
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01 Sep 2010, 17:15
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I do not think this question can be asked in the exam. These type of questions can only appear on the practice test of prep companies as brutal challenges. 
Even if it is asked in the exam, wise decision is to do random guess and save the time for further questions.
This is a perfect question for IIMCAT.
Even if it is asked in the exam, wise decision is to do random guess and save the time for further questions.
This is a perfect question for IIMCAT.
