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25 Jun 2017, 09:52
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (03:25) correct
38%
(03:36)
wrong
based on 16
sessions
History
Date
Time
Result
Not Attempted Yet
A and B had a certain number of marbles with them, not necessarily the same. First A gave 60% of his total marbles to B.
Next B gave to A 60% of the total he had then. The ratio of the number of marbles with A and B after this transaction remains
the same. What was the ratio of the total numbers of marbles with A and B initially?
(1) 2:1
(2) 3:2
(3) 5:3
(4) 4:3
(5) None of these
Next B gave to A 60% of the total he had then. The ratio of the number of marbles with A and B after this transaction remains
the same. What was the ratio of the total numbers of marbles with A and B initially?
(1) 2:1
(2) 3:2
(3) 5:3
(4) 4:3
(5) None of these
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25 Jun 2017, 10:29
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Let the number of marbles with A and B initially be \(x\) and \(y\), respectively.
After first transaction:
\(A = 0.4x\)
\(B = y + 0.6x\)
After second transaction:
\(A = 0.4x + 0.6y + 0.36x = 0.76x + 0.6y\)
\(B = 0.4y + 0.24x\)
Ratio initially \(= \frac{x}{y}\)
Current ratio \(= \frac{0.76x + 0.6y}{0.4y + 0.24x}\)
\(\frac{x}{y} = \frac{0.4x + 0.36x + 0.6y}{0.4y + 0.24x}\)
\(0.4xy + 0.24x^{2} = 0.4xy + 0.36xy + 0.6y^{2}\)
\(0.36xy + 0.6y^{2} = 0.24x^{2}\)
\(0.36 + 0.6(\frac{y}{x}) - 0.24(\frac{x}{y}) = 0\)
Let \(\frac{x}{y} = k\), then \(0.36 + \frac{0.6}{k} - 24k = 0\)
\(60 + 36k - 24k^{2} = 0\)
\(24k^{2} - 36k - 60 = 0\)
\(2k^{2} - 3k - 5 = 0\)
\(2k^{2} - 5k + 2k - 5 = 0\)
\(k(2k - 5) + 1 (2k - 5) = 0\)
\((k+1) (2k - 5)\)
Ignoring the negative factor, we get \(k = \frac{5}{2}\). So, Ans - E.
I don't think this is a GMAT problem, or may be I am using a lengthy method to solve this.
After first transaction:
\(A = 0.4x\)
\(B = y + 0.6x\)
After second transaction:
\(A = 0.4x + 0.6y + 0.36x = 0.76x + 0.6y\)
\(B = 0.4y + 0.24x\)
Ratio initially \(= \frac{x}{y}\)
Current ratio \(= \frac{0.76x + 0.6y}{0.4y + 0.24x}\)
\(\frac{x}{y} = \frac{0.4x + 0.36x + 0.6y}{0.4y + 0.24x}\)
\(0.4xy + 0.24x^{2} = 0.4xy + 0.36xy + 0.6y^{2}\)
\(0.36xy + 0.6y^{2} = 0.24x^{2}\)
\(0.36 + 0.6(\frac{y}{x}) - 0.24(\frac{x}{y}) = 0\)
Let \(\frac{x}{y} = k\), then \(0.36 + \frac{0.6}{k} - 24k = 0\)
\(60 + 36k - 24k^{2} = 0\)
\(24k^{2} - 36k - 60 = 0\)
\(2k^{2} - 3k - 5 = 0\)
\(2k^{2} - 5k + 2k - 5 = 0\)
\(k(2k - 5) + 1 (2k - 5) = 0\)
\((k+1) (2k - 5)\)
Ignoring the negative factor, we get \(k = \frac{5}{2}\). So, Ans - E.
I don't think this is a GMAT problem, or may be I am using a lengthy method to solve this.
25 Jun 2017, 10:31
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theperfectgentleman
Hi..
I.. LOGICALLY
A very straight forward method would be that ..
If after getting 60% from A, B becomes A:-
\(B+0.6A=A......B=0.4A.....10B=4A....\frac{A}{B}=\frac{10}{4}=\frac{5}{2}\)
Thus E
II..ALGEBRA
Let the ratio A:B means in total number :- 100:x..
Now after A gives 60% the number becomes, 40:(x+60)..
Now the ratio again becomes the same and since the total number of marbles is same..
The number after B gives his 60% should make 100 with A..
So 40+(x+60)*60/100=100..... (x+60)*60/100=100-40=60...
So (x+60)/100=1....X+60=100....X=40..
So ratio is 100:40=5:2
So E none of these
III... CHOICES and logical
after the first transaction A:B should become B:A then only it will return back to A:B as the transfer is 60% each time..
Now look at choices to find answer
1) 2:1..
Means he gives (2-1) or 1 and % becomes 1/2 =50%..NO
2)3:2..
%= (3-2)/3=33.33%
3) 5:3
%=(5-3)/5=40%
4)4:3..
(4-3)/4=25%
So none of them is 60%
E
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
