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If m>0, n>0, is (m+x)/(n+x) > m/n?

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If m>0, n>0, is (m+x)/(n+x) > m/n? [#permalink] New post 25 Aug 2008, 02:41
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hi,
Following is OG-11 DS, Que#143:
Q:If m>0, n>0, is (m+x)/(n+x) > m/n?

I : m<n
II : x>0
--------END OF QUESTION STEM----------
If we solve the above question stem , the last two steps are as follows:
nx > mx.......(a)
n > m........(b) [This means the comparision of n and m is independent of x]

so the question stem essentially boils down to following:
IS n > m?


Now If we consider statement labeled 'b' as deduced question stem, then A should be the answer.(As Statement I answers the question stem.)
But, If we consider statement labeled 'a' as deduced question stem, then C should be the answer.(As both statement I and II are required.

What you guys think.[am i missing something here] :)

Last edited by viveksharma on 25 Aug 2008, 03:31, edited 1 time in total.
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Re: DS : OG11-Q143 [#permalink] New post 25 Aug 2008, 03:09
viveksharma wrote:
hi,
Following is OG-11 DS, Que#143:
Q:If m>0, n>0, is (m+x)/(n+x) > m/n?

I : m<n
II : x>0
--------END OF QUESTION STEM----------
If we solve the above question stem , the last two steps are as follows:
nx > mx.......(a)
n > m........(b) [This means the comparision of n and m is independent of x]

so the question stem essentially boils down to following:
IS n > m?


Now If we consider statement labeled 'b' as deduced question stem, then A should be the answer.(As Statement I answers the question stem.)
But, If we consider statement labeled 'a' as deduced question stem, then C should be the answer.(As both statement I and II is required.

What you guys think.[am i missing something here] :)


I don't know why u r considering nx and mx here. The question is concerned with n+x and m+x.
Having said that. you need both statements to determine whether (m+x)/(n+x) > m/n.

For proper fractions, increasing numerator and denominator by same amount will give u a greater proper fraction.
3/4 < 4/5<5/6. And, if you reduce both numerator and denominator by same amount, you get a smaller proper fraction.

So you need statement 1 to determine whether m/n is a proper fraction. You need statement 2 to determine if you are increasing or decreasing the numerator and denominator.
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Re: DS : OG11-Q143 [#permalink] New post 25 Aug 2008, 03:27
bhushangiri wrote:
I don't know why u r considering nx and mx here. The question is concerned with n+x and m+x.


(m+x)/(n+x) > m/n
= mn+nx > mn+mx
= nx>mx.....(I marked this 'a' above)
= n>m......(I marked this 'b' above)

I understood whats ur logic is.but can u plz explain me the flaw in above reasoning
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Re: DS : OG11-Q143 [#permalink] New post 25 Aug 2008, 03:35
viveksharma wrote:
bhushangiri wrote:
I don't know why u r considering nx and mx here. The question is concerned with n+x and m+x.


(m+x)/(n+x) > m/n
= mn+nx > mn+mx
= nx>mx.....(I marked this 'a' above)
= n>m......(I marked this 'b' above)

I understood whats ur logic is.but can u plz explain me the flaw in above reasoning

OK i got that. to reach n>m from nx>mx, u are dividing both side by x.But since we dont know whether x is -ve or +ve,
we can't say for sure that the next step is n>m.it could be n>m(if x>0) or n<m(if x<0).So second statemnet x>0, is required.Same line of reasoning, as urs
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Re: DS : OG11-Q143 [#permalink] New post 25 Aug 2008, 04:24
viveksharma wrote:
viveksharma wrote:
bhushangiri wrote:
I don't know why u r considering nx and mx here. The question is concerned with n+x and m+x.


(m+x)/(n+x) > m/n
= mn+nx > mn+mx
= nx>mx.....(I marked this 'a' above)
= n>m......(I marked this 'b' above)

I understood whats ur logic is.but can u plz explain me the flaw in above reasoning

OK i got that. to reach n>m from nx>mx, u are dividing both side by x.But since we dont know whether x is -ve or +ve,
we can't say for sure that the next step is n>m.it could be n>m(if x>0) or n<m(if x<0).So second statemnet x>0, is required.Same line of reasoning, as urs


Precisely. When ever there is an inequality involved, you need to be careful in cross multiplying and dividing both sides. +/- change things completely
Re: DS : OG11-Q143   [#permalink] 25 Aug 2008, 04:24
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