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

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Re: M versus N [#permalink]

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New post 02 Aug 2011, 09:29
You're more than welcome, 144144. It's my job to help/teach GMAT aspirants in the real world. I am very happy to extend it to this forum now. Thank you for your appreciation!
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Re: M versus N [#permalink]

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New post 13 Aug 2011, 08:20
Is this really a 700-level problem?

When you say GMAT prep are you referring to the two free tests available?
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Re: M versus N [#permalink]

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New post 13 Aug 2011, 11:57
well i prefered pcking numbers:

S1- M<N M=2 N=4 X=2 4/6>2/4 YES BUT M=2 N=4 X=-1 1/3>1/2 NO.
S2 M=N NO M>N NO M<N YES

FROM S2 YOU ALLREADY KNOW THAT THE INEQUALITY HOLDS TRUE ONLY FOR M<N SO
S1-S2 C0MBINES WHEN X>0 AND M<N
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Re: M versus N [#permalink]

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New post 18 Aug 2011, 07:47
Too easy, don't even need a pen for this one.

Basically, when you add a +'ve constant to numerator and denominator of a +'ve fraction is gets larger, when you add a +'ve constant to numerator and denominator of a +'ve important fraction it gets smaller.

(1) m<n so m/n is a fraction, but we don't know if the constant is +'ve or negative so we don't know what adding it will do.

(2) X>0 so we know the constant is +'ve but we dont know if the fraction is proper or improper.

(1)+(2) we have a fraction and we add a +'ve constant to the numerator and denominator, therefore it gets bigger. suff.
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Re: M versus N [#permalink]

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New post 18 Aug 2011, 10:39
Well, for proper fraction (when numerator is less than denominator), if we add any same constant number both in num & den, new fraction is greater than the original fraction. But if same number is subtracted, new faction is smaller than the original one.
whereas in improper fractions (when numerator is greater than denominator), above conditions becomes exactly opposite.

So, to answer this ques, we need to know whether m>n and sign of x. Both these conditions are supplied by combining both the statements.
Re: M versus N   [#permalink] 18 Aug 2011, 10:39

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

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