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If m > 0 and n > 0, is \frac{m+x}{n+x} >

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Manager
Joined: 09 Jul 2008
Posts: 111
Location: Dallas, TX
Schools: McCombs 2011
If m > 0 and n > 0, is \frac{m+x}{n+x} > [#permalink]

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16 Nov 2008, 09:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If m > 0 and n > 0, is $$\frac{m+x}{n+x} > \frac{m}{n}$$ ?

(1) m < n
(2) x > 0

How should one approach this type of problem. For two minutes, I tried to plug in numbers and got confused. The OG solves in a different way, but my problem is the initial approach. Please help..
SVP
Joined: 17 Jun 2008
Posts: 1529
Re: Inequality , how to approach [#permalink]

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16 Nov 2008, 10:19
I think, it should be C.

In a fraction m/n where m < n, any positive numebr that is added to both numerator and denominator, will make the new fraction bigger.
VP
Joined: 05 Jul 2008
Posts: 1401
Re: Inequality , how to approach [#permalink]

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16 Nov 2008, 13:49
agree with C.

The principle is the value of a fraction approaches 1 as the numerator and denominator increase by the same amount. We need to know that m < n and x is +ve
SVP
Joined: 29 Aug 2007
Posts: 2467
Re: Inequality , how to approach [#permalink]

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16 Nov 2008, 13:56
kman wrote:
If m > 0 and n > 0, is $$\frac{m+x}{n+x} > \frac{m}{n}$$ ?

(1) m < n
(2) x > 0

How should one approach this type of problem. For two minutes, I tried to plug in numbers and got confused. The OG solves in a different way, but my problem is the initial approach. Please help..

$$\frac{m+x}{n+x} > \frac{m}{n}$$?
$$\frac{m+x}{n+x} - \frac{m}{n}$$ > 0?
$$x(n-m)$$ > 0?

(1) m < n: do not know about x.
(2) x > 0. do not know about (n-m).

togather x is +ve, (n-m) is +ve. so $$x(n-m)$$ > 0 or $$\frac{m+x}{n+x} > \frac{m}{n}$$.
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Manager
Joined: 09 Jul 2008
Posts: 111
Location: Dallas, TX
Schools: McCombs 2011
Re: Inequality , how to approach [#permalink]

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16 Nov 2008, 14:29
Thanks, I'll have to remember the rule that when a +ve number is added to a fraction, it becomes larger

Tiger: I don't understand your method. Is it a sigma ??? How?
VP
Joined: 05 Jul 2008
Posts: 1401
Re: Inequality , how to approach [#permalink]

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16 Nov 2008, 15:28
kman wrote:
Thanks, I'll have to remember the rule that when a +ve number is added to a fraction, it becomes larger

Tiger: I don't understand your method. Is it a sigma ??? How?

Here is what I think Tiger was getting at

\frac{m+x}{n+x} > \frac{m}{n}?
\frac{m+x}{n+x} - \frac{m}{n} > 0?

means (m+x) n - (n+x)m > 0

mn + nx -mn -mx > 0

x(n-m) > 0

So we have to know either both x and (n-m) are +ve or -ve

(1) and (2) taken together (n-m) and x are +ve and hence C
Re: Inequality , how to approach   [#permalink] 16 Nov 2008, 15:28
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If m > 0 and n > 0, is \frac{m+x}{n+x} >

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