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# If m>0, n>0, is (m+x)/(n+x)>m/n? 1. m<n 2.

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Senior Manager
Joined: 02 Feb 2004
Posts: 344
If m>0, n>0, is (m+x)/(n+x)>m/n? 1. m<n 2. [#permalink]

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26 May 2004, 18:27
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If m>0, n>0, is (m+x)/(n+x)>m/n?

1. m<n
2. x>0
Senior Manager
Joined: 07 Oct 2003
Posts: 350
Location: Manhattan

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26 May 2004, 19:23
my vote goes to 'C'
a doesn't work since x could be negative
b doesn't work since we don't know whether 'n' or 'm' is largest
c combines the two and we have a solution

any thoughts?
SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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26 May 2004, 20:15
A) m = 0.1 n = 0.2 m/n = 0.5 if x = 1000 then (m+x)/(n+x) = 1
so (m+x)/(n+x) > m/n
but if x = -0.1 then (m+x)/(n+x) < m/n
(Insufficient)

B) if m = n then (m+x)/(n+x) = m/n
but if m = 0.1 n = 0.2 x = 0.1 then (m+x)/(n+x) > m/n
(insufficient)

if you combine you can tell (m+x)/(n+x) > m/n

I think the answer is C.
Senior Manager
Joined: 02 Mar 2004
Posts: 327
Location: There

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26 May 2004, 21:56
Substract 1 from both sides

(m-n)/(n+x) > (m-n)/n

let g = (m-n)/(n+x) -(m-n)/n = (n-m)x/[n(n+x)]

1 and 2 each alone are not sufficient.
combined, g > 0

C it is
Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US

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27 May 2004, 05:37
C here too.

To evaluate the equation, we need to know if m>n OR if m< n. We also need to know whether x is +ve or -ve.
Senior Manager
Joined: 02 Feb 2004
Posts: 344

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27 May 2004, 06:57
If m>0, n>0, is (m+x)/(n+x)>m/n?

1. m<n
2. x>0

(m+x)/(n+x)>m/n
n(m+x)>m(n+x)
mn+nx>mn+mx
nx>mx

We can not divide both by x, because x can be negative resuting in changes of inequalities.

nx>mx this would be true if m>0, n>0, n>m & x>0

Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)

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29 May 2004, 03:45
mirhaque wrote:
If m>0, n>0, is (m+x)/(n+x)>m/n?

1. m<n
2. x>0

1 is not sufficient: x = -1, m = 1, n = 2.

2 is not sufficient: x = 1000, m = 2, n = 1.

1 and 2 are sufficient: (m+x)/(n+x)>m/n <=> (m + x)n > (n + x)m <=> n > m (since x > 0, n + x > 0, m + x > 0, n, m > 0).
Re: DS: Inequality challange-3   [#permalink] 29 May 2004, 03:45
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