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Re: Arithmetic operations and inequalities - DS [#permalink]
24 Aug 2007, 11:31

above720 wrote:

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

(1) m is less than n

(2) x is greater than 0

I get E.

Break down the equation:
(m+x) / (n+x) > m/n

if x<-n, then (n+x) is negative, you have
n*(m+x) < m*(n+x)
nx < mx and you know that x is negative, you have
n > m

if x>-n, then (n+x) is positive, you have
n*(m+x) > m*(n+x)
nx > mx
This case, you can break down further in to two equations:
if x<0, then you have n<m
if x>0, then you have n>m

In sum, you have three cases:
Case 1: if x<-n, n>m
Case 2: if -n<x<0, n<m
Case 3: if x>0, n>m

(1) Given n>m, the equation is true only for Case 1 and 3; thus, insufficient
(2) Given x>0, this is only true for Case 3; thus. insufficient.

Together, still only true for Case 1 and 3. insufficient.

Re: Arithmetic operations and inequalities - DS [#permalink]
24 Aug 2007, 11:44

lets take an example X=1/2 M=1/2 and N=5/2

then M+x=1 N+x=5/2

=2/5>1/5

take x=2 m=1 N=2

3/4>1/2

we need C here...Or am I missing something???

bkk145 wrote:

above720 wrote:

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

(1) m is less than n

(2) x is greater than 0

I get E.

Break down the equation: (m+x) / (n+x) > m/n

if x<-n, then (n+x) is negative, you have n*(m+x) < m*(n+x) nx <mx> m

if x>-n, then (n+x) is positive, you have n*(m+x) > m*(n+x) nx > mx This case, you can break down further in to two equations: if x<0, then you have n<m>0, then you have n>m

In sum, you have three cases: Case 1: if x<n>m Case 2: if -n<x<0, n<m>0, n>m

(1) Given n>m, the equation is true only for Case 1 and 3; thus, insufficient (2) Given x>0, this is only true for Case 3; thus. insufficient.

Together, still only true for Case 1 and 3. insufficient.

Re: Arithmetic operations and inequalities - DS [#permalink]
24 Aug 2007, 12:06

fresinha12 wrote:

lets take an example X=1/2 M=1/2 and N=5/2

then M+x=1 N+x=5/2

=2/5>1/5

take x=2 m=1 N=2

3/4>1/2

we need C here...Or am I missing something???

bkk145 wrote:

above720 wrote:

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

(1) m is less than n

(2) x is greater than 0

I get E.

Break down the equation: (m+x) / (n+x) > m/n

if x<-n, then (n+x) is negative, you have n*(m+x) < m*(n+x) nx <mx> m

if x>-n, then (n+x) is positive, you have n*(m+x) > m*(n+x) nx > mx This case, you can break down further in to two equations: if x<0, then you have n<m>0, then you have n>m

In sum, you have three cases: Case 1: if x<n>m Case 2: if -n<x<0, n<m>0, n>m

(1) Given n>m, the equation is true only for Case 1 and 3; thus, insufficient (2) Given x>0, this is only true for Case 3; thus. insufficient.

Together, still only true for Case 1 and 3. insufficient.

I think you guys are right. I might have confused about my cases:
I think it's C.