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If m>0 and n>0, is (m+x)/(n+x) > m/n? (1) m is less

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Manager
Joined: 05 May 2005
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If m>0 and n>0, is (m+x)/(n+x) > m/n? (1) m is less [#permalink]  23 Aug 2007, 17:44
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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
Director
Joined: 03 May 2007
Posts: 888
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 73 [0], given: 7

Re: Arithmetic operations and inequalities - DS [#permalink]  24 Aug 2007, 10:57
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

E.
1: if x is +ve, then (m+x)/(n+x) > m/n. if x is -ve, then (m+x)/(n+x) < m/n.
2. depends which one is grater m or n.

1 and 2: if m and n are in fraction and x > 1, we do not know. so
its E.
VP
Joined: 10 Jun 2007
Posts: 1464
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Kudos [?]: 142 [0], given: 0

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.
Current Student
Joined: 28 Dec 2004
Posts: 3387
Location: New York City
Schools: Wharton'11 HBS'12
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Kudos [?]: 186 [0], given: 2

I get C...

1) M <N>0

2) x is >0 ok but dont know anything about M and N

together sufficient
Current Student
Joined: 28 Dec 2004
Posts: 3387
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 14

Kudos [?]: 186 [0], given: 2

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.
VP
Joined: 10 Jun 2007
Posts: 1464
Followers: 6

Kudos [?]: 142 [0], given: 0

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.
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 6

Kudos [?]: 136 [0], given: 0

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

n*(m+x) > m*(n+x)

n*m+x*n > m*n+x*m

x*n > x*m

n > m

Note - this will be true ONLY when n+x > 0 - otherwise n < m

statement 1

m is less than n.

meaning n+x > 0

we can conclude that the stem is correct.

sufficient

statement 2

we can determine that n+x > 0 is true since x > 0 (from statement) and n > 0 (from stem) but we have no way to know if n > m.

insufficient

Current Student
Joined: 28 Dec 2004
Posts: 3387
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 14

Kudos [?]: 186 [0], given: 2

you cant do the bold..cause you dont know if X is positive or NOT.

KillerSquirrel wrote:
(m+x)/(n+x) > m/n

n*(m+x) > m*(n+x)

n*m+x*n > m*n+x*m

x*n > x*m

n > m

Note - this will be true ONLY when n+x > 0 - otherwise n <m> 0

we can conclude that the stem is correct.

sufficient

statement 2

we can determine that n+x > 0 is true since x > 0 (from statement) and n > 0 (from stem) but we have no way to know if n > m.

insufficient

VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 6

Kudos [?]: 136 [0], given: 0

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