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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   5% (low)

Question Stats: 78% (00:49) correct 22% (00:55) wrong based on 353 sessions

### HideShow timer Statistics Is $$x \gt y$$ ?

(1) $$\frac{x + y}{x - y} = 1$$

(2) $$x \gt 0$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 56304

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Official Solution:

Statements (1) and (2) combined are sufficient. From S1 it follows that $$x + y = x - y$$ or $$y = 0$$. Combining this information with $$x \gt 0$$ from S2, we conclude that $$x \gt y$$.

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Manager  B
Joined: 18 Mar 2015
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Schools: ISB '19
GMAT 1: 600 Q47 V26 GPA: 3.59

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can someone explain me more ?
my logic
x+y=x-y
y = 1/2

from statement 2, x>0 means x can 1/2 , 1 , 2

Math Expert V
Joined: 02 Sep 2009
Posts: 56304

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r19 wrote:
can someone explain me more ?
my logic
x+y=x-y
y = 1/2

from statement 2, x>0 means x can 1/2 , 1 , 2

x+y=x-y
x cancels out: y = -y
2y = 0
y = 0.
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Senior Manager  S
Joined: 08 Jun 2015
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Location: India
GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 GPA: 3.33

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The answer must be option C.

Statement 1) This gives x+y=x-y ,i.e. y=0. Not sufficient
Statement 2) x>0 . This is clearly not sufficient.

Combine the two statements y=0 and x>o .ie, x>y.

Hence the answer is option C.
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Intern  B
Joined: 17 Mar 2017
Posts: 3
Location: United States
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I don't get it? What if I come up with y=0, and I put it in the equation given, I will find that x = 1; thus x>y. That means, I can solve the question without using x>0. Can someone answer me?
Math Expert V
Joined: 02 Sep 2009
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2
linhmiu wrote:
I don't get it? What if I come up with y=0, and I put it in the equation given, I will find that x = 1; thus x>y. That means, I can solve the question without using x>0. Can someone answer me?

If you put y = 0 into (x + y)/(x - y) = 1 you'll get x/x = 1, which is true for any non-zero x, not only for x = 1. For example, try x = -1.

Hope it helps.
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I missed the case in which x= + 1 and -1 both....silly one...
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Intern  B
Joined: 09 Sep 2015
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Bunuel wrote:
r19 wrote:
can someone explain me more ?
my logic
x+y=x-y
y = 1/2

from statement 2, x>0 means x can 1/2 , 1 , 2

x+y=x-y
x cancels out: y = -y
2y = 0
y = 0.

Hi Bunuel

Cant we substitute y=0 back into the statement to find value of x ?

Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 56304

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Bunuel wrote:
r19 wrote:
can someone explain me more ?
my logic
x+y=x-y
y = 1/2

from statement 2, x>0 means x can 1/2 , 1 , 2

x+y=x-y
x cancels out: y = -y
2y = 0
y = 0.

Hi Bunuel

Cant we substitute y=0 back into the statement to find value of x ?

Thanks

Check this: https://gmatclub.com/forum/m17-184125.html#p1847599
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Manager  S
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As from a) x+y=x-y, y must be 0, as it doesn't make any difference in the equation.
from b) we get x>0, so the answer is C.
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IIMA, IIMC School Moderator V
Joined: 04 Sep 2016
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niks18 pikolo2510 pushpitkc

Quote:
x+y = x-y
Hence y =0

We could perform subtraction of x from both sides of equality even though
we do not know sign of x since we are dealing with equality and subtraction / addition.

We can not perform same step in case of inequality and division. Am I correct?
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BSchool Forum Moderator P
Joined: 05 Jul 2017
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1
niks18 pikolo2510 pushpitkc

Quote:
x+y = x-y
Hence y =0

We could perform subtraction of x from both sides of equality even though
we do not know sign of x since we are dealing with equality and subtraction / addition.

We can not perform same step in case of inequality and division. Am I correct?

Yes , you are right here.

P.S : - Just to be sure , I would like niks18 and pushpitkc to confirm the same as well
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Joined: 25 Feb 2013
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1
niks18 pikolo2510 pushpitkc

Quote:
x+y = x-y
Hence y =0

We could perform subtraction of x from both sides of equality even though
we do not know sign of x since we are dealing with equality and subtraction / addition.

We can not perform same step in case of inequality and division. Am I correct?

you can perform addition and subtraction with inequalities as well even if you don't know the value of the variable, provided the sign of inequality is SAME.

for e.g if x>y and a>b, then you can add them as x+a>y+b. but you cannot subtract the second inequality from the first because on subtracting the second inequality changes its sign as -a<-b.

but if you had x>y and d<e, then second inequality can be subtracted because in that case sign of both inequality will be same. x>y & -d>-e now add them to get
=>x-d>y-e

However you cannot multiply or divide the inequalities without knowing the value of the variable i.e. whether they are positive or negative. Re: M17-19   [#permalink] 10 Jul 2018, 09:09
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# M17-19

Moderators: chetan2u, Bunuel  