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# M17-19

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Math Expert
Joined: 02 Sep 2009
Posts: 50626

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16 Sep 2014, 00:01
00:00

Difficulty:

5% (low)

Question Stats:

78% (00:47) correct 22% (00:48) wrong based on 293 sessions

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Is $$x \gt y$$ ?

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

(2) $$x \gt 0$$

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Math Expert
Joined: 02 Sep 2009
Posts: 50626

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16 Sep 2014, 00:01
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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Joined: 18 Mar 2015
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14 Aug 2016, 09:24
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
Joined: 02 Sep 2009
Posts: 50626

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15 Aug 2016, 02:42
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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GMAT 2: 700 Q48 V38
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05 May 2017, 01:47
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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Joined: 17 Mar 2017
Posts: 3
Location: United States
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05 May 2017, 11:43
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
Joined: 02 Sep 2009
Posts: 50626

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05 May 2017, 11:51
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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10 Aug 2017, 03:22
I missed the case in which x= + 1 and -1 both....silly one...
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17 Sep 2017, 03:44
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
Joined: 02 Sep 2009
Posts: 50626

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17 Sep 2017, 03:48
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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09 Apr 2018, 23:13
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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10 Jul 2018, 06:50
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
Joined: 05 Jul 2017
Posts: 489
Location: India
GMAT 1: 700 Q49 V36
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10 Jul 2018, 07:29
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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Location: India
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10 Jul 2018, 08:09
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 &nbs [#permalink] 10 Jul 2018, 08:09
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# M17-19

Moderators: chetan2u, Bunuel

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