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M17-19

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M17-19  [#permalink]

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New post 16 Sep 2014, 01:01
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

78% (00:47) correct 22% (00:44) wrong based on 256 sessions

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New post 16 Sep 2014, 01:01
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Re: M17-19  [#permalink]

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New post 14 Aug 2016, 10:24
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

so answer is E
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New post 15 Aug 2016, 03:42
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Re: M17-19  [#permalink]

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New post 05 May 2017, 02:47
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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Re: M17-19  [#permalink]

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New post 05 May 2017, 12:43
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?
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Re: M17-19  [#permalink]

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New post 05 May 2017, 12:51
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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Re: M17-19  [#permalink]

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New post 10 Aug 2017, 04:22
I missed the case in which x= + 1 and -1 both....silly one...
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Re: M17-19  [#permalink]

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New post 17 Sep 2017, 04:44
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

so answer is E


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
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New post 17 Sep 2017, 04:48
amargad0391 wrote:
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

so answer is E


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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Re: M17-19  [#permalink]

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New post 10 Apr 2018, 00:13
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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Re: M17-19  [#permalink]

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New post 10 Jul 2018, 07:50
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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Re: M17-19  [#permalink]

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New post 10 Jul 2018, 08:29
1
adkikani wrote:
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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Re: M17-19  [#permalink]

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New post 10 Jul 2018, 09:09
1
adkikani wrote:
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?


Hi adkikani

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, 09:09
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