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Expert replies
by sykedaddy » Tue Jul 24, 2007 4:22 am
If x does NOT equal y, is (x-y)/(x+y) > 1?

(1) x>0
(2) y<0

Thanks..OA shortly
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Source: — Data Sufficiency |

by chatekar » Tue Jul 24, 2007 5:28 am
From statement [1]

x > 0, now y can be positive or negative,

assume y is positive, then (x-y) < (x+y) ==> (x-y)/(x+y) < 1
assume y is negative, then (x-y) > (x+y) ==> (x-y)/(x+y) > 1

Hence statement [1] alone is not sufficient.


From statement[2]

Following the above logic for y, we get the 2 different results and hence statement [2] alone is not sufficient.

Considering statement [1] and [2] together, we get a unique answer, so statement [1] and [2] together are sufficient to solve the problem

Whats the OA?

Thanks
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by sykedaddy » Tue Jul 24, 2007 6:06 am
I had the same thought process but the OA is E...not sure why
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by givemeanid » Tue Jul 24, 2007 6:17 am
(x-y)/(x+y) > 1?

(1) x>0
(2) y<0

x = 1/4
y = -1/2
x-y = 1/4 - (-1/2) = 3/4
x+y = 1/4 - 1/2 = -1/2
(x-y)/(x+y) = 3/4 / -1/2 = -3/2 (<1)

x = 4
y = -2
x-y = 6
x+y = 2
6/2 = 3 (>1)

NOT SUFFICIENT. Hence E.
So It Goes
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by bingojohn » Tue Aug 07, 2007 10:21 am
givemeanid wrote:(x-y)/(x+y) > 1?

(1) x>0
(2) y<0

x = 1/4
y = -1/2
x-y = 1/4 - (-1/2) = 3/4
x+y = 1/4 - 1/2 = -1/2
(x-y)/(x+y) = 3/4 / -1/2 = -3/2 (<1)

x = 4
y = -2
x-y = 6
x+y = 2
6/2 = 3 (>1)

NOT SUFFICIENT. Hence E.
I don't understand, the answer should be (B). Here is my thinking:

x-y/x+y > 1 ?
... x-y > x+y ?
... x-x > y+y ?
... 2y < 0 ?
... y < 0 ? must be answered, for the original question to be answered.

Choice (1) doesn't answer this, (2) does. Hence (B). \

Correct me if I am wrong, please.
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by givemeanid » Tue Aug 07, 2007 10:59 am
x-y/x+y > 1 ?
... x-y > x+y ?
If m/n > 1, it does not mean m > n. You DO NOT know whether n is +ve or -ve.

Remember, if you multiply (or divide) both sides of the inequality by a negative number, the sign of inequality reverses.

In this case. you do not know whether x+y is +ve or -ve. So, you cannot multiply both sides by (x+y).
So It Goes
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by bingojohn » Tue Aug 07, 2007 11:15 am
givemeanid wrote:
x-y/x+y > 1 ?
... x-y > x+y ?
If m/n > 1, it does not mean m > n. You DO NOT know whether n is +ve or -ve.

Remember, if you multiply (or divide) both sides of the inequality by a negative number, the sign of inequality reverses.

In this case. you do not know whether x+y is +ve or -ve. So, you cannot multiply both sides by (x+y).
Oops, my bad... I guess I better brush up my inequalities...

Thanks a lot givemeanid...
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by sochatte » Tue Aug 07, 2007 11:22 am
How about this?

x-y/x+y > 1 = x-y/x+y -1 > 0 = -2y/x+y > 0
so the question would be
is -2y/x+y > 0

1. x > 0
we don't know anything about y so NS

2. y < 0
we don't know anything about x , so NS

1 & 2 . if y is -ve, -2y will be +ve, however x+y can be +ve or -ve depending on the values of x and y. hence NS

Ans is E
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