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gkumar
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(x-y) / (x+y) < 1

by gkumar Wed Oct 21, 2009 10:30 am

I couldn't find this problem in the search results. I am confused by the below GPrep question:
If x is not equal to y, is (x-y)/ (x+y) < 1?
1) x>0
2) y<0

I rephrased the question to be the following:
Case 1: x+y>0
[editor: where did you get x + y > 0? the statement is just x > 0.]

Cross multiply to get: x-y < x+y => 0 < y
[editor: you can't "cross multiply" here, because you don't know the sign of the denominator.]

Case 2: x-y<0
[editor: where did you get x - y > 0? the statement is just y < 0.]

Cross multiply and switch inequality sign: x-y > x+y => 0 > y
[editor: you can't "cross multiply" here, because you don't know the sign of the denominator.]

So the rephrased question is asking: "Is y>0 or y<0?"
[editor: you have arrived at two completely contradictory questions. if this happens, you know that whatever you are doing is wrong.]

Looking at (1), X is not involved in the rephrase that contains only y, so it is irrelevant and hence insufficient.

Looking at (2), y<0 is given and answers the rephrased question of "is y<0" and the answer is yes. So I assumed B.

But the official answer is E. I realize that I could plug in numbers to get this answer, but isn't that time consuming? Why did my approach not work?
2amitprakash
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Re: (x-y) / (x+y) < 1

by 2amitprakash Wed Oct 21, 2009 5:47 pm

your rephrasing is not complete. You have assumed one condition and then simplified. So you should not forgot about the first condition.
Case 1: what about x+y>0? Under this assumption only you get y>0.

Hence in a way to tell the result or inequality in question, we need the values for x and y both. In both the options only one value is provided and hence none of the options are sufficient to answer it.
gkumar
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Re: (x-y) / (x+y) < 1

by gkumar Wed Oct 21, 2009 5:52 pm

2amitprakash, can you clarify on the incomplete rephrase? I am unclear on why Case 1 should be considered when statement (2) provides Case (2).

Does the question "Is y<0 or y>0?" require both of its cases to be true for a statement to be considered sufficient? If that's the case, then statement (2) explains only 50% or 1 of the 2 cases, and hence is insufficient. Otherwise, I am unclear.
2amitprakash
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Re: (x-y) / (x+y) < 1

by 2amitprakash Wed Oct 21, 2009 6:14 pm

Let me try to explain my way of thinking and hope it will help:
when you rephrased the question, you got 2 possible situations with original question. so if any of the provided options or a combination of options eliminates one of the possible solutions, you solved the problem.
Now in this case, you reached the simplified version y>0 or y<0 by assuming that (x+y) is < 0 or (x+y) >0. Basically you took care of the situation if the multiplication factor is -ve then you need to reverse the inequality. However, never forgot about the original assumption.
So your simplification is x+y>0 and y>0 or x+y<0 and y<0. The given options are providing information about either x or y and hence none of them are sufficient alone. Even when you combine the options, you don't get a definite answer (check the 2nd situation as with x>0 and y<0, you cannot definitely say x+y<0). If the 1st option would have been x<0 then answer will be 'C'.
RonPurewal
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Re: (x-y) / (x+y) < 1

by RonPurewal Sat Oct 24, 2009 8:30 am

there are some problems in the original post; see the editor's comments inline.

very important takeaway:
there is no such thing as "cross multiply".
what you know as "cross multiply" is actually MULTIPLICATION BY ALL DENOMINATORS.


therefore, we can't multiply this inequality unless we know the sign of (x + y).
which we don't.
even if we have both statements.

see here for more explanations related to this problem:
ds-if-x-not-equal-to-y-t7761.html