need help with method...

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need help with method...

by shipra » Thu Jun 26, 2008 10:15 pm
If x is not = y, is (x-y)/(x+y) > 1?

1. x>0
2. y<0

can sum1 pls help me with the method to do such questions...please?

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Re: need help with method...

by atlantic » Thu Jun 26, 2008 11:18 pm
[quote="shipra"]If x is not = y, is (x-y)/(x+y) > 1?

1. x>0
2. y<0

can sum1 pls help me with the method to do such questions...please?[/quote]

(x-y)/(x+y)>1 <-> x-y>x+y <-> -2y<0 <-> y>0 so (2) is SUFF

Answer B

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by floravaze » Fri Jun 27, 2008 4:28 am
i seriously doubt that the answer is B.I'd rather go with E.
Statement 1 is insufficient because it tells you nothing about y and statement 2 is also insufficient for the same reason that gives no infomation about x.Now,the two statements taken together are also insufficient coz the relative magnitudes are unknown.take values x=2,y=-1 and x-10,y=-10.The two give you 3 and o respectively.Therefore answer is 3.Hope this helps.
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by beeparoo » Fri Jun 27, 2008 7:48 am
floravaze wrote:i seriously doubt that the answer is B.I'd rather go with E.
Statement 1 is insufficient because it tells you nothing about y and statement 2 is also insufficient for the same reason that gives no infomation about x.Now,the two statements taken together are also insufficient coz the relative magnitudes are unknown.take values x=2,y=-1 and x-10,y=-10.The two give you 3 and o respectively.Therefore answer is 3.Hope this helps.
I agree with floravaze that the ans should be E.

Both statements together are INSUFF "coz the relative magnitudes are unknown".

But floravaze, your example, in which
x = 10 (I assume is what you meant to type)
y = -10
When plugged into the eq'n is (10 + 10)/(10 - 10)
This does NOT equal ZERO. It is UNDEFINED.

A better second example (to prove insufficiency) would be:
x = 1
y = -3
Then (1 + 3)/(1 - 3) = 4/(-2) = -2, which is NOT >1.

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by floravaze » Sat Jun 28, 2008 2:35 am
Thanks Beeparo
U're right about my example.it is undefinable and not zero.I really do make these stupid mistakes at times.
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