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x-y > 1?

Expert replies
Source: — Data Sufficiency |

by cramya » Sun Dec 21, 2008 12:14 pm
I will go wiht E right or wrong :-)

I initially though OF B) where I could see its telling x is positive(hope thats right) but I am not convinced on what sqrt(y) ^ 2 = y is telling (positive or negative)
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by cramya » Sun Dec 21, 2008 12:31 pm
This is my reasoning: ( still a little doubtful on stmt II)

Q: Is x> y?

Stmt I

x/y < 1/2

x = 2 y = 6
x= 4 y = -2

INSUFF

Stmt II

|x| = x whis means x is positive

sqrt(y)^2 = y could mean y is potive or negative (Brent, this is part I was a little doubtful I.E its interpretation)

INSUFF

Taken together


x = 2 y = 6 x<y
x= 4 y = -2 x>y

E)
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by Brent@GMATPrepNow » Sun Dec 21, 2008 1:08 pm
The answer is C

(2) tells us that x and y are both positive
If we apply this result to (1), we see that both x and y are positive AND y must be greater than x in order for x/y < /1/2

If x<y, then x-y will always be a negative number, and negative numbers are always less than 1 (the original question asks "is x-y >1" and the answer is no)
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by cramya » Sun Dec 21, 2008 1:35 pm
Thats a good one. I went with C) then changed to E) as I was not entriely sure on the sqrt(y)^2 part.


Could u explain the sqrt(y)^2 part some more? Have u just written it in a different way and this still means |y| = y which makes y positive?
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by Brent@GMATPrepNow » Sun Dec 21, 2008 1:52 pm
Sure thing.

If y were negative, then we wouldn't be able to find sqrt(y).
For the purposes of the GMAT, we can't find the sqrt of a negative number. So, in this case, we know that y must be positive.

Hmm, now that I have written my explanation, I don't like my original question. As it is currently written, it leaves room for some ambiguity.
While the GMAT tests only our knowledge of real numbers, the test writers are often careful to avoid contradicting other areas of math. At an early age, we learn that we can't find the square root of a negative number. Then, as we progress through mathematics, we later learn about the realm of complex (imaginary) numbers, where the sqrt(-1) = i and i^2=-1 and when we allow sqrt(y) to be a compex number then y^2 can be a negative number.

My current question allows for sqrt(y) to be an imaginary number, in which case y (and y^2) can be positive or negative. In which case the answer is E.

Had I added the proviso "x and y are real numbers" then the answer would be C, since we sqrt(negative number) is undefined and we can't take an undefined number and square it and get some real number.

I hope that helps.
Last edited by Brent@GMATPrepNow on Sun Dec 21, 2008 2:43 pm, edited 1 time in total.
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by cramya » Sun Dec 21, 2008 2:24 pm
Thanks Brent; the explanation was helpful!
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by ronniecoleman » Sun Dec 21, 2008 11:08 pm
should be C

1.. X/Y < 1/2

So any day x has to be greater than Y ..

but x-y >1 would not be true for all +ve values.

negative

2/-10 = - .10 2- (-10) = 12 > 1

now we have yes and no..

so A out

2. |x| = x
y = +ve

Club 2 &1 together and we see that x -y < 1 always
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by vittalgmat » Mon Dec 22, 2008 2:37 am
Thanks Brent,
the explanation was very helpful .
I was confused with stmt 2. Thanks for clearing this.
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Question not Viewable

by CrackGMAC » Wed Dec 24, 2008 7:56 am
Can't see question again :?:
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