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by theachiever » Mon Dec 10, 2012 12:44 am
If x is not equal to zero, is x^2+1/x>y?

1.x=y
2.y>0
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Source: — Data Sufficiency |
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by savvysatyam » Mon Dec 10, 2012 1:59 am
answer should be determined from any of the statements
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by Brent@GMATPrepNow » Mon Dec 10, 2012 7:03 am
theachiever wrote:If x is not equal to zero, is x^2+1/x>y?

1.x=y
2.y>0
From your notation, it's difficult to be certain what the actual expression is.
x^2+1/x>y could mean x^2 + (1/x) > y, or it could mean (x^2+1)/x > y

Some brackets would help to remove any doubt before we proceed with a solution.

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by theachiever » Mon Dec 10, 2012 7:59 pm
Hello Brent the expression is

(x^2+1)/x > y
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by puneetkhurana2000 » Mon Dec 10, 2012 8:45 pm
Rephrasing the equation as:- Is(x + 1/x) > y ?

From statement 1: x = y
Plugging some numbers as

1) y = 2

2 + 1/2 > 2 .... Yes

2) y = -2

-2 + (-1/2) > -2 ---> -5/2 > -2 .... No

Statement 1 Not Sufficient

From statement 2: y > 0
Plugging some numbers as

1) y = 100 , x = 50

50 + 1/50 > 100 .... No

2) y = 2 , x = 4

4 + 1/4 > 2 .... Yes

Statement 2 Not Sufficient

Together when x=y and y>0, (x + 1/x) will always be greater than x by 1/x and x is positive, so together the statements are sufficient. Answer C

Hope this helps!!!
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by The Iceman » Tue Dec 11, 2012 4:13 am
savvysatyam wrote:answer should be determined from any of the statements
Satyam, think again buddy.

Our problem is x + 1/x > y?

I=> 1/x > 0? We cannot conclude as we do not know the parity of x, whether x is +ve or -ve
II=> x + 1/x > 0? We cannot conclude as we do not know the parity of x, whether x is +ve or -ve
I+II=>x > 0 => Indeed x + 1/x is greater than y

Hence, C
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