If x is not equal to zero, is x^2+1/x>y?
1.x=y
2.y>0
1.x=y
2.y>0
Resilience is a trait that manifests in an individual when he/she faces consistent failures motivating them to succeed in the end.
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Source: Beat The GMAT — Data Sufficiency |
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Brent@GMATPrepNow
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From your notation, it's difficult to be certain what the actual expression is.theachiever wrote:If x is not equal to zero, is x^2+1/x>y?
1.x=y
2.y>0
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.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Hello Brent the expression is
(x^2+1)/x > y
(x^2+1)/x > y
Resilience is a trait that manifests in an individual when he/she faces consistent failures motivating them to succeed in the end.
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!!!
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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The Iceman
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Satyam, think again buddy.savvysatyam wrote:answer should be determined from any of the statements
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
