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Inequalities - GMATPrep
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AleksandrM
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AbhishekGakhar
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Answer should be A.
1) 2x-3y=-2= -ve. Since, x^2=+ve => 2x-3y <x^2 Sufficient
2) x>2; y>0 =>2x-3y>4 i.e. 2x-3y = 5,6,7....
Also, x>2 =>x^2>4 i.e. x^2 = 5,6,7....
Now if 2x-3y = 5 and x^2 =6 then Left side < Right side.
If 2x-3y = 6 and x^2 = 5 then LS > RS. Not sufficient.
1) 2x-3y=-2= -ve. Since, x^2=+ve => 2x-3y <x^2 Sufficient
2) x>2; y>0 =>2x-3y>4 i.e. 2x-3y = 5,6,7....
Also, x>2 =>x^2>4 i.e. x^2 = 5,6,7....
Now if 2x-3y = 5 and x^2 =6 then Left side < Right side.
If 2x-3y = 6 and x^2 = 5 then LS > RS. Not sufficient.
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missionmba
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aspirant1976
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missionmba
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AleksandrM
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The answer is D. However, I would like to know more than just the correct answer, which I already knew. I want to know WHY the answer is D and HOW you would get there. Thanks. 
(1) Sufficient because:
=> x^2 is always +ve.
=> given 2x - 3y = -2, which is -ve.
=> Since a -ve num < +ve num, we can say with certainty that 2x - 3y < x^2.
(2) Sufficient because:
=> Given x > 2, y > 0
=> Note that for all n > 2, 2n < n^2 ( you can plug numbers in)
=> Also given that y > 0, so 2n - 3y further reduces the Left hand side.
Therefore, Ans is D
=> x^2 is always +ve.
=> given 2x - 3y = -2, which is -ve.
=> Since a -ve num < +ve num, we can say with certainty that 2x - 3y < x^2.
(2) Sufficient because:
=> Given x > 2, y > 0
=> Note that for all n > 2, 2n < n^2 ( you can plug numbers in)
=> Also given that y > 0, so 2n - 3y further reduces the Left hand side.
Therefore, Ans is D
AleksandrM wrote:The answer is D. However, I would like to know more than just the correct answer, which I already knew. I want to know WHY the answer is D and HOW you would get there. Thanks.
