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Is 2X - 3Y < X^2?

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Senior | Next Rank: 100 Posts
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by eustudent » Thu Sep 03, 2009 7:40 pm
Is 2X - 3Y< X^2

1) 2X - 3Y= 2

2) X>2 and Y>0

OA will be disclosed in a little bit.
Please elaborate!

I do apologize if the problem has been posted before. I searched for it but I couldn't find it.

Thank you!
Last edited by eustudent on Fri Sep 04, 2009 7:41 am, edited 2 times in total.
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Source: — Data Sufficiency |
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Re: Is 2X - 3Y < sqrtX?

by catchyblues » Thu Sep 03, 2009 11:32 pm
I think the answer is "E"

Taking 2X - 3Y= 2, the equation is valid for both X < 0 and X >0 and hence the above inequality can't be solved.

Now Considering X > 2 and Y > 0, let X = 8 and X =1 and this implies 2X-3Y = 7 which is greater than SqrtX

but if X=3 and Y =2, 2X-3Y = 0 which will be greater than SqrtX

Considering 1 & 2 also doesn't solve the inequality...

Hence "E"
eustudent wrote:Is 2X - 3Y< SqrtX?

1) 2X - 3Y= 2

2) X>2 and Y>0

OA will be disclosed in a little bit.
Please elaborate!

I do apologize if the problem has been posted before. I searched for it but I couldn't find it.

Thank you!
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by electrico » Thu Sep 03, 2009 11:49 pm
Is 2X - 3Y< SqrtX?

1) 2X - 3Y= 2

2) X>2 and Y>0


Stmt1: Not Sufficient because Y is not known.

Stmnt2: Lets X=3, Y = 1
2x-3y = 6 - 3 =3 > sqrt(3)

Lets x= 3, y = 100

2x-3y = a negative number < sqrt(3)

So, stmnt 2 is also Not sufficient.

Take both the statment:

2x-3y = 2 => 2x and 3y will be both even.

so, solution is (4,2), (7, 4), (10,6).....

So, answer is E as the first one is conflicting with other.
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by eustudent » Fri Sep 04, 2009 7:46 am
Answer is D.

Apologies.

I saw that problem was posted before in the forum.
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by navalpike » Fri Sep 04, 2009 10:28 am
I believe that you posted the question incorrectly.
S1 should read:

1. 2x-3y = -2
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