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!
Is 2X - 3Y < X^2?
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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"
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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- Senior | Next Rank: 100 Posts
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- Joined: Tue Feb 17, 2009 11:14 pm
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- Thanked: 2 times
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.
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.