Bunuel wrote:
VeritasPrepKarishma wrote:
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?(1) 2xy < 100 --> clearly insufficient: if \(x=y=0\) then the answer will be NO but if \(x=10\) and \(y=-10\) then the answer will be YES.
(2) (x + y)^2 > 200 --> \(x^2+2xy+y^2>200\). Now, as \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) then \(x^2+y^2\geq{2xy}\)
so we can safely substitute \(2xy\) with \(x^2+y^2\) (as \(x^2+y^2\) is at least as big as \(2xy\) then the inequality will still hold true) --> \(x^2+(x^2+y^2)+y^2>200\) --> \(2(x^2+y^2)>200\) --> \(x^2+y^2>100\). Sufficient.
Answer: B.
Are you sure the OA is C?
To put it in words, think of it this way:
Is x^2 + y^2 > 100?
(2) (x + y)^2 > 200
which means: x^2 + y^2 + 2xy > 200
Now you know that 2xy is less than or equal to x^2 + y^2.
If 2xy is equal to \((x^2 + y^2)\), \((x^2 + y^2)\) will be greater than 100 since the total sum is greater than 200.
If 2xy is less than \((x^2 + y^2)\), then anyway \((x^2 + y^2)\) will be greater than 100 (which is half of 200).
So statement 2 is sufficient alone.
Thanks Bunuel and Karishma for detailed explanation... one confusion though...
from point 2) if 2xy is less than x^2 + y^2 .... lets say it is - (x^2 +y^2)... than the equation becomes 0 > 200 ? Am I missing some point
( SInce we need to know the minimum value of left side ... we should be able to substitute (x^2 + y^2 ) by 2xy..... but not 2xy by (x^2 + y^2 )