Problem 1 -Data Sufficiency

[er.twi.fb]

1. If x and y are non-negative, is (x + y)
greater than xy?
(1) x = y
(2) x + y is greater than x2 + y2

* x2 means x square.

Thanks..

[mj78ind]

Stmt 1 - make x = y in the eqtn x+y>xy, x = y,

x^2 - 2x<0, which gives us 0<x<2, but x can be greater than 2 hence insufficient

Stmt 2 - x+y>x^2 + y^2 can only be true if 0<x<1 and 0<y<1. An interesting thing to note here is a boundary condition, wherein x or y could be = 1 but the other has to be still less than 1 and the equation holds. Now, when 0<x<1 and 0<y<1,
x+y>xy holds. Hence SUFFICIENT.

Ans B

OA please?

[ayushiiitm]

1. If x and y are non-negative, is (x + y)
greater than xy?
(1) x = y
(2) x + y is greater than x2 + y2

* x2 means x square.

Thanks..

imo b

stem 1
x=y
putting in given situation
is 2x>x2
this can be true for 1, but not for 3 so insufficient

stem 2
x+y>x2+y2
add -2xy on both sides

x+y-2xy>(x-y)^2
x+y>(x-y)^2+2xy

since x and y both greater than 0

x+y is more than xy +xy + (x-y)^2

so x+y>xy
sufficient

IMOB

wats the OA

[er.twi.fb]

Stmt 1 - make x = y in the eqtn x+y>xy, x = y,

x^2 - 2x<0, which gives us 0<x<2, but x can be greater than 2 hence insufficient

Stmt 2 - x+y>x^2 + y^2 can only be true if 0<x<1 and 0<y<1. An interesting thing to note here is a boundary condition, wherein x or y could be = 1 but the other has to be still less than 1 and the equation holds. Now, when 0<x<1 and 0<y<1,
x+y>xy holds. Hence SUFFICIENT.

Ans B

OA please?

Hi.

Answer is B. But I would like to have more clear explanation for this.


Thanks..