Is x^2 - y^2 greater than x - y? (1) x - y

Question

Is x^2 - y^2 greater than x - y?

(1) x - y < 0
(2) x + y < 0

Archit3110 · Answer

target
Is x^2 - y^2 greater than x - y
x^2-y^2-x+y>0
#1
x-y<0
possible when
x=1 , y=-2
we get no to target
x=-3, y=-1
we get yes to target
insufficient
#2
x + y < 0
same case as #1 
x=1 , y=-2
we get no to target
x=-3, y=-1
we get yes to target
from 1 &2
nothing in common
option E

Kinshook · Answer

Asked: Is x^2 - y^2 greater than x - y?

Is x^2 - y^2 = (x-y)(x+y) > x-y
Is (x-y)(x+y-1) > 0

(1) x - y < 0
If x+y-1 > 0; (x-y)(x+y-1) <0
But if x+y-1 < 0; (x-y)(x+y-1) >0
NOT SUFFICIENT

(2) x + y < 0
x + y - 1 < 0
If x-y > 0; (x-y)(x+y-1) <0
But if x-y < 0; (x-y)(x+y-1) >0
NOT SUFFICIENT

(1) + (2)
(1) x - y < 0
(2) x + y < 0
x + y - 1 < 0
(x-y)(x+y-1) >0
SUFFICIENT

IMO C

sivatx2 · Answer

Is: (x^2-y^2) > (x-y)

True: (x^2-y^2) > 1
False: (x^2-y^2) <= 1

(1) alone:

x-y < 0 (or) x < y

Take x = 0, y = 1 which satisfies (1)

is (x^2-y^2) > (x-y)?
is( 0 -1) > (0-1) FALSE

Take x = -1, y = 2

is (x^2-y^2) > (x-y)?
is( 1 -4) > (-1-3)?
is -3 > -4 TRUE

One false and one true is enough to say (1) alone cannot identify that x^2 - y^2 greater than x - y.

A and D is gone.
--------------------------------------------------------------
(2) alone:

(x+y) < 0

Take x = 0, y = -1 which satisfies (2).

is (x^2-y^2) > (x-y)?

is (0-1) > 0+1?
is -1 > 1 FALSE

Take x = -1 y = 0 which satisfies (2)

is (x^2-y^2) > (x-y)?
is(1-0) > (-1-0) ?

is 1 > -1 TRUE

One false and one true is enough to say (2) alone cannot identify that x^2 - y^2 greater than x - y.

B is gone. Only C & E left.
----------------------------------------------------------------------------------------------

(1) & (2) together we have:

(x-y < 0 OR x<y) and (x+y) < 0 -----(3)

Take x = -1 and y = 0 which satisifies (3)

Then, is (x^2-y^2) > (x-y)?
is (1-0) > (-1-0)?

is 1 > -1 ? TRUE

Take x = -2 and y = -1 which satisfies (3)

Then, is (x^2-y^2) > (x-y)?

(4 -1) > -2+1 TRUE

We can't arrive a FALSE condition. => We say that (x^2-y^2) > (x-y) is always true.

Hence C is the correct answer choice.

AndrewN · Answer

Yes, I think this one needs to be reevaluated. I also answered (C) on the grounds that the quadratic can be factored into (x - y)(x + y), a difference of squares, and if both binomials are negative, as the two statements together would tell us, then it  must be true  that the positive quadratic is greater than the negative binomial.

What is the source of the question,  IdanGolan ? What is the rationale behind (E), the given OA?

- Andrew

Is x^2 - y^2 greater than x - y? (1) x - y < 0 (2) x + y < 0

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