123 is x<0

[ern5231]

x<y. Is x<0?
1) x^2>y^2 2) y>1

[Rahul@gurome]

(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.

(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.

Combining (1) and (2), we get that x < 0 always as x^2 > y^2.

The correct answer is (C).

[[email protected]]

(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.

(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.

Combining (1) and (2), we get that x < 0 always as x^2 > y^2.

The correct answer is (C).

:

if x<y or x-y<0 (or is negative), is x<0 .... hmmm

1. x^2>y^2 means x^2-y^2 > 0 or (x-y)(x+y) > 0;

given that x-y<0, it can only mean for (x-y)(x+y) to be positive, (x+y) is negative or x+y<0!!

INSUFF

2. y>1 (i.e. positive)

INSUFF

combining 1&2: given that x-y<0, we ONLY need to worry about x+y<0 (from statement 1).

if y>1, we'll have to add to it a very big negative x to get a positive (x-y)(x+y) or (x-y)(x+y) > 0!

ANS (C)

[analyst218]

(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.

(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.

Combining (1) and (2), we get that x < 0 always as x^2 > y^2.

The correct answer is (C).

well in the question it gives you X<Y

i say A..

[electrico]

IMO A.

Its already given that x<y, so point considering the case If x = 4 and y = 3.

[Rahul@gurome]

Agree. Since x < y so the example x = 4 and y = 3 is not possible. Hence, x < 0 always.

The correct answer should be (A).

[amitu]

THE first condition is x^2 > y^2 this implies that

x < -y or x > y but x > y ruled out so x < -y

if Y > 0 then X < 0

if Y < 0 then x can be anything

so we need condition 2 also

so C ?