Is x y? (1) |x| y. (2) x + y 0.

Question

Is x > y?

(1) |x| > y
(2) x + y > 0

Ejiroosa · Answer

St 1. x >y or x < - y NS

St 2 x > - y NS

Ans E

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Luckisnoexcuse · Answer

(1) if x>0 then x>y
if x<0 then x>-y
Insufficient

(2) x>-y insufficient

on combining we get x>-y..insufficient

Hence E

Supermaverick · Answer

1) x>y (when x>0) and x<-y when (x<0).....Insufficient 2) x>-y, Clearly insufficient Combining we get x > y answer C

Devbek · Answer

Why do we get x>-y not x>y?

Mo2men · Answer

Is x > y?

(1) |x| > y

Let x = 10 & y = 1...........Answer is Yes

Let x = -10 & y =1...........Answer is No

Insufficient

(2) x + y > 0

Case 1 : Let x = 10 & y = 1...........Answer is Yes

Case 2 : Let x = 10 & y = -1...........Answer is Yes

Case 3 : Let x = 1 & y = 10...........Answer is No

Case 4 : Let x = -1 & y = 10...........Answer is No

Insufficient

Combining 1 & 2

|x| - y > 0

x + y > 0
----------------- Sum both inequalities as both are in same direction

x + |x| >0 ........The only way to prove this inequality is that  x is POSITIVE . If x is negative then result is zero which is invalid.

x > 0 & |x| > y ...........then x > y

Answer: C

Another Approach:

Combine 1 & 2 together.

We need to match cases from Statement 2 that fit into Statement 1

Case 3 & 4 are NOT valid.

Case 1 & 2 are valid

based on above x > y

Answer: C

pradeepmaria · Answer

St 1: Insuff
 6,3 - Yes
-6,3 - No

St 2: Insuff
-2,4 - No
4,-2 - Yes

Combining Yes & Yes - Suff
Answer C

Madhavi1990 · Answer

For me its E is x> y? 1) |x| >y so two cases: a) x > y b) x< -y Not suff 2) x + y> 0 x> - y Not suff 1) + 2): we have 3 equations --> x> y; x < -y; x> -y So both not suff. So E OA and OE awaited!

antztheman · Answer

x> y; x < -y; x> -y. x>y is as good as x>-y

So x>y, hence C

hellosanthosh2k2 · Answer

x > y  ?

Statement 1:  |x| > y  ?
Insufficient
case 1: x = -16, y = 8
case 2: x = 16, y = 8

Statement 2:  x + y > 0 
InSufficient
case 1: x = 3, y = 2
case 2: x = 2, y = 3

(1) + (2)

adding two inequalities,  x + |x| + y > y  =>  x + |x| > 0  => x is positive
 since x is positive => statement 1 :  |x| > y  =>  x > y  => answers our question => (C)

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Is x > y? (1) |x| > y. (2) x + y > 0.

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