Is |x^2| - |y^2| = |x^2–y^2| ? (1) x 0 (2) y

Question

Is  |x^2| - |y^2| = |x^2–y^2|  ?

(1)  x > 0

(2)  y < 0

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sampriya · Answer

1) x=2 , y=0 yes
Y=0, x=2 no

2) x=1, y=-1 yes
x=0 , y= -1 no

Both 1 and 2 together

X=1, y=-1 yes 
X=1, y=-4 no

E is the correct answer

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nick1816 · Answer

|x^2| - |y^2| = |x^2–y^2| 
or
is |x|≥|y|?

Sign of x or y won't tell us anything about the magnitude of x or y

Clearly E

satya2029 · Answer

Ix^2I=lxl^2=x^2 
so question is asking whether
 x^2-y^2=Ix^2-y^2I  
or
 x^2-y^2>=0 
From 1)
No info about Y
Not sufficient
From 2)
No info about X
Not sufficient
From 1 and 2
x=4, y= -1(yes)
x=4, y=-9(No)
E:)

Manku · Answer

The LHS can be re-written as : x^2- y^2

So basically the question asks ,
whether x^2- y^2 = |x^2–y^2|
The above will hold true when |x^2–y^2|  >  0
 >  0]

Rephrasing the question stem we get : Is (x^2- y^2)  >  0 ? OR Is |x| >  |y| ?

Clearly neither of the statement by itself nor both combined can give us definitive answer to the prompt .

Lets try few values 
1 +2 : x and y are of opposite nature . 
x = 3 & y= -5 we have |x| < |y|
x= 3 & y = -2 we have |x| > |y|

Correct Ans. E

Hope it helps.

Fdambro294 · Answer

Property: |a| - |b| ≤ |a - b| the two expressions will be equal when one of two following cases are true: |a| - |b| = |a - b| will be true when: (case 1) ab > 0 and |a| ≥ |b| (i.e., the variables have the same sign AND the magnitude of a is greater than the magnitude of b, so that the left hand side is not negative) OR (case 2) |b| = 0 The wrinkle in this question in this question stem is that no matter the sign of (a) or (b), the squaring within the modulus will turn the result positive. As such, so long as the absolute value of |a| exceeds or is equal to the absolute value of |b|, the two expressions will be equal and we will get a YES. Understanding this concept and looking at the two statements, we can jump right to (s1 & s2) together: x > 0 y < 0 case 1: |x| > |y| x = 2 ; y = -1 -----> |4| - |1| = |4 - 1| = 3 YES, the expressions are equal case 2: |x| < |y| x = 1 ; y = -2 ------> |1| - |4| < |1 - 4| ----> -3 < +3 NO, the expression on the left hand side is less than that on the right hand side. *E*

Is |x^2| - |y^2| = |x^2–y^2| ? (1) x > 0 (2) y < 0

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