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Re: Which is greater |x| or y [#permalink]
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Here, x^2 = |y|. both x^2 and |y| positive. No matter x = 1 or -1 and y = 1 or -1.

Option A

x^2 is always positive.

OPtion B

Y might be positive or might be negative.

So, option D is the answer.
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Which is greater |x| or y [#permalink]
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sujoykrdatta wrote:
Carcass wrote:
\(x^2=|y|\)

Quantity A
Quantity B
\(|x|\)
\(y\)


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Kudos for the right answer and explanation



We know that: \(x^2 = |y|\)

\(=> |x|^2 = |y|\)

\(=> |x| = \sqrt{|y|}\)

Thus, we need to compare:

Quantity A: \(|x| = \sqrt{|y|}\)
Quantity B: \(y\)

The various possibilities are:

#1. If \(y < 0\): Quantity A will be greater since it is positive while Quantity B is negative

#2. If \(y = 1\) or \(y = 0\): Quantity A will be equal to Quantity B, both being zero

#3. If \(0 < y < 1\): Quantity A will be greater than Quantity B since square root of a fraction is greater than the fraction itself (ex. \(\sqrt{0.49} = 0.7 > 0.49\))

#4. If \(y > 1\): Quantity B will be greater than Quantity A

Thus, there is no relation

Answer D


Would like to point out. when square is removed from x, it becomes +sqrt(|y|) or - sqrt(|y|). not just the positive sqrt(|y|), but it can be negative also. that much information is enough to tell that the answer is D.

So in case #1 as you point out, it can actually be either greater or lesser than B
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