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# Is x^2 = y^4/(y^4)^(1/2) ?

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Intern
Joined: 17 Jan 2012
Posts: 41

Kudos [?]: 196 [1], given: 16

GMAT 1: 610 Q43 V31
Is x^2 = y^4/(y^4)^(1/2) ? [#permalink]

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28 Jan 2012, 11:53
1
KUDOS
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Difficulty:

25% (medium)

Question Stats:

78% (01:05) correct 22% (02:38) wrong based on 49 sessions

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Is x^2 = y^4/(y^4)^(1/2) ?

(1) -1 < x < 0 < y < 1 < 1-x+y
(2) |x| = |y|
[Reveal] Spoiler: OA

Kudos [?]: 196 [1], given: 16

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132618 [0], given: 12326

Re: Is X^2 = Y^4/(Y^4)^(1/2)? [#permalink]

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28 Jan 2012, 12:22
docabuzar wrote:
Is X^2 = Y^4/(Y^4)^(1/2) ?

A. -1 < x < 0 < y < 1 < 1-x+y

B. IxI = IyI

Is $$x^2=\frac{y^4}{\sqrt{y^4}}$$?

Note that: realistic GMAT question would have mentioned that $$y\neq{0}$$, as it's in denominator.

Is $$x^2=\frac{y^4}{\sqrt{y^4}}$$ --> is $$x^2=\frac{y^4}{y^2}$$? --> is $$x^2=y^2$$? --> is $$|x|=|y|$$?

(1) -1 < x < 0 < y < 1 < 1-x+y. Clearly insufficient: if $$x=-\frac{1}{2}$$ and $$y=\frac{1}{2}$$ then the answer is YES but if $$x=-\frac{1}{3}$$ and $$y=\frac{1}{2}$$ then the answer is NO.

(2) |x|=|y|. Directly gives an YES answer to the question.

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Kudos [?]: 132618 [0], given: 12326

Non-Human User
Joined: 09 Sep 2013
Posts: 15704

Kudos [?]: 282 [0], given: 0

Re: Is x^2 = y^4/(y^4)^(1/2) ? [#permalink]

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29 Mar 2017, 05:43
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Director
Joined: 12 Nov 2016
Posts: 794

Kudos [?]: 36 [0], given: 165

Re: Is x^2 = y^4/(y^4)^(1/2) ? [#permalink]

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22 May 2017, 16:25
Bunuel wrote:
docabuzar wrote:
Is X^2 = Y^4/(Y^4)^(1/2) ?

A. -1 < x < 0 < y < 1 < 1-x+y

B. IxI = IyI

Is $$x^2=\frac{y^4}{\sqrt{y^4}}$$?

Note that: realistic GMAT question would have mentioned that $$y\neq{0}$$, as it's in denominator.

Is $$x^2=\frac{y^4}{\sqrt{y^4}}$$ --> is $$x^2=\frac{y^4}{y^2}$$? --> is $$x^2=y^2$$? --> is $$|x|=|y|$$?

(1) -1 < x < 0 < y < 1 < 1-x+y. Clearly insufficient: if $$x=-\frac{1}{2}$$ and $$y=\frac{1}{2}$$ then the answer is YES but if $$x=-\frac{1}{3}$$ and $$y=\frac{1}{2}$$ then the answer is NO.

(2) |x|=|y|. Directly gives an YES answer to the question.

Bunuel I like the explanation- but also for sake of clarity I think it's important to point to others that PEMDAS isn't entirely accurate- multiplication and division are of the same rank and therefore are performed before addition or subtraction; however, two operations of the same rank and performed in the order they appear- in the case of statement we could have

1- (-.5) + .75
1.5 +.75 = 2,25>1

OR

1- (-.5) + .5 =
1.5 +.5 = 2 > 1

Kudos [?]: 36 [0], given: 165

Re: Is x^2 = y^4/(y^4)^(1/2) ?   [#permalink] 22 May 2017, 16:25
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