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

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

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New post 28 Jan 2012, 11:53
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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

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

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New post 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.

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

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New post 29 Mar 2017, 05:43
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Re: Is x^2 = y^4/(y^4)^(1/2) ? [#permalink]

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New post 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.

Answer: B.


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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