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Given that x is a negative number and 0 < y < 1, which of the followin

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Bunuel
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Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 01:54
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A
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C
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Given that x is a negative number and 0 < y < 1, which of the following is the greatest?


(A) x^2

(B) (xy)^2

(C) (x/y)^2

(D) x^2/y

(E) x^2*y
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C
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 05:31
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Let's simplify these options in terms of x^2. Consider, x^2 = a

A) x^2 = a

(B) (xy)^2 = x^2 * y^2 = a*y^2

(C) (x/y)^2 = x^2 / y^2 = a/y^2

(D) x^2/y = a/y

(E) x^2*y = a*y

Now the entire game depends on y. Since the y is between 0 and 1, it will behave opposite to a standard number greater than 1.

Let's take y = 1/2
=> y^2 = 1/4

Plug in these values of y and y^2 in given options -

(A) x^2 = a
(B) (xy)^2 = x^2 * y^2 = a*(1/4) = a/4
(C) (x/y)^2 = x^2 / y^2 = a/y^2 = a/(1/4) = 4a
(D) x^2/y = a/y = a/(1/2) = 2a
(E) x^2*y = a*y = a*(1/2) = a/2

We don't need to worry about the sign of a as x^2 is always positive.
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 02:32
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Bunuel wrote:
Given that x is a negative number and 0 < y < 1, which of the following is the greatest?


(A) x^2

(B) (xy)^2

(C) (x/y)^2

(D) x^2/y

(E) x^2


Bunuel, Both option A and E are the same.
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 06:21
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ias882000 wrote:
Can anyone tell me what is the difficulty level of this question plz?

Posted from my mobile device



Check the tag above the question. It's 600-700 Lvl
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 02:38
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Afc0892 wrote:
Bunuel wrote:
Given that x is a negative number and 0 < y < 1, which of the following is the greatest?


(A) x^2

(B) (xy)^2

(C) (x/y)^2

(D) x^2/y

(E) x^2


Bunuel, Both option A and E are the same.

_______________
Edited. Thank you.
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 05:34
neelesh1508 wrote:
Let's simplify these options in terms of x^2. Consider, x^2 = a

A) x^2 = a

(B) (xy)^2 = x^2 * y^2 = a*y^2

(C) (x/y)^2 = x^2 / y^2 = a/y^2

(D) x^2/y = a/y

(E) x^2*y = a*y

Now the entire game depends on y. Since the y is between 0 and 1, it will behave opposite to a standard number greater than 1.

Let's take y = 1/2
=> y^2 = 1/4

Plug in these values of y and y^2 in given options -

(A) x^2 = a
(B) (xy)^2 = x^2 * y^2 = a*(1/4) = a/4
(C) (x/y)^2 = x^2 / y^2 = a/y^2 = a/(1/4) = 4a
(D) x^2/y = a/y = a/(1/2) = 2a
(E) x^2*y = a*y = a*(1/2) = a/2

We don't need to worry about the sign of a as x^2 is always positive.


You don't know whether x is an integer or a fraction. So its dependent on x as well.
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Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 05:45
Afc0892 wrote:
neelesh1508 wrote:
Let's simplify these options in terms of x^2. Consider, x^2 = a

A) x^2 = a

(B) (xy)^2 = x^2 * y^2 = a*y^2

(C) (x/y)^2 = x^2 / y^2 = a/y^2

(D) x^2/y = a/y

(E) x^2*y = a*y

Now the entire game depends on y. Since the y is between 0 and 1, it will behave opposite to a standard number greater than 1.

Let's take y = 1/2
=> y^2 = 1/4

Plug in these values of y and y^2 in given options -

(A) x^2 = a
(B) (xy)^2 = x^2 * y^2 = a*(1/4) = a/4
(C) (x/y)^2 = x^2 / y^2 = a/y^2 = a/(1/4) = 4a
(D) x^2/y = a/y = a/(1/2) = 2a
(E) x^2*y = a*y = a*(1/2) = a/2

We don't need to worry about the sign of a as x^2 is always positive.


You don't know whether x is an integer or a fraction. So its dependent on x as well.


I agree. The behavior of x also depends on whether it is x<=-1 or -1>x>0. That's why no final answer - just the comparison among all given choices. :roll:
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 06:00
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Bunuel wrote:
Given that x is a negative number and 0 < y < 1, which of the following is the greatest?


(A) x^2

(B) (xy)^2

(C) (x/y)^2

(D) x^2/y

(E) x^2*y

The correct alternative choice must be the same for any PARTICULAR CASE that obeys the restrictions imposed in the question stem.

Hence, taking x = -1 and y = 1/2 we have:

(A) 1
(B) 1/4
(C) 4
(D) 2
(E) 1/2

We are sure (A), (B), (D) and (E) must be refuted.


The right answer must therefore be (C).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 06:17
Can anyone tell me what is the difficulty level of this question plz?

Posted from my mobile device
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   04 Oct 2018, 06:31
Irrespective of x, y is a number between 0 and 1 therefore will amplify value of x if it is in an inverse relation. Therefore option C is the right choice.

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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   07 Oct 2018, 19:31
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Bunuel wrote:
Given that x is a negative number and 0 < y < 1, which of the following is the greatest?


(A) x^2

(B) (xy)^2

(C) (x/y)^2

(D) x^2/y

(E) x^2*y


Let’s check the answer choices by using strategic and convenient numbers for x and y. So let’s let x = -2 and y = 1/2.

A) x^2 = (-2)^2 = 4

B) (xy)^2 = [(-2)(1/2)]^2 = (-1)^2 = 1

C) (x/y)^2 = [(-2)/(1/2)]^2 = (-4)^2 = 16

D) x^2/y = (-2)^2/(1/2) = 4/(1/2) = 8

E) x^2*y = (-2)^2*(1/2) = 4*(1/2) = 2

Alternate Solution:

Since 0 < y < 1, |x|y < |x| and thus, (xy)^2 < x^2; so A > B.

Since |x| < |x|/y, x^2 < (x/y)^2 and thus, C > A.

Notice that (x/y)^2 = (x^2/y)*(1/y). Since (1/y) > 1, (x/y)^2 > x^2/y and thus, C > D.

Finally, since 0 < y < 1, x^2*y < x^2 < (x/y)^2. Thus, C > E.

Answer: C
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink] New post   07 Oct 2023, 07:04
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Re: Given that x is a negative number and 0 < y < 1, which of the followin [#permalink]
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