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# If x and y are non-zero numbers and the value of (y)10x5 is -1, which

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If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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Updated on: 29 May 2017, 12:21
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If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

Originally posted by niteshwaghray on 29 May 2017, 10:28.
Last edited by Bunuel on 29 May 2017, 12:21, edited 1 time in total.
Edited the question.
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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29 May 2017, 12:34
4
3
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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01 Jun 2017, 01:20
1
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

Thanks Bunuel for the Solution.

But How did you do this

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

To cancel the powers the bases should be equal?
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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01 Jun 2017, 01:43
1
joepc wrote:
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

Thanks Bunuel for the Solution.

But How did you do this

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

To cancel the powers the bases should be equal?

$$(\frac{y}{x})^5=-1$$;

Take the fifths root from both sides: $$\frac{y}{x}=\sqrt[5]{(-1)}=-1$$.

Hope it's clear.
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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01 Jun 2017, 03:11
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

Bunuel, Is it too tough or is it not a GMAT type question?
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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01 Jun 2017, 06:32
understandZERO wrote:
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

Bunuel, Is it too tough or is it not a GMAT type question?

Not that tough but I don't like III option. It gives undefined value for LHS, don't remember seeing such thing in official questions.
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02 Jun 2017, 23:37
Bunuel

Can we say that, for a negative number, ODD ROOT is defined but EVEN ROOT isn't.

5THROOT(-1)=-1
SQRT(-Y^2)=NOT DEFINED.

Thank you.

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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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03 Jun 2017, 17:49
Sirakri wrote:
Bunuel

Can we say that, for a negative number, ODD ROOT is defined but EVEN ROOT isn't.

5THROOT(-1)=-1
SQRT(-Y^2)=NOT DEFINED.

Thank you.

Posted from my mobile device

Yes.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root. That is, $$\sqrt{16}=4$$, NOT +4 or -4. Even roots have only a positive value on the GMAT.

$$\sqrt{negative}=undefined$$

In contrast, the equation $$x^2=16$$ has TWO solutions, +4 and -4.

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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12 Jun 2017, 21:57
for II doesn't y=-1 and x=1 work and thus it must not be true? becomes -1>1 which is not correct, or am i missing something easy?

Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.
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Posts: 58340
Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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12 Jun 2017, 22:23
1
mdacosta wrote:
for II doesn't y=-1 and x=1 work and thus it must not be true? becomes -1>1 which is not correct, or am i missing something easy?

Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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13 Jun 2017, 16:24
Before we start point to remember is root of negative number is not defined. ---(a)
Now Numerator(N) is (root y)^10 and denominator(D) is x^5.
since N/D =-1 and N has root y ,Therefore root y is positive (from (a)) =>(root y )^10 is +ve =>y^5 is +ve ---(b)
Since N>0 ,therefore D will be <0 and D=-N =>x^5<0 =>x<0 ---(C)
also from (b) and (c) |X|=|Y| ---(d)

now lets check the options
1> |y-1|>|x-1| ; from (d) we know|x|=|y| and from (c) we know x<0 => |y-1|<|x-1| (substitute some random values for x and y and verify for better understanding) .... NOT POSSIBLE
2>y|y| > x|x| ; again from (c) and (d)
since y>x and |y|=|x| therefore y|y|> 0 and x|x| <0 hence this is correct OPTION

Looking into the options.
We have only one option (B) where we do not have 1 but have 2. Hence it will be only (B) no need to verify option 3
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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12 Jul 2017, 18:05
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

I'm confused.

(-5)^2 =25
So what if y = 25? Then, the root of y can be either 5 or -5.
Correct me if I'm wrong.
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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12 Aug 2018, 07:41
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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12 Oct 2019, 11:56
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

Asked: If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|
When y=1; x=-1; NOT TRUE

II. y|y| > x|x|
y>0; x=-y;
y|y|>(-y)|-y|
y^2>-y^2
MUST BE TRUE

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$
-y|x|<0
NOT TRUE

IMO B

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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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14 Oct 2019, 02:29
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

Dear Sir,

thank You for the elegant solution......

However, would you please clear my doubt...

Condition is x and y are non-zero numbers .... so for statement II......I AM FREE TO CHOOSE Y=(-1)................for which , it does not hold ......
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which  [#permalink]

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14 Oct 2019, 02:37
1
avikroy wrote:
Bunuel wrote:
niteshwaghray wrote:
If x and y are non-zero numbers and the value of $$\frac{(\sqrt{y})^{10}}{x^5}$$ is -1, which of the following expressions must be true?

I. |y-1| > |x-1|

II. y|y| > x|x|

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

A. I only
B. II only
C. III only
D. I, II and III
E. None of I, II and III

First of all notice for $$\sqrt{y}$$ to be defined y must be positive (the square root from a negative number is not defined for the GMAT + we are told that y is non-zero therefore y is positive).

$$\frac{(\sqrt{y})^{10}}{x^5}=-1$$;

$$\frac{y^5}{x^5}=-1$$;

$$\frac{y}{x}=-1$$;

$$y = -x$$.

Since y is positive the x is negative.

I. |y-1| > |x-1|. If y = 1 and x = -1, this won't be true. Discard.

II. y|y| > x|x|

y*y > (-y)|-y|

y^2 > -y^2.

y^2 + y^2 > 0.

Since y is non-zero, then this must be true,

III. $$\sqrt{−y|x|}=\sqrt{−xy}$$

$$\sqrt{−y|-y|}=\sqrt{−(-y)y}$$

$$\sqrt{−y^2}=\sqrt{y^2}$$

-y^2 will be negative, so the left hand side won't be defined. Thus, this also cannot be true.

P.S. Also, not very nice question.

Dear Sir,

thank You for the elegant solution......

However, would you please clear my doubt...

Condition is x and y are non-zero numbers .... so for statement II......I AM FREE TO CHOOSE Y=(-1)................for which , it does not hold ......

y cannot be -1, because even roots (such as the square root) of negative numbers are not defined on the GMAT so if y = -1, then $$\sqrt{y}=\sqrt{-1}=undefined$$, and $$\frac{(\sqrt{y})^{10}}{x^5}$$ will not equal to -1, as we are given in the stem.
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Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which   [#permalink] 14 Oct 2019, 02:37
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