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805+ (Hard)|   Absolute Values|   Algebra|      
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niteshwaghray
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which Updated on: 29 May 2017, 12:21
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.
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95% (hard)

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35% (02:17) correct 65% (02:30) wrong based on 568 sessions

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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
ShowHide Answer
Official Answer
B
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Bunuel
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 29 May 2017, 12:34
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niteshwaghray
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
Show more

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.

Answer: B.

P.S. Also, not very nice question.
General Discussion
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joepc
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 01 Jun 2017, 01:20
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Bunuel
niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.
Show more

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?
or am i missing anything? Please help
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 01 Jun 2017, 01:43
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Bunuel
niteshwaghray
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.

Answer: B.

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?
or am i missing anything? Please help
Show more

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

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

Hope it's clear.
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 01 Jun 2017, 03:11
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niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.
Show more


Bunuel, Is it too tough or is it not a GMAT type question?
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 01 Jun 2017, 06:32
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Bunuel
niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.


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

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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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 02 Jun 2017, 23:37
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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
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 03 Jun 2017, 17:49
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Sirakri
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
Show more

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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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 12 Jun 2017, 21:57
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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
niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.
Show more
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 12 Jun 2017, 22:23
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mdacosta
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
niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.
Show more

Please re-read the first sentence of the solution.
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 13 Jun 2017, 16:24
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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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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 12 Jul 2017, 18:05
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Bunuel
niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.
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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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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 12 Aug 2018, 07:41
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Bunuel
niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.
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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 12 Oct 2019, 11:56
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niteshwaghray
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
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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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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 14 Oct 2019, 02:29
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Bunuel
niteshwaghray
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.

Answer: B.

P.S. Also, not very nice question.
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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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If x and y are non-zero numbers and the value of (y)10x5 is -1, which 14 Oct 2019, 02:37
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Bunuel
niteshwaghray
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.

Answer: B.

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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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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