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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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New post Updated on: 29 May 2017, 11:21
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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, 09:28.
Last edited by Bunuel on 29 May 2017, 11:21, 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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New post 29 May 2017, 11:34
2
2
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.

Answer: B.

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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New post 01 Jun 2017, 00:20
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.

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
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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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New post 01 Jun 2017, 00:43
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.

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


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

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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New post 01 Jun 2017, 02:11
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.

Answer: B.

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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New post 01 Jun 2017, 05:32
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.

Answer: B.

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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New post 02 Jun 2017, 22:37
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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New post 03 Jun 2017, 16:49
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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New post 12 Jun 2017, 20:57
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.

Answer: B.

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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New post 12 Jun 2017, 21:23
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.

Answer: B.

P.S. Also, not very nice question.


Please re-read the first sentence of the solution.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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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New post 13 Jun 2017, 15:24
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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New post 12 Jul 2017, 17:05
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.

Answer: B.

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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New post 12 Aug 2018, 06:41
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.

Answer: B.

P.S. Also, not very nice question.
GMAT Club Bot
Re: If x and y are non-zero numbers and the value of (y)10x5 is -1, which &nbs [#permalink] 12 Aug 2018, 06:41
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