Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 14 Sep 2015
Posts: 64
Location: India
GPA: 3.41

If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
Updated on: 29 May 2017, 12:21
Question Stats:
31% (02:17) correct 69% (02:31) wrong based on 388 sessions
HideShow timer Statistics
If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true? I. y1 > x1 II. yy > xx III. \(\sqrt{−yx}=\sqrt{−xy}\) A. I only B. II only C. III only D. I, II and III E. None of I, II and III
Official Answer and Stats are available only to registered users. Register/ Login.
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.




Math Expert
Joined: 02 Sep 2009
Posts: 59147

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
29 May 2017, 12:34
niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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.
_________________




Intern
Joined: 28 Sep 2016
Posts: 18

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
01 Jun 2017, 01:20
Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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



Math Expert
Joined: 02 Sep 2009
Posts: 59147

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
01 Jun 2017, 01:43
joepc wrote: Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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.
_________________



SVP
Status: It's near  I can see.
Joined: 13 Apr 2013
Posts: 1701
Location: India
Concentration: International Business, Operations
GPA: 3.01
WE: Engineering (Real Estate)

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
01 Jun 2017, 03:11
Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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?
_________________
"Do not watch clock; Do what it does. KEEP GOING."



Math Expert
Joined: 02 Sep 2009
Posts: 59147

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
01 Jun 2017, 06:32
understandZERO wrote: Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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.
_________________



Current Student
Status: Preparing for GMAT!!
Joined: 11 Oct 2015
Posts: 126
Location: India
Concentration: Entrepreneurship, International Business
GMAT 1: 660 Q47 V34 GMAT 2: 700 Q48 V38
GPA: 3.1
WE: General Management (Entertainment and Sports)

BunuelCan 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



Math Expert
Joined: 02 Sep 2009
Posts: 59147

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
03 Jun 2017, 17:49
Sirakri wrote: BunuelCan 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 \(\sqrt[3]{64} =4\).
_________________



Manager
Joined: 05 Dec 2015
Posts: 99

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
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 nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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.



Math Expert
Joined: 02 Sep 2009
Posts: 59147

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
12 Jun 2017, 22:23
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 nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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 reread the first sentence of the solution.
_________________



Intern
Joined: 11 Aug 2013
Posts: 24
Location: India
Concentration: Finance, General Management
GPA: 3.23
WE: Information Technology (Investment Banking)

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
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> y1>x1 ; from (d) we knowx=y and from (c) we know x<0 => y1<x1 (substitute some random values for x and y and verify for better understanding) .... NOT POSSIBLE 2>yy > xx ; again from (c) and (d) since y>x and y=x therefore yy> 0 and xx <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
_________________
Thanks Sindbad  Click +1 Kudos if my post helped...



Current Student
Joined: 22 Sep 2016
Posts: 156
Location: India
GPA: 4

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
12 Jul 2017, 18:05
Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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.
_________________
Desperately need 'KUDOS' !!



Intern
Joined: 25 Jul 2018
Posts: 2

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
12 Aug 2018, 07:41
Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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.



SVP
Joined: 03 Jun 2019
Posts: 1849
Location: India

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
12 Oct 2019, 11:56
niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true? I. y1 > x1 When y=1; x=1; NOT TRUE II. yy > xx y>0; x=y; yy>(y)y y^2>y^2 MUST BE TRUE III. \(\sqrt{−yx}=\sqrt{−xy}\) yx<0 NOT TRUE IMO B Posted from my mobile device
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts." Please provide kudos if you like my post. Kudos encourage active discussions. My GMAT Resources:  Efficient LearningAll you need to know about GMAT quantTele: +911140396815 Mobile : +919910661622 Email : kinshook.chaturvedi@gmail.com



Manager
Joined: 04 Jun 2010
Posts: 87
Location: India
GPA: 3.22

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
14 Oct 2019, 02:29
Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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 nonzero numbers .... so for statement II......I AM FREE TO CHOOSE Y=(1)................for which , it does not hold ......



Math Expert
Joined: 02 Sep 2009
Posts: 59147

Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
Show Tags
14 Oct 2019, 02:37
avikroy wrote: Bunuel wrote: niteshwaghray wrote: If x and y are nonzero numbers and the value of \(\frac{(\sqrt{y})^{10}}{x^5}\) is 1, which of the following expressions must be true?
I. y1 > x1
II. yy > xx
III. \(\sqrt{−yx}=\sqrt{−xy}\)
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 nonzero 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. y1 > x1. If y = 1 and x = 1, this won't be true. Discard. II. yy > xx y*y > (y)y y^2 > y^2. y^2 + y^2 > 0. Since y is nonzero, then this must be true, III. \(\sqrt{−yx}=\sqrt{−xy}\) \(\sqrt{−yy}=\sqrt{−(y)y}\) \(\sqrt{−y^2}=\sqrt{y^2}\) 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 nonzero 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.
_________________




Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
[#permalink]
14 Oct 2019, 02:37






