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If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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Updated on: 29 May 2017, 11:21
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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
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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 nonzero numbers and the value of (y)10x5 is 1, which
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29 May 2017, 11:34
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.
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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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01 Jun 2017, 00:20
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



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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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01 Jun 2017, 00:43
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.
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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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01 Jun 2017, 02:11
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?
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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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01 Jun 2017, 05:32
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.
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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



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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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03 Jun 2017, 16:49
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 \(\sqrt[3]{64} =4\).
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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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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 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.



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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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12 Jun 2017, 21:23
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.
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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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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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> 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
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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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12 Jul 2017, 17:05
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.
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Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which
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12 Aug 2018, 06:41
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.




Re: If x and y are nonzero numbers and the value of (y)10x5 is 1, which &nbs
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