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x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 07:09
Question Stats:
48% (02:09) correct 52% (02:22) wrong based on 84 sessions
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x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true? i) \(x  y < 0\) ii) \(x < 2y\) iii) \(x^p < y^p\) A) i only B) ii only C) iii only D) i and ii only E) none of the above
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x, y and p are integers, and xyp ≠ 0. If
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Updated on: 14 Jul 2019, 09:16
Given: P^X<P^Y and XYP different than 0( could be +ve or ve)
evaluation of choices: i) x−y<0 for simplicity suppose P=2 x=3 y=4
P^X= 2^3=8< P^Y=2^4=16 xy= 34=1<0
now suppose P=2 X=3 and Y=4
P^X= 2^3= = 1/8< P^Y=2^4 = 1/16
xy=3(4)=1>0
statement 1 is excluded since for different values of xpy it gives different answers
ii) x<2y for x=3 and y=4 this equation is true, for x=3 and y=4 this equation is not true so we reject it.noting that the stem didnt mention that xyp should be positive integers so they could be both negative or positive
iii) x^p<y^p this one is true for all integer values of x,y,p whether they are positive or negtive
so correct answer is (C)
Originally posted by GeorgeKo111 on 14 Jul 2019, 08:47.
Last edited by GeorgeKo111 on 14 Jul 2019, 09:16, edited 1 time in total.



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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 09:14
GeorgeKo111 wrote: P^X= 2^3= = 1/8< P^Y=2^4 = 1/16
Be careful; (2)^4 = 1/16 (not 1/16)
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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 09:16
Quote: Be careful; (2)^4 = 1/16 (not 1/16) a typo man, thanks for pointing that out i will fix it what do you think of my solution, is it correct? or i missed something?



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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 09:17
GeorgeKo111 wrote: What do you think of my solution, is it correct? or i missed something? You missed something. Keep at it!
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x, y and p are integers, and xyp ≠ 0. If
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Updated on: 14 Jul 2019, 11:55
IMO E; check with below case x=1 and y=2 ; p = 3 x=4 and y=2 and p = 3 And P=2 x=3 y=4 condition \(p^x < p^y\) i) \(x  y < 0\) yes for +ve integers and yes for ve integers ; YES No ii) \(x < 2y\) yes for +ve integers and may or may not for ve integers ; NO iii) \(x^p < y^p\) No for +ve case ; yes for ve integers case GMATPrepNow ; hope this is correct GMATPrepNow wrote: x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x  y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only B) ii only C) iii only D) i and ii only E) none of the above
Originally posted by Archit3110 on 14 Jul 2019, 09:26.
Last edited by Archit3110 on 14 Jul 2019, 11:55, edited 5 times in total.



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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 09:37
GMATPrepNow wrote: x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x  y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only B) ii only C) iii only D) i and ii only E) none of the above Given:  \(x, y, p \in \mathbb{Z}_{>0}\)
 \(x*y*z \ne 0\)
 \(p^x<p^y\)
When inequalities contain integer exponents, one should always check the cases for both positive and negative integers. If \((x, y, p) = (5, 3, 2)\), statement (i) isn't satisfied. If \((x, y, p) = (5, 1, 2)\), statement (ii) isn't satisfied. If \((x, y, p) = (5, 6, 2)\), statement (iii) isn't satisfied. So, answer is E.
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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 11:39
Archit3110 wrote: IMO A; check with below case x=1 and y=2 ; p = 3 x=4 and y=2 and p = 3 condition \(p^x < p^y\) i) \(x  y < 0\) yes for +ve integers and yes for ve integers ; YES ii) \(x < 2y\) yes for +ve integers and may or may not for ve integers ; NO iii) \(x^p < y^p\) No for ve case ; yes for ve integers case GMATPrepNow ; hope this is correct GMATPrepNow wrote: x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x  y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only B) ii only C) iii only D) i and ii only E) none of the above I'll give you a hint: Among the 3 answers given so far (A, C, E), one is correct
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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 11:57
GMATPrepNow must say it's a very mind wrecking question.. took over 2 mins coz there was no limit on integers range had it been mentioned that integers are +ve then it would had been a bit easy .. GMATPrepNow wrote: Archit3110 wrote: IMO A; check with below case x=1 and y=2 ; p = 3 x=4 and y=2 and p = 3 condition \(p^x < p^y\) i) \(x  y < 0\) yes for +ve integers and yes for ve integers ; YES ii) \(x < 2y\) yes for +ve integers and may or may not for ve integers ; NO iii) \(x^p < y^p\) No for ve case ; yes for ve integers case GMATPrepNow ; hope this is correct GMATPrepNow wrote: x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x  y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only B) ii only C) iii only D) i and ii only E) none of the above I'll give you a hint: Among the 3 answers given so far (A, C, E), one is correct Posted from my mobile device



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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 11:58
Answer is option E?
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x, y and p are integers, and xyp ≠ 0. If
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Updated on: 14 Jul 2019, 13:37
GMATPrepNow wrote: Archit3110 wrote: IMO A; check with below case x=1 and y=2 ; p = 3 x=4 and y=2 and p = 3 condition \(p^x < p^y\) i) \(x  y < 0\) yes for +ve integers and yes for ve integers ; YES ii) \(x < 2y\) yes for +ve integers and may or may not for ve integers ; NO iii) \(x^p < y^p\) No for ve case ; yes for ve integers case GMATPrepNow ; hope this is correct I'll give you a hint: Among the 3 answers given so far (A, C, E), one is correct Look again, GMATPrepNow. Archit3110 changed their answer to E.
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Originally posted by Palladin on 14 Jul 2019, 13:32.
Last edited by Palladin on 14 Jul 2019, 13:37, edited 1 time in total.



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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 13:36
Archit3110 wrote: GMATPrepNow must say it's a very mind wrecking question.. took over 2 mins coz there was no limit on integers range had it been mentioned that integers are +ve then it would had been a bit easy .. You need to find only one set of integers to eliminate each option. Isn't hard if you know number theory. Can be done within 2:30 minutes.
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Re: x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 13:47
Palladin wrote: Archit3110 wrote: GMATPrepNow must say it's a very mind wrecking question.. took over 2 mins coz there was no limit on integers range had it been mentioned that integers are +ve then it would had been a bit easy .. You need to find only one set of integers to eliminate each option. Isn't hard if you know number theory. Can be done within 2:30 minutes. Palladin GMAT doesn't give luxury of time to solve in 2:30 mins.. Posted from my mobile device



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x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 13:58
Archit3110 wrote: Palladin wrote: Archit3110 wrote: GMATPrepNow must say it's a very mind wrecking question.. took over 2 mins coz there was no limit on integers range had it been mentioned that integers are +ve then it would had been a bit easy .. You need to find only one set of integers to eliminate each option. Isn't hard if you know number theory. Can be done within 2:30 minutes. Palladin GMAT doesn't give luxury of time to solve in 2:30 mins.. Archit3110, 2 minutes per question is an ideal rule. It cannot always be met. Sometimes you finish a question within a minute. Sometimes it takes longer than 2 minutes. More practical rule is to complete 10 questions in 20 minutes. See the 2 minutes per question as more of an average rate of question completion and less of an explicit deadline for each question.
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x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 14:28
Archit3110 wrote: GMAT doesn't give luxury of time to solve in 2:30 mins..
Surprisingly, the stats (based on 17 sessions) suggest the average time (thus far) is 1:36 100% correct?? Hmmm.
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x, y and p are integers, and xyp ≠ 0. If
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14 Jul 2019, 15:57
GMATPrepNow wrote: x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x  y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only B) ii only C) iii only D) i and ii only E) none of the above Example 1: Let p =2, x=5, y=2...........2^5 < 2^2 i) \(x  y < 0\) From example 1 above, x >y.............Eliminate A & D ii) \(x < 2y\) From example 1 above, x >2y.............Eliminate B Example 2: Let p =1, x=3, y=4...........1^3 < 1^4 iii) \(x^p < y^p\) From example 1 above......3^1 > 4^1.............Eliminate C Answer: E



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Re: x, y and p are integers, and xyp ≠ 0. If
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15 Jul 2019, 06:33
GMATPrepNow wrote: x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x  y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only B) ii only C) iii only D) i and ii only E) none of the above Two important rules: ODD exponents preserve the sign of the base. So, ( NEGATIVE)^( ODD integer) = NEGATIVEand ( POSITIVE)^( ODD integer) = POSITIVEAn EVEN exponent always yields a positive result (unless the base = 0) So, ( NEGATIVE)^( EVEN integer) = POSITIVEand ( POSITIVE)^( EVEN integer) = POSITIVE So, one solution to the inequality \(p^x < p^y\) is \(p = 1\), \(x = 7\) and \(y = 2\) Plugging those values into the inequality, we get: \((1)^7 < (1)^2\) Simplify to get: \(1 < 1\), WORKS. Now plug \(p = 1\), \(x = 7\) and \(y = 2\) into the three statements to get: i) \(7  2 < 0\) Simplify to get: \(5 < 0\) NOT true. So, statement i need not be true. ii) \(7 < 2(2)\) Simplify to get: \(7 < 4\) NOT true. So, statement ii need not be true. iii) \(7^{1} < 2^{1}\) Simplify to get: \(\frac{1}{7} < \frac{1}{2}\) This is TRUE. So, we can't (yet) conclude that statement iii need not be true.  Let's see if any other values will show that statement iii need not be true. Another solution to the inequality \(p^x < p^y\) is \(p = 1\), \(x = 1\) and \(y = 2\) Plugging those values into the inequality, we get: \((1)^1 < (1)^2\) Simplify to get: \(1 < 1\), WORKS. Now plug \(p = 1\), \(x = 1\) and \(y = 2\) into statement iii to get: iii) \(1^{1} < 2^{1}\) Simplify to get: \(\frac{1}{1} < \frac{1}{2}\) NOT true. So, statement iii need not be true. Answer: E Cheers, Brent
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Re: x, y and p are integers, and xyp ≠ 0. If
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15 Jul 2019, 15:33
I) is not true, when p<0, and x and y are positive odd integers x=3 and y=1 and p=2 II) is not true, when p<0, and x/y>2, where x and y are positive odd integers. Can consider the same case as in statement I III) is not true, when p is positive even integer, and x and y are negative integers. p=4, x=4 and y=2 GMATPrepNow wrote: x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x  y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only B) ii only C) iii only D) i and ii only E) none of the above



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Re: x, y and p are integers, and xyp ≠ 0. If
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16 Jul 2019, 05:45
GMATPrepNow wrote: Surprisingly, the stats (based on 17 sessions) suggest the average time (thus far) is 1:36 100% correct?? Hmmm. Oops. I just realized that, when I delay the appearance of the official answer by 24 hours, all responses are deemed correct until the correct answer is displayed (now the success rate is 50%) Cheers, Brent
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Re: x, y and p are integers, and xyp ≠ 0. If
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