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# x, y and p are integers, and xyp ≠ 0. If

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x, y and p are integers, and xyp ≠ 0. If  [#permalink]

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14 Jul 2019, 07:09
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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  [#permalink]

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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
x-y= 3-4=-1<0

now suppose P=-2 X=-3 and Y=-4

P^X= -2^-3= = -1/8< P^Y=-2^-4 = 1/16

x-y=-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

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  [#permalink]

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14 Jul 2019, 09:14
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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  [#permalink]

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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  [#permalink]

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14 Jul 2019, 09:17
Top Contributor
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  [#permalink]

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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  [#permalink]

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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.

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Re: x, y and p are integers, and xyp ≠ 0. If  [#permalink]

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14 Jul 2019, 11:39
Top Contributor
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  [#permalink]

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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

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Re: x, y and p are integers, and xyp ≠ 0. If  [#permalink]

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14 Jul 2019, 11:58

Posted from my mobile device
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x, y and p are integers, and xyp ≠ 0. If  [#permalink]

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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  [#permalink]

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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  [#permalink]

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14 Jul 2019, 13:47
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  [#permalink]

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14 Jul 2019, 13:58
Archit3110 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  [#permalink]

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14 Jul 2019, 14:28
Top Contributor
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  [#permalink]

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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

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Re: x, y and p are integers, and xyp ≠ 0. If  [#permalink]

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15 Jul 2019, 06:33
Top Contributor
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) = NEGATIVE
and (POSITIVE)^(ODD integer) = POSITIVE

An EVEN exponent always yields a positive result (unless the base = 0)
So, (NEGATIVE)^(EVEN integer) = POSITIVE
and (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.

Cheers,
Brent
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Re: x, y and p are integers, and xyp ≠ 0. If  [#permalink]

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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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Posts: 4331
Re: x, y and p are integers, and xyp ≠ 0. If  [#permalink]

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16 Jul 2019, 05:45
Top Contributor
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   [#permalink] 16 Jul 2019, 05:45
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