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

Question

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

GeorgeKo111 · Answer

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

so correct answer is (C)

BrentGMATPrepNow · Answer

Be careful; (-2)^-4 = 1/16 (not -1/16)

GeorgeKo111 · Answer

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?

BrentGMATPrepNow · Answer

You missed something. 
Keep at it!

Archit3110 · Answer

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

Palladin · Answer

Given:
 
  x, y, p \in \mathbb{Z}_{>0}  
  x*y*z 
e 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   .

BrentGMATPrepNow · Answer

I'll give you a hint: Among the 3 answers given so far (A, C, E), one is correct

Archit3110 · Answer

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 ..

Posted from my mobile device

Apt0810 · Answer

Answer is option E?

Posted from my mobile device

Palladin · Answer

Look again,  GMATPrepNow .  Archit3110  changed their answer to E.

Palladin · Answer

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.

Archit3110 · Answer

Palladin  GMAT doesn't give luxury of time to solve in 2:30 mins..

Posted from my mobile device

Palladin · Answer

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.

BrentGMATPrepNow · Answer

Surprisingly, the stats (based on 17 sessions) suggest the average time (thus far) is 1:36 100% correct?? Hmmm.

Mo2men · Answer

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

BrentGMATPrepNow · Answer

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 .

Answer: E

Cheers,
Brent

nick1816 · Answer

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

BrentGMATPrepNow · Answer

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

bumpbot · Answer

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

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