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Re: x, y and p are integers, and xyp ≠ 0. If [#permalink]
I got the first equation as right, this is what I did, p^y- p^x > 0, y-x>0 therefore, x-y<0

GreenlightTestPrep wrote:
GreenlightTestPrep 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.

Answer: E

Cheers,
Brent
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x, y and p are integers, and xyp ≠ 0. If [#permalink]
1
Chaithraln2499 wrote:
I got the first equation as right, this is what I did, p^y - p^x > 0, y - x >0 therefore, x-y<0


In many cases, the greater the exponent the greater the value of the power.
For example, since \(8 > 5\), it's also true that \(2^8 > 2^5\)
Similarly, since \(6 > 4\), it's also true that \(10^6 > 10^4\)
Notice that the two bases, 2 and 10, are both greater than one.

Also notice that the property does not hold when the base is less than 1.
For example, \(3 > 2\), BUT \(0.5^3 \) is less than \(0.5^2\).

So, if \(p = 0.5\), \(y = 2\) and \(x = 3\), then it's true that \(p^y - p^x > 0\), HOWEVER we can't conclude that \(x-y<0\), which means statement I is not necessarily true.
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Re: x, y and p are integers, and xyp 0. If [#permalink]
Hi Brent GreenlightTestPrep, is there another way to solve this?
As testing number can miss out easily as when I test x=2, y=3 , p=2 and I got I & III true.
Thanks Brent
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Re: x, y and p are integers, and xyp 0. If [#permalink]
Expert Reply
Aside from the explanation provided by Brent, which is amazing

x,y, and p are integers but we DO NOT have any information about

Of the three inequalities, we can have infinite results. It depends on the numbers we do have in hand

E is the only way to conceptualize this. None of the above

These questions are created specifically to waste your time that there will be a possible answer.

It is not case, though
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