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If |x – (9/2) | = 5/2 , and if y is the median of a set of p consecutive integers, where p is odd, which of the following must be true?

I. \(xyp\)is odd

II. \(xy(p^2 + p)\) is even

III. \(x^2y^2p^2\) is even

A. II only B. III only C. I and III D. II and III E. I, II, and III

x is in range of 2 and 7 so it can be even or odd y is median of set of odd integers so it can be even or odd p is given as odd using above information only II for sure can be inferred as even. other two, either can be even or odd correct response is A.

If |x – (9/2) | = 5/2 , and if y is the median of a set of p consecutive integers, where p is odd, which of the following must be true?

I. \(xyp\)is odd

II. \(xy(p^2 + p)\) is even

III. \(x^2y^2p^2\) is even

A. II only B. III only C. I and III D. II and III E. I, II, and III

x is in range of 2 and 7 so it can be even or odd y is median of set of odd integers so it can be even or odd p is given as odd using above information only II for sure can be inferred as even. other two, either can be even or odd correct response is A.

X is not between 2 and 7. X is either 2 or 7. But the rest of the explanation is correct I believe.

Re: If |x – 9/2| = 5/2, and if y is the median of a set of p [#permalink]

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21 Jul 2014, 22:25

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Re: If |x – 9/2| = 5/2, and if y is the median of a set of p [#permalink]

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31 Mar 2016, 19:11

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If |x – 9/2| = 5/2, and if y is the median of a set of p [#permalink]

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03 Oct 2016, 06:41

mads wrote:

If |x – 9/2| = 5/2, and if y is the median of a set of p consecutive integers, where p is odd, which of the following must be true?

I. xyp is odd II. xy(p^2 + p) is even III. x^2y^2p^2 is even

A. II only B. III only C. I and III D. II and III E. I, II, and III

I got to A.

from the given info, x is either 2 or 7. p is definitely odd. y can be either odd or even.

1. might be true. x can be 2, and in this case, xyp is even. C and E are out.

2. xy(p^2 +p) p is odd. p^2+p = even. so everything, regardless of x and y, will be even. B is out.

3. x can be even or odd; y can be even or odd; p is odd. since all 3 variables can be odd, there is a possibility that the number is not even. D is out

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