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

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

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18 Apr 2010, 16:27
4
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9
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Difficulty:

65% (hard)

Question Stats:

65% (02:55) correct 35% (01:58) wrong based on 506 sessions

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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
[Reveal] Spoiler: OA

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+1Kudos, if this helps

Last edited by Bunuel on 06 Jul 2013, 12:27, edited 3 times in total.
Edited the question and added the OA
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18 Apr 2010, 16:51
A - II only

P^2+P will always be even because p is odd adding an odd to an odd is even.

x can be 2 or 7 so it can be odd or even which makes it impossible for I or III to always be true
also, y can be odd or even

-ANy time an even number is in a multiple the product is even.
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18 Apr 2010, 19:12
1
KUDOS
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.
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06 Jul 2013, 12:16
einstein10 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

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.
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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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21 Jul 2014, 23:25
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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, 20:11
Hello from the GMAT Club BumpBot!

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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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14 May 2016, 23:46
given:
p is Odd
y is the median of odd numbers, hence y is Odd
x = solving the modulus you get 7 and 2. Hence x is Even or Odd

Therefore: p = O, y = O and x = O or E

Statement I = x*y*p = y*p*x = O x O x O/E = O x O/E = Odd or even. So False

Statement II = xy(p^2 + p) = O/E x O (O + O) = O/E x O(E) = O/E x E = Always Even. So true

Statement III. x^2y^2p^2 is even. E/O x O x O = Odd or even. Hence False.

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
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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, 07:41
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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If |x – 9/2| = 5/2, and if y is the median of a set of p [#permalink]

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22 Oct 2016, 02:32
x= 7 or x=2, So x can be either odd or even

|x-9/2|=5/2, we have two different cases

1/ |x-9/2|>0 => |x-9/2|=x-9/2 and x>9/2
x-9/2 = 5/2 => x = 14/2 = 7

2/ |x-9/2|<0 => |x-9/2|=-(x-9/2) = -x + 9/2 and x<9/2
-x + 9/2 = 5/2 => x = 9/2 - 5/2 = 4/2 = 2

p is odd from the question

y can be either odd or even
{1,2,3} median is 2 even
{1,2,3,4,5} median is 3 odd

I. xyp is odd
Not necessarily if x is even

II. xy(p^2+p) is even
p is odd, so p² is odd and odd + odd = enve
so, xy(p2+p) is necessarily even

III. x^2y^2p^2 is even
not necessarily if x is odd, y is odd and p is odd

If |x – 9/2| = 5/2, and if y is the median of a set of p   [#permalink] 22 Oct 2016, 02:32
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