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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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Updated on: 06 Jul 2013, 12:27
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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
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Originally posted by mads on 18 Apr 2010, 16:27.
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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Re: MGMAT question
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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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Re: MGMAT question
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18 Apr 2010, 19:12
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 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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Re: MGMAT question
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06 Jul 2013, 12:16
einstein10 wrote: 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 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
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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. Answer = A. 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



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Re: If x – 9/2 = 5/2, and if y is the median of a set of p
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03 Oct 2016, 07:41
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 A is the answer.



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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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22 Oct 2016, 02:32
x= 7 or x=2, So x can be either odd or even
x9/2=5/2, we have two different cases
1/ x9/2>0 => x9/2=x9/2 and x>9/2 x9/2 = 5/2 => x = 14/2 = 7
2/ x9/2<0 => x9/2=(x9/2) = 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
II only, answer choice A



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