We are looking at whether the variables are odd or even.

\(|x – \frac{9}{2}| = \frac{5}{2}\)........
\(x = \frac{9}{2}+ \frac{5}{2}=7\) or \(-x + \frac{9}{2}| = \frac{5}{2}......x=2\)
So x can be both even and odd

p is odd

y can be even or odd number, for example 3,4,5,6,7 and 3,4,5 ---- median or y is 5 and 4

Now let us look at the choices

I. \(xyp\) is odd
If both x and y are odd, yes, otherwise no

II. \(xy(p^2 + p)\) is even
p^2+p is odd+odd or even, so \(xy(p^2 + p)\) is always even.

III. \(x^2y^2p^2\) is even
If both x and y are odd, yes, otherwise no

only II

A

[quote="Bunuel"]If \(|x – \frac{9}{2}| = \frac{5}{2}\), 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

Solution

• \(|x-\frac{9}{2}| = \frac{5}{2}\)

o If \(x > \frac{9}{2} \)then \(x-\frac{9}{2} = \frac{5}{2}\) \(⟹x =7\)
o If \(x < \frac{9}{2}\) then \(x-\frac{9}{2} = \frac{-5}{2} \)\(⟹x =2\)
o Hence, x can be even (i.e. x =2) or odd (i.e. x =7)
• p is odd, so median of consecutive p integers will be an integer. i.e. y is an integer.

o let us take two simple examples, if p = 5 and

 Case 1: consecutive integers are \({2, 3, 4, 5, 6} \) then y =4 which is even
 Case 2: consecutive integers are \({3, 4, 5, 6, 7}\) then y = 5 which is odd.
 Hence, y can be even or odd.

Now, let us analyse the given statements:

• I. \( xyp\) is odd.

o For \(xyp\) to be odd, all three i.e. x and y and p must be odd.

 p is odd but x and y may be odd or may be even.
o So, this statement is not always true, and we can eliminate options C and E.
• II. \(xy(p^2+p) \)is even.

o We can rewrite it as, \(xy(p^2+p) = xyp(p+1)\).

 For above expression to be even any one of x , y, p and (p+1) must be even.
o Since p is odd so p+1 is even hence \(xy(p^2+p)\) is always even.
o We can eliminate answer option B.
• III. \(x^2y^2p^2\) is even.

o \(x^2y^2p^2\) to be even, at least one of them must be even.

 We know p is odd. However, x and y may be even or may not be even.
 Hence, this statement is not always true.

Thus, the correct answer is Option A.
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|x - 4.5| = 2.5 so x is a point 2.5 away from 4.5.
Then x = 2 or 7

y is the median of a set of p consecutive integers, where p is odd,
So y is median of 2, 3, 4 (y is 3)
or y is median of 3, 4, 5 (y is 4)
y may be odd or even.

I. \(xyp\) is odd
Not necessary. If x =2, this is even.

II. \(xy(p^2 + p)\) is even
p is odd so p^2 is odd too. Then, (p^2 + p) = Even because Odd + Odd = Even.
Since one term is even, the whole product becomes even. Correct.

III. \(x^2y^2p^2\) is even
Not necessary, x, y and p, all may be odd.

Answer (A)
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\(|x – \frac{9}{2}| = \frac{5}{2}…|x-4.5|=2.5\)
x>4.5 (pos): (x-4.5)=2.5…x=7…(7>4.5=valid)
x<4.5 (neg): -(x-4.5)=2.5…x=2…(2<4.5=valid)
x=2 (even) or 7 (odd)

p=odd

y=median=mean of odd number of consecutive integers
y=integer odd or even

I. \(xyp\) is odd
x=even,odd y=even,odd p=odd
xyp=eoo=even
xyp=ooo=odd

II. \(xy(p^2 + p)\) is even
pˆ2+p=oˆ2+o=o+o=even
xy(even)=eo(even)=even
xy(even)=ee(even)=even
xy(even)=oo(even)=even

III. \(x^2y^2p^2\) is even
eˆ2eˆ2oˆ2=eeo=even
oˆ2oˆ2oˆ2=ooo=odd

Ans (A)

Bunuel chetan2u when studying |x-9/2|= 5/2 I considered two cases:
1 - IF X<=0 the solution is x=2 BUT I rejected it because x has to be <=0.
2 - IF X>0 the solution is compatible and it is x=7
What is wrong with this approach?
In this question https://gmatclub.com/forum/what-is-the-product-of-all-the-solutions-of-x-2-4x-147152.html sir Bunuel rejected both solutions for X because they were outside the range.
Please help me because I know that there has to be some really silly thing that I am missing
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You are wrong in taking the values of the critical points.

Here critical point is \(\frac{9}{2}\)
1) If \(x\leq{\frac{9}{2}}\), then \(x-\frac{9}{2}\leq 0\), and \(|x-\frac{9}{2}|=\frac{9}{2}-x\)
\(\frac{9}{2}-x=5/2.........x=2\)...Valid value as \(2<\frac{9}{2}\)
2)vIf \(x>{\frac{9}{2}}\), then \(x-\frac{9}{2}>0\), and \(|x-\frac{9}{2}|=x-\frac{9}{2}\)
\(x-\frac{9}{2}=5/2.........x=7\)...Valid value as \(7>\frac{9}{2}\)

You are wrong in taking the values of the critical points.

Here critical point is \(\frac{9}{2}\)
1) If \(x\leq{\frac{9}{2}}\), then \(x-\frac{9}{2}\leq 0\), and \(|x-\frac{9}{2}|=\frac{9}{2}-x\)
\(\frac{9}{2}-x=5/2.........x=2\)...Valid value as \(2<\frac{9}{2}\)
2)vIf \(x>{\frac{9}{2}}\), then \(x-\frac{9}{2}>0\), and \(|x-\frac{9}{2}|=x-\frac{9}{2}\)
\(x-\frac{9}{2}=5/2.........x=7\)...Valid value as \(7>\frac{9}{2}\)[/quote]

chetan2u got it now, thanks!!
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