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

|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)

\(|x – \frac{9}{2}| = \frac{5}{2}…|x-4.5|=2.5\)
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

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!!

Hi GMATWhizTeam , Bunuel ,

Why can't y be 0? y is the median and if the set of consecutive numbers is say (-1,0,1), then y=0.

And then none of the options are correct. Let me know if I am making a mistake. Thanks

Check out this post for more on this concept:
https://anaprep.com/algebra-the-why-beh ... questions/

I would like to draw everybody's attention to this perspective which I'm sharing below:-

If you find the explanation to be correct and helpful then do acknowledge it and if it's not relevant or incorrect then also do comment.

Please bear with me for sometime for the long explanation.

Here I go with my explanation-

I believe this question has nothing to do with the value of 'x' initially.
It will only come into picture for a small time.
So we need to know the two values of 'x' i.e 7 & 2 one is odd and the other is even.
We can very much draw a concrete inference about all the statements I, II and III whether they are must be true (100% true) or not.

Let's first analyze statement II.
It can be re-written as xyp(p+1).
Now the question clearly says that 'p' is odd. Then (p+1) will be even for sure. If (p+1) is even then Statement II can be concluded to be even.
It means statement II is 100% must be true.
So I can easily eliminate B and C.

Now look at Statement I and III very carefully, they are of similar nature i.e, let say if 'A=xyp' is even then A^2 will be even.
Similarly if 'A' is odd then A^2 will be odd, which means Statement I and III are identical.

Now let's assume that Statement I is odd and it is true.

Then Statement III will also be odd, which means it cannot be even and hence Statement III is false.

That means Statement I and III will never be both true at the same time and hence cannot be together in the answer option.

So option E can be eliminated.

If you are having trouble in understanding the above explanation why E is eliminated, then let me give you a solid reason to check I and III.

I hope the reasoning about Statement I and III being identical is clear to you all, because I'll be checking for statement I and applying the deduction to Statement III as well.

Let's analyze Statement I -

Its says 'xyp' - remember I told you that we will not be taking the value of 'x' into consideration initially. .

'p' is odd (given in the question).
Now 'y' can be both even and odd.
Let say 'y' is even then 'xyp' will be even. Statement I is false.
If 'xyp' is even then (xyp)^2 should also be even. In this case Statement III will be true.

It means that I and III will never be both true at the same time and hence will never occur together in the answer. Hence option-E was eliminated.

Similarly when you take 'y' to be odd ,and now comes 'x' into picture for this short duration, and 'x' to be odd then 'xyp' will be odd which translates into Statement I as true and Statement III as false.
Hence we cannot say with 100% surety that either of I or III is correct.

So only II is 100% true and option-A is the correct answer.

Posted from my mobile device

In that case Statement II will be 0 and zero is an even number, hence Statement II is correct and for all other values also Statement II is always even.
Hence Statement II is always even and 100% true for all cases.

But for Statement III it will be 0, and in this case III will be true, but for other cases III can be odd. So, it cannot said with certainty that III is always true.

Posted from my mobile device

Thanks "sumit99kr" for the detailed explanation.