mgmat 700 -800 mean and median

Hi Pradeep,

|x - 4.5| = 2.5 means that x-4.5=2.5 or x-4.5=-2.5. So x is either 2 or 7.

Since p is odd, the median of a set of p consecutive integer will be the middle integer. In other words, y is an integer. y could be even or odd. So to summarize.

x = 2 or 7
p = odd integer
y = integer

What must be true?

I. If x = 2, then this will be false.

II.Since p is odd, p^2+p is odd+odd=even. Thus the product of (p^2+p) and any other integer is even. II must be true.

III. We know that p is odd and x can be odd, so x^2p^2 can be odd. If y is also odd, then this statement will be false.


Only II must be true. The answer is A
-Patrick

pradeepkaushal9518 wrote:If |x – 9/2 | =5/2, and if y is the median of a set of p consecutive integers, where p isodd, which of the following must be true?

I. xyp is odd

II. xy(p2 + p) is even

III. x2y2p2 is even
II only
III only
I and III
II and III
I, II, and III

|x – 9/2 | =5/2
if x>9/2
then x= 7
if x<9/2
x=2
y is median of p consecutive integers.
p is odd
Let us look at options:
I says xyp is odd. x can be either 7 or 2. Hence, we do not know if its odd or even. y can be odd or even. p is odd.
not useful.
CE out.
same problem exists with III statement.
BD out.

Answer should be A.
Let us confirm this:
we know p is odd.
So p^2 will be odd.
odd + odd = even
so p^2 +p is even.
even *(even or odd) = even
so II is correct.

hope this helps!!

I just want to confirm my understanding here.
Before we break it down, we know that X and Y can both be positive and negative.

If we jump to I. - xyp is odd
Here, we know that xyp must be odd. However, because X and Y can be positive or negative we can not say that it is odd - it is insufficient.
Nonetheless, fact I is telling us that xyp must be odd. so does X, Y, and P have to be odd?
Perhaps, I'm getting my questions mixed up...?

(II. xy(p^2 + p) is even --- I get it.
III. x^2y^2p^2 is even --- Same with I, where X and Y can be positive or negative so they are insufficient.)

I see, here you are trying to prove for I., II. and III. I see what I did wrong.

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(p² + p) is even

III. x²y²p² is even

II only
III only
I and III
II and III
I, II, and III

Case 1: No signs changed
x - 9/2 = 5/2
x = 14/2
x = 7.

Case 2: Signs changed on one side
x - 9/2 = -5/2
x = 4/2
x = 2.

Try to prove that the statements DON'T have to be true.
Let p=1, implying that there is only one integer in the set.
Let the set = {1}.
Since the median of the set is 1, y=1.

Statement I: xyp is odd
If x=2, y=1 and p=1, then xyp = 2*1*1 = 2, which is NOT odd.
Since statement I does not have to be true, eliminate C and E.

Statement II: xy(p² + p) is even
If x=2, y=1 and p=1, then xy(p² + p) =(2)(1)(1² + 1) = 4, which is even.
If x=7, y=1 and p=1, then xy(p² + p) =(7)(1)(1² + 1) = 14, which is even.
Hold onto statement II.

Statement III: x²y²p² is even
If x=2, y=1 and p=1, then x²y²p²= 2²1²1² = 4, which is even.
If x=7, y=1 and p=1, then x²y²p²= 7²1²1² = 49, which is NOT even.
Since statement III does not have to be true, eliminate B and D.

The correct answer is A.