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

Question

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

Are You Up For the Challenge: 700 Level Questions

chetan2u · Accepted Answer

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

GMATWhizTeam · Answer

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  .

KarishmaB · Answer

|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)

exc4libur · Answer

|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)

Michele4 · Answer

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

chetan2u · Answer

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}

sashankagarwal78 · Answer

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

KarishmaB · Answer

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

sumit99kr · Answer

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

sumit99kr · Answer

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

akshitaa_goyal · Answer

can someone explain why the value of x as 2 is acceptable when we only considered x to be negative when opening the modulus in that scenario, shouldn't we reject if we get negative x in this case? I'm so confused

Bunuel · Answer

|x – 9/2| = 5/2

x – 9/2 = 5/2 or x – 9/2 = -5/2

x = 7 or x = 2
 
 10. Absolute Value 
     Theory   
 Math: Absolute value (Modulus) 
 Absolute Value: Tips and hints 
 Properties of Absolute Values on the GMAT 
 Absolute modulus : A better understanding 
 GMAT Question Patterns: Abolute Value 
 The Reason Behind Absolute Value Questions on the GMAT 
     Questions   
 Inequality and absolute value questions from my collection 
 The E-GMAT Question Series on ABSOLUTE VALUE 
 DS Questions 
 PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

ManishaPrepMinds · Answer

Solution:

Given |x - 9/2| = 5/2:

Solve for x:
Case 1: x = 7
Case 2: x = 2
Given y is the median of p consecutive integers with p odd, y is an integer.

Evaluate Statements:

Statement I: xyp is odd

If x = 7 (odd), xyp is odd; if x = 2 (even), xyp is even.
Not necessarily true.
Statement II: xy(p^2 + p) is even

Since p is odd, p^2 + p is even. Thus, xy(p^2 + p) is always even.
True.
Statement III: x^2y^2p^2 is even

If x = 2, x^2y^2p^2 is even; otherwise, it depends on x and y.
Not necessarily true.
Conclusion: Only Statement II is always true.

Final Answer: II only

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards