Two sets each contain 3 consecutive positive integers. None of the int

Question

Two sets each contain 3 consecutive positive integers. None of the integers included in one set are included in the other set. The two sets are combined to form a set of 6 ordered positive integers. If the median of the combined set is also an integer, which of the following could be the range of the integers in the new set?

I. 15
II. 16
III. 17

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

I. 15
II. 16
III. 17

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

rajatchopra1994 · Answer

Explanation:

The new set will contain 6 integers. 
Median will be the average of 3rd & 4th integer, Sum of 3rd & 4th integer has to be even for median to be an integer.

if 3rd and 4th integer are odd, then 1st and 6th integer will be odd too, therefore range will be even
If 3rd and 4th integer are even, then 1st and 6th integer will be even too, therefore range will be even

So only 16 is the option.

IMO-B

manikmehta95 · Answer

Two sets each contain 3 consecutive positive integers. None of the integers included in one set are included in the other set. The two sets are combined to form a set of 6 ordered positive integers. If the median of the combined set is also an integer, which of the following could be the range of the integers in the new set?

I. 15
II. 16
III. 17

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

I. 15
II. 16
III. 17

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

let 1st set = {(x-1), x, (x+1)}
2nd set = {(y-1), y, (y+1)}

Average of both sets = \frac{(3*x + 3*y)}{6} 
According to question (x + y) has 2 as a factor. Therefore, either both are even or both are odd.

Range = |x-y| + 2 
from the above expression we see that is always even in both cases.

Ans. B

sambitspm · Answer

Kindly see the attachment. The link between set1 and set2 is important to evaluate.
B

hacorus · Answer

This is how I did it:

Range = max value - lowest value.
We have 2 sets of 3 consecutive positive integers
Median is an integer so the sets are both starting with odd numbers or bot starting with even numbers.

could we have a set with a range of 15? 1-2-3-14-15-16 has a range of 15, however median is not an integer (8.5)
could we have a set with a range of 16? 1-2-3-15-16-17 has a range of 16, and a median of 9 so it works.
could we have a set with a range of 17? 1-2-3-16-17-18 has a range of 17, however median is not an integer (9.5)

Answer is B, II only.

AnirudhaS · Answer

think about it this way, how do we get an integer for a set of even number of integers - well the sum of the middle elements must add up to EVEN
How can we achieve an EVEN sum? 2 cases -

Case I: Both middle terms are odd, for e.g. {1,2,3,5,6,7}
If both middle terms are odd, both extreme terms must be odd as well (as 6 terms in total)
Range=ODD-ODD=EVEN

Case II: Both middle terms are EVEN, for e.g. {2,3,4,6,7,8}
If both middle terms are EVEN, both extreme terms must be EVEN as well (as 6 terms in total)
Range=EVEN-EVEN=EVEN

Only even range is 16, hence the answer.

Answer:  B

vipulshahi · Answer

S1 = { n,n+1,n+2}
S2 = { m,m+1, m+3}

Given: Median of combined set is integer

Required: Range of combined set

Solution: After combining the two sets S1 & S2,
S3 = { n,n+1,n+2,m, m+1, m+2} 
median is integer
it means   / 2 = integer
Thus, (n+2) + (m) = even

Possibility (1) : (n+2) & (m) both are even
it means n & m+2 are even 
then range is even.

Possibility (2) : (n+2) & (m) both are odd
it means n & m+2 are odd
even then also range is even

Among given option , only 16 is even.

Answer : B

Papist · Answer

In order for the median to be an integer, the middle two digits must either both be even or both be odd. By extension, the first and last digit must either both be even or both be odd. Two numbers can only produce an odd solution (i.e. range) when they one is even and the other is odd, so both odd options can be eliminated.

Neha2404 · Answer

IMO -B

Assume 
Set1 -a,a+1, a+2
Set-2 -b,b+1,b+2
Set -3. (a,a+1, a+2,b,b+1,b+2)

Median of the set is =(a+2+b)/2
Median will be integer only if total of (a+b) is even

And we will get even Number only
Which is 16

Posted from my mobile device

IanStewart · Answer

Just in case this sentence is confusing for any test takers: sets are, by definition, unordered. If something is in order, it's a sequence, not a set. So it doesn't make sense to talk about a set of "ordered positive integers", and the word "ordered" should be deleted from the question (it doesn't serve any purpose anyway).

I won't solve the problem since there are a half-dozen good solutions above.

mSKR · Answer

e.g.:

SetA: 2,3,4

SetB: 15,16,17

Range = 17-2 = 15
but median should be an integer.
4+15= 19/2= 9.5 - not an integer

It means the range should be even number
it means only 16 is possible answer among available options

II- B

Oppenheimer1945 · Answer

Set 1={odd, even, odd}, set 2={odd, even, odd}

or

Set 1={even, odd, even}, set 2={even, odd, even}

Then only median of new set is even.

Combining the sets. Range is either odd-odd or even-even=even . hence only 16

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Two sets each contain 3 consecutive positive integers. None of the int

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Two sets each contain 3 consecutive positive integers. None of the int

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