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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
1
set-1: 12,13,14
set-2: 16,17,18

new set: 12,13,14,16,17,18

median: (14+16)/2=15
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
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souravp94 wrote:
set-1: 12,13,14
set-2: 16,17,18

new set: 12,13,14,16,17,18

median: (14+16)/2=15


Wrong. Please read the question carefully.
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
1
Official Explanation:

In this question, each of the two sets contains 3 consecutive positive integers and the two sets have no common numbers. We can easily imagine how these two sets could be, for example, one option would be 3, 4, 5 and 9, 10, 11.

When the two sets are combined to form a new set of 6 ordered numbers, all numbers are distinct. This is because if one number appears twice, this would mean that either the original set that the number comes from does not include consecutive integers or the two original sets have common numbers. Neither of these is possible according to the information given in this question.

We know that the median of this ordered list of 6 numbers is the average of the 3d and 4th integer which are not equal as we explained above.

This is the first hard part of this question:

How must the 3d and 4th integers relate so that their average is also an integer?

The average of two integers is also an integer if

a) the two integers are equal which is not possible according to the information given or

b) the two integers differ by an even number 2, 4, 6,...

Does b) make sense? Yes, because if you have an integer x and another integer that is say

x + 8, then their sum is always a multiple of 2 because you write it as

x + x + 8 = 2x + 8 = 2(x+4)

And thus it is divided by 2 and gives an integer. And this is the case for any even number that you can have in 8’s place above.

Also you can intuitively check b) by taking random pairs of integers that they differ by an even number. For example, 5 and 7 which differ by 2 have average equal to 6 because 5 + 7 = 12 and \(\frac{12}{2}\) = 6. Or 12 and 22 which differ by 10 have average equal to 17 because 12 + 22 = 34 and \(\frac{34}{2}\) = 17

So, the difference of the 3d and 4th integers is an even.

And how are we going to somehow relate this result to the range of the data that we finally want to find?

Here is the second tricky part of the question.

Consider the following cases:

1) If the 3d number is even then the 4th number is also an even since they differ by an even number. But since 1st, 2nd, and 3d are consecutive then 1st is also even. And since the 4th, 5th and 6th are consecutive, then the 6th number is also an even. The range is the difference of the 1st number from the 6th number and thus a difference of two even numbers which gives an even.

2) If the 3d number is odd then the 4th number is also an odd since they differ by an even number. But since 1st, 2nd, and 3d are consecutive then 1st is also odd. And since the 4th, 5th and 6th are consecutive, then the 6th number is also an odd. The range is the difference of the 1st number from the 6th number and thus a difference of two odd numbers which gives an even.

Therefore, we conclude that the range must always be an even number and the correct answers are all choices that contain even numbers (A), (C), and (E).
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
If CLUB X HAS 40 members and CLub Y has 28 members.If 8 people are members of both the clubs how many people are members of at least one of the two clubs?

CAN SOMEONE EXPLAIN THIS ??
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
Expert Reply
Sir

you cannot post a question along with an open discussion which is completely unrelated to.....

Read the rules of the board https://gre.myprepclub.com/forum/rules-for ... -1083.html

Thank you for your collaboration

Regards
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
2
Solution:

Let the set of 1st consecutive integer be: x-1,x,x+1
Let the set of the other consecutive integer be: y-1,y,y+1

Now, the first thing we are said here is, the median of the new set is an integer:
New set={x-1,x,x+1,y-1,y,y+1}
Therefore, the median will be \(\frac{x+1+y-1}{2}\)=\(\frac{x+y}{2}\)

This simply means that x+1 & y-1 both are either even number or both are odd.
Lets consider both are odd then the new set will have number {odd, even,odd,odd,even,odd} and when we add them we get even(Even times odd yields even value)
Lets consider both are even te the new set will be {even,odd,even, even, odd, even} similarly this will also yield an even value.

Now, to the final part Odd-Odd=Even & Even-Even=Even
Thus, the range should be even value.

Option A, C & E satisfies.


P.s. Knowing basics like the sum of odd and even number etc can not only save time but also helps you to get accurate answers.

Hope this helps.
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
Consider set 1 as 1,4,9 and set 2 as 2,8,18. When these sets are combined, the median would be 4+8/2 which is 6 which is an integer. In this case, the range is 18-1 which is 17. Therefore shouldn't F also be considered as an option?
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
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vangapalliamulya wrote:
Consider set 1 as 1,4,9 and set 2 as 2,8,18. When these sets are combined, the median would be 4+8/2 which is 6 which is an integer. In this case, the range is 18-1 which is 17. Therefore shouldn't F also be considered as an option?


Consecutive numbers are numbers that follow each other in order. They have a difference of 1 between every two numbers.
i.e. n, (n+1), (n+2), .....

Consecutive even numbers are even numbers that follow each other in order. They have a difference of 2 between every two numbers.

Consecutive odd numbers are odd numbers that follow each other in order. They have a difference of 1 between every two numbers.

Let's take your example: 1, 4, 9 are consecutive what?? They form an A.P!

Also, the second set: 2, 8, 18 neither forms an A.P nor are consecutive!

8 - 2 = 6
18 - 8 = 10
They don't have a common difference.
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
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rx10 wrote:
Great question:

Two sets with 3 consecutive positive integers.

Given : All the no in both the sets are distinct. & when combined the median is integer.

Now, Take set A {x , y , z } & set B {p , q , r} & consider z < p

Combining { x , y , z , p , q , r }
median will be \(\frac{z+p}{2}\) , which is an integer hence both must be even

z & p are even. all are consecutive integers so x , z , p , r = even & y , q are odd

Range = r - p = even - even = even no.

Answer A C E


rx10 Thank you the explanation.
I was taking the standard way i.e x, x+1 etc.
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
rx10 wrote:
Great question:

Two sets with 3 consecutive positive integers.

Given : All the no in both the sets are distinct. & when combined the median is integer.

Now, Take set A {x , y , z } & set B {p , q , r} & consider z < p

Combining { x , y , z , p , q , r }
median will be \(\frac{z+p}{2}\) , which is an integer hence both must be even

z & p are even. all are consecutive integers so x , z , p , r = even & y , q are odd

Range = r - p = even - even = even no.



Answer A C E


what if 1,2,3 and 16,17,18 we take
then the range will be 17
??
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
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Hey drasticgre

It is given that the median is an integer.

With \(1,2,3\) & \(16,17,18\) the median will be \(\frac{3+16}{2}\) = not an integer


drasticgre wrote:
what if 1,2,3 and 16,17,18 we take
then the range will be 17
??
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
drasticgre wrote:
what if 1,2,3 and 16,17,18 we take
then the range will be 17
??


ohh, I missed it. thank you.
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
rx10 wrote:
Great question:

Two sets with 3 consecutive positive integers.

Given : All the no in both the sets are distinct. & when combined the median is integer.

Now, Take set A {x , y , z } & set B {p , q , r} & consider z < p

Combining { x , y , z , p , q , r }
median will be \(\frac{z+p}{2}\) , which is an integer hence both must be even

z & p are even. all are consecutive integers so x , z , p , r = even & y , q are odd

Range = r - p = even - even = even no.

Answer A C E



Although I agree with your final solution, you must also consider that both z & p could be odd (for example, 3 and 5 average to 4).
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Re: Two sets each contain 3 consecutive positive integers. None [#permalink]
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