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Two sets each contain 3 consecutive positive integers. None
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Updated on: 23 Feb 2021, 02:29

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Question Stats:

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?

Indicate all such values.

A. \(12\)

B. \(13\)

C. \(14\)

D. \(15\)

E. \(16\)

F. \(17\)

_________________

Indicate all such values.

A. \(12\)

B. \(13\)

C. \(14\)

D. \(15\)

E. \(16\)

F. \(17\)

_________________

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Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
11 Feb 2020, 04:36

6

huda wrote:

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?

Indicate all such values.

A. \(12\)

B. \(13\)

C. \(14\)

D. \(15\)

E. \(16\)

F. \(17\)

G. \(18\)

E. \(20\)

Indicate all such values.

A. \(12\)

B. \(13\)

C. \(14\)

D. \(15\)

E. \(16\)

F. \(17\)

G. \(18\)

E. \(20\)

Show: ::

LATER

The new set will have 6 elements. Median will be the average of 3rd and 4th element (after arranging in the ascending order).

Therefore the sum of 3rd and 4th element has to be even in order to generate a integer median. There are two possible cases

1. 3rd and 4th element are odd, which implies that the 1st and 6th element are odd. Therefore range is even.

For example: {1,2,3,11,12,13} Median is 7 and range is 12

2. 3rd and 4th element are even, which implies that the 1st and 6th element are even. Therefore range is even

For Example: {2,3,4,12,13,14} Median is 8 and Range is 12

Therefore all the options, which have an even number can be an answer

_________________

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If you like the solution, give a Kudo

Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
22 Feb 2021, 20:02

5

1

Bookmarks

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

_________________

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

_________________

Want to crush the GRE? One stop SHOP

Debrief - And it's done with 328

Which exam to go for? - GRE v/s GMAT Diagnostic Test

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Which exam to go for? - GRE v/s GMAT Diagnostic Test

Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
11 Feb 2020, 03:17

1

set-1: 12,13,14

set-2: 16,17,18

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

median: (14+16)/2=15

set-2: 16,17,18

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

median: (14+16)/2=15

Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
11 Feb 2020, 04:23

1

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

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]
11 Feb 2020, 17:06

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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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]
30 Sep 2020, 10:05

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 ??

CAN SOMEONE EXPLAIN THIS ??

Re: Two sets each contain 3 consecutive positive integers. None
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30 Sep 2020, 12:41

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

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

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Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
25 Feb 2021, 09:24

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.

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.

Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
03 Mar 2021, 23:13

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]
04 Mar 2021, 00:22

1

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.

_________________

I hope this helps!

Regards:

Karun Mendiratta

Founder and Quant Trainer

Prepster Education, Delhi, India

https://www.instagram.com/prepster_education/

Regards:

Karun Mendiratta

Founder and Quant Trainer

Prepster Education, Delhi, India

https://www.instagram.com/prepster_education/

Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
27 Oct 2021, 06:49

1

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

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.

Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
11 Dec 2021, 02:19

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

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]
11 Dec 2021, 03:01

2

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

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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

??

then the range will be 17

??

_________________

Debrief - And it's done with 328

Which exam to go for? - GRE v/s GMAT Diagnostic Test

Re: Two sets each contain 3 consecutive positive integers. None
[#permalink]
17 Dec 2021, 23:51

drasticgre wrote:

what if 1,2,3 and 16,17,18 we take

then the range will be 17

??

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]
05 Sep 2022, 09:33

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

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).

gmatclubot

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