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# What is the average of eleven consecutive integers? 1)

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VP
Joined: 20 Sep 2005
Posts: 1017
What is the average of eleven consecutive integers? 1) [#permalink]

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26 Mar 2006, 22:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the average of eleven consecutive integers?
1) average of first 9 integers is 7
2) average of last 9 integers is 9
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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27 Mar 2006, 00:31
Hallo face,
Consecutive integers are:
N, N+1, N+2, N+10 then
from A) 9N+36=63 N=3
from B) 9N+54=81 N=3
so D) seems correct IMO
Regards
SVP
Joined: 14 Dec 2004
Posts: 1689

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27 Mar 2006, 01:27
"D" it is
SVP
Joined: 05 Apr 2005
Posts: 1710

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27 Mar 2006, 14:08
lhotseface wrote:
What is the average of eleven consecutive integers?
1) average of first 9 integers is 7
2) average of last 9 integers is 9

D too.
VP
Joined: 20 Sep 2005
Posts: 1017

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27 Mar 2006, 14:55
Fellow GMATers,

What if it is a decreasing sequence....
N, N-1 , N-2.....
SVP
Joined: 14 Dec 2004
Posts: 1689

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27 Mar 2006, 21:42
lhotseface wrote:
Fellow GMATers,

What if it is a decreasing sequence....
N, N-1 , N-2.....

I think still we can determine!

The numbers are arranged in reverse order, so it can still be determined. Am I right?
VP
Joined: 20 Sep 2005
Posts: 1017

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27 Mar 2006, 23:10
Here is where I differ...kindly point out the flaw..I can't spot one

If the numbers are x, x+1, x+2.....then statement I gives us....
(9x + 36)/9 = 7 => x = 3
and the average of 11 numbers is ( 11(3) + 55 )/11 = 8.

If the numbers are x,x-1,x-2.....then statement I gives us....
(9x - 36)/9 = 7 => x = 11
and the average of 11 numbers is ( 11(11) - 55 )/11 = 6.

Thus, we get two different values and hence A is INSUFF IMHO and similarly B is INSUFF.

vivek123 wrote:
lhotseface wrote:
Fellow GMATers,

What if it is a decreasing sequence....
N, N-1 , N-2.....

I think still we can determine!

The numbers are arranged in reverse order, so it can still be determined. Am I right?
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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28 Mar 2006, 01:05
possible sets
3,4,5,6,7,8,9,10,11,12,13
11,10,9,8,7,6,5,4,3,2,1
The average for first 9 terms for both is 7 so A) is insufficient and as you mentioned B) is also insufficient by the same reasoning.
Then from both statements togehter seems that only one set of numbers can be determined.
Good reasoning Ihotseface
SVP
Joined: 14 Dec 2004
Posts: 1689

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28 Mar 2006, 01:47
lhotseface wrote:
Here is where I differ...kindly point out the flaw..I can't spot one

Actually, I meant the same, that the value can still be determined, thus D can't be the answer. Sorry for putting it in wrong way
VP
Joined: 29 Apr 2003
Posts: 1403

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31 Mar 2006, 06:32
if u represent the set of nos as

n, n+1... n+10 its qte easy..

ans is D
Senior Manager
Joined: 08 Jun 2004
Posts: 495
Location: Europe

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31 Mar 2006, 09:08
lhotseface wrote:
Fellow GMATers,

What if it is a decreasing sequence....
N, N-1 , N-2.....

As I know in such type of Q GMAT always use increasing order.
Am I right?
Manager
Joined: 04 Jan 2006
Posts: 58

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31 Mar 2006, 15:45
Can anybody explain this line pls?

(9x + 36)/9 = 7 => x = 3

My expln: (Assuming increasing order)
stmt 1: Since the nos are consecutive 7 is the 5th element. For 11 numbers, the average should be the 6th number. That is 8.

Stmt2:
Same logic as above
Intern
Joined: 03 Mar 2006
Posts: 19

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31 Mar 2006, 19:52
Consecutive integers are integers n1 and n2, such that n2-n1=1 , i.e., n2 follows immediately after n1.

If the Sequence is in decreasing order.. then n2 - n1 = -1... so by definition they are not consecutive.

If I'm wrong, somebody please let me know.
Manager
Joined: 13 Dec 2005
Posts: 224
Location: Milwaukee,WI

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31 Mar 2006, 21:25
I concurr with RAO ... the defination of CONSECUTIVE INTEGER is the integers that follows in sequence,each number being one greater than the previous number ,represented by n,n+1,n+2 .....where n is any number ...

Hence the answer for the question is D .
31 Mar 2006, 21:25
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