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If the eleven consecutive integers are listed from least to

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If the eleven consecutive integers are listed from least to  [#permalink]

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New post 09 Jun 2009, 07:05
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If the eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of eleven integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 09 Jun 2009, 12:37
joyseychow wrote:
If the eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of eleven integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9


i think the answer should be D

1)let x= sum of first 9 int.

x/9=x, x=63

sum of consect int = (first term + last term)/2 * #terms

so (a+b)/2 *9=63
9(a+b)=126
a+b=14

since these are consec. int., b-a=8, b=a+8

a+a+8=14, 2a=6, a=3

integers are 3-13, avg=8

2) you can do it the same way
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 09 Jun 2009, 19:41
D.
I did not calculated actually but i assumed numbers to be
n,n+1,............n+10

Stmt. 1
It is suff. to give us n.So you can calculate the av. of all 11.

Stmt. 2
It is suff. to give us n.So you can calculate the av. of all 11.
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 13 Jun 2009, 02:54
mdfrahim wrote:
D.
I did not calculated actually but i assumed numbers to be
n,n+1,............n+10

Stmt. 1
It is suff. to give us n.So you can calculate the av. of all 11.

Stmt. 2
It is suff. to give us n.So you can calculate the av. of all 11.


I believe you can only apply n, n+1, n+2....so on, when you know that the consecutive integers are evenly spaced by 1. Coincidentally, the eleven consecutive integers here fulfills that criteria.
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 06 Apr 2011, 07:12
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We can solve the problem without much calculations.

In case consecutive numbers mean=median

So for 11 consecutive integers 6th number is the median so it will be the mean.

St1. the mean of first 9 integers is 7, which tells that the 5th number is 7, you know 5th number so you know 6th number is 8.

So avg of 11 numbers is 8.

SUFFICIENT.

St.2 In this list of 9 numbers mean is 9 so median = 9.
this is 7th number of original list so 6th number is 8. Which is the median = mean.

SUFFICIENT

Ans D.
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 08 Apr 2011, 23:12
Average of odd number of cons. numbers is the same as median and is equals to the middle number.

(1) The average of the first nine integers is 7
Using above theory, 5th digit is 7, so avg of 11 numbers is 6th digit which is 8. Sufficient

(2) The average of the last nine integers is 9
Using above theory, 7th digit is 9, so avg of 11 numbers is 6th digit which is 8. Sufficient.

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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 28 Aug 2011, 11:29
Baten80 wrote:
We can solve the problem without much calculations.

In case consecutive numbers mean=median

So for 11 consecutive integers 6th number is the median so it will be the mean.

St1. the mean of first 9 integers is 7, which tells that the 5th number is 7, you know 5th number so you know 6th number is 8.

So avg of 11 numbers is 8.

SUFFICIENT.

St.2 In this list of 9 numbers mean is 9 so median = 9.
this is 7th number of original list so 6th number is 8. Which is the median = mean.

SUFFICIENT

Ans D.


Please help me understand statement 2 better....shouldn't median be 5th number??
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 28 Aug 2011, 11:43
DeeptiM wrote:
Please help me understand statement 2 better....shouldn't median be 5th number??


(2) The average of the last nine integers is 9

Let each asterisk symbolize one integer.
Given: ****9****

Now, prefix two stars because there are total 11 integers. Last 9, we already represented before. 2 more needs to come before at the beginning.
******9****

These are consecutive integers. 6 consecutive numbers before 9 and 4 consecutive numbers after 9.
3,4,5,6,7,8,9,10,11,12,13

We have our set. Check the middle integer; that's the mean: 8
Sufficient.
**************************

For consecutive numbers;
The mean=median=middle term, if the number of integers is odd. Here, there are 11 terms, so means will be the middle term. 6th term.

The mean=median=Average of two middle terms, if the number of integers is even. So, we don't need to consider this case. If there were 10 consecutive numbers, the mean would be:

Average of 5th and 6th term; (5th+6th)/2
****************************************
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 28 Aug 2011, 12:04
Fluke...can't thank u enough for all the support....
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 30 Aug 2011, 19:36
Let x,x+1,.....x+10 be the sequence of consecutive integer sequence.

average = (x+x+1+......x+10)/11 ?

=> x=?

1. Sufficient

(x+x+1+.....x+8)/9 = 7

we can find out x from the above expression. hence sufficient.

2. Sufficient
(x+2......x+10)/9= 9

we can find out x from the above expression. hence sufficient.

Answer is D.
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 31 Aug 2011, 00:37
Nice explanation everyone. Yes its D.
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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New post 02 Sep 2011, 07:36
joyseychow wrote:
If the eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of eleven integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9


Will you please Cite Official Answer? I am looking for good explanation.
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If the eleven consecutive integers are listed from least to  [#permalink]

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New post 15 Jul 2016, 00:05
joyseychow wrote:
If the eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of eleven integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9


For odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question
(1) The average of the first nine integers is 7
7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7
we will have {3,4,5,6,7,8,9,10,11}
now we can add last two consecutive integer after 11, they will be 12,13
our new set will become = {3,4,5,6,7,8,9,10,11,12,13}
again since the number of total elements in the set is odd, Mean will simply be the middle value = 8
SUFFICIENT

(2) The average of the last nine integers is 9
Again number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right
Middle value will be
{5,6,7,8,9,10,11,12,13}
Add 3,4 at the start of the set
new set = {3,4,5,6,7,8,9,10,11,12,13}
Mean will be 8
Sufficient

ANSWER IS D
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Re: If the eleven consecutive integers are listed from least to  [#permalink]

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Re: If the eleven consecutive integers are listed from least to &nbs [#permalink] 17 Jul 2018, 04:57
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