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What is the average (arithmetic mean) of eleven consecutive

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What is the average (arithmetic mean) of eleven consecutive [#permalink] New post 23 Apr 2010, 17:05
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What is the average (arithmetic mean) of eleven consecutive integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
[Reveal] Spoiler: OA
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink] New post 23 Apr 2010, 17:11
Some how i got E

1. (63+ X10 + X11) / 11 = ?? ..sitill missing two numbers? so insuff??
2. (X1+X2 +81)/ 11 = ?? Still missing two numbers so Insuff???

and combining 1 and 2 still missing some info ? .. Sorry friends, i need help here.
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink] New post 24 Apr 2010, 05:10
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zz0vlb wrote:
What is the average ( arithmetic mean ) of eleven consecutive integers?

1.The avg of first nine integers is 7
2. The avg of the last nine integers is 9
[Reveal] Spoiler:
D



friends, please explain this to me.


Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case mean=median=x_6.

(1) x_1+x_2+...+x_9=63 --> there can be only one set of 9 consecutive integers to total 63. Sufficient.

If you want to calculate: (x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63 --> x_6=8.

OR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> x_5*9=63 --> x_5=7 --> x_6=7+1=8

(2) x_3+x_4+...+x_{11}=81 --> there can be only one set of 9 consecutive integers to total 81. Sufficient.

If you want to calculate: (x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81 --> x_6=8.

OR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> x_7*9=81 --> x_7=9 --> x_6=9-1=8


Answer: D.
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink] New post 24 Apr 2010, 06:11
Thanks for clarifying this Bunuel. +1Kudos to you.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink] New post 12 Aug 2013, 03:00
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink] New post 12 Aug 2013, 03:39
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?

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


Here is a neat little trick for such kind of problems:

Will be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly \frac{d}{2} units \to The new Average = 4-\frac{1}{2} = 3.5

Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.

Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.


Back to the problem:

From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.

From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.

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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink] New post 12 Aug 2013, 22:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?

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



As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1

1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient

And is D
Re: What is the average (arithmetic mean ) of eleven consecutive   [#permalink] 12 Aug 2013, 22:15
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