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What is the average of eleven consecutive integers? 1)

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What is the average of eleven consecutive integers? 1) [#permalink] New post 25 Dec 2005, 19:24
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D
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What is the average of eleven consecutive integers?
1) average of first 9 integers is 7
2) average of last 9 integers is 9
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Re: DS - GMATPrep [#permalink] New post 25 Dec 2005, 19:56
MBAHopeful 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


Answer is D

The key here is 11 consecutive integers..

1st term = x
2nd = x+2
3rd = x + 3
....
...
11th = x + 10

(1) So the total sum for the first 9 integers is 63 ==> 9x + 36 = 63 ==> x = 3

We can find out the first term is 3, last term is 11. We can find out the average.

(2) So the total sum for the last 9 integers is 81 ==> 9x + 54 = 81 ==> x = 3

We can find out the first term is 3, last term is 11. We can find out the average.
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Re: DS - GMATPrep [#permalink] New post 25 Dec 2005, 22:36
MBAHopeful 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


The consecutive integers could either be of the increasing sequence or the decreasing sequence.

From stmt 1, we get (9x+36)/9 = 7 if it is an increasing sequence.
solving we get x = 3.
If it is a decreasing sequence we get (9x-36/9 = 7 solving we get x = 11.
So insuff.

From stmt2 we get (9x+54)/9 = 9 solving we get x = 3 , when the sequence is increasing.
9x-54/9 = 9 we get x = 15. So insuff.

Combining both we see x =3 is common, and the sequence is increasing.

So my take is C.
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Re: DS - GMATPrep [#permalink] New post 25 Dec 2005, 23:08
krisrini wrote:
MBAHopeful 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


The consecutive integers could either be of the increasing sequence or the decreasing sequence.

From stmt 1, we get (9x+36)/9 = 7 if it is an increasing sequence.
solving we get x = 3.
If it is a decreasing sequence we get (9x-36/9 = 7 solving we get x = 11.
So insuff.

From stmt2 we get (9x+54)/9 = 9 solving we get x = 3 , when the sequence is increasing.
9x-54/9 = 9 we get x = 15. So insuff.

Combining both we see x =3 is common, and the sequence is increasing.

So my take is C.


I neglected to include decreasing sequence. :oops:

However, I think you may have made a mistake in your working.

In the decreasing sequence, (1) it should be 9x - 54 = 63 ==> x = 13

So the first term is still 3.

Similar working for (2)
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Re: DS - GMATPrep [#permalink] New post 26 Dec 2005, 00:03
TeHCM wrote:
krisrini wrote:
MBAHopeful 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


The consecutive integers could either be of the increasing sequence or the decreasing sequence.

I neglected to include decreasing sequence. :oops:

However, I think you may have made a mistake in your working.

In the decreasing sequence, (1) it should be 9x - 54 = 63 ==> x = 13

So the first term is still 3.

Similar working for (2)


I am not able to find the mistake.

Increasing seq-> x,x+1,x+2,x+3,x+4,x+5,X+6,X+7,X+8,X+9,X+10.
Decreasing seq-> x,X-1,X-2,X-3,X-4,X-5,X-6,X-7,X-8,X-9,X-10
From stmt1 we get 2 equn
a) For increasing 9x+36 = 63 => 9x = 27 => x= 3
b) For decreasing 9x - 36 = 63 => 9x = 99 => X = 11

From stmt we get 2 equn
a) For incresing 9x+54=81=> 9x= 27 => x = 3
b) For decreasing 9x - 54 = 81=> 9x = 135 => x = 15

So comibing it is still 3.

If you have time can you tell me where am i making a mistake.

Thanks a ton.

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 [#permalink] New post 26 Dec 2005, 02:04
I didnt get the increasing or decreasing sequence...

Since the 11 integers are consecutive, let's assume the 6th integer to be X
The sequence becomes
(X-5,X-4,X-3,X-2,X-1,X,X+1,X+2,X+2,X+4,X+5)

From Statement I:
X-1=7 from which rest all numbers can be figured......sufficient
From Statement II:
X+1=9 from which rest all numbers can be figured......sufficient

Hence D
  [#permalink] 26 Dec 2005, 02:04
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