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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]

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New post 23 Apr 2010, 18:05
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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
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What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 24 Apr 2010, 06:10
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What is the average (arithmetic mean) of eleven consecutive integers?

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

(1) The average of the first nine integers is 7 --> \(x_1+x_2+...+x_9=63\) --> 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) The average of the last nine integers is 9 --> \(x_3+x_4+...+x_{11}=81\) --> 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: What is the average (arithmetic mean ) of eleven consecutive  [#permalink]

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New post 12 Aug 2013, 04:39
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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: If 11 consecutive integers are listed from least to  [#permalink]

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New post 08 Mar 2013, 05:51
so its not possible to have a list of numbers with positive and negative numbers?
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Re: If 11 consecutive integers are listed from least to  [#permalink]

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New post 08 Mar 2013, 06:02
fozzzy wrote:
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.

so its not possible to have a list of numbers with positive and negative numbers?


How it is possible? From both statements it follows that the set is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
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Re: What is the average (arithmetic mean ) of eleven consecutive  [#permalink]

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New post 12 Aug 2013, 23:15
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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
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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 09 Mar 2015, 19:15
I considered following approach

if the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A
if largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55

Now as per statement 1 , average of first 9 numbers is 7 i.e sum =63
sum of 11 numbers =63+x+9+x+10----->B

Equating A& B
11X+55=63+X+9+10 ,which can be solved to get x=3

statement I is sufficient
similar approach for Statement II
11X-55=8+2X-19 ,can be solved to get X=13

statement 2 is sufficient

OA=D
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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 09 Mar 2015, 19:26
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Hi All,

When you look at this question, if you find yourself unsure of where to "start", it might help to break down everything that you know into small pieces:

1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place "in line", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?)

2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7.

What would have to happen for a group of consecutive integers to have an average of 7?

Here are some examples:

7

6, 7, 8

5, 6, 7, 8, 9

Notice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern.

With 9 total terms, that means there has to be 4 above and 4 below:

3, 4, 5, 6,.......7.......8, 9, 10, 11

Now we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT

With this same approach, we can deal with Fact 2.

The key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to "play around" with a question if you're immediately certain about how to handle it.

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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 29 Jun 2016, 03:10
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wow such complex explanations for such a simple problem?

given :

11 consec integers

let them be x,x+1,x+2,...,x+10

Q: what is their mean?

mean is (11x+55)/11 = x+5.
Q becomes what is x+5

1) mean first 9 is 7.

so (9x+36)/9 = x+4 = 7 , so x+5 =8 ,--> sufficient A or D

2) mean of last 9 is 9.

so (9x+54)/9 = x+6= ---> x+5=8, sufficient . so D

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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 15 Jul 2016, 00:10
zz0vlb wrote:
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


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: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 20 Dec 2016, 17:37
Nice Official Question>
Here is my solution to this one =>
Set of consecutive integers =>
N
N+1
N+2
.
.
.
N+10

AP series with D=2
Hence Mean = Median = Average of the first and the last terms= N+5

So we just ned the value of N

Statement 1
N
N+1
.
.
N+8

Mean => N+4=7
Hence N+5=> 8
So the mean of the original set will be 8
Hence Sufficient

Statement 2-->
p
p+2
p+3
.
.
p+8
Mean = p+4=9
p=5
Hene p+8=>13
So N+10=>13
Hence N+5=>8
Hence the mean of the original data set must be 8

Hence Sufficient

Hence D

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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 26 Jan 2017, 05:52
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The key is that the integers are consecutive. So, if we can determine any one of the 11 (and know where it falls), we can answer the question.

(1) The average of the first 9 consecutive integers is 7.

We know that avg = sum of terms / # of terms.

So, 7 = sum of terms/9
sum of terms = 63.

Well, there's only going to be one set of 9 consecutive integers that add up to 63. If we can determine the first 9, we can certainly determine the last 2: sufficient.

(2) The average of the last 9 terms is 9.

Exact same reasoning as (1): sufficient.

Both (1) and (2) are sufficient: choose (D).
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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 13 Mar 2017, 05:39
The trick here is to catch the phase "11 consecutive integers". We know that an odd set of consecutive integers have the same median and mean {i.e. set 1,2,3 has a median and mean of 2}.

Based on this we can say the same for the statements:

s1) Represent the first 9 integers as: A+B+C+D+E+F+G+H+I. If the mean of this set is 7 then the median is also 7 so we found that the 7th number in the total set. We know they are consecutive and therefore we could count forward and backward to get the unknown numbers. Therefore statement is sufficient.

s2) Same concept as above. Statement is sufficient.
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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 24 Mar 2018, 07:32
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zz0vlb wrote:
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


There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."

Since the consecutive integers are equally-spaced, their mean and median will be equal.

Target question: What is the average of eleven consecutive integers?

Statement 1: The average of the first nine integers is 7.
This also tells us that the MEDIAN of the first nine integers is 7.
In other words, the MIDDLEMOST value is 7.
This means, the first nine integers are 3, 4, 5, 6, 7, 8, 9, 10, 11
So, ALL 11 integers must be 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
Since we've identified all 11 integers, we can DEFINITELY find their average.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The average of the last night integers is 9
This also tells us that the MEDIAN of the last nine integers is 9.
In other words, the MIDDLEMOST value is 9.
This means, the last nine integers are 5, 6, 7, 8, 9, 10, 11, 12, 13
So, ALL 11 integers must be 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
Since we've identified all 11 integers, we can DEFINITELY find their average.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 13 Jun 2018, 08:09
How do you all know 'consecutive' means consecutive increasing rather than decreasing?
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Re: What is the average (arithmetic mean) of eleven consecutive  [#permalink]

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New post 13 Jun 2018, 19:20
Hi shiv19911,

Functionally, it's easiest for most people to think about consecutive numbers in terms of 'least to greatest.' With the way the question is phrased, it makes no difference whether you think of the integers as increasing or decreasing. However, the two facts refer to the fact that the 'first nine' has an average of 7 and the 'last nine' has an average of 9. Both groups have the same number of terms (re: nine), and since the second grouping has a HIGHER average, the sequence must be increasing.

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Re: What is the average (arithmetic mean) of eleven consecutive &nbs [#permalink] 13 Jun 2018, 19:20
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