What is the average (arithmetic mean) of eleven consecutive

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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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.

D.
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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

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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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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1: Sufficient: Let first number be n...therefore we are tying to find n + (n+1)....+(n+10)/11. Statement one says that n + (n+1)...+(n+8)/9=7. Therefore, n + (n+1)...+(n+8)=63. Therefore, 9n+36=63 and n=3. Can use value of n to find the avg of the n and the next 10 integers.

2. Sufficient. Similarly, statement two says that (n+10)+ (n+9)....+(n+2)/9=9. Therefore, (n+10)+ (n+9)....+(n+2)=81. Can solve for n and use value of n to find the avg of n and the next 10 integers.

Answer D.

Brunel!
Can't the consecutive integers be decreasing. I mean it can happen that the first integer in the sequence is the highest and , therefore, the last one the lowest. In such case, the answer must be C.

HI tanuj3,

What you're describing IS possible, but that type of specific language would be stated (re: "sequence X is a series of decreasing consecutive integers....").

Here, the two Facts do NOT logically match with the idea that the consecutive integers were decreasing though....If that were the case, then the average of the "first" integers would have to be GREATER than the average of the "last" integers. The two Facts state the OPPOSITE of that, so the sequence must be INCREASING.

GMAT assassins aren't born, they're made,
Rich

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

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).

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

RELATED VIDEO

How do you all know 'consecutive' means consecutive increasing rather than decreasing?

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.

GMAT assassins aren't born, they're made,
Rich

Hi,

So if we can find any term and know its place value we can determine the series/ mean / media/ average of a evenly spaced series of n terms.
here are my two cents for this question. If it doe not strike you that mean= median if the series is equally spaced then here is the alternate solution.

Statement 1:
We are told that average of first 9 terms is 7

So if we consider this series of 9 terms ,can we use this to find the the first term.
We know that
Sum = n\((\frac{First term +last Term}{2})\)

Sum = n\((\frac{a_{1} +a_{1} +(n-1)d}{2})\)

We know that sum of 9 terms is 63 , n = 9, d=1
We can figure out \(a_{1}\), then accordingly we can determine all the terms and we can know the average.

Just in case if you want to know what the first term is \(a_{1}\),= 3



Statement 2:
We are told that average of last 9 terms is 9

So if we consider this series of 9 terms ,can we use this to find the the first term( which will be third term of our original series) .
We know that
Sum = n\((\frac{First term +last Term}{2})\)

Sum = n\((\frac{a_{1} +a_{1} +(n-1)d}{2})\)

We know that sum of 9 terms is 81 , n = 9, d=1
We can figure out \(a_{1}\), then accordingly we can determine all the terms and we can know the average.

Just in case if you want to know what the first term is \(a_{3}\),= 5


So D
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just put the formula for consecutive avg integer :

1st term + last term / 2 = avg

let us assume : n is the 1 st term and so 9 th is n+8 as it is consecutive

for statement 1 :

{ n + ( n + 8 ) } / 2 = 7

for statement 2 :

{ (n + 2) + ( n + 10 ) } / 2 = 9

so it's Clear D .

n= 3 .

Need help. Am I to assume the consecutive 11 integers are evenly spaced by 1 unit? Why not assume its spaced by 2 units? (Please note, I know it's much easier to space by 1 unit and I did this when solving this problem and got it correct My objective here is to help me deeply understand the question).

For instance, statement 1:
(1) The objective of question is to determine the value of the 6th term of the consecutive 11 integers, provided the avg=median concept
(2) Assuming consectuive integers are spaced by 2 units (rather than 1); the SUM FORMULA of the first 9 integers could be written as (n) + (n+2) + (n+4) + ... + (n+16) = 9n + 72
(3) Plug SUM FORMULA into AVG FORMULA to find the value of n --> (9n + 72)/9=7 --> (9n + 72)=63 --> n=-1
(4) The 6th term is defined as (n+14) --> (-1)+10 = 9

The value for the answer I found when spacing by 2 units is 9, as opposed to spacing by 1 and getting the answer value of 8.


.... Bunuel post states, Mean(=median of first 9 terms=5th term)*# of terms=63 --> x5∗9=63x5∗9=63 --> x5=7x5=7 --> x6=7+1=8, which seems soundproof to me. Nonetheless, I'd love to understand how to select the unit spacing.
Simply put, why can there be only one, unique set of 9 consecutive integers to total 63? Am I to assume the spacing is by 1 unit? and if so, how do I know that when taking the test? Many thanks and I appreciate your help.

When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x-3, x-2, x-1, x, x+1, x+2, ....

For example:

-7, -6, -5 are consecutive integers.

2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers.

3, 5, 7 ARE NOT consecutive integers, they are consecutive odd integers.

So, not all evenly spaced sets represent consecutive integers.
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Bunuel thank you, thank, thank you! So simple yet I cannot forget this. Thank you!