If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the 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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- sharmin.karim
- Course Students
**Posts:**18**Joined:**Thu Jul 31, 2008 1:35 pm

If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the eleven integers?

1. The average of the first nine integers is 7

2. The average of the last nine integers is 9

1. The average of the first nine integers is 7

2. The average of the last nine integers is 9

- nitin_prakash_khanna
- Students
**Posts:**68**Joined:**Sat Aug 08, 2009 1:16 am

The only fundamental we need to remember is that for a equally spaced set , if the total number of items are odd then mean = median.

Consecutive numbers are a special case of equally spaced set.

So for 11 conscutive 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.

Consecutive numbers are a special case of equally spaced set.

So for 11 conscutive 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.

- Ben Ku
- ManhattanGMAT Staff
**Posts:**817**Joined:**Sat Nov 03, 2007 7:49 pm

Nitin did a great job of explaining it.

Sharmin: it would be helpful to write a bit of explanation about how you got to your answer, or where you got stuck. THen we can respond to your confusion directly.

The key to this problem is knowing that when we have an odd number of consecutive integers, the mean = median. So the question can be rephrased: "what is the 6th integer"?

Statement (1) is a set of the first nine integers, so the mean = median = 7 = 5th integer. Well, the 6th integer then is 8, so (1) is sufficient.

If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the eleven integers?

You can use the same reasoning to determine that (2) is also sufficient.

Sharmin: it would be helpful to write a bit of explanation about how you got to your answer, or where you got stuck. THen we can respond to your confusion directly.

The key to this problem is knowing that when we have an odd number of consecutive integers, the mean = median. So the question can be rephrased: "what is the 6th integer"?

Statement (1) is a set of the first nine integers, so the mean = median = 7 = 5th integer. Well, the 6th integer then is 8, so (1) is sufficient.

If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the eleven integers?

You can use the same reasoning to determine that (2) is also sufficient.

Ben Ku

Instructor

ManhattanGMAT

Instructor

ManhattanGMAT

- mohit_cs2002
- Forum Guests
**Posts:**2**Joined:**Sun Nov 22, 2009 8:00 am

good job

I might me asking a silly question. If no interval of the consecutive numbers is mentioned, we can assume as 1???

Can we call the series as consecutive no.s : 2,5,8,11 (interval 3)

I might me asking a silly question. If no interval of the consecutive numbers is mentioned, we can assume as 1???

Can we call the series as consecutive no.s : 2,5,8,11 (interval 3)

- Ben Ku
- ManhattanGMAT Staff
**Posts:**817**Joined:**Sat Nov 03, 2007 7:49 pm

Hi Mohit,

Yes, "consecutive integers" means the numbers are separated by one. "consecutive even" or "consecutive odd" means the numbers are separated by two.

Yes, "consecutive integers" means the numbers are separated by one. "consecutive even" or "consecutive odd" means the numbers are separated by two.

Ben Ku

Instructor

ManhattanGMAT

Instructor

ManhattanGMAT

- mindadze
- Students
**Posts:**14**Joined:**Wed Dec 31, 1969 8:00 pm

Ben Ku wrote:Nitin did a great job of explaining it.

Sharmin: it would be helpful to write a bit of explanation about how you got to your answer, or where you got stuck. THen we can respond to your confusion directly.

The key to this problem is knowing that when we have an odd number of consecutive integers, the mean = median. So the question can be rephrased: "what is the 6th integer"?

Statement (1) is a set of the first nine integers, so the mean = median = 7 = 5th integer. Well, the 6th integer then is 8, so (1) is sufficient.

If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the eleven integers?

You can use the same reasoning to determine that (2) is also sufficient.

Hi,

Know this is an old post but I am very curious. Question to Ben Ku:

Why should we have only ODD number of consecutive integers to say that mean=median in a consecutive list of integers?

I think it works for EVEN number of consecutive integers as well. E.g. 1,2,3,4,5,6 (in total 6 integers).

Mean=(1+6)/2=3.5 [in consecutive list of integers, mean equals (1st number+last number)/2]

Median=(3+4)/2=3.5

Thanks for clarifying.

- RonPurewal
- ManhattanGMAT Staff
**Posts:**17384**Joined:**Tue Aug 14, 2007 8:23 am

mindadze wrote:Know this is an old post but I am very curious. Question to Ben Ku:

Why should we have only ODD number of consecutive integers to say that mean=median in a consecutive list of integers?

I think it works for EVEN number of consecutive integers as well. E.g. 1,2,3,4,5,6 (in total 6 integers).

Mean=(1+6)/2=3.5 [in consecutive list of integers, mean equals (1st number+last number)/2]

Median=(3+4)/2=3.5

Thanks for clarifying.

that's true as well.

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- rachelhong2012
- Course Students
**Posts:**52**Joined:**Wed Dec 31, 1969 8:00 pm

If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the eleven integers?

1. The average of the first nine integers is 7

2. The average of the last nine integers is 9

x x+1 x+2 x+3 x+4 x+5 x+6 x+7 x+8 x+9 x+10

average= sum of these numbers (all in terms of x)/11

if we know x, we can solve for average

1. since everything is in terms of x, if we know the relationship between one variable and one constant, we can solve for the variable x, suff

try it yourself

(x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8)/9=7

you can solve for x

2. same idea, when we only have one variable and one constant, we can solve for the variable x. suff.

D

1. The average of the first nine integers is 7

2. The average of the last nine integers is 9

x x+1 x+2 x+3 x+4 x+5 x+6 x+7 x+8 x+9 x+10

average= sum of these numbers (all in terms of x)/11

if we know x, we can solve for average

1. since everything is in terms of x, if we know the relationship between one variable and one constant, we can solve for the variable x, suff

try it yourself

(x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8)/9=7

you can solve for x

2. same idea, when we only have one variable and one constant, we can solve for the variable x. suff.

D

- RonPurewal
- ManhattanGMAT Staff
**Posts:**17384**Joined:**Tue Aug 14, 2007 8:23 am

that's a legitimate algebraic way to do the problem, yes.

however, it's also important to develop a conceptual intuition for this one, because that kind of algebra will get really annoying if there are even a few more integers than the eleven that we have here.

here's what i'm getting at: these are CONSECUTIVE integers. therefore, if we know any of the following --

* the position of the first integer

* the position of the last integer

* the position of ANY particular integer (e.g., "the fourth integer is -2")

-- then we know the position of the entire sequence.

this is an observation that generalizes effortlessly to any number of consecutive integers, even if there are thousands or millions of them.

however, it's also important to develop a conceptual intuition for this one, because that kind of algebra will get really annoying if there are even a few more integers than the eleven that we have here.

here's what i'm getting at: these are CONSECUTIVE integers. therefore, if we know any of the following --

* the position of the first integer

* the position of the last integer

* the position of ANY particular integer (e.g., "the fourth integer is -2")

-- then we know the position of the entire sequence.

this is an observation that generalizes effortlessly to any number of consecutive integers, even if there are thousands or millions of them.

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- jeffrey.k.l.ho
- Course Students
**Posts:**9**Joined:**Mon Dec 13, 2010 7:31 am

Hi there,

I was doing this question and someone how I recall somewhere about a rule/concept that I would like clarification on.

If an n is given (n being the number of consecutive integers) and the sum for these consecutive integers is given, there is only ONE set of consecutive integers that will fit the sum?

for instance, given n=3 sum = 12, there is only one set of consecutive integers that will fit this, which is 3, 4, 5.

I was doing this question and someone how I recall somewhere about a rule/concept that I would like clarification on.

If an n is given (n being the number of consecutive integers) and the sum for these consecutive integers is given, there is only ONE set of consecutive integers that will fit the sum?

for instance, given n=3 sum = 12, there is only one set of consecutive integers that will fit this, which is 3, 4, 5.

- tim
- ManhattanGMAT Staff
**Posts:**5532**Joined:**Tue Sep 11, 2007 9:08 am**Location:**Southwest Airlines, seat 21C

there is never more than one set that fits the bill, but in some cases there is no such set of numbers. consider n=4 and sum=12 for instance..

Tim Sanders

Manhattan GMAT Instructor

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- nowwithgmat
- Forum Guests
**Posts:**27**Joined:**Sun May 20, 2012 8:26 am

for 11 consecutive integer average is

let n first term

and n+10 is the last i.e 11th term

so ave: n+n+10/2=n+5

now st-1

first 9 integer average is 7 so

n+n+8/2=7 >>>>>n+4=7 >>>>n =3

so suff..

and st-2

ave of last 9 integer is

n+2+n+10 /2=9 >>> n=3

suff

so D winn...

let n first term

and n+10 is the last i.e 11th term

so ave: n+n+10/2=n+5

now st-1

first 9 integer average is 7 so

n+n+8/2=7 >>>>>n+4=7 >>>>n =3

so suff..

and st-2

ave of last 9 integer is

n+2+n+10 /2=9 >>> n=3

suff

so D winn...

- RonPurewal
- ManhattanGMAT Staff
**Posts:**17384**Joined:**Tue Aug 14, 2007 8:23 am

tim wrote:there is never more than one set that fits the bill, but in some cases there is no such set of numbers. consider n=4 and sum=12 for instance..

this is correct.

on the other hand, remember that you won't have to worry about this kind of situation on data sufficiency.

each DS statement will have either ...

... exactly one solution, or

... multiple solutions.

there will be no impossible statements in DS.

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