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# If m is a positive odd integer, what is the average (arithme

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If m is a positive odd integer, what is the average (arithme  [#permalink]

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Updated on: 02 Jul 2013, 00:30
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If m is a positive odd integer, what is the average (arithmetic mean) of a certain set of m integers

(1) The integers in the set are consecutive multiples of 3.
(2) The median of the set of integers is 33

Originally posted by rigger on 24 Oct 2005, 22:54.
Last edited by Bunuel on 02 Jul 2013, 00:30, edited 1 time in total.
Edited the question and added the OA
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Re: If m is a positive odd integer, what is the average (arithme  [#permalink]

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02 Jul 2013, 00:37
6
If m is a positive odd integer, what is the average (arithmetic mean) of a certain set of m integers

(1) The integers in the set are consecutive multiples of 3. Clearly insufficient. Consider: {0, 3, 6} and {3, 6, 9}. Not sufficient.

From this statement, though we can deduce that since the set is evenly spaced then its mean equals to its median.

(2) The median of the set of integers is 33. Also, insufficient. Consider: {31, 33, 35} and {31, 33, 133}. Not sufficient.

(1)+(2) From (1) we have that mean=median and from (2) we have that meidan=33, thus mean=median=33. Sufficient.

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Joined: 22 Aug 2005
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Updated on: 24 Oct 2005, 23:21
C.
I am not sure about ans. I could infer following from the question.
(1)
the numbers in the set are:
3, 6, 9,...,3m, m is odd > 0

avg = 3(1+2+3+...+m) / m = 3 m (m + 1) / 2m or 3(m+1)/2
insufficient.

(2) insufficient. median doesn't tell anything about mean.

together

as m is odd, median will be n/2 + 1 the member in the set(sorted)
so there are 21 numbers in the set(3,6,....,33,.....,63)
mean = 3 * 22 / 2 = 33

Originally posted by duttsit on 24 Oct 2005, 23:09.
Last edited by duttsit on 24 Oct 2005, 23:21, edited 1 time in total.
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24 Oct 2005, 23:18
1
1
in a set of consecutive numbers the mean is equal to the median. hence, C)...
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25 Oct 2005, 09:48
IF M is ODD and we have M integers in a set, shouldnt Median=mean
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01 Jul 2011, 15:29
FN wrote:
IF M is ODD and we have M integers in a set, shouldnt Median=mean

2,3,99: median = 3. mean != 3.

2,3,4: median = mean = 3

The rule applies when integers are consecutive (or in this case multiples)

1: ns
2: ns
1+2: suff

C

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Updated on: 11 Jul 2011, 03:34
duttsit wrote:
C.
I am not sure about ans. I could infer following from the question.
(1)
the numbers in the set are:
3, 6, 9,...,3m, m is odd > 0

avg = 3(1+2+3+...+m) / m = 3 m (m + 1) / 2m or 3(m+1)/2
insufficient.

(2) insufficient. median doesn't tell anything about mean.

together

as m is odd, median will be n/2 + 1 the member in the set(sorted)
so there are 21 numbers in the set(3,6,....,33,.....,63)
mean = 3 * 22 / 2 = 33

Is my approach right?
Lets says the numbers are 30 , 33 , 36 then the median is 33 but the mean = 33

if the numbers are 27 , 30 , 33,36, 39 the median is 33 but the mean is 33.. There can be any number of odd positive numbers in the set M ....

Originally posted by siddhans on 11 Jul 2011, 03:26.
Last edited by siddhans on 11 Jul 2011, 03:34, edited 1 time in total.
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11 Jul 2011, 03:32
1
siddhans wrote:
duttsit wrote:
C.
I am not sure about ans. I could infer following from the question.
(1)
the numbers in the set are:
3, 6, 9,...,3m, m is odd > 0

avg = 3(1+2+3+...+m) / m = 3 m (m + 1) / 2m or 3(m+1)/2
insufficient.

(2) insufficient. median doesn't tell anything about mean.

together

as m is odd, median will be n/2 + 1 the member in the set(sorted)
so there are 21 numbers in the set(3,6,....,33,.....,63)
mean = 3 * 22 / 2 = 33

How can we say that the answer is C?

Lets says the numbers are 30 , 33 , 36 then the median is 33 but the mean = 33

if the numbers are 27 , 30 , 33,36, 39 the median is 33 but the mean is 35..Where does it say we need to start with 3 ... There can be any number of odd psotive numbers in the set M .... I am confused why is the OA C

(27 + 30 + 33 + 36 + 39) / 5 = 33

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Joined: 29 Jan 2011
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11 Jul 2011, 03:35
pike wrote:
siddhans wrote:
duttsit wrote:
C.
I am not sure about ans. I could infer following from the question.
(1)
the numbers in the set are:
3, 6, 9,...,3m, m is odd > 0

avg = 3(1+2+3+...+m) / m = 3 m (m + 1) / 2m or 3(m+1)/2
insufficient.

(2) insufficient. median doesn't tell anything about mean.

together

as m is odd, median will be n/2 + 1 the member in the set(sorted)
so there are 21 numbers in the set(3,6,....,33,.....,63)
mean = 3 * 22 / 2 = 33

How can we say that the answer is C?

Lets says the numbers are 30 , 33 , 36 then the median is 33 but the mean = 33

if the numbers are 27 , 30 , 33,36, 39 the median is 33 but the mean is 35..Where does it say we need to start with 3 ... There can be any number of odd psotive numbers in the set M .... I am confused why is the OA C

(27 + 30 + 33 + 36 + 39) / 5 = 33

Posted from my mobile device

My bad I figured out after i posted ...I edited but you replied before i posted my edited reply
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Posts: 759
Re: If m is a positive odd integer, what is the average  [#permalink]

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02 Jul 2013, 00:19
1
Statement 1 says

multiples of 3

so it can be 3,6,9 or 3,6,9,12,15

different mean's insufficient

Statement 2 says

33 is the middle value but we have no other info insufficient

Combined we know that 33 is the middle value so there are 10 terms on either side of 33

1st value is 3
11th value is 33
21st value is 63

---------33-------

So we can find out mean = median = 33 sufficient
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Joined: 13 Jun 2013
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02 Jul 2013, 12:12
1
christoph wrote:
in a set of consecutive numbers the mean is equal to the median. hence, C)...

Oops I made the classic mistake of unconciously carrying over data from statement 1 to my assessment of statement 2.

I had what you said in my mind, saw a median given in statement 2 and concluded that it is equal to the mean and chose B.

However you need statement 1 to know it is a evenly spaced set (for which the same rule holds) and thus answer choice C.

You do not need any calculations for this. All you need to know is that:

A) it is an evenly spaced set
B) the value of the median
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Joined: 07 Apr 2015
Posts: 165
Re: If m is a positive odd integer, what is the average (arithme  [#permalink]

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09 Jul 2015, 03:56
christoph wrote:
in a set of consecutive numbers the mean is equal to the median. hence, C)...

Oops I made the classic mistake of unconciously carrying over data from statement 1 to my assessment of statement 2.

I had what you said in my mind, saw a median given in statement 2 and concluded that it is equal to the mean and chose B.

However you need statement 1 to know it is a evenly spaced set (for which the same rule holds) and thus answer choice C.

You do not need any calculations for this. All you need to know is that:

A) it is an evenly spaced set
B) the value of the median

And I did exactly the same, just carried over information and therefore assumed that Mean = Median should apply to statement 2.
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Re: If m is a positive odd integer, what is the average (arithme  [#permalink]

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20 Oct 2015, 19:12
Bunuel wrote:
If m is a positive odd integer, what is the average (arithmetic mean) of a certain set of m integers

(1) The integers in the set are consecutive multiples of 3. Clearly insufficient. Consider: {0, 3, 6} and {3, 6, 9}. Not sufficient.

From this statement, though we can deduce that since the set is evenly spaced then its mean equals to its median.

(2) The median of the set of integers is 33. Also, insufficient. Consider: {31, 33, 35} and {31, 33, 133}. Not sufficient.

(1)+(2) From (1) we have that mean=median and from (2) we have that meidan=33, thus mean=median=33. Sufficient.

Answer and solution apart, is there a contradiction between question stem and (1)?
m is a positive odd integer
Clearly, then such sets as {0, 3, 6} and {3, 6, 9} are not possible
Further, since sets are of consecutive multiples of 3, it can have only 1 element e.g. {3} or {9}
I'm only interested in knowing whether my interpretation is right since I agree with the solution given by you. Thanks in advance for your comments!
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Joined: 02 Sep 2009
Posts: 51280
Re: If m is a positive odd integer, what is the average (arithme  [#permalink]

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20 Oct 2015, 20:12
1
rahulsnh wrote:
Bunuel wrote:
If m is a positive odd integer, what is the average (arithmetic mean) of a certain set of m integers

(1) The integers in the set are consecutive multiples of 3. Clearly insufficient. Consider: {0, 3, 6} and {3, 6, 9}. Not sufficient.

From this statement, though we can deduce that since the set is evenly spaced then its mean equals to its median.

(2) The median of the set of integers is 33. Also, insufficient. Consider: {31, 33, 35} and {31, 33, 133}. Not sufficient.

(1)+(2) From (1) we have that mean=median and from (2) we have that meidan=33, thus mean=median=33. Sufficient.

Answer and solution apart, is there a contradiction between question stem and (1)?
m is a positive odd integer
Clearly, then such sets as {0, 3, 6} and {3, 6, 9} are not possible
Further, since sets are of consecutive multiples of 3, it can have only 1 element e.g. {3} or {9}
I'm only interested in knowing whether my interpretation is right since I agree with the solution given by you. Thanks in advance for your comments!

No, your interpretation is not right.

Given: m is a positive odd integer.
Question: what is the average of a set of m integers.

So, we need to know the average of a set which has m integers in it (the number of element in the set is m, which is a positive odd integer). It does NOT mean that the set consists of positive odd integers.
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Re: If m is a positive odd integer, what is the average (arithme  [#permalink]

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15 Jul 2016, 09:31
rigger wrote:
If m is a positive odd integer, what is the average (arithmetic mean) of a certain set of m integers

(1) The integers in the set are consecutive multiples of 3.
(2) The median of the set of integers is 33

(1) The integers in the set are consecutive multiples of 3.
{3,6,9,12,15} or {3,6} or {-99,-96,-93,-90} ==> insufficient
(2) The median of the set of integers is 33
{1,23,33,55,99} or {1,2,3,4,5,} or {-88,33,1098}==> Insuficient

Marge both:- Sufficient
{first term=3,..........middle term=33,...........last term=63}

so the Median is 33; since the set contains consecutive multiples of 3 therefore mean = median = 33

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Re: If m is a positive odd integer, what is the average (arithme  [#permalink]

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07 Feb 2017, 19:41
M is a +ve Odd Integer {1,3,5,7,9,11...} What is Avg of {M}?
Rephrase the Question as What is M?

1) M:- {1,3,5,7,9,11..}
M:-{3,6} {3,6,9,12,15}.... Insuff to Find the Average

2) Median of set is 33. Insuff

Using 1 and 2
M( 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,....66,69}
Median is 33.
Since M is +ve odd integer,M should be odd.
If M is odd, then the middle integer will be the median

So if Median is 33, then M should be 21. { 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63}

Hence C is Suff
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Re: If m is a positive odd integer, what is the average (arithme  [#permalink]

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