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# M22-09

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:16
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61% (00:29) correct 39% (00:36) wrong based on 125 sessions

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What is the average (arithmetic mean) age of all students of the city's schools?

(1) The average (arithmetic mean) age of the students in each of the city's schools is 12

(2) The number of students in each of the city's schools is 70
[Reveal] Spoiler: OA

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Math Expert
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16 Sep 2014, 01:16
Official Solution:

Statement (1) by itself is sufficient. If $$N1, N2, ..., Nk$$ are the respective school populations, the mean is $$\frac{12N1 + ... + 12Nk}{N1 + ... + Nk} = 12$$.

Statement (2) by itself is insufficient. S2 says nothing about the age of the students.

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05 Mar 2016, 03:55
I think this is a poor-quality question and I don't agree with the explanation. the question asks for mean not average
Math Expert
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05 Mar 2016, 04:11
deospaima wrote:
I think this is a poor-quality question and I don't agree with the explanation. the question asks for mean not average

hi deospaima,
average age is the same as MEAN age..
there are three terms in STATISTICS..
MEAN= this is same as AVERAGE
MEDIAN= the central value when the numbers, here ages, are arranged in ascending order..
MODE= the number with max repetition..

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11 Aug 2017, 03:49
can someone provide a detailed explanation.
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11 Aug 2017, 04:00
vinnativ wrote:
can someone provide a detailed explanation.

You can find more solutions here: https://gmatclub.com/forum/what-is-the- ... 83863.html
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04 Jan 2018, 08:31
Let me take a stab at providing a simple explanation. The terms average or mean are used interchangeably here.
Consider a set "A" with 3 numbers { 4, 4, 4}. The mean (or average) of this set is obviously 4.

Now lets extra extrapolate this a little bit.
Lets assume that we have 3 sets B, C, D.
Set B = 1, 3, 8 ==> Average = 4
Set C = 4, 4, 4 ==> Average = 4
Set D = 2, 6, 4 ==> Average = 4

When we "mix" sets with the same average (or mean), the average of the resultant set doesn't change.
Lets "mix" sets B, C. The average of the resultant set is { 1, 3, 8, 4,4,4} is still 4
Lets "mix" sets C, D. The average of the resultant set { 4,4,4, 2, 6, 4, } is still 4.

Let's get back to {A} and assume that instead of containing 3 numbers, now A contained 3 subsets, with each subset having its own numbers.
So putting all of this together, if A comprised of 3 sets { {B}, {C}, {D} } and each of the sets {B}, {C}, {D} have the same average, then the average of {A} will continue to remain the same. .

vinnativ wrote:
can someone provide a detailed explanation.

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05 Jan 2018, 07:04
+1 for A. Avg age of all students is sum of age for all students divided by total number of students.

Statement 1 : Sufficient
Statement 2 : Not sufficient

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01 Feb 2018, 16:50
Hi --

How come we wouldn't need to know how many students were in each of the city's schools and get a weighted average? Is it because the average is the same across all the schools?

Thanks!
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01 Feb 2018, 21:11
ncplagge wrote:
Hi --

How come we wouldn't need to know how many students were in each of the city's schools and get a weighted average? Is it because the average is the same across all the schools?

Thanks!

Yes. If the average is the same for several groups then the overall average would be the average of any of the groups. For example, if the average age of group A is 12 years and the average age of group B is also 12 years, then the overall average age will also be 12 years irrespective of the number of people in the groups. Or using the formula: $$\frac{12N_1 + ... + 12N_k}{N_1 + ... + N_k} = \frac{12(N_1 + ... + N_k)}{N_1 + ... + N_k}=12$$ (notice that N1, N2, ..., Nk are the number of students in the schools and all of them are reduced).
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Re: M22-09   [#permalink] 01 Feb 2018, 21:11
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# M22-09

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