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There are 10 employees in an office. The table shows how man

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Senior Manager
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There are 10 employees in an office. The table shows how man  [#permalink]

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New post 17 Nov 2010, 12:57
7
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A
B
C
D
E

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Question Stats:

71% (02:47) correct 29% (02:38) wrong based on 271 sessions

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There are 10 employees in an office. The table shows how many employees have 0, 1, 2 or 3 pets. If the office manager were included in the table, the average (arithmetic mean) number of pets per person would equal the median number of pets per person. How many pets does the office manager have?


# of pets # of employees
0 ---------------2
1-------- ------3
2--------------2
3--------------3

A. 3
B. 4
C. 5
D. 6
E. 7

Edit: Formatting

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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 17 Nov 2010, 13:16
2
3
shrive555 wrote:
There are 10 employees in an office. The table shows how many employees have 0, 1, 2 or 3 pets. If the office manager were included in the table, the average (arithmetic mean) number of pets per person would equal the median number of pets per person. How many pets does the office manager have?


# of pets # of employees
0 ---------------2
1-------- ------3
2--------------2
3--------------3

3
4
5
6
7


So we have the following set {0, 0, 1, 1, 1, 2, 2, 3, 3, 3, x}, 11 terms represent 11 employees (including office manager) and x represents the number of pets office manager has.

Now, \(mean=\frac{0+0+1+1+1+2+2+3+3+3+x}{11}=\frac{16+x}{11}\) and \(median=6_{th} \ term\) (median equals to the middle term in the set with odd number of terms when arranged in ascending or descending order), so in ANY case it must be an integer --> as given that mean=median, then \(mean=\frac{16+x}{11}\) must be an integer --> \(x=6\) (in this case \(mean=\frac{16+x}{11}=2\) and median= 6th term = 2 also).

Answer: D.
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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 17 Nov 2010, 13:27
Thanks alot B.
B why Arithmetic Mean isn't included in the gmat math book ? isn't any thread that gives insight/tips/pitfalls on Mean/average
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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 18 Nov 2010, 07:28
Quite straightforward if you know your formulas. Took me some time to solve. I think I need to practice more. Thanks for the question!
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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 13 Sep 2013, 13:17
There are 10 employees in an office, not counting the office manager. The table shows how many employees have 0, 1, 2 or 3 pets. If the office manager also were included in the table, the average (arithmetic mean) number of pets per person would equal the median number of pets per person. How many pets does the office manager have?

a) 3
b) 4
c) 5
d) 6
e) 7
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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 13 Sep 2013, 13:20
AkshayChittoria wrote:
There are 10 employees in an office, not counting the office manager. The table shows how many employees have 0, 1, 2 or 3 pets. If the office manager also were included in the table, the average (arithmetic mean) number of pets per person would equal the median number of pets per person. How many pets does the office manager have?

a) 3
b) 4
c) 5
d) 6
e) 7


Merging similar topics.
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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 10 Feb 2014, 13:47
This one's pretty easy because the mean 16+x/11 is an integer only when x = 6. And since we have an odd number of members then the median is always an integer, given that all terms in the given set are integers. Therefore, only D works

Hope its clear
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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 09 Jan 2016, 12:02
let's list everything:
0 0 1 1 1 2 2 3 3 3
ok, so if we add new number, then the median would be either 1 or 2.
the total sum is 3+4+9=16.
new number = x.

16+x/11 = either 1 or 2.
16+x=11 - impossible to be negative, thus, the median, and the average has to be 2.
16+x/11=2 -> 16+x=22 -> x=6.


took me slightly over 1 minute to answer.

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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 12 Feb 2016, 19:38
In a multi-voting system, voters can vote for more than one candidate. Two candidates A and B are contesting the election. 100 voters voted for A. Fifty out of 250 voters voted for both candidates. If each voter voted for at least one of the two candidates, then how many candidates voted only for B?

(A) 50
(B) 100
(C) 150
(D) 200
(E) 250
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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 12 Feb 2016, 20:09
leve wrote:
In a multi-voting system, voters can vote for more than one candidate. Two candidates A and B are contesting the election. 100 voters voted for A. Fifty out of 250 voters voted for both candidates. If each voter voted for at least one of the two candidates, then how many candidates voted only for B?

(A) 50
(B) 100
(C) 150
(D) 200
(E) 250


Hi leve,
you should have posted the Q separately..
otherwise..
100 voters voted for A...
50 voted for both A and B, but this is already a part of 100 above..
each of 250 voted for A or B..
so remaining 250-100 voted for B only..
ans C 150

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Re: There are 10 employees in an office. The table shows how man  [#permalink]

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New post 14 Mar 2018, 20:20
Hi All,

From the table, we know that there are 10 people and 16 pets. By including the manager, we'll have 11 people, so the median will be the 6th number in line and will be an INTEGER.

We're told that once the manager's pets are included, the AVERAGE number of pets will EQUAL the MEDIAN number of pets. This significantly limits the possibilities, since the average will have to be an INTEGER as well.

X = number of manager's pets

Average = (16 + X)/11 = integer

Since we can't have a negative number of pets, X could be 6, 17, 26, etc. With the given answer choices, the only match is 6. Once you include that value, you'll see that both the average and the median are 2.

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Re: There are 10 employees in an office. The table shows how man &nbs [#permalink] 14 Mar 2018, 20:20
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