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Yesterday each of the 35 members of a certain task force [#permalink]

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21 Oct 2009, 08:56

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Yesterday each of the 35 members of a certain task force spent some time working on project P. The graph shows the number of hours and the number of members who spent that number of hours working on project P yesterday. What was the median number of hours that the members of the task force spent working on project P yesterday?

Re: 35 members of a certain task force [#permalink]

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21 Oct 2009, 10:12

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mendelay wrote:

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13996-medium.jpg

Yesterday each of the 35 members of a certain task force spent some time working on project P. The graph shows the number of hours and the number of members who spent that number of hours working on project P yesterday. What was the median number of hours that the members of the task force spent working on project P yesterday?

A. 2 B. 3 C. 4 D. 5 E. 6

We have set consisting of 35 terms (terms=number of members): 9+4+1+2+1+8+10=35. Median of a set, with odd number of terms, would be the middle term, so 18th term. Values of terms: 9 terms=1, 4 terms=2, 1 term=3, 2 terms=4 , 1 term=5, 8 terms=6 and 10 terms=7 --> 18th term is 6.

Re: Yesterday each of the 35 members of a certain task force [#permalink]

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21 Sep 2013, 15:49

A little unorthodox approach, but so far it has worked for these types of problems. I set up the number of people that worked that many hours in a number line:

9 - 4 - 1 - 2 - 1 - 8 - 10

As Bunuel said we have 35 terms so the middle, 18th term, will be the median.

I then start crossing off from either end until I find the "1" remaining:

9 4 1 2 1 8 10 0 4 1 2 1 8 1 0 3 1 2 1 8 0 0 0 1 2 1 5 0 0 0 0 2 1 4 0 0 0 0 0 1 2 0 0 0 0 0 0 1 0 - Therefore it's the sixth term, in this case "6 hours"....hope it makes sense, just a minor shortcut over writing out all 35 terms.

Re: Yesterday each of the 35 members of a certain task force [#permalink]

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27 Nov 2014, 15:42

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Re: Yesterday each of the 35 members of a certain task force [#permalink]

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27 Nov 2014, 21:27

Expert's post

Nwsmith11 wrote:

A little unorthodox approach, but so far it has worked for these types of problems. I set up the number of people that worked that many hours in a number line:

9 - 4 - 1 - 2 - 1 - 8 - 10

As Bunuel said we have 35 terms so the middle, 18th term, will be the median.

I then start crossing off from either end until I find the "1" remaining:

9 4 1 2 1 8 10 0 4 1 2 1 8 1 0 3 1 2 1 8 0 0 0 1 2 1 5 0 0 0 0 2 1 4 0 0 0 0 0 1 2 0 0 0 0 0 0 1 0 - Therefore it's the sixth term, in this case "6 hours"....hope it makes sense, just a minor shortcut over writing out all 35 terms.

Or, just start adding the number of members till you reach 18 (since you are already given in the question that there are 35 members who worked)

9+4+1+2+1 = 17 So the next fellow who worked for 6 hrs is the 18th member and lies at the median i.e. median is 6.

Re: Yesterday each of the 35 members of a certain task force [#permalink]

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19 Jun 2016, 15:36

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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