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# In a company 60% of the employees earn less than $50,000  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Senior Manager Joined: 22 Sep 2005 Posts: 281 Followers: 1 Kudos [?]: 31 [0], given: 1 In a company 60% of the employees earn less than$50,000 [#permalink]  01 Sep 2009, 14:26
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In a company 60% of the employees earn less than $50,000 a year, 60% of the employees earn more than$40,000 a year, 11% of the employees earn $43,000 a year and 5% of the employees earn$49,000 a year. What is the median salary for the company?
a) 43.000
b) 45.500
c) 46.000
d) 49.000
e) 50.000
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Re: In a company 60% of the employees earn less than $50,000 [#permalink] 01 Sep 2009, 15:10 2 This post received KUDOS In a company 60% of the employees earn less than$50,000 a year, 60% of the employees earn more than $40,000 a year, 11% of the employees earn$43,000 a year and 5% of the employees earn $49,000 a year. What is the median salary for the company? 60% of the employees earn less than$50,000 a year
=> 40% earn greater than 50,000 a year.

60% of the employees earn more than $40,000 a year => 40% earn less than 40,000 a year.. Let there are 100 employess then, to calculate median we need salary of 50th employee and 51th employee. Then, median = (salary of 50th employee + salary of 51th employee)/2 40 people<40000$40,000 20 people $50,000 40 people > 50,000 --------------------------*----------------*---------------------- Salary of 11 people = 43,000 Salary of 5 people = 49,000 whtever, be the case, 50th and 51th salary would be 43,000 and 43,000 Hence, median = \frac{(2*43000)}{2} Manager Joined: 12 Aug 2009 Posts: 107 Followers: 3 Kudos [?]: 12 [0], given: 2 Re: In a company 60% of the employees earn less than$50,000 [#permalink]  01 Sep 2009, 19:01

50 and 51 employee will be each 43K. hence median = (43K+43k)/2
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Re: In a company 60% of the employees earn less than $50,000 [#permalink] 20 Oct 2009, 18:37 Opening this thread again.. Absolutly did'nt grasp the concept guys..Please explain.. In a company 60% of the employees earn less than$50,000 a year, 60% of the employees earn more than $40,000 a year, 11% of the employees earn$43,000 a year and 5% of the employees earn $49,000 a year. What is the median salary for the company? Quote: 60% of the employees earn less than$50,000 a year
=> 40% earn greater than 50,000 a year.

60% of the employees earn more than $40,000 a year => 40% earn less than 40,000 a year.. Let there are 100 employess then, to calculate median we need salary of 50th employee and 51th employee. Then, median = (salary of 50th employee + salary of 51th employee)/2 I understood till here..after that it all went above my head:( _________________ Math Expert Joined: 02 Sep 2009 Posts: 15109 Followers: 2532 Kudos [?]: 15592 [1] , given: 1563 Re: In a company 60% of the employees earn less than$50,000 [#permalink]  20 Oct 2009, 19:32
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Expert's post
tejal777 wrote:
Absolutly did'nt grasp the concept guys..Please explain..

In a company 60% of the employees earn less than $50,000 a year, 60% of the employees earn more than$40,000 a year, 11% of the employees earn $43,000 a year and 5% of the employees earn$49,000 a year. What is the median salary for the company?

Quote:
60% of the employees earn less than $50,000 a year => 40% earn greater than 50,000 a year. 60% of the employees earn more than$40,000 a year
=> 40% earn less than 40,000 a year..

Let there are 100 employess
then, to calculate median we need salary of 50th employee and 51th employee.

Then, median = (salary of 50th employee + salary of 51th employee)/2

I understood till here..after that it all went above my head:(

Let's consider this problem in the following way:
1. We have set of 100 terms: a1, a2, ..a100.
2. 60 terms are <50
3. 60 terms are >40
4. 11 terms =43
5. 5 terms =49.
Q what is median of the set?

Clearly as we have even number of terms (100) in the set, the median would be (a50+a51)/2. So we should determine a50 and a51.

From 2 and 3 we can conclude that 20 terms: from a41 to a60 are in the range 40-50. Or 40<a41<=a42<=...<=a60<50. From this it's already clear that median will be <50. Plus we know 16 terms from this 20 terms (a41-a60): 11 terms=43 and 5 terms=49.

4 terms out of 20 (a41-a60) are unknown: they can be in the range 40-43, 43-49 or 49-50 (all or any combination of them in any of these three ranges). BUT they doesn't matter because in ANY case a50 and a51 would be 43. You can not move 11 terms, which are 43, in the range of a41 to a60, so that a50 and a51 not to be 43.

a50=43, a51=43 --> median=43

Well I don't know how I managed to clear this problem, but this is the way how I solved it.
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Re: In a company 60% of the employees earn less than $50,000 [#permalink] 07 Oct 2010, 07:18 Hi Bunuel, I am not able to understand this problem You are saying From 2 and 3 we can conclude that 20 terms: from a41 to a60 are in the range 40-50 how it is so? Math Expert Joined: 02 Sep 2009 Posts: 15109 Followers: 2532 Kudos [?]: 15592 [0], given: 1563 Re: In a company 60% of the employees earn less than$50,000 [#permalink]  07 Oct 2010, 07:26
Expert's post
prashantbacchewar wrote:
Hi Bunuel,

I am not able to understand this problem

You are saying From 2 and 3 we can conclude that 20 terms: from a41 to a60 are in the range 40-50 how it is so?

Total 100 terms --> 60(<50)+60(>40()=120 --> 20 overlap in the range 40-50.
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Re: In a company 60% of the employees earn less than $50,000 [#permalink] 07 Oct 2010, 07:54 Got it Bunuel.. Thanks a lot Manager Joined: 26 Aug 2010 Posts: 73 Location: India Followers: 3 Kudos [?]: 28 [1] , given: 18 Re: In a company 60% of the employees earn less than$50,000 [#permalink]  08 Oct 2010, 15:38
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Nice Question!

During my first read, I realized that it is a venn diagram question. So, I solved it this way..
Attachment:

Qus.jpg [ 10.5 KiB | Viewed 1177 times ]

60(<50k) + 60 (>40k) - (Intersection between two categories) = 100

Given Three Categories:
no (I.): Earning < 40k (which amounts to first 40 people). {This is I. only region}
no (II.): Earning between 40 k and 50k (which amounts to 41st to 60 people) {This is Intersection}
no (III.): Earning > 50k (which amounts to 61st to 100th people) {This is III. only region}

We are interested in no (II.) as there are already first 40 people in no (I.).

Also, given : 11 people earn 43k. They can be either 41st to 52nd people or 50th to 60th people. {of course when put in ascending order of their salaries}.

So, In any case..50 and 51 people will be earning 43K. Hence, OA: A.
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Re: In a company 60% of the employees earn less than $50,000 [#permalink] 19 May 2011, 00:30 good concept used here. _________________ Visit -- http://www.sustainable-sphere.com/ Promote Green Business,Sustainable Living and Green Earth !! Senior Manager Joined: 12 Oct 2009 Posts: 274 Schools: Columbia, INSEAD, RSM, LBS Followers: 2 Kudos [?]: 25 [0], given: 4 Re: In a company 60% of the employees earn less than$50,000 [#permalink]  19 May 2011, 19:43
This is how id did it.

Well the Median should be between the overlap portion of the sets.

Now in this overlap section of 20 % (or say 20 values for the sake of ease) the avg of the 11th and 10th value is the Median of the whole set.

11% earn 43,000
5% earn 49,000
remaining 4%

Now even if these 4% earn less than 43,000( above 40,000) or more than 49,000 ( less than 50,000)
the 10th and the 11th term would still be 43,000

For example

say these 4 values are
42,000 each
then the set of 20 values would be
42000 42000 42000 42000 43000 _ _ _ _ 10th Term 11th term _ _ _ _ 49000 49000 49000 49000 49000

Hence under all probabilities the 10th and the 11th term would be 43000 and hence would be the Median of the entire set
Attachments

Median Question.pdf [39.73 KiB]