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There are 50 employees in the office of ABC Company. Of

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There are 50 employees in the office of ABC Company. Of [#permalink] New post 23 Nov 2012, 07:36
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There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses and 1 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses?

A. 0
B. 9
C. 10
D. 11
E. 26
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Nov 2012, 07:44, edited 1 time in total.
Renamed the topic and edited the question.
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Re: There are 50 employees in the office of ABC Company. Of [#permalink] New post 23 Nov 2012, 08:05
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cv3t3l1na wrote:
There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses and 1 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses?

A. 0
B. 9
C. 10
D. 11
E. 26


50 = 22 + 15 + 14 - 9 - 1*2 + x

x = 10

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Re: There are 50 employees in the office of ABC Company. Of [#permalink] New post 23 Nov 2012, 09:39
Expert's post
MacFauz wrote:
cv3t3l1na wrote:
There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses and 1 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses?

A. 0
B. 9
C. 10
D. 11
E. 26


50 = 22 + 15 + 14 - 9 - 1*2 + x

x = 10

Kudos Please... If my post helped.


Hii MacFauz.
I got confused with the formula applied.
Isn't it Total-neither=A+B+C-A(intersection)B-B(intersection)C-C(intersection)A+A(intersection)B(intersection)C ?
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Re: There are 50 employees in the office of ABC Company. Of [#permalink] New post 23 Nov 2012, 09:53
Expert's post
Marcab wrote:
MacFauz wrote:
cv3t3l1na wrote:
There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses and 1 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses?

A. 0
B. 9
C. 10
D. 11
E. 26


50 = 22 + 15 + 14 - 9 - 1*2 + x

x = 10

Kudos Please... If my post helped.


Hii MacFauz.
I got confused with the formula applied.
Isn't it Total-neither=A+B+C-A(intersection)B-B(intersection)C-C(intersection)A+A(intersection)B(intersection)C ?


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Re: There are 50 employees in the office of ABC Company. Of [#permalink] New post 20 Dec 2013, 02:08
50 Employees. Counting every "different" attendand to the courses we have:

Accounting: 22
Finance: 15
Marketing: 14

Which would add up to 51 "different" attendands, which is not possible.
Now 9 have taken exactly 2 courses, which means that there are 9 less "different" attendands. Say that 9 of the Finance attentands also attended Accounting.
51-9= 42

1 Person has taken all three courses. As above, we subtract him from the number of "different" attendands. Since this time the person took all three courses, we have to substract him two times.
42-2= 40.

This tells us that we had 40 "different" attendands and 10 employees who didn't take any courses.

Hope it helps.

Happy Holidays!
Re: There are 50 employees in the office of ABC Company. Of   [#permalink] 20 Dec 2013, 02:08
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