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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]

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New post 23 Nov 2012, 08:36
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Question Stats:

72% (01:16) correct 28% (01:48) wrong based on 320 sessions

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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, 08:44, 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]

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New post 23 Nov 2012, 09:05
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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]

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New post 23 Nov 2012, 10:39
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]

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New post 23 Nov 2012, 10:53
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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 ?


Check ADVANCED OVERLAPPING SETS PROBLEMS

Hope it helps.
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Re: There are 50 employees in the office of ABC Company. Of [#permalink]

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New post 20 Dec 2013, 03:08
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.

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Re: There are 50 employees in the office of ABC Company. Of [#permalink]

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New post 23 Feb 2015, 19:45
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 ?

Kudos [?]: 8 [0], given: 62

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Re: There are 50 employees in the office of ABC Company. Of [#permalink]

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New post 23 Feb 2015, 22:28
Hi Bingo123,

In this variation on the question, you have to account for the employees who have taken none of the courses. You can either subtract that group from the total OR add that group to the sum of the other groups. The net effect on the calculation will be the same.

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Re: There are 50 employees in the office of ABC Company. Of [#permalink]

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New post 24 Feb 2015, 01:11
Refer diagram below:

Attachment:
overla.png
overla.png [ 7.37 KiB | Viewed 5255 times ]


We require to find value of x (yellow shaded region)

Given that a+b+c = 9

Setting up the equation

22 + 13 - (a+c) + c + 14 - (b+c) + x = 50

x - 9 = 1

x = 10

Answer = C
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Re: There are 50 employees in the office of ABC Company. Of [#permalink]

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New post 25 Feb 2015, 19:24
Hello Rich,

Thanks for you response. However, my point is not about the employees who have taken none of the courses.

Instead, for about the employees who have taken all the courses. ( -1*2 in the below equation).

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

x = 10

As per my knowledge, we should add (1 time) the employees who have taken all the courses. However, what i see in the solution is that we are subtracting (2 times) the same. This is completely against the logic and formula.

Kindly help to explain. Thanks.

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Re: There are 50 employees in the office of ABC Company. Of [#permalink]

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Hi Bingo123,

If someone has taken ALL 3 courses, then that person has been counted 3 times (once in accounting, once in finance and once in marketing), but each person is only supposed to counted once in total. As such, you have to subtract 2 of those "counts" from the calculation.

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There are 50 employees in the office of ABC Company. Of [#permalink]

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New post 26 Feb 2015, 19:11
Hello Rich,

Thanks for your response.I now understand my mistake of ignoring the word "EXACTLY" in 2 courses.

The established formula of 3 sets takes it as " At least 2 courses" which is why it adds"All Courses" once.

Thanks.

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There are 50 employees in the office of ABC Company. Of   [#permalink] 26 Feb 2015, 19:11
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