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

Question

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

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

MacFauz · Accepted Answer

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

x = 10

Kudos Please... If my post helped.

PareshGmat · Answer

Refer diagram below: overla.png 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

Marcab · Answer

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 ?

Bunuel · Answer

Check  ADVANCED OVERLAPPING SETS PROBLEMS

Hope it helps.

unceldolan · Answer

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!

Bingo123 · Answer

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 ?

EMPOWERgmatRichC · Answer

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.

GMAT assassins aren't born, they're made,
Rich

Bingo123 · Answer

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.

EMPOWERgmatRichC · Answer

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.

GMAT assassins aren't born, they're made,
Rich

Bingo123 · Answer

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.

Kinshook · Answer

22 + 15 + 14 - 9 - 1*2 = 40
None = 50 - 40 = 10

IMO C

Posted from my mobile device

VIGHNESHKAMATH · Answer

This is direct formula application. Formula is as follows:

N (UNIVERSAL SET) = N(A)+N(B)+N(C)- N (EXACTLY 2 SET CASES)- 2 X (EXACTLY 3 SET CASES) + N (NEITHER OF THE 3 SETS)

Thank You
Vighnesh

Karmoker · Answer

You can do this by using a venn diagram rather than formula

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

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