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

Refer diagram below:



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

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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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!

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 ?

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

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.

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

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.

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

IMO C

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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

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