Can someone please solve this question? I have the answer too but didn't really understand how have they solved it ![Confused :?]()
Question: In an office, where working in at least one department is mandatory, 78% of the employees are in operations, 69% are in finance, 87% are in HR.
What are the maximum and minimum percentages of employees that could have been working in all three departments??
Solution: There is a formula for such questions.
We have 3 departments with 78%m 69% and 87%.
Now, 100 - 78 = 22
100 - 69 = 31
100 - 87 = 13
So, minimum for working in all three = 100 - (22+31+12) = 34
maximium is simple as 13 percent is required to make 100 with 87. So take this 11 from 78=78-11=67 & 2 from 69=67
minimum is to reduce overlap=78+69+87-200=34
Let I be the % of employees working in at least one department, so I = 100
II be the % of employees working in at at least two departments
III be the % of employees working in all three departments
I ≥ II ≥ III
=> I + II + III = 78 + 69 + 87 = 234
Now, to mimimize III put I and II 100
=> III(min) = 234 - 200 = 34
To maximize III, put II = III (as II ≥ III) and I has to 100
=> 2*III(max) = 234 - 100
=> III(max) = 67

Question: In an office, where working in at least one department is mandatory, 78% of the employees are in operations, 69% are in finance, 87% are in HR.
What are the maximum and minimum percentages of employees that could have been working in all three departments??
Solution: There is a formula for such questions.
We have 3 departments with 78%m 69% and 87%.
Now, 100 - 78 = 22
100 - 69 = 31
100 - 87 = 13
So, minimum for working in all three = 100 - (22+31+12) = 34
maximium is simple as 13 percent is required to make 100 with 87. So take this 11 from 78=78-11=67 & 2 from 69=67
minimum is to reduce overlap=78+69+87-200=34
Let I be the % of employees working in at least one department, so I = 100
II be the % of employees working in at at least two departments
III be the % of employees working in all three departments
I ≥ II ≥ III
=> I + II + III = 78 + 69 + 87 = 234
Now, to mimimize III put I and II 100
=> III(min) = 234 - 200 = 34
To maximize III, put II = III (as II ≥ III) and I has to 100
=> 2*III(max) = 234 - 100
=> III(max) = 67