­A certain company has the following policy regarding membership in co

This is a lovely answer. Thank you

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MartyMurray KarishmaB
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Short and simple..

If each employee was part of exactly one company, then total employee would have been 40+40+40 or 120. But there are 110 employees, so these 10 extra, 120-110, must get adjusted in employees being part of more than one company.

Least value of X: This would happen when the extra are adjusted in the overlap area of all three companies, that is the individual is part of all three companies. So each individual would account for 3 employees, one each in A, B and C. Thereby, that employ is accounting for two more employees.
If 2 employees are catered for by one employee, then 10 employees will get catered by 10/2 or 5 employees.

Maximum value of X: This would happen when the extra are adjusted in the overlap area of exactly two companies, that is the individual is part of only two companies. So each individual would account for 2 employees, one each in A and B or A and C or B and C. Thereby, that employ is accounting for one more employee.
If 1 employee is catered for by one employee, then 10 employees will get catered by 10/1 or 10 employees.
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chetan2u  The highlighted part was not completely clear to me. Will you be kind enough to  exemplify it and  break it down for me a bit ? 

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­Say there was one employee each in sections A, B and C of company X, but the total number of employee in X was only one.

How would that happen: It would be possible only when the same employee is part of the three sections. Here, three(1+1+1) employees in total in A, B and C is catered by just one employee in the overlap region 'All three'. So, this one employee in X accounts for the increase 3-1 or 2 employee. 
Relate it to a Venn Diagram and the overlapping Zone and it will be easier to comprehend.

­This is a great solution    
I had gone by the options (if not approaching the unique elements) and this is how I solved it.

Firstly, we have to understand that members can be common either in all 3 groups or in 2 groups.

If they are common in all 3 groups, then naturally the number of members overall will be less. Let's check the options.
If the number of common members is 3, then 3+((40-3)*3)=114. Wrong.
If the number of common members is 5, then 5+((40-5)*3)=110. Correct.
You can similarly calculate for other options, none will fit.

Let's now find the greatest number of overall members. Here, common members will be present in 2 groups.
If the number of common members is 3, then 3+((40-3)*2)+40=117. Wrong.
If the number of common members is 5, then 5+((40-5)*2)+40=115. Wrong.
If the number of common members is 10, then 10+((40-10)*2)+40=110. Correct.­