A certain company has the following policy regarding membership in co

Question

A certain company has the following policy regarding membership in committees A, B, C: each of the committees have exactly 40 members, and each employee must be a member of at least one of the committees. The company currently has 110 employees.

Let X be the current number of employees who are members of more than one of the committees. Based on the current total number of employees, select the least value of X that is compatible with the policy and select the greatest value of X that is compatible with the policy.

HeyBarbie · Accepted Answer

Let X be the number of employees that joined all two committees. Let Y be the number of employees that joined three committees.

Recall the formula for number of unique elements:
(Total # in A) + (Total # in B) + (Total # in C) - (# in groups of exactly 2) - 2*(# in groups of exactly 3) + (# in neither)

Applying this formula,
40+40+40-X-2(Y)+0=110

To solve for min and max, 
(1) Assume X=0, 120-0-2(Y)=110
Y=5
(2) Assume Y=0, 120-X-2(0)=110
X=10

Therefore, Min is 5, Max is 10

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chetan2u · Answer

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.

sayan640 · Answer

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 ?

chetan2u · Answer

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.

samarpan.g28 · Answer

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 .

kabirgandhi · Answer

Have a logical doubt:

When we have three groups, A, B, and C, and want to minimise the total number of people in two or more of those groups.

Let's say out of 110 people, 40 are a part of group A, leaving us with 110-40=70 members in total in the group unassigned.

Now, we assign B to 40 out of the remaining 70, which leaves us with 70-40=30.

Now, to assign C to at least 40 out of the remaining 30, we can assign it to the remaining 30 people who are unassigned, which would leave us with 10 yet to be assigned.

And since all the people have been assigned to Group A, B, or C, we can minimise the total number by having these 10 people already assigned to A/B and assigned to C.

Therefore, Min Value should be 10.

This is the approach I used to solve, prevalent among set problems with least/greatest intersection value. Can anyone help understand where I went wrong?

KarishmaB   Bunuel

KarishmaB · Answer

Nothing wrong with that but we are not taking overlap of all 3 in that case. If only overlap of 2 is allowed then 10 people must be in the overlap region since None = 0.

But when all 3 overlap is also allowed, we can reduce the number of people in the overlap region. Think in terms of instances and people. We have 110 people and 120 instances. We can distribute 110 instances to 110 people - one each since everyone should be on at least one committee. Now 10 instances are remaining to be distributed. You can give one each to 10 people or 2 each to 5 people. So minimum overlap is 5 and maximum is 10.

Ak284 · Answer

in the question, X is number of employees in more than one committee. So employees who joined 2 committees and 3 committees, right? in the formula, do we need to multiply by 2?

cheshire · Answer

When finding the maximum value of X we use 2 because it allows the maximum amount of people to be in 2 or more committees. We multiply by 3 when we are looking for the minimum amount of people in 2 or more committees.

Prak0709 · Answer

Sorry if its a bit lame but why do we multiply 3 member groups by 2?

Bunuel · Answer

Because someone who is in three committees gets counted three times in 40 + 40 + 40, but should be counted only once as an employee.

So we must remove the extra counts. Being counted three times instead of once means there are 2 extra counts to subtract.

That is why each person in all three committees is multiplied by 2 in the formula.

More on this here:  ADVANCED OVERLAPPING SETS PROBLEMS

Usernamevisible · Answer

750-Level Thinking (Target: 30–45 seconds): This is a set-counting/inclusion-exclusion problem. The key is to stop thinking about people and start thinking about committee seats.

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Step 1: Total committee memberships. Each committee has 40 members, so the total number of committee seats occupied is 40 + 40 + 40 = 120. However, the company has only 110 employees.

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Core Insight (the 750-level trigger): Every employee who belongs to exactly one committee contributes 1 seat. Every employee who belongs to multiple committees contributes extra seats. Therefore, the excess seats are 120 − 110 = 10. This is the entire problem.

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Let x2 be the number of people in exactly 2 committees and x3 be the number of people in exactly 3 committees. Then X = x2 + x3, since X represents the number of employees who belong to more than one committee. A person in 2 committees creates 1 extra seat, while a person in 3 committees creates 2 extra seats. Therefore, x2 + 2x3 = 10. This equation captures the whole problem.

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Least possible X: We want as few people as possible creating those 10 extra memberships, so we maximize the number of triple-members. Since each triple-member contributes 2 extra seats, we can have x3 = 5 and x2 = 0, giving X = 5. This is feasible because 5 people belonging to all three committees contribute 15 memberships in total, leaving each committee needing 35 additional unique members. Adding 35 unique members to each committee gives 35 + 35 + 35 + 5 = 110 employees. Therefore, Xmin = 5.

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Greatest possible X: We now want as many people as possible creating those 10 extra seats, so nobody should belong to all three committees. Instead, every overlapping employee should belong to exactly two committees. Thus, x2 = 10 and x3 = 0, giving X = 10. This is feasible because the 10 double-members create exactly 10 extra memberships, and the remaining 100 employees each belong to exactly one committee. Total memberships are therefore 100(1) + 10(2) = 120, which satisfies the requirement. Hence, Xmax = 10.

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Answer: Least value = 5, Greatest value = 10.

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750-Scorer Under-1-Minute Thought Process: Total seats = 120. Employees = 110. Therefore, there are exactly 10 extra memberships. A double-member contributes +1 extra, while a triple-member contributes +2 extras. To minimize X, maximize triple-members: 10 ÷ 2 = 5. To maximize X, make everyone a double-member: 10 ÷ 1 = 10. Therefore, the answer is (5, 10). A strong test-taker should finish this in about 20–30 seconds.

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Pattern Recognition Code (Memorize): Whenever you see fixed group sizes, a total number of people, and the possibility of multiple memberships, immediately compute: Extra memberships = Total seats − People. Then ask: how many extra seats does each overlapping person contribute? For minimum overlap, maximize the contribution per person (triples, quadruples, etc.). For maximum overlap, minimize the contribution per person (doubles, pairs, etc.). This is one of the fastest and most powerful counting shortcuts on difficult set-counting problems.

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Similar 750-Level Drills:

Drill 1:  Three clubs have 50 members each. There are 140 students, and every student belongs to at least one club. Find the minimum and maximum number belonging to multiple clubs. (Answer: 5 and 10.)

Drill 2:  Four committees have 30 members each. There are 100 employees, and everyone belongs to at least one committee. What is the minimum number belonging to multiple committees? Think: total seats = 120, extras = 20, and one person can contribute at most 3 extras. Final Answer: (Min, Max) = (7, 20).

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GMAT Takeaway: The elite pattern is Seats − People = Overlap Units. Once you know the number of overlap units, the extrema become simple: maximize the number of extra seats contributed per person to minimize the number of overlapping people, and minimize the number of extra seats contributed per person to maximize the number of overlapping people. This converts what looks like a messy Venn-diagram problem into a 20-second arithmetic problem.

A certain company has the following policy regarding membership in co

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A certain company has the following policy regarding membership in co