Monkton University has established a board which reviews all

Question

Monkton University has established a board which reviews all applications for experiments to be conducted by students or faculty. Researchers submit proposals to one or more committees, each of which is responsible for certain kinds of research subjects, and the committees’ task is to ensure that the proposed research both complies with all laws and meets the university’s standards for ethical experimentation. A proposal must gain approval from all committees for which the research parameters apply.

Board members must sit on a minimum of one committee and a maximum of three; each committee must have a minimum of 4 members. Committees 1 and 2 must reach a unanimous vote in order to approve a proposal. Committees 3 and 4 may approve a proposal with no more than one “no” vote. Committees 5 and 6 must have an odd number of members; a simple majority is sufficient to approve a proposal.

What is the minimum number of board members necessary in order to staff all six committees according to the given rules?

âƒ� 6
âƒ� 8
âƒ� 9
âƒ� 15
âƒ� 24

This is an IR question from MGMAT test. Could someone help to explain a bit the solution from MGMAT?

To compute the minimum number of board members, we must determine how many people we need to staff each of the 6 committees at the minimum level. The second paragraph tells us that each committee must have a minimum of 4 members; there are 6 committees, so it is tempting to say that we must need a minimum of 24 board members. Each board member, however, is allowed to sit on up to 3 committees. In addition, committees 5 and 6 must have an odd number of members – and 4 is not an odd number. Therefore, committees 5 and 6 must actaully have a minimum of 5 members.

The minimum number of committee members needed is then 4 + 4 + 4 + 4 + 5 + 5 = 26 members.

Remember also that each committee member is allowed to sit on 3 committees. 26/3 = 8 2/3, which means that we need at least 9 people to staff the 6 committees.

The correct answer is C.

According to the given solution, we just have to calculate the minimum number of committee members needed, and then divide it by 3. Is this approach applicable for all cases or just for the case in question (which requires a minimum of 4 members for each committee)? I mean, let's say in the extreme case, if we adjust the number of members for each commmittee (each of the first 5 committees has 1 member, the 6th committee has 21 members) such that the total number remains the same (= 26 members), then the answer could not be just 26/3 = 8 2/3, right? I don't quite understand the logic of the solution, why it could be so and when it is applicable, I do think that I am not confident to use it if I have to face it in the real test

Thanks for your help

ceilidh.erickson · Answer

According to the given solution, we just have to calculate the minimum number of committee members needed, and then divide it by 3. Is this approach applicable for all cases or just for the case in question (which requires a minimum of 4 members for each committee)? I mean, let's say in the extreme case, if we adjust the number of members for each committee (each of the first 5 committees has 1 member, the 6th committee has 21 members) such that the total number remains the same (= 26 members), then the answer could not be just 26/3 = 8 2/3, right? I don't quite understand the logic of the solution, why it could be so and when it is applicable, I do think that I am not confident to use it if I have to face it in the real test

You're right: if 5 of the 6 committees had only 1 member each and the 6th had 21, we couldn't simply divide 26/3. In this case, the 6th committee by itself would need 21 members (though any of those 21 could be on the other 1-person committees).

The reason that this strategy works in this case is that each individual committee requires fewer people than the total divided by 3.

To your larger question, though - how can we find a rule that would apply in every context? Well... we shouldn't! I think it would take more time and effort to memorize a slew of different formulas for different scenarios, especially because you're unlikely to see the exact same question types again.

Take each problem on a case-by-case basis. On this problem, I personally would have worked backwards from the answer choices, starting with the smallest:

- Can I make 6 work? 6 people onto 6 committees, each person can sit on 3 committees... that only gets us up to 6 committees of exactly 3. Doesn't work.

- How about 8 people? people: A, B, C, D, E, F, G, H
1: ABCD
2: BCDE
3: CDEF
4: EFGH
5: FGHA
6: GHAB
It works for 6 committees of 4... but committees 5 and 6 need a minimum of 5 people. So 1 more person would do it.

The answer must be 9.

Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

BTG Expert Team · Answer

What we see: a Multi-Source Reasoning question about staffing six committees. We need the smallest number of board members that can fill every committee under the rules. Step 1. Pull out only the rules that set sizes. What we see: each member sits on 1 to 3 committees; each committee has at least 4 members; committees 5 and 6 must have an odd number of members. The voting rules (unanimous, one "no" allowed, simple majority) do not change any size. What it means: committees 1 to 4 need at least 4 members each. Committees 5 and 6 need an odd number that is at least 4, so the smallest is 5. What we do and why: we ignore the voting rules. They are there to slow us down. Step 2. Count the seats. What we see: 4 + 4 + 4 + 4 + 5 + 5. What it means: at least 26 seats must be filled. What we do and why: we now ask how few people can cover 26 seats. Step 3. Use the 3-committee limit. What we see: one person can fill at most 3 seats. What it means: 8 people fill at most $8 \times 3 = 24$ seats, which is less than 26. So 8 or fewer can never work. 9 people can fill up to 27 seats, which is enough on paper. What we do and why: we still have to show 9 really works, because one person cannot take two seats on the same committee. Step 4. Build it with 9 people. Number the people 1 to 9. Committee 1: 1, 2, 3, 4 Committee 2: 5, 6, 7, 8 Committee 3: 9, 1, 2, 3 Committee 4: 4, 5, 6, 7 Committee 5: 8, 9, 1, 2, 3 Committee 6: 4, 5, 6, 7, 8 What we see: every committee has its own different people, the sizes are 4, 4, 4, 4, 5, 5, person 9 sits on 2 committees and everyone else on 3. What it means: all rules hold with 9 people. The minimum is 9. The trap: 24 (choice E). It comes from 6 committees times 4 members, which forgets two things: one person can sit on 3 committees, and committees 5 and 6 need 5 members, not 4. Faster way: seats needed divided by the most seats one person can hold, rounded up: $26 \div 3 = 8.67$, so 9. This shortcut works here because no single committee needs more than 9 people; always check that before you trust it. Takeaway: for "fewest people to fill the spots," count the total spots, divide by the most spots one person can take, round up, then build one real example to prove it works. Answer: C

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