Monkton University has established a board which reviews all applicati

Question

Monkton University has established a board that reviews all applications for experiments to be conducted by students or faculty. Researchers submit proposals to one or more committees of the board, each of which is responsible for certain 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 5 members. Committee 1 must reach a unanimous vote in order to approve a proposal. Committees 2 and 3 may approve a proposal with no more than one “no” vote. Approval for Committee 4 requires a simple majority vote.

Committee 1: humans (all ages)
Committee 2: all other animals (e.g., rats, worms)
Committee 3: all other living organisms (e.g., bacteria, plant matter)
Committee 4: research that does not involve living organismsThree proposals were recently approved by the applicable committees.

1. A doctoral candidate was given permission to conduct a study on pain management in humans using medication as well as alternative techniques such as massage therapy and meditation.
2. A zoology professor gained approval to conduct a series of studies on the swarming behaviors of various animals when predators are in the vicinity.
3. A botany professor and her students received permission to catalog all living organisms native to a nearby nature preserve that usually does not grant access to people.

One proposal was rejected.

4. An undergraduate student was refused permission to study the physiological effects of illegal drug use in human subjects. Statement 1: Yes. Tab 2 indicates that the doctoral candidate will be studying pain management in humans, so committee 1 would be required to approve the study.

Statement 2: No. The first paragraph of the first tab states that the task of each committee is to ensure that the proposed research complies with all laws and meets the university’s standards for ethical experimentation. The other information in both tabs does not indicate why this proposal was rejected; it is possible that the experiment was entirely ethical but failed to comply with a particular law.

Statement 3: No. According to tab 2, the botany professor’s study will catalog all living organisms native to a nearby nature preserve that usually does not permit access by people. At the least, then, this study would require approval from committees 2 (animals) and 3 (other organisms). However, committee 4 is for projects that don’t involve living organisms, so committee 4’s approval is not required.

A research study involving interactions between people (of any age) and their animal pets would need to be approved by Committees 1 and 2. Both committees must have at least 5 members, but Committee 1 needs unanimous approval, while Committee 2 can approve with up to one “no” vote. The minimum number of “yes” votes would occur when both committees have the minimum number of members, resulting in 5 “yes” votes from Committee 1, and 4 “yes” votes from Committee 2, for a minimum of 9 “yes” votes.

The correct answer is (D).

The correct answer is (D).

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 actually 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. \(\frac{26}{3}\) = 8 \(\frac{2}{3}\), which means that we need at least 9 people to staff the 6 committees.

The correct answer is C.

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.  \frac{26}{3}  = 8  \frac{2}{3} , which means that we need at least 9 people to staff the 6 committees.

The correct answer is C.

Statement 1: No. According to tab 2, the undergraduate’s proposal was for a study involving human subjects; this falls under committee 1. Approval in this committee requires a unanimous vote; therefore, a single “no” vote to would be enough to reject the proposal.

Statement 2: Yes. According to tab 2, the botany professor’s proposal targets “all living organisms native to a nature preserve.” Committees 1 through 3 all deal with living organisms, but the nature preserve “usually does not permit access to people,” so no people are living in the preserve. The relevant committees, then, must be committees 2 and 3. Both committees can approve with no more than a single “no” vote in each. If the minimum number of members on each committee is 5 and no more than 1 person in each committee may vote “no”, then there must be at least 8 “yes” votes.

Statement 3: No. Committee 4 handles proposals for research that does not involve living organisms. The minimum number of members on any committee is 5, and on committee 4, approval is achieved with a simple majority (over half), so a minimum of only 3 “yes” votes is required.

Kurtosis · Answer

Hi,

We have to find the minimum number of board members.

We know that a person can repeat in a maximum of 3 committees. So when minimizing the total number of members we count each member 3 times. Hence we divide the total by 3. You have to understand this part and extrapolate this logic in the exam depending upon the question.

Hope it helps.

thuyduong91vnu · Answer

Hi Vyshak, Thanks for your response

Yes I see the approach is applicable for the case in question ( which requires (1) a minimum of 4 members for each committee, and (2) odd number of members for committee 5 and 6 ), but I still wonder could it be so for all other cases. Let's say in other cases ( in which we do not have to meet 2 requirements above ), 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 OR each of the first 5 committees has 2 member, the 6th committee has 16 members) such that the total number remains the same (= 26 members), then the answer could not be just \frac{26}{3} = 8 \frac{2}{3} , right? My question is quite confusing, isn't it :-D

I just try to imagine another case and then apply the solution to see whether it works for all cases. Thanks for your help!

Kurtosis · Answer

Hi thuyduong91vnu , You are still not getting my point. What does the question stem ask? --> Minimum number of board members necessary in order to staff all six committees Given: 1) Board members must sit on a minimum of one committee and a maximum of three. 2) each committee must have a minimum of 4 members 3) Committees 5 and 6 must have an odd number of members Now how do you minimize the number of members? --> By using the same member in 3 different committees --> By doing this we are counting each member 3 times. So we have to divide the total by 3. If each member could represent 4 committees then dividing by 4 will provide the answer, provided the question stem asks to minimize the number of members. Now your question: adjust the number of members for each commmittee (each of the first 5 committees has 1 member, the 6th committee has 21 members OR each of the first 5 committees has 2 member, the 6th committee has 16 members) --> You have changed the conditions now. Let's take the case where 5 committees have 2 members and 5th committee has 16 members. Total number = 26 Let A, B, C, D, E and F be the different committees --> Committee A, B and C can be formed by using just 2 distinct members. D, E can be formed by using just 2 members. The members in D and E can represent one more committee F. So F will 14 distinct members Total distinct members = (2 * 6)/3 + 4 --> 12/3 + 14 = 18 In the earlier case, the minimum number of each committee was 4: So assume A, B, C, D were made up of 4 members each and E, F were made of 5 members each A, B, C can be formed by just 4 members D can be formed 4 members E, F will be formed by reusing the 4 members in D and adding 1 new member Total distinct members = (4 * 6)/3 + 1 --> 24/3 + 1 = 9 So it all depends on how many times a member can be reused. A logic cannot be generalized to work in all scenarios. Hope I am clear now and I answered your question

jabhatta2 · Answer

Hi  KarishmaB  - for Q4, part 2, specifically

Do you think committee 1 has to be looked at when choosing an answer for this one ?

I thought "YES" because perhaps the   nature preserve   itself has people living on the land itself

The   nature preserve   may have people living on it but dont allow  other  people to come in contact with it.

Per the OA solution however, to answer this question -- one has to look at committee 2 + committee 3 only

KarishmaB · Answer

Yes, it specifies that "people" are not granted permission to enter. It implies that committee 1 is not involved. 
In committees 2 and 3 at least 5 members will be there and there will not be more than 1 no vote in each. Hence, at least 8 yes votes were received. 
Even if you did think that people could also be involved, then also the answer is "yes". Because 8 yes votes would be received from committees 2 and 3 themselves. Committee 1 would have led to more "yes" votes, certainly not fewer.

Heema22 · Answer

KarishmaB

In question 2 can you please make me understand why can't we take committees as -

Committee 1 - A B C D E (Board members can sit on a maximum of 3 com)
Committee 2 - A B C D E (Board members can sit on a maximum of 3 com)
Committee 3 - A B C D E (Board members can sit on a maximum of 3 com)
Committee 4 - F G H I J (Since Board members must sit on a minimum of one committee and a maximum of three; each committee must have a minimum of 5 members)

For agreement of Com 1 and Com 2 - minimum numbers of YES votes will be 5 (ABCDE since both coms have same members) 
compared to 9 (ABCDE and FGHI when both coms have different members and you need 5 + 4 different votes)

Can you please make me understand which understanding of mine is wrong here???

KarishmaB · Answer

You can but how many 'Yes' VOTES do you need? Committee 1 and committee 2 both should accept the proposal. Committee 1 must give 5 Yes votes and committee 2 must give at least 4 Yes votes. So minimum number of votes obtained (not people saying Yes) will be 9. Same person serving on both committees must give 2 Yes votes.

MT1302 · Answer

Hello  Sajjad1994 ,

I think so that you need to  edit Question no. 3 and it's correct answer  because as per the information present in the passage we are having only 4 committee members and the question stem is asking about 6 committees (for which we have no clue).

So as per the information we are having 4 committees. Total Members = (5 * 4) i.e. 20 members and one committee member can board max. 3 committees thereby 20 / 3 = 6. 66... approximately equal to minimum 6 members(A) as per the answer choices present.

Sajjad1994 · Answer

Official Explanation

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

Difficulty Level:   Very Hard

Explanation

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 actually 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. \(\frac{26}{3}\) = 8 \(\frac{2}{3}\), which means that we need at least 9 people to staff the 6 committees.

The correct answer is C.

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
 \frac{26}{3} =8\frac{2}{3} , which means that we need at least 9 people to staff the 6 committees.

The correct answer is C.

captain0612 · Answer

For question 3, why are we considering the board members per comittee as 4? I could not find any information that says every comittee consists of a minimum of 4 board members.

Apeksha2101 · Answer

I don't know how you're inferring this but in the passage, it is explicitly given that a minimum 5 members of each committee must be there.

Deepa444 · Answer

Sajjad1994   KarishmaB   chetan2u  is the passage incomplete? because there is nothing given about committee 5 and 6 and also its given that min no. of members in each must be 5 unlike the solutions posted.

chetan2u · Answer

Yes, you are correct. Something is missing in the question. The official explanation for Q.3. gives minimum number to be 4, while the question talks of minimum number to be 5. Even the number of committees does not match.

The question in it's present form requires to be archived and that is what we would do till the time exact wordings are posted.

Thank you.

bumpbot · Answer

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Yes	No
		The doctoral candidate’s proposal was reviewed by committee 1.
		The undergraduate student’s proposal was rejected because it failed to meet the university’s ethical standards.
		The botany professor’s proposal required approval from all four committees.

Yes	No
		At least two people rejected the undergraduate student’s proposal.
		The botany professor’s proposal received at least 8 “yes” votes.
		A minimum of 4 “yes” votes is required for any research that does not involve living organisms.

Monkton University has established a board which reviews all applicati

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