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# Monkton University has established a board which reviews all applicati

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Re: Monkton University has established a board which reviews all applicati [#permalink]
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thuyduong91vnu wrote:
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 I just try to imagine another case and then apply the solution to see whether it works for all cases.

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.

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Re: Monkton University has established a board which reviews all applicati [#permalink]
Hi KarishmaB - for Q4, part 2, specifically

Quote:

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
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Re: Monkton University has established a board which reviews all applicati [#permalink]
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jabhatta2 wrote:
Hi KarishmaB - for Q4, part 2, specifically

Quote:

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

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.
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Monkton University has established a board which reviews all applicati [#permalink]
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???
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Re: Monkton University has established a board which reviews all applicati [#permalink]
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Paumilpatel wrote:
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???

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.
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Monkton University has established a board which reviews all applicati [#permalink]
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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.
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Re: Monkton University has established a board which reviews all applicati [#permalink]
MihirTandon wrote:

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.

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

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Re: Monkton University has established a board which reviews all applicati [#permalink]
­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.
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Re: Monkton University has established a board which reviews all applicati [#permalink]
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Kurtosis wrote:
thuyduong91vnu wrote:
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.

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.