Thread: Possible Number Committees

A class of 43 students is divided into committees in such a way that each student serves on exactly one committee. Each committee must have at least 3 members and at most 5 members. The possible number of committees that can be formed is

(a) at least 8 and at most 14
(b) at least 8 and at most 15
(c) at least 9 and at most 14
(d) at least 9 and at most 15

I want to know how you find the answer. The answer is one of the choices above.

It can't be choices (a) or (b), because if you have 8 committees with a maximum of 5 students each, then you would have 40 students total, which is not enough.

It can't be choices (b) or (d), because if you have 15 committees with a minimum of 3 students each, then you would have 45 students total, which is too many.

So the answer is (c).


01

Originally Posted by yeongil

It can't be choices (a) or (b), because if you have 8 committees with a maximum of 5 students each, then you would have 40 students total, which is not enough.

It can't be choices (b) or (d), because if you have 15 committees with a minimum of 3 students each, then you would have 45 students total, which is too many.

So the answer is (c).


01

Yes, that's what the math books says. I just wanted to know how it's done.

Thanks

Originally Posted by sharkman

A class of 43 students is divided into committees in such a way that each student serves on exactly one committee. Each committee must have at least 3 members and at most 5 members. The possible number of committees that can be formed is

(a) at least 8 and at most 14
(b) at least 8 and at most 15
(c) at least 9 and at most 14
(d) at least 9 and at most 15

I want to know how you find the answer. The answer is one of the choices above.

First look at the least number of members which will give you the most committees:

3 divides 43 14 times with one left over. So you put the extra in any one
of the committees with three.

Now look at the greatest number of members which will give you the least committess:

5 divides 43 8 times with 3 left over. You can't add these extras to one
of the existing committees because they already have the max number.
So you create one more committee for the three. 9 total.

So you have 9 min and 14 max.

It's just division and then handling the remainder.

Originally Posted by mikeyS

First look at the least number of members which will give you the most committees:

3 divides 43 14 times with one left over. So you put the extra in any one
of the committees with three.

Now look at the greatest number of members which will give you the least committess:

5 divides 43 8 times with 3 left over. You can't add these extras to one
of the existing committees because they already have the max number.
So you create one more committee for the three. 9 total.

So you have 9 min and 14 max.

It's just division and then handling the remainder.

Thank you for providing more details about the answer to this question.

Originally Posted by mikeyS

First look at the least number of members which will give you the most committees:

3 divides 43 14 times with one left over. So you put the extra in any one
of the committees with three.

Now look at the greatest number of members which will give you the least committess:

5 divides 43 8 times with 3 left over. You can't add these extras to one
of the existing committees because they already have the max number.
So you create one more committee for the three. 9 total.

So you have 9 min and 14 max.

It's just division and then handling the remainder.

Thanks for the detailed reply.