Annual Meet Puzzle

The annual meet was attended by 20 persons. The persons who attended the meet have worked in eight different firms – M, BC, K, A, BA, D, PC and E. The following table provides information about the count of persons who have worked in exactly one, two, three, four, five and six of the given eight firms. At least one person worked in each of the given eight firms.

Person count   1 4 5 2 6 2
Firm Count     1 2 3 4 5 6


The information about the number of persons who worked in a particular firm.The table provides provides information about only four of the eight firms but does not specify them.

Firm 4 -9
Firm 3 -13
Firm 2 -8
Firm 1 -11


If there is no firm in which less than 5 people have worked, then at most how many persons could have worked in a particular firm?

There are exactly X firms in which exactly Y persons have worked. What could be the maximum possible value of X?

For at most how many firms, the number of persons who have worked in it is more than those who have worked for Firm 4?

Source: www.imsindia.com

• For the second and third questions do you continue to use the assumption that "there is no firm in which less than 5 people have worked"? – hexomino Oct 29 '18 at 15:33
• No, those are independent questions. – sam Oct 29 '18 at 15:34

For the first table, define

The person count $$p_m$$ corrsponding to the firm count $$m$$.

Then, let us define the firm-person count as

$$\displaystyle \sum_{m=1}^6 m p_m = (1\times 1) + (2\times 4) + (3 \times 5) + (4 \times 2) + (5 \times 6) + (6\times 2) = 74$$ This should result in the same value as when we add up the number of people to have worked in each firm, in each case we count each firm placement for each person exactly once. In other words, $$\displaystyle \sum_{j=1}^8 \text{Firm } j = 74 \Rightarrow \text{Firm } 5 + \text{Firm }6 +\text{Firm }7 +\text{Firm }8 = 33$$

Question 1

If there is no firm in which less than $$5$$ people have worked, then any three of the four remaining firms must have had $$15$$ placements. From the formula above, this means that, at most, ($$33 - 15 =$$) 18 people have worked at a particular firm.

Question 2

I think the maximum possible value of $$X$$ is $$4$$.
$$33$$ is not divisible by $$4$$ so we could not make $$X=5$$ but for example, we could have $$\text{Firm }5 = 9,\,\,\,\, \text{Firm }6 = 9,\,\,\,\, \text{Firm }7 = 9, \,\,\,\, \text{Firm }8 = 6$$ which would make $$4$$ firms with nine people each.

Question 3

I think the answer here is $$5$$
We already know that more people have worked at Firms $$1$$ and $$3$$ than at Firm $$4$$. For the remaining firms, it is not possible to have all four with more people than Firm $$4$$ as $$\frac{33}{4} < 9$$ but we could, for example, have $$\text{Firm }5 = 10,\,\,\,\, \text{Firm }6 = 10,\,\,\,\, \text{Firm }7 = 10, \,\,\,\, \text{Firm }8 = 3$$

• Thanks for your explanation. I have a doubt here that in the beginning it is given that there are 8 firms but the values of m are1,2,3,4,5,6 and that adds upto to 21 firms .That contradicts with the statements that there 8 firms. Can you pls explain this part? – sam Oct 29 '18 at 15:52
• All this table says is that there is no person that works for more than 6 firms. The relevant bit is that the numbers on top add up to 20 which means every worker is counted at least once. Did you not write this question? You should really provide attribution for questions that are not yours. – hexomino Oct 29 '18 at 16:35