Dividing 45 plus x

Question

Two fishermen were fishing. The first fisherman caught 18 fish, and the second 27 (all fish are the same size). When they had fried all 45 fish, a stranger joined them and all the fish were divided equally into three parts. The stranger paid the fishermen \$90 for one equal share. How should the fishermen divide these \$90 among themselves so that the division is fair? I think this is a ' trick question ' since \$90 was given as a fair share it means 90 IS the fair share ( Fishermen )

Answer 1

I believe it is 90$ and a trick question . If anyone provides a number theoretical solution I would be very grateful . If it's not 90 then we have (45 + x )/3 =9o where x is fish caught by stranger .

Answer 2

The "trick" is not actually a trick, but realising that no matter what subset of the $45$ fish are given to the stranger, the amount of money will be split out at ratio of $27 : 18 \to 3 : 2$, (due to the pre-factor of each fish is same.)

This question targets the relevance of total work contributed and its subsequent income, not the splitting of the fish from each fishermen. Since fisherman 2 did more work than fisherman 1, he will get paid more.

Another factor that plays well into this is that the fish are fried and mixed together, then split equally, hence out the $15$ fish given to the stranger, $9$ fish came from fisherman $2$ and $6$ fish came from fisherman $1$. See that fisherman $2$ has contributed more fish in the equal splitting than fisherman $1$ (both before splitting and thus after).

Hence, fisherman $2$ is paid $\$54$ and fisherman $1$ is paid $\$36$.

Answer 3

I do not agree with either answer posted.

Forget about the fish being fried, mixed together so that they are indistinguishable, and then eaten. !

Instead think that the first and second fisherman sold $15$ fish to the stranger after reserving $15$ each for themselves.

Fisherman $1$ must have sold $18-15 = 3$ fish,
and fisherman $2$ must have sold $27-15= 12$ fish,

and @ $\large\frac{90}{15} = 6$\$ per fish,
fisherman $1$ should get $18$ \$,
and fisherman $2$ should get $72$ \$

It is a "trick" question only in the sense that one is likely to wrongly divide the money between them in the proportion of fish caught.