Strange Venn diagrams question

This is not a venn diagram question.

First of all, I would say this is not even a proper GMAT question.

But this is a Rate, Time, Distance problem.

Here is my solution:

J came last: 24 times.
Therefore, A and M came 1st or 2nd or draw 24 times.
A arrived before M 1/8 times: 48/8 = 6
M arrived before A 1/6 times: 48/6 = 8
total different ways when they don't have draw = 8+6 =14
Rest of the time they must have draw. i.e. draw = 48-14 = 34

Now we need to find minimum no. of times when J has to come first.
We can find this if we maximize the no. when J cannot become 1st.
This can happen if he comes 2nd, i.e. between A and M.
It can happen maximum of 14 times. (This can happen when A and M have a draw all the times when J came last)
Therefore, J can come first at least 10 times.