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The problem gives 1 child ticket for every 4 adult ticket.
It also gives 3 senior tickets for every 4 adult tickets.
Thus, for every 4 adult tickets the problem gives 1 child ticket and 3 seniors tickets.
So, we have some number of groups, each containing 4 adult, 1 child and 3 senior tickets.
In all, each such group has 4+1+3 = 8 tickets; hence, the number of groups is 72/8=9.
From this point, there is one step to get the answer.
ANSWER. There were 9 child tickets, 3*9 = 27 senior tickets and 4*9 = 36 adult tickets.
Solved.
The method which I used to solve the problem, is a "grouping method".
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I am sorry . . .
I didn't make a diagram or a table as you asked.
I simply solved the problem from the beginning to the end in a way on how it should be presented to a 2-nd or 3-rd or 4-th grade students.
