Let's represent adults with A, Pensioners with P and Children with C.
A=3.50, P=1.00 and C= 0.85
Let's take a random example. 5A+20P+75C amounts to a cost of 101.25.
Since the result is too big, we must reduce the number of A and add to the numbers of P and C.
4A+21P+75C=98.75
Note that since the value is lower than the desired end result, we must decrease the number of lower value objects and replace them with objects of higher value.
4A+22P+74C= 98.90
Here we are exchanging C for P in order to get close to 100. Note that using the exchange C for A will take the desired result over 100.
It would be useful to have some rules:
P for A=+2.50
A for P= -2.50
P for C=-0.15
C for P=+0.15
A for C=+2.65
C for A= -2.65
If the student follow these rules of exchange, he will find the answer is 3A+47P+50C=100
END