We can buy the fruit in two ways:
(88c, 2o, 10w) or (84c, 11o, 5w)
Method:
These are Linear Diophantine equations in 3 variables, so we need to deduce additional constraints based on the fact that the solutions are positive integers.
0.5x + 3y + 5z = 100
and
x + y + z = 100
where x,y,z are the number of Cherries, Oranges, and Watermelons respectively. Clearly, since there is at least one of each fruit,
based on the first equation, x is between 2 to 92, y is in the range [1,31] and z[1,19] because, for example, 20 watermelons would cost Rs. 100 and leave no room for other fruits.
To eliminate x, we can multiply the first equation by 2 and subtract the second
x + 6y + 10z = 200
x + y + z = 100
-------------------
5y + 9z = 100
Dividing by 5,
y = 20 - 9z/5
Now, knowing that y is a positive integer, the 9z/5 term must produce an integer, and that integer must also be less than 20. That means z must be divisible by 5. But the only such values z can take in the range [1,19] are (5, 10, 15). 15 clearly produces a negative y, so the solutions for z are 5 and 10, producing y as 11 and 2.
Finally subtract y+z from 100 to get x as 88 and 84.