Kata Day 35: Going to cinema

Problem Description: My friend John likes to go to the cinema. He can choose between system A and system B.

System A : buy a ticket (15 dollars) every time
System B : buy a card (500 dollars) and every time buy a ticket the price of which is 0.90 times the price he paid for the previous one.
Example:

If John goes to the cinema 3 times:

System A : 15 * 3 = 45
System B : 500 + 15 * 0.90 + (15 * 0.90) * 0.90 + (15 * 0.90 * 0.90) * 0.90
John wants to know how many times he must go to the cinema so that the final result of System B, when rounded up to the next dollar, will be cheaper than System A.

The function movie has 3 parameters: card (price of the card), ticket (normal price of a ticket), perc (fraction of what he paid for the previous ticket) and returns the first n such that

ceil(price of System B) < price of System A.
More examples:

movie(500, 15, 0.9) should return 43
    (with card the total price is 634, with tickets 645)
movie(100, 10, 0.95) should return 24
    (with card the total price is 235, with tickets 240)

Level:7kyu

It is a Kata involving basic implementation of maths. We have to find the number of tickets where the total cost involved in system B comes out to be less than the cost involved in system A. We start from 1 and then keep increasing the number of tickets by 1 and keep comparing the value in two systems.

Whenever the value of cost involved in system B is found less than system A we break out of the loop and return the value of number of tickets  as our answer to this Kata.

Happy New Year!!

Link to Kata: Going to cinema



Here is a link to the code: Kata Day 35

Please mention your suggestions or your doubts in the comment below.

Happy coding. :)