Today's challenge is to help John make an important financial decision!
John likes to go to the cinema, but he wants to find the most cost-effective way to go. He can buy a ticket for $15, or he can buy a membership card for $500. Every time he uses the membership card, the ticket will be 0.9 times the price he paid for the last one.
Ex. If John goes to the cinema three times:
A : 15 * 3 = 45
B : 500 + 15 * 0.90 + (15 * 0.90) * 0.90 + (15 * 0.90 * 0.90) * 0.90 ( = 536.5849999999999, no rounding for each ticket)
Create a function
movie with three 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 B) < price of A
movie(500, 15, 0.9)should return
(with card the total price is 634, with tickets 645)
movie(100, 10, 0.95) should return
(with card the total price is 235, with tickets 240)
Want to propose a challenge idea for a future post? Email firstname.lastname@example.org with your suggestions!