Danial went to a fruit stall that sells apples, mangoes, and papayas. Each apple costs $3$ RM ,each mango costs $4$ RM , and each papaya costs $5$ RM . He bought at least one of each fruit, and paid exactly $50$ RM. What is the maximum number of fruits that he could have bought?
Problem
Source:
Tags: algebra
Mathrocks314
19.09.2021 19:53
12 apples (36 RM) + 1 mango (4 RM) + 2 papayas (10RM) = 15 fruit costing a grand total of 50
OofPirate
19.09.2021 20:15
We need at least 1 of each fruit: $1$ apple = 3RM $1$ mango = 4RM $1$ papaya =5RM They add up to 12. 50-12=38. Put the maximum apples inside, 12 more apples, which is worth 36 RM. There are no fruits with the value 2RM, so we subtract 1 from the 13 apples to get 5RM left. The papaya fits. Hence, we can buy a maximum of 15 fruits in which 12 apples, 1 mango, and 2 papayas are bought.
Mathrocks314
19.09.2021 20:17
Exactly! Of course you could buy 11 apples, 3 mangos, and 1 papaya with the same results.