Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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+ | == Video Solution == | ||
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+ | https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2 | ||
==See also== | ==See also== |
Revision as of 15:29, 22 April 2021
Problem 1
Ike and Mike go into a sandwich shop with a total of to spend. Sandwiches cost each and soft drinks cost each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy?
Solution 1
We know that the sandwiches cost dollars. Guessing will bring us to multiplying by 6, which gives us . Since they can spend they have dollars left. Since sodas cost dollar each, they can buy 3 sodas, which makes them spend Since they bought 6 sandwiches and 3 sodas, they bought a total of items. Therefore, the answer is
- SBose
Solution 2 (Using Algebra)
Let be the number of sandwiches and be the number of sodas. We have to satisfy the equation of In the question, it states that Ike and Mike buys as many sandwiches as possible. So, we drop the number of sodas for a while. We have: We don't want a remainder so the maximum number of sandwiches is . The total money spent is . The number of dollar left to spent on sodas is dollars. dollars can buy sodas leading us to a total of items. Hence, the answer is
- by interactivemath
Video Solution
https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.