Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2 | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2 | ||
+ | |||
+ | The Learning Royal: https://youtu.be/IiFFDDITE6Q | ||
==See also== | ==See also== |
Revision as of 12:15, 1 January 2022
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
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2
The Learning Royal: https://youtu.be/IiFFDDITE6Q
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.