Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
− | We know that | + | We know that the sandwiches cost <math>4.50</math> dollars. Guessing will bring us to multiplying <math>4.50</math> by 6, which gives us <math>27.00</math>. Since they can spend <math>30.00</math> they have <math>3</math> dollars left. Since sodas cost <math>1.00</math> dollar each, they can buy 3 sodas, which makes them spend <math>30.00</math> Since they bought 6 sandwiches and 3 sodas, they bought a total of <math>9</math> items. Therefore, the answer is <math>\boxed{D = 9 }</math> |
==Solution 2 (Using Algebra)== | ==Solution 2 (Using Algebra)== |
Revision as of 19:42, 28 October 2020
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
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
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.