Difference between revisions of "2019 AMC 8 Problems/Problem 1"

(Problem 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
 
We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of <math>\$4.50</math> that is less than <math>\$30.</math> This number is <math>6.</math>
 
 
Therefore, they can buy <math>6</math> sandwiches for <math>\$4.50\cdot6=\$27.</math> They spend the remaining money on soft drinks, so they buy <math>30-27=3</math> soft drinks.
 
 
Combining the items, Mike and Ike buy <math>6+3=9</math> soft drinks.
 
 
The answer is <math>\boxed{\textbf{(D)  }9}.</math>
 
 
Https://www.youtube.com/watch?v=5i69xiEF-pk&list=PLOMzQDaUOdNtpWQbYUyAhcghw0qWx6wDA-Video Solution(also for first 10 problems)
 
  
 
==See Also==
 
==See Also==

Revision as of 14:06, 3 September 2020

Problem 1

Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each.

Solution 1

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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

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