Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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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. | 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> | + | Combining the items, Mike and Ike buy <math>6+3=9</math> soft drinks. |
| − | The answer is <math>\boxed{\textbf{( | + | The answer is <math>\boxed{\textbf{(D) }9}.</math> |
==See Also== | ==See Also== | ||
Revision as of 10:57, 3 July 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 the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?
Solution 1
We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of that is less than 
This number is 
Therefore, they can buy 
sandwiches for 
They spend the remaining money on soft drinks, so they buy 
soft drinks.
Combining the items, Mike and Ike buy 
soft drinks.
The answer is 
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 | ||
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