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 11:57, 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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