2019 AMC 8 Problems/Problem 1

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. 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?

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Solution 1

We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of $$4.50$ that is less than $$30.$ This number is $6.$

Therefore, they can buy $6$ sandwiches for $$4.50\cdot6=$27.$ They spend the remaining money on soft drinks, so they buy $30-27=3$ soft drinks.

Combining the items, Mike and Ike buy $6+3=9$ soft drinks.

The answer is $\boxed{\textbf{(D) }9}.$

Solution 2

To double check your work, you can realize that $4.5*2=9$ , so $27$ is the largest number of dollars you can spend on sandwiches. Since $4.5*2=9, 4.5*6=27$ , so you have $3$ dollars left. $3+6=$ $\boxed{\textbf{(D) }9}.$ ~heeeeeeeeeheeeee

2019 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2019 AMC 8 Problems/Problem 1

Contents

Problem 1

Solution 1

Solution 2

See Also