2019 AMC 8 Problems/Problem 1

Solution 1

We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\boxed{D = 9 }$

- SBose

Solution 2 (Using Algebra)

Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of $4.50s+d=30$ 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: $4.50s=30$ $s=\frac{30}{4.5}$ $s=6R30$ We don't want a remainder so the maximum number of sandwiches is $6$ . The total money spent is $6\cdot 4.50=27$ . The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items. Hence, the answer is $\boxed{(D) = 9}$

-by interactivemath

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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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

Solution 1

Solution 2 (Using Algebra)

See also