2019 AMC 8 Problems/Problem 1

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
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

[[David C-100%

Page

Toolbox

Search

2019 AMC 8 Problems/Problem 1

Solution 2 (Using Algebra)

See also