Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 8 Problems/Problem 1"

(→‎Problem 1)
(→‎Solution 1)
Line 1: Line 1:
−
== Solution 1 ==
 
−
We know that the sandwiches cost <math>4.50</math> dollars. Guessing will bring us to multiplying <math>4.50</math> by 6, which gives us <math>27.00</math>. Since they can spend <math>30.00</math> they have <math>3</math> dollars left. Since sodas cost <math>1.00</math> dollar each, they can buy 3 sodas, which makes them spend <math>30.00</math>  Since they bought 6 sandwiches and 3 sodas, they bought a total of <math>9</math> items. Therefore, the answer is <math>\boxed{D = 9 }</math>
 
−
 
−
- SBose
 
−
 
 
== Solution 2 (Using Algebra) ==
 
== Solution 2 (Using Algebra) ==
 
Let <math>s</math> be the number of sandwiches and <math>d</math> be the number of sodas. We have to satisfy the equation of
 
Let <math>s</math> be the number of sandwiches and <math>d</math> be the number of sodas. We have to satisfy the equation of

Revision as of 18:27, 6 November 2020

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_1&oldid=136805"