2019 AMC 8 Problems/Problem 1
Solution 2 (Using Algebra)
Let 
be the number of sandwiches and 
be the number of sodas. We have to satisfy the equation of
![\[4.50s+d=30\]](http://latex.artofproblemsolving.com/9/c/8/9c8e776fe1543c6342e6785f6d11ae6e2cc34420.png)
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:
![\[4.50s=30\]](http://latex.artofproblemsolving.com/f/4/d/f4d8f91964baca1af9c6f70895e783fb024ab326.png)
![\[s=\frac{30}{4.5}\]](http://latex.artofproblemsolving.com/0/4/6/046b85b7444cc380805e25771bb8a2c6817be185.png)
![\[s=6R30\]](http://latex.artofproblemsolving.com/3/0/7/3074101700fc53c9ebcf78fb3a3adc42a794a5b5.png)
We don't want a remainder so the maximum number of sandwiches is 
.
The total money spent is 
.
The number of dollar left to spent on sodas is 
dollars.
dollars can buy sodas leading us to a total of
items.
Hence, the answer is 
-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. 
