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

Revision as of 18:27, 6 November 2020

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

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-== 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
-<cmath>4.50s+d=30</cmath>
-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:
-<cmath>4.50s=30</cmath>
-<cmath>s=\frac{30}{4.5}</cmath>
-<cmath>s=6R30</cmath>
-We don't want a remainder so the maximum number of sandwiches is <math>6</math>.
-The total money spent is <math>6\cdot 4.50=27</math>.
-The number of dollar left to spent on sodas is <math>30-27=3</math> dollars.
-<math>3</math> dollars can buy <math>3</math> sodas leading us to a total of
-<math>6+3=9</math> items.
-Hence, the answer is <math>\boxed{(D) = 9}</math>
--by interactivemath
 ==See also==
 {{AMC8 box|year=2019|before=First Problem|num-a=2}}