Difference between revisions of "2019 AMC 8 Problems/Problem 1"
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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== | ==See also== | ||
{{AMC8 box|year=2019|before=First Problem|num-a=2}} | {{AMC8 box|year=2019|before=First Problem|num-a=2}} | ||
[[David C-100% | [[David C-100% |
Revision as of 17:26, 7 November 2020
Solution 2 (Using Algebra)
Let be the number of sandwiches and be the number of sodas. We have to satisfy the equation of 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: 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 |
[[David C-100%