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

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== Solution 2 (Using Algebra) ==
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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
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<cmath>4.50s+d=30</cmath>
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In the question, it states that Ike and Mike buys as many sandwiches as possible.
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So, we drop the number of sodas for a while.
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We have:
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<cmath>4.50s=30</cmath>
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<cmath>s=\frac{30}{4.5}</cmath>
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<cmath>s=6R30</cmath>
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We don't want a remainder so the maximum number of sandwiches is <math>6</math>.
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The total money spent is <math>6\cdot 4.50=27</math>.
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The number of dollar left to spent on sodas is <math>30-27=3</math> dollars.
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<math>3</math> dollars can buy <math>3</math> sodas leading us to a total of
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<math>6+3=9</math> items.
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Hence, the answer is <math>\boxed{(D) = 9}</math>
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-by interactivemath
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==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 18:26, 7 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 (ProblemsAnswer KeyResources)
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%