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

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== Solution 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>
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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{\textbf{(D) }9}</math>.
  
 
- SBose
 
- SBose
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So, we drop the number of sodas for a while.
 
So, we drop the number of sodas for a while.
 
We have:  
 
We have:  
<cmath>4.50s=30</cmath>
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<cmath>\begin{align*}
<cmath>s=\frac{30}{4.5}</cmath>
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4.50s&=30 \\
<cmath>s=6R30</cmath>
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s&=\frac{30}{4.5} \\
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s&=6R3
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\end{align*}</cmath>
 
We don't want a remainder so the maximum number of sandwiches is <math>6</math>.
 
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 total money spent is <math>6\cdot 4.50=27</math>.
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<math>3</math> dollars can buy <math>3</math> sodas leading us to a total of  
 
<math>3</math> dollars can buy <math>3</math> sodas leading us to a total of  
 
<math>6+3=9</math> items.  
 
<math>6+3=9</math> items.  
Hence, the answer is <math>\boxed{(D) = 9}</math>
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Hence, the answer is <math>\boxed{\textbf{(D) }9}</math>.
  
 
- by interactivemath
 
- by interactivemath
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== Video Solution ==
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2
  
 
==See also==
 
==See also==

Revision as of 11:16, 13 June 2021

Problem 1

Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy?

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Solution 1

We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$. Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\boxed{\textbf{(D) }9}$.

- SBose

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: \begin{align*} 4.50s&=30 \\ s&=\frac{30}{4.5} \\ s&=6R3 \end{align*} 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 $\boxed{\textbf{(D) }9}$.

- by interactivemath

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2

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

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