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

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==Problem 1==
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==Problem==
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Ike and Mike go into a sandwich shop with a total of <math>\$30.00</math> to spend. Sandwiches cost <math>\$4.50</math> each and soft drinks cost <math>\$1.00</math> each. Ike and Mike plan to buy as many sandwiches as they can,
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and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
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many items will they buy?
  
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Ike and Mike go into a sandwich shop with a total of <math>\$30.00</math> to spend. Sandwiches cost <math>\$4.50</math> each and soft drinks cost <math>\$1.00</math> each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?
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<math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10</math>
  
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<math>\textbf{(A) }6\qquad\textbf{(B) }5\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }2</math>
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== Solution 1 ==
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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 soft drinks cost <math>1.00</math> dollar each, they can buy 3 soft drinks, which makes them spend <math>30.00</math>  Since they bought 6 sandwiches and 3 soft drinks, they bought a total of <math>9</math> items. Therefore, the answer is <math>\boxed{\textbf{(D) }9}</math>.
  
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==Solution 1==
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- SBose
  
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We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of <math>\$4.50</math> that is less than <math>\$30.</math> This number is <math>6.</math>  
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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>\begin{align*}
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4.50s&=30 \\
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s&=\frac{30}{4.5} \\
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s&=6R3
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\end{align*}</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{\textbf{(D) }9}</math>.
  
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Therefore, they can buy <math>6</math> sandwiches for <math>\$4.50\cdot6=\$27.</math> They spend the remaining money on soft drinks, so they buy <math>30-27=3</math> soft drinks.
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- by interactivemath
  
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Combining the items, Mike and Ike buy <math>6+3=9</math> soft drinks.
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== Video Solution ==
  
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The answer is <math>\boxed{\textbf{(D) }9}.</math>
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==Video Solution by Math-X (First fully understand the problem!!!)==
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https://youtu.be/IgpayYB48C4?si=2nxxlhOYNmBnEhj5
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~Math-X
  
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==See Also==
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{{AMC8 box|year=2019|before=First Problem|num-a=2}}
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The Learning Royal: https://youtu.be/IiFFDDITE6Q
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== Video Solution 2 ==
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=Puzy1HAlAKk&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=2
 +
 
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==Video Solution 3==
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https://youtu.be/Y7wzKYaSOhI
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 +
~savannahsolver
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/WggTFT9FhNg
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
 +
 
 +
==See also==
 +
{{AMC8 box|year=2019|before = First Problem|num-a=2}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:29, 9 November 2024

Problem

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 soft drinks cost $1.00$ dollar each, they can buy 3 soft drinks, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 soft drinks, 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

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=2nxxlhOYNmBnEhj5 ~Math-X


The Learning Royal: https://youtu.be/IiFFDDITE6Q

Video Solution 2

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

Video Solution 3

https://youtu.be/Y7wzKYaSOhI

~savannahsolver

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/WggTFT9FhNg

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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

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