Difference between revisions of "2017 AMC 10A Problems/Problem 14"

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Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was <math>A</math> dollars. The cost of his movie ticket was <math>20\%</math> of the difference between <math>A</math> and the cost of his soda, while the cost of his soda was <math>5\%</math> of the difference between <math>A</math> and the cost of his movie ticket. To the nearest whole percent, what fraction of <math>A</math> did Roger pay for his movie ticket and soda?
 
Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was <math>A</math> dollars. The cost of his movie ticket was <math>20\%</math> of the difference between <math>A</math> and the cost of his soda, while the cost of his soda was <math>5\%</math> of the difference between <math>A</math> and the cost of his movie ticket. To the nearest whole percent, what fraction of <math>A</math> did Roger pay for his movie ticket and soda?
  
<math> \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%</math>
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<math> \mathrm{\textbf{(A)} \ }9\%\qquad \mathrm{\textbf{(B)} \ } 19\%\qquad \mathrm{\textbf{(C)} \ } 22\%\qquad \mathrm{\textbf{(D)} \ } 23\%\qquad \mathrm{\textbf{(E)} \ }25\%</math>
  
 
==Solution==
 
==Solution==
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<math>s = \frac{4}{99}A</math><br><br>
 
<math>s = \frac{4}{99}A</math><br><br>
  
Since we want to find what fraction <math>A</math> did Roger pay for his movie ticket and soda, we add <math>m</math> and <math>s</math> to get:<br>
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Since we want to find what fraction of <math>A</math> did Roger pay for his movie ticket and soda, we add <math>m</math> and <math>s</math> to get:<br>
  
<math>\frac{19}{99}A + {4}{99}A = {\qquad \mathrm{(D) \ } 23\%\qquad \mathrm}</math>
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<math>\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}</math>
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==Solution 2==
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We have two equations from the problem:
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<math>5M=A-S</math> and
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<math>20S=A-M</math>
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If we replace <math>A</math> with <math>100</math> we get a system of equations, and the sum of the values of <math>M</math> and <math>S</math> is the percentage of <math>A</math>.
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Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>.
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Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math> which is <math>(D)</math>.
 +
 
 +
-Harsha12345
 +
 
 +
==Solution 3==
 +
WLOG let <math>A=20.</math>
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Let <math>m</math> be the price of the movie ticket.
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Let <math>s</math> be the price of the soda.
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Thus,
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\begin{align*}
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m &=\frac{1}{5}\left(20-s\right) \\
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s &= \frac{1}{20}\left(20-m\right)
 +
\end{align*}
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Simplifying, we have
 +
\begin{align*}
 +
5m &= 20 - s \\
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20s &= 20-m
 +
\end{align*}
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 +
Multiplying the first equation by <math>4</math> and adding them, we have
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<cmath>m+s = \frac{100 - 4s - m}{20}</cmath>
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Finding <math>m</math> and <math>s</math> is straightforward from there.
 +
 
 +
~mathboy282
 +
 
 +
==Video Solution==
 +
https://youtu.be/s4vnGlwwHHw
 +
 
 +
https://youtu.be/zY726PV6XU8
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 00:29, 7 September 2021

Problem

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?

$\mathrm{\textbf{(A)} \ }9\%\qquad \mathrm{\textbf{(B)} \ } 19\%\qquad \mathrm{\textbf{(C)} \ } 22\%\qquad \mathrm{\textbf{(D)} \ } 23\%\qquad \mathrm{\textbf{(E)} \ }25\%$

Solution

Let $m$ = cost of movie ticket
Let $s$ = cost of soda

We can create two equations:

$m = \frac{1}{5}(A - s)$

$s  = \frac{1}{20}(A - m)$

Substituting we get:

$m = \frac{1}{5}(A - \frac{1}{20}(A - m))$

which yields:
$m = \frac{19}{99}A$

Now we can find s and we get:

$s = \frac{4}{99}A$

Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:

$\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}$

Solution 2

We have two equations from the problem: $5M=A-S$ and $20S=A-M$ If we replace $A$ with $100$ we get a system of equations, and the sum of the values of $M$ and $S$ is the percentage of $A$. Solving, we get $S=\frac{400}{99}$ and $M=\frac{1900}{99}$. Adding, we get $\frac{2300}{99}$, which is closest to $23$ which is $(D)$.

-Harsha12345

Solution 3

WLOG let $A=20.$ Let $m$ be the price of the movie ticket. Let $s$ be the price of the soda. Thus, \begin{align*} m &=\frac{1}{5}\left(20-s\right) \\ s &= \frac{1}{20}\left(20-m\right) \end{align*} Simplifying, we have \begin{align*} 5m &= 20 - s \\ 20s &= 20-m \end{align*}

Multiplying the first equation by $4$ and adding them, we have \[m+s = \frac{100 - 4s - m}{20}\]

Finding $m$ and $s$ is straightforward from there.

~mathboy282

Video Solution

https://youtu.be/s4vnGlwwHHw

https://youtu.be/zY726PV6XU8

~savannahsolver

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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