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>
+
<math> \textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) } 23\%\qquad \textbf{(E) } 25\%</math>
  
 
==Solution==
 
==Solution==
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Let <math>s</math>  = cost of soda
 
Let <math>s</math>  = cost of soda
  
We can create two equations:<br>
+
We can create two equations:
  
<math>m = \frac{1}{5}(A - s)</math><br><br>
+
<cmath>m = \frac{1}{5}(A - s)</cmath>
<math>s  = \frac{1}{20}(A - m)</math><br>
+
<cmath>s  = \frac{1}{20}(A - m)</cmath>
  
 
Substituting we get: <br>
 
Substituting we get: <br>
  
<math>m = \frac{1}{5}(A - \frac{1}{20}(A - m))</math> <br><br>
+
<cmath>m = \frac{1}{5}(A - \frac{1}{20}(A - m))</cmath>
 
which yields:<br>
 
which yields:<br>
<math>m = \frac{19}{99}A</math><br>
+
<cmath>m = \frac{19}{99}A</cmath>
  
 
Now we can find s and we get:<br>
 
Now we can find s and we get:<br>
  
<math>s = \frac{4}{99}A</math><br><br>
+
<cmath>s = \frac{4}{99}A</cmath>
  
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>
+
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:
  
<math>\frac{19}{99}A + \frac{4}{99}A = \boxed{\textbf{(D)}\ 23}</math>
+
<cmath>\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}</cmath>
 +
 
 +
==Solution 2==
 +
We have two equations from the problem:
 +
<math>5M=A-S</math> and
 +
<math>20S=A-M</math>
 +
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>.
 +
Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>.
 +
Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math> which is <math>\boxed{\textbf{(D)}\ 23\%}</math>.
 +
 
 +
-Harsha12345
 +
 
 +
==Solution 4==
 +
Let <math>m</math> be the price of a movie ticket and <math>s</math> be the price of a soda.
 +
 
 +
Then,
 +
 
 +
<cmath>m=\frac{A-s}{5}</cmath>
 +
and
 +
<cmath>s=\frac{A-m}{20}</cmath>
 +
Then, we can turn this into
 +
<cmath>5m=A-s</cmath>
 +
<cmath>20s=A-m</cmath>
 +
 
 +
Subtracting and getting rid of A, we have <math>20s-5m=-m+s \rightarrow 19s=4m</math>. Assume WLOG that <math>s=4</math>, <math>m=19</math>, thus making a solution for this equation. Substituting this into the 1st equation, we get <math>A=99</math>. Hence, <math>\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{\textbf{(D)}\ 23\%}</math>
 +
 
 +
~MrThinker
 +
==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 15:54, 4 July 2023

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?

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

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 $\boxed{\textbf{(D)}\ 23\%}$.

-Harsha12345

Solution 4

Let $m$ be the price of a movie ticket and $s$ be the price of a soda.

Then,

\[m=\frac{A-s}{5}\] and \[s=\frac{A-m}{20}\] Then, we can turn this into \[5m=A-s\] \[20s=A-m\]

Subtracting and getting rid of A, we have $20s-5m=-m+s \rightarrow 19s=4m$. Assume WLOG that $s=4$, $m=19$, thus making a solution for this equation. Substituting this into the 1st equation, we get $A=99$. Hence, $\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{\textbf{(D)}\ 23\%}$

~MrThinker

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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