2017 AMC 10A Problems/Problem 14

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
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png