# 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

~savannahsolver

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