# Difference between revisions of "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?

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

## 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

~savannahsolver