text: # 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?
  
  $\mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%$
  
  ## 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 $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:
  
  $\frac{19}{99}A + {4}{99}A = \boxed{\textbf{(D)}\ 23\%\qquad \mathrm}$ (Error compiling LaTeX. ! Missing } inserted.)
  
   2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 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
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