Difference between revisions of "2017 AMC 10A Problems/Problem 14"

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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:<br>
  
<math>\frac{19}{99}A + \frac{4}{99}A = \boxed{\textbf{(D)}\ 23}</math>
+
<math>\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23%}</math>
  
 
==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}}

Revision as of 19:39, 8 February 2017

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 of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:

$\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23%}$ (Error compiling LaTeX. ! File ended while scanning use of \boxed.)

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

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