2017 AMC 10A Problems/Problem 14

Revision as of 18:38, 8 February 2017 by Roverav (talk | contribs) (Solution)

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:

$\frac{19}{99}A + {4}{99}A = {\qquad \mathrm{(D) \ } 23\%\qquad \mathrm}$ (Error making remote request. No response to HTTP request)

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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