2017 AMC 10A Problem 14

Attempt Problem 14 of the 2017 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger’s allowance was AA dollars. The cost of his movie ticket was 20%20\% of the difference between AA and the cost of his soda, while the cost of his soda was 5%5\% of the difference between AA and the cost of his movie ticket. To the nearest whole percent, what fraction of AA did Roger pay for his movie ticket and soda?
9%9\%
19%19\%
22%22\%
23%23\%
25%25\%
Answer: D
Concepts:percentagesystem of equations
Small hint:
Let the ticket and soda costs be variables
Big hint:
Turn the two percent statements into a two-equation linear system
Solution:
Let tt be the cost of the ticket and ss be the cost of the soda. Then we get the following equations. t=A−s5 t = \dfrac{A - s}{5} s=A−t20 s = \dfrac{A - t}{20}
Cross-multiplying the first equation gives us 5t=A−s.5t = A - s. Substituting in the expression for ss yields 5t=A−A−t20. 5t = A - \dfrac{A - t}{20}. Solving yields 5t=A−A−t20100t=20A−A+t99t=19At=19A99. \begin{aligned} 5t &= A - \dfrac{A - t}{20} \\ 100t &= 20A - A + t \\ 99t &= 19A \\ t &= \dfrac{19A}{99}. \end{aligned}
This also gives us s=A−19A9920=4A99. s = \dfrac{A - \frac{19A}{99}}{20} = \dfrac{4A}{99}.
Adding together the costs gives us 19A99+4A99=23A99≈23%. \dfrac{19A}{99} + \dfrac{4A}{99} = \dfrac{23A}{99} \approx 23 \%.
Thus, D is the correct answer.
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