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 | + | <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 20:39, 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 dollars. The cost of his movie ticket was of the difference between and the cost of his soda, while the cost of his soda was of the difference between and the cost of his movie ticket. To the nearest whole percent, what fraction of did Roger pay for his movie ticket and soda?
Solution
Let = cost of movie ticket
Let = cost of soda
We can create two equations:
Substituting we get:
which yields:
Now we can find s and we get:
Since we want to find what fraction of did Roger pay for his movie ticket and soda, we add and 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 (Problems • Answer Key • Resources) | ||
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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