Difference between revisions of "2017 AMC 10A Problems/Problem 14"
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<math>\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}</math> | <math>\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | We have two equations from the problem. | ||
| + | 5M=A-S | ||
| + | 20S=A-M | ||
| + | If we replace A with 100 we get a system of equations, and the sum of the values of M and S is the percentage of A. | ||
| + | Solving, we get S=400/99 and M=1900/99. | ||
| + | Adding we get 2300/99, which is closest to 23. | ||
| + | -Harsha12345 | ||
==See Also== | ==See Also== | ||
Revision as of 10:55, 30 June 2019
Contents
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:
Solution 2
We have two equations from the problem. 5M=A-S 20S=A-M If we replace A with 100 we get a system of equations, and the sum of the values of M and S is the percentage of A. Solving, we get S=400/99 and M=1900/99. Adding we get 2300/99, which is closest to 23. -Harsha12345
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
