Difference between revisions of "2017 AMC 10A Problems/Problem 14"
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Let <math>s</math> = cost of soda | Let <math>s</math> = cost of soda | ||
| − | We can create two equations: | + | We can create two equations: |
| − | < | + | <cmath>m = \frac{1}{5}(A - s)</cmath> |
| − | < | + | <cmath>s = \frac{1}{20}(A - m)</cmath> |
Substituting we get: <br> | Substituting we get: <br> | ||
| − | < | + | <cmath>m = \frac{1}{5}(A - \frac{1}{20}(A - m))</cmath> |
which yields:<br> | which yields:<br> | ||
| − | < | + | <cmath>m = \frac{19}{99}A</cmath> |
Now we can find s and we get:<br> | Now we can find s and we get:<br> | ||
| − | < | + | <cmath>s = \frac{4}{99}A</cmath> |
| − | 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: | + | 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: |
| − | < | + | <cmath>\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}</cmath> |
==Solution 2== | ==Solution 2== | ||
Revision as of 08:46, 3 September 2022
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:
![\[m = \frac{1}{5}(A - s)\]](http://latex.artofproblemsolving.com/c/2/2/c225bfe998b71a7dfb2713ba8bcda44671f41eb3.png)
![\[s = \frac{1}{20}(A - m)\]](http://latex.artofproblemsolving.com/7/5/d/75d93a0482e185d50835ef040cd4c9f05a47cd87.png)
Substituting we get:
![\[m = \frac{1}{5}(A - \frac{1}{20}(A - m))\]](http://latex.artofproblemsolving.com/2/3/4/234f13fb1a857e5e12b6b63e8e0fd26606023db0.png)
which yields:
![\[m = \frac{19}{99}A\]](http://latex.artofproblemsolving.com/c/b/f/cbfad79ad2eab599b87d92603ba7d070d7858a3d.png)
Now we can find s and we get:
![\[s = \frac{4}{99}A\]](http://latex.artofproblemsolving.com/a/2/1/a212bfb55c0fc18e3d650bd713be3bfe86181961.png)
Since we want to find what fraction of did Roger pay for his movie ticket and soda, we add and to get:
![\[\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}\]](http://latex.artofproblemsolving.com/c/0/a/c0aef4fb2cf37753e4197badb9f069a4e2fbc5ba.png)
Solution 2
We have two equations from the problem:
and
If we replace with 
we get a system of equations, and the sum of the values of 
and 
is the percentage of .
Solving, we get 
and 
.
Adding, we get 
, which is closest to 
which is 
.
-Harsha12345
Solution 3
WLOG let 
Let be the price of the movie ticket.
Let be the price of the soda.
Thus,
Simplifying, we have
Multiplying the first equation by 
and adding them, we have
![\[m+s = \frac{100 - 4s - m}{20}\]](http://latex.artofproblemsolving.com/4/4/1/441be36349a4ffd83050a4b3700d2768b828af44.png)
Finding and is straightforward from there.
~mathboy282
Video Solution
~savannahsolver
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. 
