Difference between revisions of "2017 AMC 10A Problems/Problem 14"
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Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was <math>A</math> dollars. The cost of his movie ticket was <math>20\%</math> of the difference between <math>A</math> and the cost of his soda, while the cost of his soda was <math>5\%</math> of the difference between <math>A</math> and the cost of his movie ticket. To the nearest whole percent, what fraction of <math>A</math> did Roger pay for his movie ticket and soda? | Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was <math>A</math> dollars. The cost of his movie ticket was <math>20\%</math> of the difference between <math>A</math> and the cost of his soda, while the cost of his soda was <math>5\%</math> of the difference between <math>A</math> and the cost of his movie ticket. To the nearest whole percent, what fraction of <math>A</math> did Roger pay for his movie ticket and soda? | ||
− | <math> \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%</math> | + | <math> \mathrm{\textbf{(A)} \ }9\%\qquad \mathrm{\textbf{(B)} \ } 19\%\qquad \mathrm{\textbf{(C)} \ } 22\%\qquad \mathrm{\textbf{(D)} \ } 23\%\qquad \mathrm{\textbf{(E)} \ }25\%</math> |
==Solution== | ==Solution== | ||
Line 23: | Line 23: | ||
<math>s = \frac{4}{99}A</math><br><br> | <math>s = \frac{4}{99}A</math><br><br> | ||
− | Since we want to find what fraction <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 + {4}{99}A = {\ | + | <math>\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}</math> |
+ | |||
+ | ==Solution 2== | ||
+ | We have two equations from the problem: | ||
+ | <math>5M=A-S</math> and | ||
+ | <math>20S=A-M</math> | ||
+ | If we replace <math>A</math> with <math>100</math> we get a system of equations, and the sum of the values of <math>M</math> and <math>S</math> is the percentage of <math>A</math>. | ||
+ | Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>. | ||
+ | Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math> which is <math>(D)</math>. | ||
+ | |||
+ | -Harsha12345 | ||
+ | |||
+ | ==Solution 3== | ||
+ | WLOG let <math>A=20.</math> | ||
+ | Let <math>m</math> be the price of the movie ticket. | ||
+ | Let <math>s</math> be the price of the soda. | ||
+ | Thus, | ||
+ | \begin{align*} | ||
+ | m &=\frac{1}{5}\left(20-s\right) \\ | ||
+ | s &= \frac{1}{20}\left(20-m\right) | ||
+ | \end{align*} | ||
+ | Simplifying, we have | ||
+ | \begin{align*} | ||
+ | 5m &= 20 - s \\ | ||
+ | 20s &= 20-m | ||
+ | \end{align*} | ||
+ | |||
+ | Multiplying the first equation by <math>4</math> and adding them, we have | ||
+ | <cmath>m+s = \frac{100 - 4s - m}{20}</cmath> | ||
+ | |||
+ | Finding <math>m</math> and <math>s</math> is straightforward from there. | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/s4vnGlwwHHw | ||
+ | |||
+ | https://youtu.be/zY726PV6XU8 | ||
+ | |||
+ | ~savannahsolver | ||
==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}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 00:29, 7 September 2021
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: 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, \begin{align*} m &=\frac{1}{5}\left(20-s\right) \\ s &= \frac{1}{20}\left(20-m\right) \end{align*} Simplifying, we have \begin{align*} 5m &= 20 - s \\ 20s &= 20-m \end{align*}
Multiplying the first equation by and adding them, we have
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 |
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