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Difference between revisions of "2017 AMC 10A Problems/Problem 14"

(solution 3)
(→‎Solution 3)
Line 42: Line 42:
 
Let <math>s</math> be the price of the soda.
 
Let <math>s</math> be the price of the soda.
 
Thus,  
 
Thus,  
−
\begin{align*}
+
<cmath>\begin{align*}
 
m &=\frac{1}{5}\left(20-s\right) \\  
 
m &=\frac{1}{5}\left(20-s\right) \\  
 
s &= \frac{1}{20}\left(20-m\right)
 
s &= \frac{1}{20}\left(20-m\right)
−
\end{align*}
+
\end{align*}</cmath>
 
Simplifying, we have  
 
Simplifying, we have  
−
\begin{align*}
+
<cmath>\begin{align*}
 
5m &= 20 - s \\
 
5m &= 20 - s \\
 
20s &= 20-m
 
20s &= 20-m
−
\end{align*}
+
\end{align*}</cmath>
  
 
Multiplying the first equation by <math>4</math> and adding them, we have
 
Multiplying the first equation by <math>4</math> and adding them, we have

Revision as of 10:02, 27 July 2022

Problem

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?

$\mathrm{\textbf{(A)} \ }9\%\qquad \mathrm{\textbf{(B)} \ } 19\%\qquad \mathrm{\textbf{(C)} \ } 22\%\qquad \mathrm{\textbf{(D)} \ } 23\%\qquad \mathrm{\textbf{(E)} \ }25\%$

Solution

Let $m$ = cost of movie ticket
Let $s$ = cost of soda

We can create two equations:

$m = \frac{1}{5}(A - s)$

$s = \frac{1}{20}(A - m)$

Substituting we get:

$m = \frac{1}{5}(A - \frac{1}{20}(A - m))$

which yields:
$m = \frac{19}{99}A$

Now we can find s and we get:

$s = \frac{4}{99}A$

Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:

$\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}$

Solution 2

We have two equations from the problem: $5M=A-S$ and $20S=A-M$ 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 $S=\frac{400}{99}$ and $M=\frac{1900}{99}$. Adding, we get $\frac{2300}{99}$, which is closest to $23$ which is $(D)$.

-Harsha12345

Solution 3

WLOG let $A=20.$ Let $m$ be the price of the movie ticket. Let $s$ 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 $4$ and adding them, we have \[m+s = \frac{100 - 4s - m}{20}\]

Finding $m$ and $s$ is straightforward from there.

~mathboy282

Video Solution

https://youtu.be/s4vnGlwwHHw

https://youtu.be/zY726PV6XU8

~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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