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

Latest revision as of 02:06, 18 December 2024

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?

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

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 $\boxed{\textbf{(D)}\ 23\%}$ .

~Harsha12345

Solution 3

Let $m$ be the price of a movie ticket and $s$ be the price of a soda.

Then,

$m=\frac{A-s}{5}$ and $s=\frac{A-m}{20}$ Then, we can turn this into $5m=A-s$ $20s=A-m$

Subtracting and getting rid of A, we have $20s-5m=-m+s \rightarrow 19s=4m$ . Assuming WLOG (without loss of generality) that $s=4$ , $m=19$ , thus making a solution for this equation. Substituting this into the 1st equation, we get $A=99$ . Hence, $\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{\textbf{(D)}\ 23\%}$

~MrThinker

Video Solutions

i) https://youtu.be/s4vnGlwwHHw

ii) https://youtu.be/zY726PV6XU8

~savannahsolver

@@ Line 35: / Line 35: @@
 Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math> which is <math>\boxed{\textbf{(D)}\ 23\%}</math>.
--Harsha12345
+~Harsha12345
-==Solution 4==
+==Solution 3==
 Let <math>m</math> be the price of a movie ticket and <math>s</math> be the price of a soda.
@@ Line 49: / Line 49: @@
 <cmath>20s=A-m</cmath>
-Subtracting and getting rid of A, we have <math>20s-5m=-m+s \rightarrow 19s=4m</math>. Assume WLOG that <math>s=4</math>, <math>m=19</math>, thus making a solution for this equation. Substituting this into the 1st equation, we get <math>A=99</math>. Hence, <math>\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{\textbf{(D)}\ 23\%}</math>
+Subtracting and getting rid of A, we have <math>20s-5m=-m+s \rightarrow 19s=4m</math>. Assuming WLOG (without loss of generality) that <math>s=4</math>, <math>m=19</math>, thus making a solution for this equation. Substituting this into the 1st equation, we get <math>A=99</math>. Hence, <math>\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{\textbf{(D)}\ 23\%}</math>
 ~MrThinker
-==Video Solution==
-https://youtu.be/s4vnGlwwHHw
-https://youtu.be/zY726PV6XU8
+==Video Solutions==
+i) https://youtu.be/s4vnGlwwHHw
+ii) https://youtu.be/zY726PV6XU8
 ~savannahsolver

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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