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

(→‎Solution)
m (→‎Solution 2)
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==Solution 2==
 
==Solution 2==
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We have two equations from the problem.
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We have two equations from the problem:
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5M=A-S
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<math>5M=A-S</math> and
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20S=A-M
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<math>20S=A-M</math>
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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.
+
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>.
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Solving, we get S=400/99 and M=1900/99.
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Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>.
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Adding we get 2300/99, which is closest to 23.
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Adding we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math>.
 +
 
 
-Harsha12345
 
-Harsha12345
  

Revision as of 20:31, 7 October 2019

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 $23$.

-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

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