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

(Solution)
(Solution)
Line 7: Line 7:
 
Let <math>m</math> = cost of movie ticket<br>
 
Let <math>m</math> = cost of movie ticket<br>
 
Let <math>s</math>  = cost of soda
 
Let <math>s</math>  = cost of soda
 +
 +
We can create two equations:<br>
 +
 +
<math>m = \frac{1}{5}(A - s)</math><br>
 +
<math>s  = \frac{1}{20}(A - m)</math><br>
 +
 +
Substituting we get: <br>
  
 
==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}}

Revision as of 18:25, 8 February 2017

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?

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

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:

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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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