https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Roverav&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-02T00:10:40Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_16&diff=83049 2017 AMC 10A Problems/Problem 16 2017-02-08T22:47:43Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> There are &lt;math&gt;10&lt;/math&gt; horses, named Horse &lt;math&gt;1&lt;/math&gt;, Horse &lt;math&gt;2&lt;/math&gt;, . . . , Horse &lt;math&gt;10&lt;/math&gt;. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse &lt;math&gt;k&lt;/math&gt; runs one lap in exactly &lt;math&gt;k&lt;/math&gt; minutes. At time &lt;math&gt;0&lt;/math&gt; all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time &lt;math&gt;S &gt; 0&lt;/math&gt;, in minutes, at which all &lt;math&gt;10&lt;/math&gt; horses will again simultaneously be at the starting point is &lt;math&gt;S=2520&lt;/math&gt;. Let &lt;math&gt;T &gt; 0&lt;/math&gt; be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of &lt;math&gt;T?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Stratergy: find the LCM <br /> <br /> We know that Horses &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; will all meet at the starting line in 12 minutes. Therefore, the answer is: &lt;br&gt;<br /> &lt;math&gt;1 + 2 = \boxed{\textbf{(B)}\ 3}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=83045 2017 AMC 10A Problems/Problem 11 2017-02-08T22:44:57Z <p>Roverav: /* Visualizing the Region */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all point in three-dimensional space within 3 units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume 216&lt;math&gt;\pi&lt;/math&gt;. What is the length &lt;math&gt;\textit{AB}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within &lt;math&gt;3&lt;/math&gt; units of a point would be a sphere with radius &lt;math&gt;3&lt;/math&gt;. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal &lt;math&gt;216 \pi&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{4 \pi }{3}3^3+9 \pi x=216&lt;/math&gt;<br /> <br /> Where &lt;math&gt;x&lt;/math&gt; is equal to the length of our line segment.<br /> <br /> We isolate &lt;math&gt;x&lt;/math&gt;. This comes out to be &lt;math&gt;\boxed{\textbf{(D)}\ 20}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> To envision what the region must look like, we simplify the problem to finding all points within &lt;math&gt;3&lt;/math&gt; units from a point. This is a sphere. To account for the line, we drag the sphere's center across the line, sweeping out the desired volume. As stated above, this is a cylinder with two hemispheres on both ends.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83042 2017 AMC 10A Problems/Problem 14 2017-02-08T22:41:21Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;&lt;br&gt;<br /> <br /> Now we can find s and we get:&lt;br&gt;<br /> <br /> &lt;math&gt;s = \frac{4}{99}A&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> <br /> Since we want to find what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda, we add &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; to get:&lt;br&gt;<br /> <br /> &lt;math&gt;\frac{19}{99}A + \frac{4}{99}A = \boxed{\textbf{(D)}\ 23}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83041 2017 AMC 10A Problems/Problem 14 2017-02-08T22:40:53Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;&lt;br&gt;<br /> <br /> Now we can find s and we get:&lt;br&gt;<br /> <br /> &lt;math&gt;s = \frac{4}{99}A&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> <br /> Since we want to find what fraction &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda, we add &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; to get:&lt;br&gt;<br /> <br /> &lt;math&gt;\frac{19}{99}A + \frac{4}{99}A = \boxed{\textbf{(D)}\ 23}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83038 2017 AMC 10A Problems/Problem 14 2017-02-08T22:38:56Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;&lt;br&gt;<br /> <br /> Now we can find s and we get:&lt;br&gt;<br /> <br /> &lt;math&gt;s = \frac{4}{99}A&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> <br /> Since we want to find what fraction &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda, we add &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; to get:&lt;br&gt;<br /> <br /> &lt;math&gt;\frac{19}{99}A + {4}{99}A = {\qquad \mathrm{(D) \ } 23\%\qquad \mathrm}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83036 2017 AMC 10A Problems/Problem 14 2017-02-08T22:37:40Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;&lt;br&gt;<br /> <br /> Now we can find s and we get:&lt;br&gt;<br /> <br /> &lt;math&gt;s = \frac{4}{99}A&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> <br /> Since we want to find what fraction &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda, we add &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; to get:&lt;br&gt;<br /> <br /> &lt;math&gt;\frac{19}{99}A + {4}{99}A = \boxed{\textbf{(D)}\ 23\%\qquad \mathrm}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83034 2017 AMC 10A Problems/Problem 14 2017-02-08T22:36:37Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;&lt;br&gt;<br /> <br /> Now we can find s and we get:&lt;br&gt;<br /> <br /> &lt;math&gt;s = \frac{4}{99}A&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> <br /> Since we want to find what fraction &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda, we add &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; to get:&lt;br&gt;<br /> <br /> &lt;math&gt;\frac{19}{99}A + {4}{99}A = \boxed{\textbf{(D)}\ 23%}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83033 2017 AMC 10A Problems/Problem 14 2017-02-08T22:35:43Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;&lt;br&gt;<br /> <br /> Now we can find s and we get:&lt;br&gt;<br /> <br /> &lt;math&gt;s = \frac{4}{99}A&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> <br /> Since we want to find what fraction &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda, we add &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; to get:&lt;br&gt;<br /> <br /> $\frac{19}{99}A + {4}{99}A = \boxed{\textbf{(D)}\ 23}<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83032 2017 AMC 10A Problems/Problem 14 2017-02-08T22:35:19Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;&lt;br&gt;<br /> <br /> Now we can find s and we get:&lt;br&gt;<br /> <br /> &lt;math&gt;s = \frac{4}{99}A&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> <br /> Since we want to find what fraction &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda, we add &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; to get:&lt;br&gt;<br /> <br />$\frac{19}{99}A + {4}{99}A = \boxed{\textbf{(D)}\ 23%}<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83027 2017 AMC 10A Problems/Problem 14 2017-02-08T22:30:59Z <p>Roverav: </p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - \frac{1}{20}(A - m))&lt;/math&gt; &lt;br&gt;&lt;br&gt;<br /> which yields:&lt;br&gt;<br /> &lt;math&gt;m = \frac{19}{99}A&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83008 2017 AMC 10A Problems/Problem 14 2017-02-08T22:25:33Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=83006 2017 AMC 10A Problems/Problem 14 2017-02-08T22:25:00Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> We can create two equations:&lt;br&gt;<br /> <br /> &lt;math&gt;m = \frac{1}{5}(A - s)&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt;s = \frac{1}{20}(A - m)&lt;/math&gt;&lt;br&gt;<br /> <br /> Substituting we get: &lt;br&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=82997 2017 AMC 10A Problems/Problem 14 2017-02-08T22:21:15Z <p>Roverav: /* Solution */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket&lt;br&gt;<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_14&diff=82991 2017 AMC 10A Problems/Problem 14 2017-02-08T22:19:23Z <p>Roverav: /* Problem */</p> <hr /> <div>==Problem==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }9\%\qquad \mathrm{(B) \ } 19\%\qquad \mathrm{(C) \ } 22\%\qquad \mathrm{(D) \ } 23\%\qquad \mathrm{(E) \ }25\%&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; = cost of movie ticket<br /> Let &lt;math&gt;s&lt;/math&gt; = cost of soda<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Roverav