https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Elipticmodulo&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-22T00:33:02Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_6&diff=76344 2014 AMC 10A Problems/Problem 6 2016-02-17T00:51:04Z <p>Elipticmodulo: /* Solution 4 */</p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #4]] and [[2014 AMC 10A Problems|2014 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We need to multiply &lt;math&gt;b&lt;/math&gt; by &lt;math&gt;\frac{d}{a}&lt;/math&gt; for the new cows and &lt;math&gt;\frac{e}{c}&lt;/math&gt; for the new time, so the answer is &lt;math&gt;b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> <br /> We plug in &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=4&lt;/math&gt;, &lt;math&gt;d=5&lt;/math&gt;, and &lt;math&gt;e=6&lt;/math&gt;. Hence the question becomes &quot;2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?&quot;<br /> <br /> If 2 cows give 3 gallons of milk in 4 days, then 2 cows give &lt;math&gt;\frac{3}{4}&lt;/math&gt; gallons of milk in 1 day, so 1 cow gives &lt;math&gt;\frac{3}{4\cdot2}&lt;/math&gt; gallons in 1 day. This means that 5 cows give &lt;math&gt;\frac{5\cdot3}{4\cdot2}&lt;/math&gt; gallons of milk in 1 day. Finally, we see that 5 cows give &lt;math&gt;\frac{5\cdot3\cdot6}{4\cdot2}&lt;/math&gt; gallons of milk in 6 days. Substituting our values for the variables, this becomes &lt;math&gt;\frac{dbe}{ac}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We see that the the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is &lt;math&gt;\dfrac{ac}{b}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;g&lt;/math&gt; be the answer to the question. We have &lt;math&gt;\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> The problem specifics &quot;rate,&quot; so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days &lt;cmath&gt;\implies\text{rate}=\dfrac{b}{ac}&lt;/cmath&gt;<br /> <br /> Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days &lt;cmath&gt;\implies\dfrac{bde}{ac}\implies\boxed{\textbf{(A)}}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2014|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_6&diff=76343 2014 AMC 10A Problems/Problem 6 2016-02-17T00:49:27Z <p>Elipticmodulo: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #4]] and [[2014 AMC 10A Problems|2014 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We need to multiply &lt;math&gt;b&lt;/math&gt; by &lt;math&gt;\frac{d}{a}&lt;/math&gt; for the new cows and &lt;math&gt;\frac{e}{c}&lt;/math&gt; for the new time, so the answer is &lt;math&gt;b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> <br /> We plug in &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=4&lt;/math&gt;, &lt;math&gt;d=5&lt;/math&gt;, and &lt;math&gt;e=6&lt;/math&gt;. Hence the question becomes &quot;2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?&quot;<br /> <br /> If 2 cows give 3 gallons of milk in 4 days, then 2 cows give &lt;math&gt;\frac{3}{4}&lt;/math&gt; gallons of milk in 1 day, so 1 cow gives &lt;math&gt;\frac{3}{4\cdot2}&lt;/math&gt; gallons in 1 day. This means that 5 cows give &lt;math&gt;\frac{5\cdot3}{4\cdot2}&lt;/math&gt; gallons of milk in 1 day. Finally, we see that 5 cows give &lt;math&gt;\frac{5\cdot3\cdot6}{4\cdot2}&lt;/math&gt; gallons of milk in 6 days. Substituting our values for the variables, this becomes &lt;math&gt;\frac{dbe}{ac}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We see that the the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is &lt;math&gt;\dfrac{ac}{b}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;g&lt;/math&gt; be the answer to the question. We have &lt;math&gt;\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> The problem specifics &quot;rate,&quot; so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days &lt;cmath&gt;\implies\text{rate}=\dfrac{b}{ac}&lt;/cmath&gt;<br /> <br /> Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days &lt;cmath&gt;\implies\dfrac{bde}{ac}\implies\boxed{A}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2014|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_6&diff=76342 2014 AMC 10A Problems/Problem 6 2016-02-17T00:48:43Z <p>Elipticmodulo: /* Solution 3 */</p> <hr /> <div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #4]] and [[2014 AMC 10A Problems|2014 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We need to multiply &lt;math&gt;b&lt;/math&gt; by &lt;math&gt;\frac{d}{a}&lt;/math&gt; for the new cows and &lt;math&gt;\frac{e}{c}&lt;/math&gt; for the new time, so the answer is &lt;math&gt;b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> <br /> We plug in &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=4&lt;/math&gt;, &lt;math&gt;d=5&lt;/math&gt;, and &lt;math&gt;e=6&lt;/math&gt;. Hence the question becomes &quot;2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?&quot;<br /> <br /> If 2 cows give 3 gallons of milk in 4 days, then 2 cows give &lt;math&gt;\frac{3}{4}&lt;/math&gt; gallons of milk in 1 day, so 1 cow gives &lt;math&gt;\frac{3}{4\cdot2}&lt;/math&gt; gallons in 1 day. This means that 5 cows give &lt;math&gt;\frac{5\cdot3}{4\cdot2}&lt;/math&gt; gallons of milk in 1 day. Finally, we see that 5 cows give &lt;math&gt;\frac{5\cdot3\cdot6}{4\cdot2}&lt;/math&gt; gallons of milk in 6 days. Substituting our values for the variables, this becomes &lt;math&gt;\frac{dbe}{ac}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We see that the the number of cows is inversely proportional to the number of days and directly proportional to the gallons of milk. So our constant is &lt;math&gt;\dfrac{ac}{b}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;g&lt;/math&gt; be the answer to the question. We have &lt;math&gt;\dfrac{de}{g}=\dfrac{ac}{b}\implies gac=bde\implies g=\dfrac{bde}{ac}\implies\boxed{ \textbf{(A)}\ \frac{bde}{ac}}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> The problem specifics &quot;rate,&quot; so it would be wise to first find the rate at which cows produce milk. We can find the rate of work/production by finding the gallons produced by a single cow in a single day. To do this, we divide the amount produced by the number of cows and number of days &lt;cmath&gt;\implies\text{rate}=\dfrac{b}{ac}&lt;/cmath&gt;<br /> <br /> Now that we have the gallons produced by a single cow in a single day, we simply multiply that rate by the number of cows and the number of days &lt;cmath&gt;\implies\dfrac{bde}{ac}\implies\boxed{A}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2014|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_7&diff=75937 2011 AMC 12A Problems/Problem 7 2016-02-15T01:46:38Z <p>Elipticmodulo: /* Solution */</p> <hr /> <div>== Problem ==<br /> A majority of the &lt;math&gt;30&lt;/math&gt; students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than &lt;math&gt;1&lt;/math&gt;. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt;17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 7 \qquad<br /> \textbf{(B)}\ 11 \qquad<br /> \textbf{(C)}\ 17 \qquad<br /> \textbf{(D)}\ 23 \qquad<br /> \textbf{(E)}\ 77 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The total cost of the pencils can be found by &lt;math&gt;(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})&lt;/math&gt;. As the cost is &lt;math&gt;17.71&lt;/math&gt; dollars, or &lt;math&gt;1771&lt;/math&gt; cents, the cost of the pencils must divide the total cost. Scanning the answer choices, they all unfortunately divide into &lt;math&gt;1771&lt;/math&gt;, so we have to start from the beginning (although this does give us some indication that we are on the right track, because MAA doesn't want to make things too easy for us).<br /> <br /> Therefore, since &lt;math&gt;1771&lt;/math&gt; is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: &lt;math&gt;7, 11, 23&lt;/math&gt;. Since neither &lt;math&gt;(C)&lt;/math&gt; nor &lt;math&gt;(E)&lt;/math&gt; are any of these factors, they can be eliminated immediately, leaving &lt;math&gt;(A)&lt;/math&gt;, &lt;math&gt;(B)&lt;/math&gt;, and &lt;math&gt;(D)&lt;/math&gt;.<br /> <br /> Beginning with &lt;math&gt;(A) 7&lt;/math&gt;, we see that the number of pencils purchased by each student must be either &lt;math&gt;11&lt;/math&gt; or &lt;math&gt;23&lt;/math&gt;. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.<br /> <br /> Continuing with &lt;math&gt;(B) 11&lt;/math&gt;, we can conclude that the only case that fulfills the restrictions are that there are &lt;math&gt;23&lt;/math&gt; students who each purchased &lt;math&gt;7&lt;/math&gt; such pencils, so the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;. We can apply the same logic to &lt;math&gt;(E)&lt;/math&gt; as we applied to &lt;math&gt;(A)&lt;/math&gt;, if one wants to make doubly sure.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_7&diff=75936 2011 AMC 12A Problems/Problem 7 2016-02-15T01:46:15Z <p>Elipticmodulo: /* Solution */</p> <hr /> <div>== Problem ==<br /> A majority of the &lt;math&gt;30&lt;/math&gt; students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than &lt;math&gt;1&lt;/math&gt;. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt;17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 7 \qquad<br /> \textbf{(B)}\ 11 \qquad<br /> \textbf{(C)}\ 17 \qquad<br /> \textbf{(D)}\ 23 \qquad<br /> \textbf{(E)}\ 77 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The total cost of the pencils can be found by &lt;math&gt;(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})&lt;/math&gt;. As the cost is &lt;math&gt;17.71&lt;/math&gt; dollars, or &lt;math&gt;1771&lt;/math&gt; cents, the cost of the pencils must divide the total cost. Scanning the answer choices, they all unfortunately divide into &lt;math&gt;1771&lt;/math&gt;, so we have to start from the beginning (although this does give us some indication that we are on the right tack, because MAA doesn't want to make things too easy for us).<br /> <br /> Therefore, since &lt;math&gt;1771&lt;/math&gt; is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: &lt;math&gt;7, 11, 23&lt;/math&gt;. Since neither &lt;math&gt;(C)&lt;/math&gt; nor &lt;math&gt;(E)&lt;/math&gt; are any of these factors, they can be eliminated immediately, leaving &lt;math&gt;(A)&lt;/math&gt;, &lt;math&gt;(B)&lt;/math&gt;, and &lt;math&gt;(D)&lt;/math&gt;.<br /> <br /> Beginning with &lt;math&gt;(A) 7&lt;/math&gt;, we see that the number of pencils purchased by each student must be either &lt;math&gt;11&lt;/math&gt; or &lt;math&gt;23&lt;/math&gt;. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.<br /> <br /> Continuing with &lt;math&gt;(B) 11&lt;/math&gt;, we can conclude that the only case that fulfills the restrictions are that there are &lt;math&gt;23&lt;/math&gt; students who each purchased &lt;math&gt;7&lt;/math&gt; such pencils, so the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;. We can apply the same logic to &lt;math&gt;(E)&lt;/math&gt; as we applied to &lt;math&gt;(A)&lt;/math&gt;, if one wants to make doubly sure.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_7&diff=75935 2011 AMC 12A Problems/Problem 7 2016-02-15T01:45:36Z <p>Elipticmodulo: /* Solution */</p> <hr /> <div>== Problem ==<br /> A majority of the &lt;math&gt;30&lt;/math&gt; students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than &lt;math&gt;1&lt;/math&gt;. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt;17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 7 \qquad<br /> \textbf{(B)}\ 11 \qquad<br /> \textbf{(C)}\ 17 \qquad<br /> \textbf{(D)}\ 23 \qquad<br /> \textbf{(E)}\ 77 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The total cost of the pencils can be found by &lt;math&gt;(\text{students}*\text{pencils purchased by each}*\text{price of each pencil})&lt;/math&gt;. As the cost is &lt;math&gt;17.71 dollars&lt;/math&gt;, or &lt;math&gt;1771&lt;/math&gt; cents, the cost of the pencils must divide the total cost. Scanning the answer choices, they all unfortunately divide into &lt;math&gt;1771&lt;/math&gt;, so we have to start from the beginning (although this does give us some indication that we are on the right tack, because MAA doesn't want to make things too easy for us).<br /> <br /> Therefore, since &lt;math&gt;1771&lt;/math&gt; is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: &lt;math&gt;7, 11, 23&lt;/math&gt;. Since neither &lt;math&gt;(C)&lt;/math&gt; nor &lt;math&gt;(E)&lt;/math&gt; are any of these factors, they can be eliminated immediately, leaving &lt;math&gt;(A)&lt;/math&gt;, &lt;math&gt;(B)&lt;/math&gt;, and &lt;math&gt;(D)&lt;/math&gt;.<br /> <br /> Beginning with &lt;math&gt;(A) 7&lt;/math&gt;, we see that the number of pencils purchased by each student must be either &lt;math&gt;11&lt;/math&gt; or &lt;math&gt;23&lt;/math&gt;. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.<br /> <br /> Continuing with &lt;math&gt;(B) 11&lt;/math&gt;, we can conclude that the only case that fulfills the restrictions are that there are &lt;math&gt;23&lt;/math&gt; students who each purchased &lt;math&gt;7&lt;/math&gt; such pencils, so the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;. We can apply the same logic to &lt;math&gt;(E)&lt;/math&gt; as we applied to &lt;math&gt;(A)&lt;/math&gt;, if one wants to make doubly sure.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_7&diff=75934 2011 AMC 12A Problems/Problem 7 2016-02-15T01:45:13Z <p>Elipticmodulo: /* Solution */</p> <hr /> <div>== Problem ==<br /> A majority of the &lt;math&gt;30&lt;/math&gt; students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than &lt;math&gt;1&lt;/math&gt;. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt;17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 7 \qquad<br /> \textbf{(B)}\ 11 \qquad<br /> \textbf{(C)}\ 17 \qquad<br /> \textbf{(D)}\ 23 \qquad<br /> \textbf{(E)}\ 77 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The total cost of the pencils can be found by &lt;math&gt;\text{students}*\text{pencils purchased by each}*\text{price of each pencil}&lt;/math&gt;. As the cost is &lt;math&gt;17.71 dollars&lt;/math&gt;, or &lt;math&gt;1771&lt;/math&gt; cents, the cost of the pencils must divide the total cost. Scanning the answer choices, they all unfortunately divide into &lt;math&gt;1771&lt;/math&gt;, so we have to start from the beginning (although this does give us some indication that we are on the right tack, because MAA doesn't want to make things too easy for us).<br /> <br /> Therefore, since &lt;math&gt;1771&lt;/math&gt; is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: &lt;math&gt;7, 11, 23&lt;/math&gt;. Since neither &lt;math&gt;(C)&lt;/math&gt; nor &lt;math&gt;(E)&lt;/math&gt; are any of these factors, they can be eliminated immediately, leaving &lt;math&gt;(A)&lt;/math&gt;, &lt;math&gt;(B)&lt;/math&gt;, and &lt;math&gt;(D)&lt;/math&gt;.<br /> <br /> Beginning with &lt;math&gt;(A) 7&lt;/math&gt;, we see that the number of pencils purchased by each student must be either &lt;math&gt;11&lt;/math&gt; or &lt;math&gt;23&lt;/math&gt;. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.<br /> <br /> Continuing with &lt;math&gt;(B) 11&lt;/math&gt;, we can conclude that the only case that fulfills the restrictions are that there are &lt;math&gt;23&lt;/math&gt; students who each purchased &lt;math&gt;7&lt;/math&gt; such pencils, so the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;. We can apply the same logic to &lt;math&gt;(E)&lt;/math&gt; as we applied to &lt;math&gt;(A)&lt;/math&gt;, if one wants to make doubly sure.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_7&diff=75933 2011 AMC 12A Problems/Problem 7 2016-02-15T01:44:51Z <p>Elipticmodulo: /* Solution */</p> <hr /> <div>== Problem ==<br /> A majority of the &lt;math&gt;30&lt;/math&gt; students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than &lt;math&gt;1&lt;/math&gt;. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt;17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 7 \qquad<br /> \textbf{(B)}\ 11 \qquad<br /> \textbf{(C)}\ 17 \qquad<br /> \textbf{(D)}\ 23 \qquad<br /> \textbf{(E)}\ 77 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The total cost of the pencils can be found by &lt;math&gt;\text{students}*\text{pencils purchased by each}*\text{price of each pencil}&lt;/math&gt;. As the cost is &lt;math&gt;&lt;/math&gt;17.71&lt;math&gt;, or &lt;/math&gt;1771&lt;math&gt; cents, the cost of the pencils must divide the total cost. Scanning the answer choices, they all unfortunately divide into &lt;/math&gt;1771&lt;math&gt;, so we have to start from the beginning (although this does give us some indication that we are on the right tack, because MAA doesn't want to make things too easy for us).<br /> <br /> Therefore, since &lt;/math&gt;1771&lt;math&gt; is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: &lt;/math&gt;7, 11, 23&lt;math&gt;. Since neither &lt;/math&gt;(C)&lt;math&gt; nor &lt;/math&gt;(E)&lt;math&gt; are any of these factors, they can be eliminated immediately, leaving &lt;/math&gt;(A)&lt;math&gt;, &lt;/math&gt;(B)&lt;math&gt;, and &lt;/math&gt;(D)&lt;math&gt;.<br /> <br /> Beginning with &lt;/math&gt;(A) 7&lt;math&gt;, we see that the number of pencils purchased by each student must be either &lt;/math&gt;11&lt;math&gt; or &lt;/math&gt;23&lt;math&gt;. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.<br /> <br /> Continuing with &lt;/math&gt;(B) 11&lt;math&gt;, we can conclude that the only case that fulfills the restrictions are that there are &lt;/math&gt;23&lt;math&gt; students who each purchased &lt;/math&gt;7&lt;math&gt; such pencils, so the answer is &lt;/math&gt;\boxed{B}&lt;math&gt;. We can apply the same logic to &lt;/math&gt;(E)&lt;math&gt; as we applied to &lt;/math&gt;(A)$, if one wants to make doubly sure.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_7&diff=75932 2011 AMC 12A Problems/Problem 7 2016-02-15T01:44:06Z <p>Elipticmodulo: /* Solution */</p> <hr /> <div>== Problem ==<br /> A majority of the &lt;math&gt;30&lt;/math&gt; students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than &lt;math&gt;1&lt;/math&gt;. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was &lt;math&gt;17.71&lt;/math&gt;. What was the cost of a pencil in cents?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 7 \qquad<br /> \textbf{(B)}\ 11 \qquad<br /> \textbf{(C)}\ 17 \qquad<br /> \textbf{(D)}\ 23 \qquad<br /> \textbf{(E)}\ 77 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The total cost of the pencils can be found by &lt;math&gt;\text{#students}*\text{#pencils purchased by each}*\text{price of each pencil}&lt;/math&gt;. As the cost is &lt;math&gt;&lt;/math&gt;17.71&lt;math&gt;, or &lt;/math&gt;1771&lt;math&gt; cents, the cost of the pencils must divide the total cost. Scanning the answer choices, they all unfortunately divide into &lt;/math&gt;1771&lt;math&gt;, so we have to start from the beginning (although this does give us some indication that we are on the right tack, because MAA doesn't want to make things too easy for us).<br /> <br /> Therefore, since &lt;/math&gt;1771&lt;math&gt; is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: &lt;/math&gt;7, 11, 23&lt;math&gt;. Since neither &lt;/math&gt;(C)&lt;math&gt; nor &lt;/math&gt;(E)&lt;math&gt; are any of these factors, they can be eliminated immediately, leaving &lt;/math&gt;(A)&lt;math&gt;, &lt;/math&gt;(B)&lt;math&gt;, and &lt;/math&gt;(D)&lt;math&gt;.<br /> <br /> Beginning with &lt;/math&gt;(A) 7&lt;math&gt;, we see that the number of pencils purchased by each student must be either &lt;/math&gt;11&lt;math&gt; or &lt;/math&gt;23&lt;math&gt;. However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.<br /> <br /> Continuing with &lt;/math&gt;(B) 11&lt;math&gt;, we can conclude that the only case that fulfills the restrictions are that there are &lt;/math&gt;23&lt;math&gt; students who each purchased &lt;/math&gt;7&lt;math&gt; such pencils, so the answer is &lt;/math&gt;\boxed{B}&lt;math&gt;. We can apply the same logic to &lt;/math&gt;(E)&lt;math&gt; as we applied to &lt;/math&gt;(A)$, if one wants to make doubly sure.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=6|num-a=8|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75931 2011 AMC 12A Problems/Problem 12 2016-02-15T01:08:56Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Think of the blue arrow as the power boat and the red arrow as the raft in the following three diagrams, which represent different time intervals of the problem.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> Thinking about the distance covered as their distances with respect to each other, they are &lt;math&gt;0&lt;/math&gt; distance apart in the first diagram when they haven't started to move yet, some distance &lt;math&gt;d&lt;/math&gt; apart in the second diagram when the power boat reaches &lt;math&gt;B&lt;/math&gt;, and again &lt;math&gt;0&lt;/math&gt; distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of &lt;math&gt;d&lt;/math&gt; on the way there, and again cover a distance of &lt;math&gt;d&lt;/math&gt; on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; denote the speed of the power boat (only the power boat, not factoring in current) and &lt;math&gt;r&lt;/math&gt; denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the boat travels at a velocity of &lt;math&gt;b+r&lt;/math&gt;, and on the way back, travels at a velocity of &lt;math&gt;-(b-r)=r-b&lt;/math&gt;, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; becomes &lt;math&gt;(r+b)-r=b&lt;/math&gt;, and on the way back it becomes &lt;math&gt;(r-b)-r=-b&lt;/math&gt;. Since the boat's velocities with respect to the raft are exact opposites, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;-b&lt;/math&gt;, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br /> <br /> From this, we have that the boat travels a distance &lt;math&gt;d&lt;/math&gt; at rate &lt;math&gt;b&lt;/math&gt; with respect to the raft both on the way to &lt;math&gt;B&lt;/math&gt; and on the way back. Thus, using &lt;math&gt;\dfrac{distance}{speed}=time&lt;/math&gt;, we have &lt;math&gt;\dfrac{2d}{b}=9\text{ hours}&lt;/math&gt;, and to see how long it took to travel half the distance, we have &lt;math&gt;\dfrac{d}{b}=4.5\text{ hours}\implies\boxed{\textbf{D}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75930 2011 AMC 12A Problems/Problem 12 2016-02-15T01:07:49Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> Think of the blue arrow as the power boat and the red arrow as the raft. Thinking about the distance covered as their distances with respect to each other, they are &lt;math&gt;0&lt;/math&gt; distance apart in the first diagram when they haven't started to move yet, some distance &lt;math&gt;d&lt;/math&gt; apart in the second diagram when the power boat reaches &lt;math&gt;B&lt;/math&gt;, and again &lt;math&gt;0&lt;/math&gt; distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of &lt;math&gt;d&lt;/math&gt; on the way there, and again cover a distance of &lt;math&gt;d&lt;/math&gt; on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; denote the speed of the power boat (only the power boat, not factoring in current) and &lt;math&gt;r&lt;/math&gt; denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the boat travels at a velocity of &lt;math&gt;b+r&lt;/math&gt;, and on the way back, travels at a velocity of &lt;math&gt;-(b-r)=r-b&lt;/math&gt;, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; becomes &lt;math&gt;(r+b)-r=b&lt;/math&gt;, and on the way back it becomes &lt;math&gt;(r-b)-r=-b&lt;/math&gt;. Since the boat's velocities with respect to the raft are exact opposites, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;-b&lt;/math&gt;, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br /> <br /> From this, we have that the boat travels a distance &lt;math&gt;d&lt;/math&gt; at rate &lt;math&gt;b&lt;/math&gt; with respect to the raft both on the way to &lt;math&gt;B&lt;/math&gt; and on the way back. Thus, using &lt;math&gt;\dfrac{distance}{speed}=time&lt;/math&gt;, we have &lt;math&gt;\dfrac{2d}{b}=9\text{ hours}&lt;/math&gt;, and to see how long it took to travel half the distance, we have &lt;math&gt;\dfrac{d}{b}=4.5\text{ hours}\implies\boxed{\textbf{D}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75929 2011 AMC 12A Problems/Problem 12 2016-02-15T01:04:40Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> Think of the blue arrow as the power boat and the red arrow as the raft. Thinking about the distance covered as their distances with respect to each other, they are &lt;math&gt;0&lt;/math&gt; distance apart in the first diagram when they haven't started to move yet, some distance &lt;math&gt;d&lt;/math&gt; apart in the second diagram when the power boat reaches &lt;math&gt;B&lt;/math&gt;, and again &lt;math&gt;0&lt;/math&gt; distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of &lt;math&gt;d&lt;/math&gt; on the way there, and again cover a distance of &lt;math&gt;d&lt;/math&gt; on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; denote the speed of the power boat (only the power boat, not factoring in current) and &lt;math&gt;r&lt;/math&gt; denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the boat travels at a velocity of &lt;math&gt;b+r&lt;/math&gt;, and on the way back, travels at a velocity of &lt;math&gt;-(b-r)=r-b&lt;/math&gt;, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; becomes &lt;math&gt;(r+b)-r=b&lt;/math&gt;, and on the way back it becomes &lt;math&gt;(r-b)-r=-b&lt;/math&gt;. Since the boat's velocities with respect to the raft are exact opposites, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;-b&lt;/math&gt;, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br /> <br /> From this, we have that the boat travels a distance &lt;math&gt;d&lt;/math&gt; at rate &lt;math&gt;b&lt;/math&gt; with respect to the raft both on the way to &lt;math&gt;B&lt;/math&gt; and on the way back. Thus, using &lt;math&gt;\dfrac{distance}{speed}=time&lt;/math&gt;, we have &lt;math&gt;\dfrac{2d}{b}=9\text{ hours}&lt;/math&gt;, and to see how long it took to travel half the distance, we have &lt;math&gt;\dfrac{d}{b}=4.5\text{ hours}\implies\boxed{\textbf{D}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75928 2011 AMC 12A Problems/Problem 12 2016-02-15T01:04:09Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> Think of the blue arrow as the power boat and the red arrow as the raft. Thinking about the distance covered as their distances with respect to each other, they are &lt;math&gt;0&lt;/math&gt; distance apart in the first diagram when they haven't started to move yet, some distance &lt;math&gt;d&lt;/math&gt; apart in the second diagram when the power boat reaches &lt;math&gt;B&lt;/math&gt;, and again &lt;math&gt;0&lt;/math&gt; distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of &lt;math&gt;d&lt;/math&gt; on the way there, and again cover a distance of &lt;math&gt;d&lt;/math&gt; on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; denote the speed of the power boat (only the power boat, not factoring in current) and &lt;math&gt;r&lt;/math&gt; denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the boat travels at a velocity of &lt;math&gt;b+r&lt;/math&gt;, and on the way back, travels at a velocity of &lt;math&gt;-(b-r)=r-b&lt;/math&gt;, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; becomes &lt;math&gt;(r+b)-r=b&lt;/math&gt;, and on the way back it becomes &lt;math&gt;(r-b)-r=-b&lt;/math&gt;. Since the boat's velocities with respect to the raft are exact opposites, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;-b&lt;/math&gt;, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br /> <br /> From this, we have that the boat travels a distance &lt;math&gt;d&lt;/math&gt; at rate &lt;math&gt;b&lt;/math&gt; with respect to the raft both on the way to &lt;math&gt;B&lt;/math&gt; and on the way back. Thus, using &lt;math&gt;\dfrac{distance}{speed}=time&lt;/math&gt;, we have &lt;math&gt;\dfrac{2d}{b}=9\text{ hours}&lt;/math&gt;, and to see how long it took to travel half the distance, we have &lt;math&gt;\dfrac{d}{b}=4.5\text{ hours}\implies\boxed{\textbf{E}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75927 2011 AMC 12A Problems/Problem 12 2016-02-15T01:03:44Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> Think of the blue arrow as the power boat and the red arrow as the raft. Thinking about the distance covered as their distances with respect to each other, they are &lt;math&gt;0&lt;/math&gt; distance apart in the first diagram when they haven't started to move yet, some distance &lt;math&gt;d&lt;/math&gt; apart in the second diagram when the power boat reaches &lt;math&gt;B&lt;/math&gt;, and again &lt;math&gt;0&lt;/math&gt; distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of &lt;math&gt;d&lt;/math&gt; on the way there, and again cover a distance of &lt;math&gt;d&lt;/math&gt; on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; denote the speed of the power boat (only the power boat, not factoring in current) and &lt;math&gt;r&lt;/math&gt; denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the boat travels at a velocity of &lt;math&gt;b+r&lt;/math&gt;, and on the way back, travels at a velocity of &lt;math&gt;-(b-r)=r-b&lt;/math&gt;, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; becomes &lt;math&gt;(r+b)-r=b&lt;/math&gt;, and on the way back it becomes &lt;math&gt;(r-b)-r=-b&lt;/math&gt;. Since the boat's velocities with respect to the raft are exact opposites, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;-b&lt;/math&gt;, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br /> <br /> From this, we have that the boat travels a distance &lt;math&gt;d&lt;/math&gt; at rate &lt;math&gt;b&lt;/math&gt; with respect to the raft both on the way to &lt;math&gt;B&lt;/math&gt; and on the way back. Thus, using &lt;math&gt;\dfrac{distance}{speed}=time&lt;/math&gt;, we have &lt;math&gt;\dfrac{2d}{b}=9\text{ hours}&lt;/math&gt;, and to see how long it took to travel half the distance, we have &lt;math&gt;\dfrac{d}{b}=4.5text{ hours}\implies\boxed{\textbf{E}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75926 2011 AMC 12A Problems/Problem 12 2016-02-15T01:01:08Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> Think of the blue arrow as the power boat and the red arrow as the raft. Thinking about the distance covered as their distances with respect to each other, they are &lt;math&gt;0&lt;/math&gt; distance apart in the first diagram when they haven't started to move yet, some distance &lt;math&gt;d&lt;/math&gt; apart in the second diagram when the power boat reaches &lt;math&gt;B&lt;/math&gt;, and again &lt;math&gt;0&lt;/math&gt; distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of &lt;math&gt;d&lt;/math&gt; on the way there, and again cover a distance of &lt;math&gt;d&lt;/math&gt; on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; denote the speed of the power boat (only the power boat, not factoring in current) and &lt;math&gt;r&lt;/math&gt; denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the boat travels at a velocity of &lt;math&gt;b+r&lt;/math&gt;, and on the way back, travels at a velocity of &lt;math&gt;-(b-r)=r-b&lt;/math&gt;, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; becomes &lt;math&gt;(r+b)-r=b&lt;/math&gt;, and on the way back it becomes &lt;math&gt;(r-b)-r=-b&lt;/math&gt;. Since the boat's velocities with respect to the raft are exact opposites, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;-b&lt;/math&gt;, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br /> <br /> From this, we have that the boat travels a distance &lt;math&gt;d&lt;/math&gt; at rate &lt;math&gt;b&lt;/math&gt; with respect to the raft both on the way to &lt;math&gt;B&lt;/math&gt; and on the way back. Thus, using &lt;math&gt;\dfrac{distance}{speed}=time&lt;/math&gt;, we have &lt;math&gt;\dfrac{2d}{b}=9 hours&lt;/math&gt;, and to see how long it took to travel half the distance, we have &lt;math&gt;\dfrac{d}{b}=4.5 hours\implies\boxed{\textbf{E}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75925 2011 AMC 12A Problems/Problem 12 2016-02-15T00:59:45Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> Think of the blue arrow as the power boat and the red arrow as the raft. Thinking about the distance covered as their distances with respect to each other, they are &lt;math&gt;0&lt;/math&gt; distance apart in the first diagram when they haven't started to move yet, some distance &lt;math&gt;d&lt;/math&gt; apart in the second diagram when the power boat reaches &lt;math&gt;B&lt;/math&gt;, and again &lt;math&gt;0&lt;/math&gt; distance apart in the third diagram when they meet. Therefore, with respect to each other, the boat and the raft cover a distance of &lt;math&gt;d&lt;/math&gt; on the way there, and again cover a distance of &lt;math&gt;d&lt;/math&gt; on when drawing closer. This makes sense, because from the 1st diagram to the second, the raft moves in the same direction as the boat, while from the 2nd to the 3rd, the boat and raft move in opposite directions.<br /> <br /> Let &lt;math&gt;b&lt;/math&gt; denote the speed of the power boat (only the power boat, not factoring in current) and &lt;math&gt;r&lt;/math&gt; denote the speed of the raft, which, as given by the problem, is also equal to the speed of the current. Thus, from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the boat travels at a velocity of &lt;math&gt;b+r&lt;/math&gt;, and on the way back, travels at a velocity of &lt;math&gt;-(b-r)=r-b&lt;/math&gt;, since the current aids the boat on the way there, and goes against the boat on the way back. With respect to the raft then, the boat's velocity from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; becomes &lt;math&gt;(r+b)-r=b&lt;/math&gt;, and on the way back it becomes &lt;math&gt;(r-b)-r=-b&lt;/math&gt;. Since the boat's velocities with respect to the raft are exact opposites, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;-b&lt;/math&gt;, we therefore know that the boat and raft travel apart from each other at the same rate that they travel toward each other.<br /> <br /> From this, we have that the boat travels a distance &lt;math&gt;d&lt;/math&gt; at rate &lt;math&gt;b&lt;/math&gt; with respect to the raft both on the way to &lt;math&gt;B&lt;/math&gt; and on the way back. Thus, using &lt;math&gt;\frac_{distance}{speed}=time&lt;/math&gt;, we have &lt;math&gt;\frac_{2d}{b}=9 hours&lt;/math&gt;, and to see how long it took to travel half the distance, we have &lt;math&gt;\frac_{d}{b}=4.5 hours\implies\boxed{\textbf{E}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75923 2011 AMC 12A Problems/Problem 12 2016-02-15T00:37:15Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((.3,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75922 2011 AMC 12A Problems/Problem 12 2016-02-15T00:36:59Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((1,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75921 2011 AMC 12A Problems/Problem 12 2016-02-15T00:36:47Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((1.1,4.4),dir(360),blue);<br /> arrow((1.1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75920 2011 AMC 12A Problems/Problem 12 2016-02-15T00:36:19Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((1.2,4.4),dir(360),blue);<br /> arrow((1,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75919 2011 AMC 12A Problems/Problem 12 2016-02-15T00:35:53Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((1.8,4.4),dir(360),blue);<br /> arrow((1.8,4.2),dir(180),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75918 2011 AMC 12A Problems/Problem 12 2016-02-15T00:35:26Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.6,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((1.8,4.4),dir(180),blue);<br /> arrow((1.8,4.2),dir(360),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75917 2011 AMC 12A Problems/Problem 12 2016-02-15T00:34:55Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((2.8,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((1.8,6.4),dir(180),blue);<br /> arrow((1.8,6.2),dir(360),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75916 2011 AMC 12A Problems/Problem 12 2016-02-15T00:34:17Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((3,5.4),dir(360),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((1.8,5.4),dir(180),blue);<br /> arrow((-1.8,5.2),dir(360),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75915 2011 AMC 12A Problems/Problem 12 2016-02-15T00:33:21Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((3.5,5.4),dir(180),blue);<br /> arrow((-0.5,5.2),dir(360),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((3.5,5.4),dir(180),blue);<br /> arrow((-0.5,5.2),dir(360),red);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75914 2011 AMC 12A Problems/Problem 12 2016-02-15T00:32:32Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> arrow((3.5,5.4),dir(180),blue);<br /> arrow((-0.5,5.2),dir(180),red);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75911 2011 AMC 12A Problems/Problem 12 2016-02-15T00:30:57Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> arrow((-2.5,6.4),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75910 2011 AMC 12A Problems/Problem 12 2016-02-15T00:30:34Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> arrow((-2.5,6),dir(180),blue);<br /> arrow((-2.5,6.2),dir(180),red);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75909 2011 AMC 12A Problems/Problem 12 2016-02-15T00:30:04Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6.2);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> arrow((-2.5,4),dir(180),blue);<br /> arrow((-2.5,4.2),dir(180),red);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75908 2011 AMC 12A Problems/Problem 12 2016-02-15T00:29:47Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> arrow((-2.5,4),dir(180),blue);<br /> arrow((-2.5,4.2),dir(180),red);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75907 2011 AMC 12A Problems/Problem 12 2016-02-15T00:29:01Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> arrow((-2.5,3),dir(180),green);<br /> arrow((-2.5,3.2),dir(180),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75906 2011 AMC 12A Problems/Problem 12 2016-02-15T00:28:35Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> arrow((-3,3),dir(180),green);<br /> arrow((-3,3.2),dir(180),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75905 2011 AMC 12A Problems/Problem 12 2016-02-15T00:27:43Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> arrow((-2,3),dir(380),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75904 2011 AMC 12A Problems/Problem 12 2016-02-15T00:27:24Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> arrow((-2,3),dir(180),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75903 2011 AMC 12A Problems/Problem 12 2016-02-15T00:27:10Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> arrow((-1,3),dir(180),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75902 2011 AMC 12A Problems/Problem 12 2016-02-15T00:26:44Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> arrow((-3,3),dir(180),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75901 2011 AMC 12A Problems/Problem 12 2016-02-15T00:26:18Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> arrow((3,0),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75900 2011 AMC 12A Problems/Problem 12 2016-02-15T00:25:56Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> arrow((3,0),dir(180),green);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75899 2011 AMC 12A Problems/Problem 12 2016-02-15T00:23:53Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> A=(-3,5);<br /> B=(3,5);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75898 2011 AMC 12A Problems/Problem 12 2016-02-15T00:23:22Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((-3,3),blue);<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,8);<br /> B=(3,8);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75897 2011 AMC 12A Problems/Problem 12 2016-02-15T00:22:30Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((0,0),blue);<br /> A=(-3,6);<br /> B=(3,6);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,8);<br /> B=(3,8);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75896 2011 AMC 12A Problems/Problem 12 2016-02-15T00:21:29Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> dot((0,0),blue);<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75895 2011 AMC 12A Problems/Problem 12 2016-02-15T00:18:07Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);<br /> label(&quot;$A$&quot;,A,NE);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75894 2011 AMC 12A Problems/Problem 12 2016-02-15T00:17:33Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(-3,4);<br /> B=(3,4);<br /> draw(A--B);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75893 2011 AMC 12A Problems/Problem 12 2016-02-15T00:16:59Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(3,4);<br /> B=(3,-4);<br /> draw(A--B);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75892 2011 AMC 12A Problems/Problem 12 2016-02-15T00:16:30Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> pair A, B;<br /> A=(3,4);<br /> B=(3,-4);<br /> draw(O--Z);&lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75891 2011 AMC 12A Problems/Problem 12 2016-02-15T00:15:37Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> pair A, B;<br /> A=(3,4);<br /> B=(3,-4);<br /> &lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75890 2011 AMC 12A Problems/Problem 12 2016-02-15T00:15:02Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> A=(3,4);<br /> B=(3,-4);<br /> &lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75889 2011 AMC 12A Problems/Problem 12 2016-02-15T00:14:47Z <p>Elipticmodulo: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm,8cm);<br /> pair A, B;<br /> O=(3,4);<br /> Z=(3,-4);<br /> &lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_12&diff=75888 2011 AMC 12A Problems/Problem 12 2016-02-15T00:10:59Z <p>Elipticmodulo: /* Solution */</p> <hr /> <div>== Problem ==<br /> A power boat and a raft both left dock &lt;math&gt;A&lt;/math&gt; on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock &lt;math&gt;B&lt;/math&gt; downriver, then immediately turned and traveled back upriver. It eventually met the raft on the river 9 hours after leaving dock &lt;math&gt;A.&lt;/math&gt; How many hours did it take the power boat to go from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 3 \qquad<br /> \textbf{(B)}\ 3.5 \qquad<br /> \textbf{(C)}\ 4 \qquad<br /> \textbf{(D)}\ 4.5 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Since the speed of the river is not specified, the outcome of the problem must be independent of this speed. We may thus trivially assume that the river has a speed of &lt;math&gt;0&lt;/math&gt;. In this case, when the powerboat travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, the raft remains at &lt;math&gt;A&lt;/math&gt;. Thus the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; takes the same time as the trip from &lt;math&gt;B&lt;/math&gt; to the raft. Since these times are equal and sum to &lt;math&gt;9&lt;/math&gt; hours, the trip from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; must take half this time, or &lt;math&gt;4.5&lt;/math&gt; hours. The answer is thus &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> What's important in this problem is to consider everything in terms of the power boat and the raft, since that is how the problem is given to us. Let us denote the speed of the boat as &lt;math&gt;b&lt;/math&gt; and the speed of the raft at &lt;math&gt;r&lt;/math&gt;. dot((20,0));<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=11|num-a=13|ab=A}}<br /> {{MAA Notice}}</div> Elipticmodulo