https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Carpedatdiem&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-08T19:03:57Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_15&diff=84686 2017 AMC 10A Problems/Problem 15 2017-03-13T02:22:00Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> Chloé chooses a real number uniformly at random from the interval &lt;math&gt;[0, 2017]&lt;/math&gt;. Independently, Laurent chooses a real number uniformly at random from the interval &lt;math&gt;[0, 4034]&lt;/math&gt;. What is the probability that Laurent's number is greater than Chloé's number?<br /> <br /> &lt;math&gt; \mathrm{\textbf{(A)} \ }\frac{1}{2}\qquad \mathrm{\textbf{(B)} \ } \frac{2}{3}\qquad \mathrm{\textbf{(C)} \ } \frac{3}{4}\qquad \mathrm{\textbf{(D)} \ } \frac{5}{6}\qquad \mathrm{\textbf{(E)} \ }\frac{7}{8}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Denote &quot;winning&quot; to mean &quot;picking a greater number&quot;.<br /> There is a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance that Laurent chooses a number in the interval &lt;math&gt;[2017, 4034]&lt;/math&gt;. In this case, Chloé cannot possibly win, since the maximum number she can pick is &lt;math&gt;2017&lt;/math&gt;. Otherwise, if Laurent picks a number in the interval &lt;math&gt;[0, 2017]&lt;/math&gt;, with probability &lt;math&gt;\frac{1}{2}&lt;/math&gt;, then the two people are symmetric, and each has a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of winning. Then, the total probability is &lt;math&gt;\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can use geometric probability to solve this.<br /> Suppose a point &lt;math&gt;(x,y)&lt;/math&gt; lies in the &lt;math&gt;xy&lt;/math&gt;-plane. Let &lt;math&gt;x&lt;/math&gt; be Chloe's number and &lt;math&gt;y&lt;/math&gt; be Laurent's number. Then obviously we want &lt;math&gt;y&gt;x&lt;/math&gt;, which basically gives us a region above a line. We know that Chloe's number is in the interval &lt;math&gt;[0,2017]&lt;/math&gt; and Laurent's number is in the interval &lt;math&gt;[0,4034]&lt;/math&gt;, so we can create a rectangle in the plane, whose length is &lt;math&gt;2017&lt;/math&gt; and whose width is &lt;math&gt;4034&lt;/math&gt;. Drawing it out, we see that it is easier to find the probability that Chloe's number is greater than Laurent's number and subtract this probability from &lt;math&gt;1&lt;/math&gt;. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line &lt;math&gt;y&gt;x&lt;/math&gt;, which is &lt;math&gt;\frac{2017 \cdot 2017}{2}&lt;/math&gt;. Instead of bashing this out we know that the rectangle has area &lt;math&gt;2017 \cdot 4034&lt;/math&gt;. So the probability that Laurent has a smaller number is &lt;math&gt;\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}&lt;/math&gt;. Simplifying the expression yields &lt;math&gt;\frac{1}{4}&lt;/math&gt; and so &lt;math&gt;1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_12&diff=84657 2017 AMC 10A Problems/Problem 12 2017-03-12T00:08:04Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be a set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3,~x+2,&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\\qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\\qquad\textbf{(D)}\ \text{a triangle} \qquad\textbf{(E)}\ \text{three rays with a common endpoint}&lt;/math&gt;<br /> <br /> ==Solution==<br /> If the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;x+2&lt;/math&gt;, then &lt;math&gt;x=1&lt;/math&gt;. Also, &lt;math&gt;y-4\le 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y \le 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;x=1&lt;/math&gt; where &lt;math&gt;y \le 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt;. This is a ray with an endpoint of &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Similar to the process above, we assume that the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt;. Solving the equation &lt;math&gt;3=y-4&lt;/math&gt; then &lt;math&gt;y=7&lt;/math&gt;. Also, &lt;math&gt;x+2\le 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x\le1&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=7&lt;/math&gt; where &lt;math&gt;x\le 1&lt;/math&gt; is also part of &lt;math&gt;S&lt;/math&gt;. This is another ray with the same endpoint as the above ray: &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;x+2&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are the two equal values, then &lt;math&gt;x+2=y-4&lt;/math&gt;. Solving the equation for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y=x+6&lt;/math&gt;. Also &lt;math&gt;3\le y-4&lt;/math&gt; because &lt;math&gt;y-4&lt;/math&gt; is one way to express the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y\ge 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=x+6&lt;/math&gt; where &lt;math&gt;y\ge 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt; like the other two rays. The lowest possible value that can be achieved is also &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;S&lt;/math&gt; is made up of three rays with common endpoint &lt;math&gt;(1, 7)&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\textbf{(E) }\text{three rays with a common endpoint}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=84656 2017 AMC 10A Problems/Problem 11 2017-03-12T00:05:51Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all point in three-dimensional space within 3 units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume 216&lt;math&gt;\pi&lt;/math&gt;. What is the length &lt;math&gt;\textit{AB}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Solution==<br /> In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within &lt;math&gt;3&lt;/math&gt; units of a point would be a sphere with radius &lt;math&gt;3&lt;/math&gt;. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal &lt;math&gt;216 \pi&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is equal to the length of our line segment.<br /> <br /> Solving, we find that &lt;math&gt;x = \boxed{\textbf{(D)}\ 20}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> <br /> http://i.imgur.com/cwNt293.png<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_10&diff=84655 2017 AMC 10A Problems/Problem 10 2017-03-11T23:57:06Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> Joy has &lt;math&gt;30&lt;/math&gt; thin rods, one each of every integer length from &lt;math&gt;1&lt;/math&gt; cm through &lt;math&gt;30&lt;/math&gt; cm. She places the rods with lengths &lt;math&gt;3&lt;/math&gt; cm, &lt;math&gt;7&lt;/math&gt; cm, and &lt;math&gt;15&lt;/math&gt; cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> ==Solution==<br /> The triangle inequality generalizes to all polygons, so &lt;math&gt;x &lt; 3+7+15&lt;/math&gt; and &lt;math&gt;x+3+7&gt;15&lt;/math&gt; to get &lt;math&gt;5&lt;x&lt;25&lt;/math&gt;. Now, we know that there are &lt;math&gt;19&lt;/math&gt; numbers between &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; exclusive, but we must subtract &lt;math&gt;2&lt;/math&gt; to account for the 2 lengths already used that are between those numbers, which gives &lt;math&gt;19-2=\boxed{\textbf{(B)}\ 17}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_8&diff=84653 2017 AMC 10A Problems/Problem 8 2017-03-11T22:57:04Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> At a gathering of &lt;math&gt;30&lt;/math&gt; people, there are &lt;math&gt;20&lt;/math&gt; people who all know each other and &lt;math&gt;10&lt;/math&gt; people who know no one. People who know each other a hug, and people who do not know each other shake hands. How many handshakes occur?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Each one of the ten people has to shake hands with all the &lt;math&gt;20&lt;/math&gt; other people they don’t know. So &lt;math&gt;10\cdot20 = 200&lt;/math&gt;. From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands, or &lt;math&gt;\binom{10}{2} = 45&lt;/math&gt;. Thus the answer is &lt;math&gt;200 + 45 = \boxed{\textbf{(B)}\ 245}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can also use complementary counting. First of all, &lt;math&gt;\dbinom{30}{2}=435&lt;/math&gt; handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from &lt;math&gt;435&lt;/math&gt; to find the handshakes. Hugs only happen between the 20 people who know each other, so there are &lt;math&gt;\dbinom{20}{2}=190&lt;/math&gt; hugs. &lt;math&gt;435-190= \boxed{\textbf{(B)}\ 245}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_21&diff=84556 2006 AMC 10A Problems/Problem 21 2017-03-09T16:52:04Z <p>Carpedatdiem: /* Solution */</p> <hr /> <div>== Problem ==<br /> How many four-[[digit]] [[positive integer]]s have at least one digit that is a 2 or a 3? <br /> <br /> &lt;math&gt;\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad&lt;/math&gt;<br /> <br /> == Solution ==<br /> Since we are asked for the number of positive 4-digit [[integer]]s with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits. <br /> <br /> The total number of 4-digit integers is &lt;math&gt;9 \cdot 10 \cdot 10 \cdot 10 = 9000&lt;/math&gt;, since we have 10 choices for each digit except the first (which can't be 0).<br /> <br /> Similarly, the total number of 4-digit integers without any 2 or 3 is &lt;math&gt;7 \cdot 8 \cdot 8 \cdot 8 =\boxed{3584}&lt;/math&gt;.<br /> <br /> Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their [[decimal representation]] is &lt;math&gt;9000-3584=5416 \Longrightarrow \mathrm{(E)} &lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_7&diff=84514 2017 AMC 10A Problems/Problem 7 2017-03-08T23:51:19Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%&lt;/math&gt;<br /> ==Solution==<br /> <br /> Let &lt;math&gt;j&lt;/math&gt; represent how far Jerry walked, and &lt;math&gt;s&lt;/math&gt; represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, <br /> &lt;math&gt;j = 2&lt;/math&gt;<br /> Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus,<br /> &lt;math&gt;s = \sqrt{2} = 1.414...&lt;/math&gt;<br /> We can then take <br /> &lt;math&gt;\frac{j-s}{j} \approx \frac{2 - 1.4}{2}=0.3 \implies \boxed{\textbf{(A)}\ 30\%}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_6&diff=84504 2017 AMC 10A Problems/Problem 6 2017-03-08T22:29:00Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which one of these statements necessarily follows logically?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{If Lewis did not receive an A, then he got all of the multiple choice questions wrong.}\\\qquad\textbf{(B)}\ \text{If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}\\\qquad\textbf{(C)}\ \text{If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A.}\\\qquad\textbf{(D)}\ \text{If Lewis received an A, then he got all of the multiple choice questions right.}\\\qquad\textbf{(E)}\ \text{If Lewis received an A, then he got at least one of the multiple choice questions right.}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Rewriting the given statement: &quot;if someone got all the multiple choice questions right on the upcoming exam then he or she would receive an A on the exam.&quot;<br /> If that someone is Lewis the statement becomes: &quot;if Lewis got all the multiple choice questions right, then he got an A on the exam.&quot;<br /> The contrapositive: &quot;If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong (did not get all of them right)&quot; must also be true leaving B as the correct answer. B is also equivalent to the contrapositive of the original statement, which implies that it must be true, so the answer is &lt;math&gt;\boxed{\textbf{(B)}\text{ If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_5&diff=84503 2017 AMC 10A Problems/Problem 5 2017-03-08T22:25:31Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the two real numbers be &lt;math&gt;x,y&lt;/math&gt;. We are given that &lt;math&gt;x+y=4xy,&lt;/math&gt; and dividing both sides by &lt;math&gt;xy&lt;/math&gt;, &lt;math&gt;\frac{x}{xy}+\frac{y}{xy}=4.&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.&lt;/cmath&gt;<br /> <br /> Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.<br /> <br /> <br /> ==Solution 2==<br /> Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction.<br /> See for yourself. And by looking into fractions, we immediately see that &lt;math&gt;\frac{1}{3}&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; would fit the rule. &lt;math&gt;\boxed{\textbf{(C)} 4}.&lt;/math&gt;<br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=84500 2017 AMC 10A Problems/Problem 4 2017-03-08T22:20:29Z <p>Carpedatdiem: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Mia is helping her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Every &lt;math&gt;30&lt;/math&gt; seconds &lt;math&gt;3-2=1&lt;/math&gt; toys are put in the box, so after &lt;math&gt;27\cdot30&lt;/math&gt; seconds there will be &lt;math&gt;27&lt;/math&gt; toys in the box. Mia's mom will then put &lt;math&gt;3&lt;/math&gt; toys into to the box and we have our total amount of time to be &lt;math&gt;27\cdot30+30=840&lt;/math&gt; seconds, which equals &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Though Mia's mom places &lt;math&gt;3&lt;/math&gt; toys every &lt;math&gt;30&lt;/math&gt; seconds, Mia takes out &lt;math&gt;2&lt;/math&gt; toys right after. Therefore, after &lt;math&gt;30&lt;/math&gt; seconds, the two have collectively placed &lt;math&gt;1&lt;/math&gt; toy into the box. Therefore by &lt;math&gt;13.5&lt;/math&gt; minutes, the two would have placed &lt;math&gt;27&lt;/math&gt; toys into the box. Therefore, at &lt;math&gt;14&lt;/math&gt; minutes, the two would have placed &lt;math&gt;30&lt;/math&gt; toys into the box. Though Mia may take &lt;math&gt;2&lt;/math&gt; toys out right after, the number of toys in the box first reaches &lt;math&gt;30&lt;/math&gt; by &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> ==See also==<br /> <br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=84499 2017 AMC 10A Problems/Problem 4 2017-03-08T22:19:17Z <p>Carpedatdiem: /* Problem */</p> <hr /> <div>==Problem==<br /> Mia is helping her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Every &lt;math&gt;30&lt;/math&gt; seconds &lt;math&gt;3-2=1&lt;/math&gt; toys are put in the box, so after &lt;math&gt;27\cdot30&lt;/math&gt; seconds there will be &lt;math&gt;27&lt;/math&gt; toys in the box. Mia's mom will then put &lt;math&gt;3&lt;/math&gt; toys into to the box and we have our total amount of time to be &lt;math&gt;27\cdot30+30=840&lt;/math&gt; seconds, which equals &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Though Mia's mom places &lt;math&gt;3&lt;/math&gt; toys every &lt;math&gt;30&lt;/math&gt; seconds, Mia takes out &lt;math&gt;2&lt;/math&gt; toys right after. Therefore, after &lt;math&gt;30&lt;/math&gt; seconds, the two have collectively placed &lt;math&gt;1&lt;/math&gt; toy into the box. Thereforeby &lt;math&gt;13.5&lt;/math&gt; minutes, the two would have placed &lt;math&gt;27&lt;/math&gt; toys into the box. Therefore, at &lt;math&gt;14&lt;/math&gt; minutes, the two would have placed &lt;math&gt;30&lt;/math&gt; toys into the box. Though Mia may take &lt;math&gt;2&lt;/math&gt; toys out right after, the number of toys in the box first reaches &lt;math&gt;30&lt;/math&gt; by &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> ==See also==<br /> <br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Carpedatdiem https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_2&diff=84496 2017 AMC 10A Problems/Problem 2 2017-03-08T22:12:23Z <p>Carpedatdiem: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> Pablo buys popsicles for his friends. The store sells single popsicles for &lt;math&gt;\$1&lt;/math&gt; each, &lt;math&gt;3&lt;/math&gt;-popsicle boxes for &lt;math&gt;\$2&lt;/math&gt; each, and &lt;math&gt;5&lt;/math&gt;-popsicle boxes for &lt;math&gt;\$3&lt;/math&gt;. What is the greatest number of popsicles that Pablo can buy with &lt;math&gt;\$8&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;\$3&lt;/math&gt; boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying &lt;math&gt;2&lt;/math&gt;, we have &lt;math&gt;\$2&lt;/math&gt; left. We cannot buy a third &lt;math&gt;\$3&lt;/math&gt; box, so we opt for the &lt;math&gt;\$2&lt;/math&gt; box instead (since it has a higher popsicles/dollar ratio than the &lt;math&gt;\\$1&lt;/math&gt; pack). We're now out of money. We bought &lt;math&gt;5+5+3=13&lt;/math&gt; popsicles, so the answer is &lt;math&gt;\boxed{\textbf{(D) }13}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Carpedatdiem