https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Nezha33&feedformat=atom AoPS Wiki - User contributions [en] 2021-11-28T00:44:03Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_17&diff=136372 2018 AMC 8 Problems/Problem 17 2020-11-01T22:36:55Z <p>Nezha33: /* Solution */</p> <hr /> <div>==Problem 17==<br /> Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is &lt;math&gt;2&lt;/math&gt; miles, which is &lt;math&gt;10,560&lt;/math&gt; feet, and Bella covers &lt;math&gt;2 \tfrac{1}{2}&lt;/math&gt; feet with each step. How many steps will Bella take by the time she meets Ella?<br /> <br /> &lt;math&gt;\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since Ella rides 5 times as fast as Bella, Ella rides at a rate of &lt;math&gt;\frac{25}{2}&lt;/math&gt; or &lt;math&gt;12 \tfrac{1}{2}&lt;/math&gt;. Together, they move &lt;math&gt;15&lt;/math&gt; feet towards each other every unit. Dividing &lt;math&gt;10560&lt;/math&gt; by &lt;math&gt;15&lt;/math&gt; to find the number of steps Bella takes results in the answer of &lt;math&gt;\boxed{\textbf{(A) }704}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since Ella rides 5 times faster than Bella, the ratio of their speeds is 5:1. This means that Bella travels 1/6 of the way, and 1/6 of 10560 feet is 1760 feet. Bella also walks 2.5 feet in a step, and 1760 divided by 2.5 is &lt;math&gt;\boxed{\textbf{(A) }704}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> We know that Ella's speed is &lt;math&gt;12.5&lt;/math&gt; st/s (steps per second) and Bella's speed is &lt;math&gt;2.5&lt;/math&gt; st/s. Let the speeds in which they are walking and riding at be the slope of two equations, one for Bella and one for Ella. We make &lt;math&gt;s&lt;/math&gt; the amount of steps she takes and &lt;math&gt;d&lt;/math&gt; distance from their houses. Since we are finding where they meet, we need to solve for the intersections of the two equations for Ella and Bella. The equations we get are:<br /> <br /> &lt;cmath&gt;d = -12.5s + 10560&lt;/cmath&gt;<br /> &lt;cmath&gt;d = 2.5s&lt;/cmath&gt;<br /> <br /> Subtracting both equations from each other gives us:<br /> <br /> &lt;cmath&gt;-15s + 10560 = 0&lt;/cmath&gt;<br /> <br /> Solving that equation grants us:<br /> <br /> &lt;cmath&gt;s = 704&lt;/cmath&gt;<br /> <br /> Since we are finding the number of steps they take until they meet each other, we can stop there. The solution is <br /> <br /> &lt;math&gt;\boxed{\textbf{(A) }704}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_16&diff=136371 2018 AMC 8 Problems/Problem 16 2020-11-01T22:36:12Z <p>Nezha33: /* Solution */</p> <hr /> <div>==Problem 16==<br /> Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?<br /> <br /> &lt;math&gt;\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just &lt;math&gt;5&lt;/math&gt; factorial. Now we multiply this product by &lt;math&gt;2!&lt;/math&gt; because there are &lt;math&gt;2&lt;/math&gt; factorial ways to arrange the Arabic books within themselves, and &lt;math&gt;4!&lt;/math&gt; ways to arrange the Spanish books within themselves. Multiplying all these together, we have &lt;math&gt;2!\cdot 4!\cdot 5!=\boxed{\textbf{(C) }5760.}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_3&diff=136370 2018 AMC 8 Problems/Problem 3 2020-11-01T22:34:21Z <p>Nezha33: /* Solution */</p> <hr /> <div>==Problem 3==<br /> Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The five numbers which cause people to leave the circle are &lt;math&gt;7, 14, 17, 21,&lt;/math&gt; and &lt;math&gt;27.&lt;/math&gt;<br /> <br /> The most straightforward way to do this would be to draw out the circle with the people, and cross off people as you count.<br /> <br /> Assuming the six people start with &lt;math&gt;1&lt;/math&gt;, Arn counts &lt;math&gt;7&lt;/math&gt; so he leaves first. Then Cyd counts &lt;math&gt;14&lt;/math&gt;, as there are &lt;math&gt;7&lt;/math&gt; numbers to be counted from this point. Then Fon, Bob, and Eve, count, &lt;math&gt;17,&lt;/math&gt; &lt;math&gt;21,&lt;/math&gt; and &lt;math&gt;27&lt;/math&gt; respectively, so last one standing is Dan. Hence, the answer would be &lt;math&gt;\boxed{\text{(D) Dan}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_2&diff=136369 2018 AMC 8 Problems/Problem 2 2020-11-01T22:33:57Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 2==<br /> What is the value of the product&lt;cmath&gt;\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> ==Solution==<br /> By adding up the numbers in each of the 6 parentheses, we have: <br /> <br /> &lt;math&gt;\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}&lt;/math&gt;.<br /> <br /> Using telescoping, most of the terms cancel out diagonally. We are left with &lt;math&gt;\frac{7}{1}&lt;/math&gt; which is equivalent to &lt;math&gt;7&lt;/math&gt;. Thus the answer would be &lt;math&gt;\boxed{\textbf{(D) }7}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2018|num-b=1|num-a=3}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_1&diff=136368 2018 AMC 8 Problems/Problem 1 2020-11-01T22:33:09Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 1==<br /> An amusement park has a collection of scale models, with a ratio &lt;math&gt; 1: 20&lt;/math&gt;, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its duplicate to the nearest whole number?<br /> <br /> &lt;math&gt;\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> You can see that since the ratios of real building's heights to the model building's height is 1:20. We also know that the U.S Capitol is 289 feet in real life so to find the height of the model, we divide by 20. That gives us 14.45 which rounds to 14. So the answer is &lt;math&gt;\boxed{\textbf{(A)}14}&lt;/math&gt;. That is how I did it ~avamarora.<br /> <br /> ==Solution 2==<br /> <br /> <br /> You can just do &lt;math&gt;\frac{289}{20}&lt;/math&gt; and round your answer to get &lt;math&gt;\boxed{\textbf{(A)}14}&lt;/math&gt;.<br /> It is basically Solution 1 without the ratio calculation, which might not be necessary.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2018|before=First Problem|num-a=2}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_14&diff=136254 2019 AMC 8 Problems/Problem 14 2020-10-31T16:35:33Z <p>Nezha33: /* Solution 5 */</p> <hr /> <div>==Problem 14==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\text{Day }1&lt;/math&gt; to &lt;math&gt;\text{Day }2&lt;/math&gt; denote a day where one coupon is redeemed and the day when the second coupon is redeemed. <br /> <br /> If she starts on a &lt;math&gt;\text{Monday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Thursday}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(A)}\ \text{Monday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Tuesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(B)}\ \text{Tuesday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Wednesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Saturday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Saturday}&lt;/math&gt; to &lt;math&gt;\text{Tuesday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Tuesday}&lt;/math&gt; to &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> And on &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her last coupon. <br /> <br /> <br /> No sunday occured thus &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt; is correct. <br /> <br /> <br /> Checking for the other options, <br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(D)}\ \text{Thursday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Friday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> <br /> Checking for the other options gave us negative results, thus the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Let <br /> <br /> &lt;math&gt;Sunday \equiv 0 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Monday \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Tuesday \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Wednesday \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Thursday \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Friday \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Saturday \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;10 \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;20 \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;30 \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;40 \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;50 \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;60 \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> Which clearly indicates if you start form a &lt;math&gt;x \equiv 3 \pmod{7}&lt;/math&gt; you will not get a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Any other starting value may lead to a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Which means our answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 3 ==<br /> Like Solution 2, let the days of the week be numbers&lt;math&gt;\pmod 7&lt;/math&gt;. &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are coprime, so continuously adding &lt;math&gt;3&lt;/math&gt; to a number&lt;math&gt;\pmod 7&lt;/math&gt; will cycle through all numbers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. If a string of 6 numbers in this cycle does not contain &lt;math&gt;0&lt;/math&gt;, then if you minus 3 from the first number of this cycle, it will always be &lt;math&gt;0&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. ~~SmileKat32<br /> <br /> == Solution 4 ==<br /> Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. <br /> ~~ gorefeebuddie<br /> Note: This only works when 7 and 3 are relatively prime.<br /> <br /> == Solution 5 ==<br /> Let Sunday be Day 0, Monday be Day 1, Tuesday be Day 2, and so forth. We see that Sundays fall on Day &lt;math&gt;n&lt;/math&gt;, where n is a multiple of seven. If Isabella starts using her coupons on Monday (Day 1), she will fall on a Day that is a multiple of seven, a Sunday (her third coupon will be &quot;used&quot; on Day 21). Similarly, if she starts using her coupons on Tuesday (Day 2), Isabella will fall on a Day that is a multiple of seven (Day 42). Repeating this process, if she starts on Wednesday (Day 3), Isabella will first fall on a Day that is a multiple of seven, Day 63 (13, 23, 33, 43, 53 are not multiples of seven), but on her seventh coupon, of which she only has six. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\text{ Wednesday}}&lt;/math&gt;.<br /> <br /> == Solution 6 ==<br /> Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=13|num-a=15}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_14&diff=136253 2019 AMC 8 Problems/Problem 14 2020-10-31T16:31:48Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 14==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\text{Day }1&lt;/math&gt; to &lt;math&gt;\text{Day }2&lt;/math&gt; denote a day where one coupon is redeemed and the day when the second coupon is redeemed. <br /> <br /> If she starts on a &lt;math&gt;\text{Monday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Thursday}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(A)}\ \text{Monday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Tuesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(B)}\ \text{Tuesday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Wednesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Saturday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Saturday}&lt;/math&gt; to &lt;math&gt;\text{Tuesday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Tuesday}&lt;/math&gt; to &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> And on &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her last coupon. <br /> <br /> <br /> No sunday occured thus &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt; is correct. <br /> <br /> <br /> Checking for the other options, <br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(D)}\ \text{Thursday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Friday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> <br /> Checking for the other options gave us negative results, thus the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Let <br /> <br /> &lt;math&gt;Sunday \equiv 0 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Monday \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Tuesday \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Wednesday \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Thursday \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Friday \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Saturday \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;10 \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;20 \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;30 \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;40 \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;50 \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;60 \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> Which clearly indicates if you start form a &lt;math&gt;x \equiv 3 \pmod{7}&lt;/math&gt; you will not get a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Any other starting value may lead to a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Which means our answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 3 ==<br /> Like Solution 2, let the days of the week be numbers&lt;math&gt;\pmod 7&lt;/math&gt;. &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are coprime, so continuously adding &lt;math&gt;3&lt;/math&gt; to a number&lt;math&gt;\pmod 7&lt;/math&gt; will cycle through all numbers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. If a string of 6 numbers in this cycle does not contain &lt;math&gt;0&lt;/math&gt;, then if you minus 3 from the first number of this cycle, it will always be &lt;math&gt;0&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. ~~SmileKat32<br /> <br /> == Solution 4 ==<br /> Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. <br /> ~~ gorefeebuddie<br /> Note: This only works when 7 and 3 are relatively prime.<br /> <br /> == Solution 5 ==<br /> Let Sunday be Day 0, Monday be Day 1, Tuesday be Day 2, and so forth. We see that Sundays fall on Day &lt;math&gt;n&lt;/math&gt;, where n is a multiple of seven. If Isabella starts using her coupons on Monday (Day 1), she will fall on a Day that is a multiple of seven, a Sunday (her third coupon will be &quot;used&quot; on Day 21). Similarly, if she starts using her coupons on Tuesday (Day 2), Isabella will fall on a Day that is a multiple of seven (Day 42). Repeating this process, if she starts on Wednesday (Day 3), Isabella will first fall on a Day that is a multiple of seven, Day 63 (13, 23, 33, 43, 53 are not multiples of seven), but on her eleventh coupon, of which she only has ten. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\text{ Wednesday}}&lt;/math&gt;.<br /> <br /> == Solution 6 ==<br /> Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=13|num-a=15}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_13&diff=136252 2019 AMC 8 Problems/Problem 13 2020-10-31T16:31:05Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 13==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that the only positive 2-digit palindromes are multiples of 11, namely &lt;math&gt;11, 22, \ldots, 99&lt;/math&gt;. Since &lt;math&gt;N&lt;/math&gt; is the sum of 2-digit palindromes, &lt;math&gt;N&lt;/math&gt; is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so &lt;math&gt;N=110&lt;/math&gt; is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as &lt;math&gt;110=77+22+11&lt;/math&gt;. Then &lt;math&gt;N = 110&lt;/math&gt;, and the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;1+1+0 = \boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Associated video - https://www.youtube.com/watch?v=bOnNFeZs7S8<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=12|num-a=14}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_12&diff=136251 2019 AMC 8 Problems/Problem 12 2020-10-31T16:29:40Z <p>Nezha33: /* Note */</p> <hr /> <div>==Problem==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> [[File:2019AMC8Prob12.png]<br /> <br /> ==Solution 1==<br /> &lt;math&gt;B&lt;/math&gt; is on the top, and &lt;math&gt;R&lt;/math&gt; is on the side, and &lt;math&gt;G&lt;/math&gt; is on the right side. That means that (image &lt;math&gt;2&lt;/math&gt;) &lt;math&gt;W&lt;/math&gt; is on the left side. From the third image, you know that &lt;math&gt;P&lt;/math&gt; must be on the bottom since &lt;math&gt;G&lt;/math&gt; is sideways. That leaves us with the back, so the back must be &lt;math&gt;A&lt;/math&gt;. The front is opposite of the back, so the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.~heeeeeeeheeeee<br /> <br /> ==Solution 2==<br /> Looking closely we can see that all faces are connected with &lt;math&gt;R&lt;/math&gt; except for &lt;math&gt;A&lt;/math&gt;. Thus the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.<br /> <br /> It is A, just draw it out!<br /> ~phoenixfire<br /> <br /> ==Solution 3==<br /> Associated video - https://www.youtube.com/watch?v=K5vaX_EzjEM<br /> <br /> Video Solution- https://youtu.be/Lw8fSbX_8FU ( Also explains problems 11-20)<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=11|num-a=13}}<br /> Only two of the cubes are required to solve the problem.</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&diff=136250 2019 AMC 8 Problems/Problem 11 2020-10-31T16:27:04Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 11==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eighth graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of students taking both a math and a foreign language class.<br /> <br /> By P-I-E, we get &lt;math&gt;70 + 54 - x&lt;/math&gt; = &lt;math&gt;93&lt;/math&gt;. <br /> <br /> Solving gives us &lt;math&gt;x = 31&lt;/math&gt;.<br /> <br /> But we want the number of students taking only a math class.<br /> <br /> Which is &lt;math&gt;70 - 31 = 39&lt;/math&gt;.<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 39}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Solution 2==<br /> We have &lt;math&gt;70 + 54 = 124&lt;/math&gt; people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that &lt;math&gt;31&lt;/math&gt; people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get &lt;math&gt;70 - 31 = \boxed{\textbf{D} \, 39}&lt;/math&gt; -fath2012<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> draw(circle((-0.5,0),1));<br /> draw(circle((0.5,0),1));<br /> label(&quot;$\huge{x}$&quot;, (0, 0));<br /> label(&quot;$70-x$&quot;, (-1, 0));<br /> label(&quot;$54-x$&quot;, (1, 0));<br /> &lt;/asy&gt;<br /> <br /> We know that the sum of all three areas is &lt;math&gt;93&lt;/math&gt;<br /> So, we have: <br /> &lt;cmath&gt;93 = 70-x+x+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 70+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 124 - x&lt;/cmath&gt;<br /> &lt;cmath&gt;-39=-x&lt;/cmath&gt;<br /> &lt;cmath&gt;x=39&lt;/cmath&gt;<br /> <br /> We are looking for the number of students in only math. This is &lt;math&gt;70-x&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; with &lt;math&gt;31&lt;/math&gt;, our answer is &lt;math&gt;\boxed{39}&lt;/math&gt;.<br /> <br /> -mathnerdnair<br /> <br /> ==Solution 4==<br /> Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU, (also explains problems 11-20)<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&diff=136249 2019 AMC 8 Problems/Problem 11 2020-10-31T16:26:02Z <p>Nezha33: /* Problem 11 */</p> <hr /> <div>==Problem 11==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eighth graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of students taking both a math and a foreign language class.<br /> <br /> By P-I-E, we get &lt;math&gt;70 + 54 - x&lt;/math&gt; = &lt;math&gt;93&lt;/math&gt;. <br /> <br /> Solving gives us &lt;math&gt;x = 31&lt;/math&gt;.<br /> <br /> But we want the number of students taking only a math class.<br /> <br /> Which is &lt;math&gt;70 - 31 = 39&lt;/math&gt;.<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 39}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Solution 2==<br /> We have &lt;math&gt;70 + 54 = 124&lt;/math&gt; people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that &lt;math&gt;31&lt;/math&gt; people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get &lt;math&gt;70 - 31 = \boxed{\textbf{D} \, 39}&lt;/math&gt; -fath2012<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> draw(circle((-0.5,0),1));<br /> draw(circle((0.5,0),1));<br /> label(&quot;$\huge{x}$&quot;, (0, 0));<br /> label(&quot;$70-x$&quot;, (-1, 0));<br /> label(&quot;$54-x$&quot;, (1, 0));<br /> &lt;/asy&gt;<br /> <br /> We know that the sum of all three areas is &lt;math&gt;93&lt;/math&gt;<br /> So, we have: <br /> &lt;cmath&gt;93 = 70-x+x+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 70+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 124 - x&lt;/cmath&gt;<br /> &lt;cmath&gt;-39=-x&lt;/cmath&gt;<br /> &lt;cmath&gt;x=39&lt;/cmath&gt;<br /> <br /> We are looking for the number of students in only math. This is &lt;math&gt;70-x&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; with &lt;math&gt;31&lt;/math&gt;, our answer is &lt;math&gt;\boxed{39}&lt;/math&gt;.<br /> <br /> -mathnerdnair<br /> <br /> ==Solution 4==<br /> Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU, (also explains problems 11-20)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_9&diff=136248 2019 AMC 8 Problems/Problem 9 2020-10-31T16:23:19Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 9==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Using the formula for the volume of a cylinder, we get Alex, &lt;math&gt;\pi108&lt;/math&gt;, and Felicia, &lt;math&gt;\pi216&lt;/math&gt;. We can quickly notice that &lt;math&gt;\pi&lt;/math&gt; cancels out on both sides, and that Alex's volume is &lt;math&gt;1/2&lt;/math&gt; of Felicia's leaving &lt;math&gt;1/2 = \boxed{1:2}&lt;/math&gt; as the answer. <br /> <br /> ~aopsav<br /> <br /> ==Solution 2==<br /> <br /> Using the formula for the volume of a cylinder, we get that the volume of Alex's can is &lt;math&gt;3^2\cdot12\cdot\pi&lt;/math&gt;, and that the volume of Felicia's can is &lt;math&gt;6^2\cdot6\cdot\pi&lt;/math&gt;. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get &lt;math&gt;\frac{1}{2}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; <br /> <br /> lol this is something no one should be able to do.-(Algebruh123)2020<br /> <br /> ==Solution 3==<br /> <br /> The ratio of the numbers is &lt;math&gt;1/2&lt;/math&gt;. Looking closely at the formula &lt;math&gt;r^2 * h * \pi&lt;/math&gt;, we see that the &lt;math&gt;r * h * \pi&lt;/math&gt; will cancel, meaning that the ratio of them will be &lt;math&gt;\frac{1(2)}{2(2)}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B)}\ 1:2}&lt;/math&gt; <br /> <br /> -Lcz<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_8&diff=136247 2019 AMC 8 Problems/Problem 8 2020-10-31T16:20:25Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 8==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> After Gilda gives &lt;math&gt;20&lt;/math&gt;% of the marbles to Pedro, she has &lt;math&gt;80&lt;/math&gt;% of the marbles left. If she then gives &lt;math&gt;10&lt;/math&gt;% of what's left to Ebony, she has &lt;math&gt;(0.8*0.9)&lt;/math&gt; = &lt;math&gt;72&lt;/math&gt;% of what she had at the beginning. Finally, she gives &lt;math&gt;25&lt;/math&gt;% of what's left to her brother, so she has &lt;math&gt;(0.75*0.72)&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(E)}\ 54}&lt;/math&gt;. of what she had in the beginning left.<br /> <br /> ==Solution 2==<br /> Suppose Gilda has 100 marbles. <br /> <br /> Then she gives Pedro 20% of 100 = 20, she remains with 80 marbles.<br /> <br /> Out of 80 marbles she gives 10% of 80 = 8 to Ebony. <br /> <br /> Thus she remains with 72 marbles. <br /> <br /> Then she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54. <br /> <br /> And &lt;math&gt;\frac{54}{100}&lt;/math&gt;=54%=&lt;math&gt;\boxed{\textbf{(E)}\ 54}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> Solution 3 (Only if you have lots of time do it this way)<br /> Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So the only option that is greater than 100% - 55% is &lt;math&gt;\boxed{\textbf{(E)}\ 54}&lt;/math&gt;.<br /> <br /> ==See also== <br /> {{AMC8 box|year=2019|num-b=7|num-a=9}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_7&diff=136016 2019 AMC 8 Problems/Problem 7 2020-10-29T00:55:25Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 7==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; for the scores on the last two tests. &lt;cmath&gt;\frac{76+94+87+x+y}{5} = 81,&lt;/cmath&gt; &lt;cmath&gt;\frac{257+x+y}{5} = 81.&lt;/cmath&gt; We can now cross multiply to get rid of the denominator. &lt;cmath&gt;257+x+y = 405,&lt;/cmath&gt; &lt;cmath&gt;x+y = 148.&lt;/cmath&gt; Now that we have this equation, we will assign &lt;math&gt;y&lt;/math&gt; as the lowest score of the two other tests, and so: &lt;cmath&gt;x = 100,&lt;/cmath&gt; &lt;cmath&gt;y=48.&lt;/cmath&gt; Now we know that the lowest score on the two other tests is &lt;math&gt;\boxed{48}&lt;/math&gt;.&lt;math&gt;\frac{3}{4}&lt;/math&gt;<br /> <br /> ~ aopsav<br /> <br /> ==Solution 2==<br /> Right now, she scored &lt;math&gt;76, 94,&lt;/math&gt; and &lt;math&gt;87&lt;/math&gt; points, with a total of &lt;math&gt;257&lt;/math&gt; points. She wants her average to be &lt;math&gt;81&lt;/math&gt; for her &lt;math&gt;5&lt;/math&gt; tests so she needs to score &lt;math&gt;405&lt;/math&gt; points in total. She needs to score a total of &lt;math&gt;(405-257) <br /> 148&lt;/math&gt; points in her &lt;math&gt;2&lt;/math&gt; tests. So the minimum score she can get is when one of her &lt;math&gt;2&lt;/math&gt; scores is &lt;math&gt;100&lt;/math&gt;. So the least possible score she can get is &lt;math&gt;\boxed{\textbf{(A)}\ 48}&lt;/math&gt;. &lt;math&gt;3.1415926&lt;/math&gt;<br /> <br /> <br /> Note: You can verify that &lt;math&gt;\boxed{48}&lt;/math&gt; is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer.<br /> <br /> ==Solution 3==<br /> We can compare each of the scores with the average of &lt;math&gt;81&lt;/math&gt;:<br /> &lt;math&gt;76&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;-5&lt;/math&gt;,<br /> &lt;math&gt;94&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+13&lt;/math&gt;,<br /> &lt;math&gt;87&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+6&lt;/math&gt;,<br /> &lt;math&gt;100&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;19&lt;/math&gt;;<br /> <br /> So the last one has to be &lt;math&gt;-33&lt;/math&gt; (since all the differences have to sum to &lt;math&gt;0&lt;/math&gt;), which corresponds to &lt;math&gt;81-33 = \boxed{48}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=6|num-a=8}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_6&diff=136015 2019 AMC 8 Problems/Problem 6 2020-10-29T00:54:23Z <p>Nezha33: /* Also See */</p> <hr /> <div>== Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> draw((0,4)--(3,4));<br /> draw((0,8)--(3,5));<br /> draw((8,8)--(5,5));<br /> draw((4,8)--(4,5));<br /> draw((4,0)--(4,3));<br /> draw((0,0)--(3,3));<br /> draw((8,0)--(5,3));<br /> draw((5,4)--(8,4));<br /> &lt;/asy&gt;<br /> Lines of symmetry go through point &lt;math&gt;P&lt;/math&gt;, and there are &lt;math&gt;8&lt;/math&gt; directions the lines could go, and there are &lt;math&gt;4&lt;/math&gt; dots at each direction.&lt;math&gt;\frac{4\times8}{80}=\boxed{\textbf{(C)} \frac{2}{5}}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=5|num-a=7}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=136013 2019 AMC 8 Problems/Problem 5 2020-10-29T00:52:50Z <p>Nezha33: /* See Also */</p> <hr /> <div>== Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> <br /> [[File:2019_AMC_8_-4_Image_1.png|900px]]<br /> <br /> [[File:2019_AMC_8_-4_Image_2.png|600px]]<br /> <br /> ==Solution 1==<br /> First, the tortoise walks at a constant rate, ruling out &lt;math&gt;(D)&lt;/math&gt;<br /> Second, when the hare is resting, the distance will stay the same, ruling out &lt;math&gt;(E)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt;.<br /> Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> Therefore, the answer &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt; is the only one left.<br /> <br /> &lt;math&gt;\phantom{Note to the original author of this solution: &quot;we shouldn't be able to edit&quot; is incorrect (if its definition is what I think it is), because I was able to edit. Also, I deleted that, (but did nothing else) }&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> First, we know that the rabbit beats the tortoise in the first half of the race. So he is going to be ahead of the tortoise. We also know, while he rested, he didn't move. The only graph portraying that is going to be &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;. This is our answer. ~bobthefam<br /> <br /> Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&amp;t=2s<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_2&diff=136012 2019 AMC 8 Problems/Problem 2 2020-10-29T00:50:56Z <p>Nezha33: /* See also */</p> <hr /> <div>=Problem 2=<br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution 1==<br /> <br /> We know that the length of the shorter side of the 3 identical rectangles are all 5 so we can use that by seeing that the longer side of the right rectangle is the same as 2 of the shorter sides of the other 2 left rectangles. This means that &lt;math&gt;2\cdot{5}\ = 10&lt;/math&gt; which is the longer side of the right rectangle, and because all the rectangles are congruent, we see that each of the rectangles have a longer side of 10 and a shorter side of 5. Now the bigger rectangle has a shorter length of 10(because the shorter side of the bigger rectangle is the bigger side of the shorter rectangle, which is 10) and so the bigger side of the bigger rectangle is the bigger side of the smaller rectangle + the smaller side of the smaller rectangle, which is &lt;math&gt;10 + 5 = 15&lt;/math&gt; . Thus, the area is &lt;math&gt;15\cdot{10}\ = 150&lt;/math&gt; for choice &lt;math&gt;\boxed{\textbf{(E)}\ 150}&lt;/math&gt; ~~Saksham27<br /> <br /> ==Solution 2==<br /> Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is &lt;math&gt;5 \cdot 2 = 10&lt;/math&gt;. So the area of the identical rectangles is &lt;math&gt;5 \cdot 10 = 50&lt;/math&gt;. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is &lt;math&gt;50 \cdot 3 = \boxed{\textbf{(E)}\ 150}&lt;/math&gt;. ~~fath2012<br /> <br /> ==Solution 3==<br /> We see that if the short sides are 5, the long side has to be &lt;math&gt;5\cdot2=10&lt;/math&gt; because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle &lt;math&gt;ABCD&lt;/math&gt;)<br /> is &lt;math&gt;10+5=15&lt;/math&gt; because long side + short side of the small rectangle is &lt;math&gt;15&lt;/math&gt;. The short side of rectangle &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;10&lt;/math&gt; because it is the long side of the short rectangle. Multiplying &lt;math&gt;15&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt; together gets us &lt;math&gt;15\cdot10&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(E)}\ 150}&lt;/math&gt;.<br /> ~~mathboy282<br /> <br /> Video Solution (also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&amp;t=2s<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=1|num-a=3}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_4&diff=136011 2019 AMC 8 Problems/Problem 4 2020-10-29T00:50:13Z <p>Nezha33: /* See Also */</p> <hr /> <div>== Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> <br /> == Solution 1 ==<br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,5)--(12,0));<br /> draw((12,0)--(0,-5));<br /> draw((0,-5)--(-12,0));<br /> draw((0,0)--(12,0));<br /> draw((0,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,-5));<br /> dot((-12,0));<br /> dot((0,5));<br /> dot((12,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(12,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> label(&quot;E&quot;,(0,0),SW);<br /> &lt;/asy&gt;<br /> <br /> A rhombus has sides of equal length. Because the perimeter of the rhombus is &lt;math&gt;52&lt;/math&gt;, each side is &lt;math&gt;\frac{52}{4}=13&lt;/math&gt;. In a rhombus, diagonals are perpendicular and bisect each other, which means &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt; = &lt;math&gt;\overline{EC}&lt;/math&gt;.<br /> <br /> Consider one of the right triangles:<br /> <br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,5));<br /> dot((-12,0));<br /> dot((0,5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;E&quot;,(0,0),SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\overline{AB}&lt;/math&gt; = &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt;. Using Pythagorean theorem, we find that &lt;math&gt;\overline{BE}&lt;/math&gt; = &lt;math&gt;5&lt;/math&gt;.<br /> &quot;You may recall the famous pythagorean triple, (5, 12, 13), that's how I did it&quot;-Zack2008<br /> <br /> Thus the values of the two diagonals are &lt;math&gt;\overline{AC}&lt;/math&gt; = &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; = &lt;math&gt;10&lt;/math&gt;.<br /> The area of a rhombus is = &lt;math&gt;\frac{d_1\cdot{d_2}}{2}&lt;/math&gt; = &lt;math&gt;\frac{24\cdot{10}}{2}&lt;/math&gt; = &lt;math&gt;120&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 120}&lt;/math&gt; ~phoenixfire<br /> <br /> Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&amp;t=2s<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=3|num-a=5}}<br /> <br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=136010 2019 AMC 8 Problems/Problem 3 2020-10-29T00:48:25Z <p>Nezha33: /* See Also */</p> <hr /> <div>==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> ==Solution 1==<br /> Each one is in the form &lt;math&gt;\frac{x+4}{x}&lt;/math&gt; so we are really comparing &lt;math&gt;\frac{4}{11}, \frac{4}{15},&lt;/math&gt; and &lt;math&gt;\frac{4}{13}&lt;/math&gt; where you can see &lt;math&gt;\frac{4}{11}&gt;\frac{4}{13}&gt;\frac{4}{15}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We take a common denominator:<br /> &lt;cmath&gt;\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> ~ dolphin7 - I took your idea and made it an explanation.<br /> <br /> ==Solution 3==<br /> When &lt;math&gt;\frac{x}{y}&gt;1&lt;/math&gt; and &lt;math&gt;z&gt;0&lt;/math&gt;, &lt;math&gt;\frac{x+z}{y+z}&lt;\frac{x}{y}&lt;/math&gt;. Hence, the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> ~ ryjs<br /> <br /> This is also similar to Problem 20 on the AMC 2012.<br /> <br /> ==Solution 4(probably won't use this solution)==<br /> We use our insane mental calculator to find out that &lt;math&gt;\frac{15}{11} \approx 1.36&lt;/math&gt;, &lt;math&gt;\frac{19}{15} \approx 1.27&lt;/math&gt;, and &lt;math&gt;\frac{17}{13} \approx 1.31&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ~~ by an insane math guy<br /> <br /> ==Solution 5==<br /> Suppose each fraction is expressed with denominator &lt;math&gt;2145&lt;/math&gt;: &lt;math&gt;\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}&lt;/math&gt;. Clearly &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}} The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_1&diff=136009 2019 AMC 8 Problems/Problem 1 2020-10-29T00:42:55Z <p>Nezha33: /* Solution 1 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We know that the sandwiches cost &lt;math&gt;4.50&lt;/math&gt; dollars. Guessing will bring us to multiplying &lt;math&gt;4.50&lt;/math&gt; by 6, which gives us &lt;math&gt;27.00&lt;/math&gt;. Since they can spend &lt;math&gt;30.00&lt;/math&gt; they have &lt;math&gt;3&lt;/math&gt; dollars left. Since sodas cost &lt;math&gt;1.00&lt;/math&gt; dollar each, they can buy 3 sodas, which makes them spend &lt;math&gt;30.00&lt;/math&gt; Since they bought 6 sandwiches and 3 sodas, they bought a total of &lt;math&gt;9&lt;/math&gt; items. Therefore, the answer is &lt;math&gt;\boxed{D = 9 }&lt;/math&gt;<br /> <br /> ==Solution 2 (Using Algebra)==<br /> Let &lt;math&gt;s&lt;/math&gt; be the number of sandwiches and &lt;math&gt;d&lt;/math&gt; be the number of sodas. We have to satisfy the equation of<br /> &lt;cmath&gt;4.50s+d=30&lt;/cmath&gt;<br /> In the question, it states that Ike and Mike buys as many sandwiches as possible. <br /> So, we drop the number of sodas for a while.<br /> We have: <br /> &lt;cmath&gt;4.50s=30&lt;/cmath&gt;<br /> &lt;cmath&gt;s=\frac{30}{4.5}&lt;/cmath&gt;<br /> &lt;cmath&gt;s=6R30&lt;/cmath&gt;<br /> We don't want a remainder so the maximum number of sandwiches is &lt;math&gt;6&lt;/math&gt;.<br /> The total money spent is &lt;math&gt;6\cdot 4.50=27&lt;/math&gt;.<br /> The number of dollar left to spent on sodas is &lt;math&gt;30-27=3&lt;/math&gt; dollars.<br /> &lt;math&gt;3&lt;/math&gt; dollars can buy &lt;math&gt;3&lt;/math&gt; sodas leading us to a total of <br /> &lt;math&gt;6+3=9&lt;/math&gt; items. <br /> Hence, the answer is &lt;math&gt;\boxed{(D) = 9}&lt;/math&gt;<br /> <br /> -by interactivemath<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|before=First Problem|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Nezha33 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_5&diff=135430 2019 AMC 8 Problems/Problem 5 2020-10-20T00:02:49Z <p>Nezha33: /* Also see */</p> <hr /> <div>== Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> <br /> [[File:2019_AMC_8_-4_Image_1.png|900px]]<br /> <br /> [[File:2019_AMC_8_-4_Image_2.png|600px]]<br /> <br /> ==Solution 1==<br /> First, the tortoise walks at a constant rate, ruling out &lt;math&gt;(D)&lt;/math&gt;<br /> Second, when the hare is resting, the distance will stay the same, ruling out &lt;math&gt;(E)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt;.<br /> Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out &lt;math&gt;(A)&lt;/math&gt;.<br /> Therefore, the answer &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt; is the only one left.<br /> <br /> &lt;math&gt;\phantom{Note to the original author of this solution: &quot;we shouldn't be able to edit&quot; is incorrect (if its definition is what I think it is), because I was able to edit. Also, I deleted that, (but did nothing else) }&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> First, we know that the rabbit beats the tortoise in the first half of the race. So he is going to be ahead of the tortoise. We also know, while he rested, he didn't move. The only graph portraying that is going to be &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;. This is our answer. ~bobthefam<br /> <br /> Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&amp;t=2s<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Nezha33