2011 AMC 12B Problems Solution

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2006年AMC8试题及答案

Copyright © 2016 Art of Problem Solving Mindy made three purchases for dollars, dollars, and dollars. What was her total, tothe nearest dollar?The three prices round to , , and , which has a sum of2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byFirstQuestion Followed by Problem 21 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_1&oldid=55964"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingOn the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn'tanswer the last 5. What is his score?As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly, .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 1Followed by Problem 31 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_2&oldid=55965"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingElisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now, she canfinish 12 laps in 24 minutes. By how many minutes has she improved her lap time?When Elisa started, she finished a lap in minutes. Now, she finishes a lap is minutes.The difference is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 2Followed by Problem 41 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_3&oldid=55966"SolutionSee AlsoCopyright © 2016 Art of Problem Solving Initially, a spinner points west. Chenille moves it clockwise revolutions and then counterclockwiserevolutions. In what direction does the spinner point after the two moves?If the spinner goes clockwise revolutions and then counterclockwise revolutions, it ultimately goes counterclockwise which brings the spinner pointing.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 3Followed by Problem 51 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_4&oldid=55967"SolutionSee AlsoPoints and are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?Drawing segments and , the number of triangles outside square is the same as the number of triangles inside the square. Thus areas must be equal so the area of is half the area of thelarger square which is .If the side length of the larger square is , the side length of the smaller square is . Therefore the area of the smaller square is , half of the larger square's area, .Thus, the area of the smaller square in the picture is .Copyright © 2016 Art of Problem Solving The letter T is formed by placing twoinch rectangles next to each other, as shown. What is theperimeter of the T, in inches?If the two rectangles were seperate, the perimeter would be . It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 5Followed by Problem 71 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_6&oldid=55969"SolutionSee AlsoCopyright © 2016 Art of Problem Solving Circle has a radius of . Circle has a circumference of . Circle has an area of . List thecircles in order from smallest to largest ing the formulas of circles, and , we find that circle has a radius of and circle has a radius of . Thus, the order from smallest to largest radius is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 6Followed by Problem 81 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21• 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_7&oldid=55970"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingThe table shows some of the results of a survey by radiostation KAMC. What percentage of the males surveyedlisten to the station?Filling out the chart, it becomes Thus, the percentage of males surveyed that listen to the station is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 7Followed by Problem 91 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_8&oldid=55971"SolutionSee AlsoCopyright © 2016 Art of Problem Solving What is the product of?By telescoping, it's easy to see the sum becomes .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 8Followed by Problem 101 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_9&oldid=55972"SolutionSee AlsoJorge's teacher asks him to plot all the ordered pairs of positive integers for which is the width and is the length of a rectangle with area 12. What should his graph look like?Copyright © 2016 Art of Problem SolvingThe length of the rectangle will relate invertly to the width, specifically using the theorem . Theonly graph that could represent a inverted relationship is . (The rest are linear graphs thatrepresent direct relationships)2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 9Followed by Problem 111 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_10&oldid=55973"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingHow many two-digit numbers have digits whose sum is a perfect square?There is integer whose digits sum to : .There are integers whose digits sum to : .There are integers whose digits sum to : .There are integers whose digits sum to :.Two digits cannot sum toor any greater square since the greatest sum of digits of a two-digit number is.Thus, the answer is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 10Followed by Problem 121 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_11&oldid=80944"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingAntonette gets on a 10-problem test, on a 20-problem test and on a 30-problem test. Ifthe three tests are combined into one 60-problem test, which percent is closest to her overall score?Adding them up gets. The overall percentage correct would be.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 11Followed by Problem 131 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_12&oldid=55975"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingCassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time inthe morning do they meet?If Cassie leavesan hour earlier then Brian, when Brian starts, the distance between them will be. Every hour, they will getmiles closer., so 2 hours from 9:00AM is when they meet, which is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 12Followed by Problem 141 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_13&oldid=55976"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingProblems 14, 15 and 16 involve Mrs. Reed's English assignment.A Novel AssignmentThe students in Mrs. Reed's English class are reading the same -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in seconds and Chandra reads a page inseconds.If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?The information is the same for Problems 14,15, and 16. Therefore, we shall only use the information we need. All we need for this problem is that there's 760 pages, Bob reads a page in 45 seconds and Chandrareads a page in 30 seconds. A lot of people will find how long it takes Bob to read the book, how long it takes Chandra to read the book, and then find the seconds. However, if we just set up the expression, we can find an easier way.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 13Followed by Problem 151 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_14&oldid=55977"SolutionSee AlsoLet be the number of pages that Chandra reads.Distribute theAdd to both sidesDivide both sides by to make it easier to solveDivide both sides bySolution 2Bob and Chandra read at a rate of seconds per page, respectively. Simplifying that gets us Bob reads pages for every pages that Chandra reads. Therefore Chandra should read of the book. =$\boxed{\textbf{(C)} 456}See AlsoCopyright © 2016 Art of Problem SolvingProblems 14, 15 and 16 involve Mrs. Reed's English assignment.A Novel AssignmentThe students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have toread?The amount of pages Bob, Chandra, and Alice would read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading in the same amount of time because the ratio of pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 15Followed by Problem 171 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_16&oldid=55979"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingJeff rotates spinners ,andand adds the resulting numbers. What is the probability that his sum isan odd number?In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even +Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner contains only even numbers.Therefore we must work with Even + Even + Odd and spinner will give us one of our even numbers. We also see that spinner only contains odd, so spinner must give us our odd nmber. We still need one evennumber from spinner . There is only 1 even number: . Since spinning the required numbers are automatic onthe other spinners, we only have to find the probability of spinning a in spinner , which clearly is2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 16Followed by Problem 181 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24• 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_17&oldid=55980"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingA cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?The surface area of the cube is . Each of the eight black cubes has 3 faces on the outside,making black faces. Therefore there are white faces. To find the probability,we evaluate.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 17Followed by Problem 191 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 •25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_18&oldid=55981"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingTriangleis an isosceles triangle with . Pointis the midpoint of bothand, and is 11 units long. Triangleis congruent to triangle . What is the length of?Since triangleis congruent to triangleand ,. Since,. Because pointis the midpoint of,.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded by Problem 18Followed by Problem 201 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19• 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_19&oldid=55982"SolutionSee AlsoA singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how manygames did Monica win?Since there are 6 players, a total ofgames are played. So far,games finished (one person won from each game), so Monica needs to win.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 19Followed byProblem 211 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 •20• 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'sAmerican Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_20&oldid=81054"SolutionSee AlsoCopyright © 2016 Art of Problem SolvingCopyright © 2016 Art of Problem Solving An aquarium has a rectangular base that measures cm by cm and has a height ofcm. The aquarium is filled with water to a depth of cm. A rock with volume is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?The water level will rise cm for every . Since is of , the waterwill rise 2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 20Followed by Problem 221 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22• 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_21&oldid=55984"SolutionSee AlsoThree different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process toobtain a number in the top cell. What is the difference between the largest and smallest numbers possible inthe top cell?If the lower cells contain and , then the second row will contain and , and the top cell will contain . To obtain the smallest sum, place in the center cell and and in the outer ones. The top number will be . For the largest sum, place in the center cell and and in the outer ones. This top number will be . The difference is .2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 21Followed by Problem 231 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 •22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_22&oldid=55985"SolutionSee AlsoThe counting numbers that leave a remainder of 4 when divided by 6 are The counting numbers that leave a remainder of 3 when divided by 5 are So 28 isthe smallest possible number of coins that meets both conditions. Because , there areIf there were two more coins in the box, the number of coins would be divisible by both 6 and 5. The smallest number that is divisible by 6 and 5 is , so the smallest possible number of coins in the box isand the remainder when divided by 7 is .The problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Copyright © 2016 Art of Problem Solving In the multiplication problem below , , , and are different digits. What is?, so . Therefore, and , so.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 23Followed by Problem 251 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 •24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_24&oldid=68712"SolutionSee AlsoCopyright © 2016 Art of Problem Solving Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides areprime numbers. What is the average of the hidden prime numbers?Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be anodd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is . Thus, the first card's hidden number is , and the last card's hidden number is .Since the sum of the hidden primes is, the average of the primes is.2006 AMC 8 (Problems • Answer Key • Resources(/Forum/resources.php?c=182&cid=42&year=2006))Preceded byProblem 24Followed by Last Problem1 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 •19 • 20 • 21 • 22 • 23 • 24 • 25All AJHSME/AMC 8 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America ()'s American Mathematics Competitions ().Retrieved from "/wiki/index.php?title=2006_AMC_8_Problems/Problem_25&oldid=55988"SolutionSee Also。

2004 AMC 12A Problems and Solution

2004 AMC 12A Problems and Solution Problem 1Alicia earns dollars per hour, of which is deducted to pay local taxes. Howmany cents per hour of Alicia's wages are used to pay local taxes?Solution20 dollars is the sam e as 2000 cents, and of 2000 iscents .Problem 2On the AMC 12, each correct answer is worth points, each incorrect answer isworth points, and each problem left unanswered is worth points. If Charlynleaves of the problems unanswered, how m any of the rem aining problemsmust she answer correctly in order to score at least ?SolutionShe gets points for the problems she didn't answer. She must getproblems right to score at least 100.Problem 3For how m any ordered pairs of positive integers is ?SolutionEvery integer value of leads to an integer solution for Since must be positive,Also, Since must be positive,This leaves values for y, which m ean there are solutions to theequationProblem 4Bertha has daughters and no sons. Som e of her daughters have daughters, andthe rest have none. Bertha has a total of daughters and granddaughters, and nogreat-granddaughters. How many of Bertha's daughters and grand-daughters have no children?SolutionSince Bertha has 6 daughters, Bertha has granddaughters, of whichnone have daughters. Of Bertha's daughters, have daughters, sodo not have daughters.Therefore, of Bertha's daughters and granddaughters, do not havedaughters .ORDraw a tree diagram and see that the answer can be found in the sum of 6 + 6 granddaughters, 5 + 5 daughters, and 4 more daughters.Problem 5The graph of the line is shown. Which of the following is true?SolutionIt looks like it has a slope of and is shifted up.Problem 6Let , , , ,and . Which of the following is the largest?SolutionAfter comparison, is the largest.Problem 7A gam e is played with tokens according to the following rules. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The gam e ends when some player r uns out oftokens. Players , and start with , and tokens, respectively. Howmany rounds will there be in the game?SolutionLook at a set of 3 rounds, where the players have , , and tokens. Each of the players will gain two tokens from the others and give away 3 tokens, so overall, each player will lose 1 token.Therefore, after 12 sets of 3 rounds, or 36 rounds, the players will have 3, 2, and 1 tokens, repectively. After 1 m ore round, player will give away his last 3 tokensand the gam e will stop .Problem 8In the overlapping triangles and sharing common side ,and are right angles, , , , and andintersect at . What is the difference between the areas of and?SolutionSolution 1Since and , . By alternate interior angles and AA~,we find that , with side length ratio . Their heights also havethe sam e ratio, and since the two heights add up to , we have thatand . Subtracting the areas,.Solution 2Let represent the area of figure . Note thatand ..Problem 9A com pany sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars would increase sales. If the diam eter of the jars is increased bywithout altering the volume, by what percent m ust the height be decreased?SolutionWhen the diam eter is increased by , it is increased by , so the area of the baseis increased by .To keep the volum e the sam e, the height m ust be of the original height,which is a reduction .Problem 10The sum of consecutive integers is . What is their m edian?SolutionThe m edian of a sequence is the m iddle number of the sequence when t he sequence is arranged in order. Since the integers are consecutive, the m edian is also themean, so the m edian is .Problem 11The average value of all the pennies, nickels, dimes, and quarters in Paula's purse iscents. If she had one m ore quarter, the average value would be cents. Howmany dimes does she have in her purse?SolutionSolution 1Let the total value (in cents) of the coins Paula has originally be , and the numberof coins she has be . Then and . Substitutingyields . It is easy to see now that Paulahas 3 quarters, 1 nickel, so she has dimes.Solution 2If the new coin was worth 20 cents, adding it would not change the m ean at all. The additional 5 cents raise the m ean by 1, thus the new number of coins m ust be 5.Therefore initially there were 4 coins worth a total of cents. As in the previous solution, we conclude that the only way to get 80 cents using 4 coins is 25+25+25+5.Problem 12Let and . Points and are on the line , andand intersect at . What is the length of ?SolutionThe equation of can be found using points to be. Similarily, has the equation. These two equations intersectthe line at and . Using the distance formula or righttriangles, the answer is .Problem 13Let be the set of points in the coordinate plane, where each of and maybe , , or . How many distinct lines pass through at least two m embers of ?SolutionSolution 1Let's count them by cases:▪Case 1: The line is horizontal or vertical, clearly .▪Case 2: The line has slope , with through and additional ones one unit above or below those. These total .▪Case 3: The only remaining lines pass through two points, a vertex and a non-vertex point on the opposite side. Thus we have each vertex pairing up with twopoints on the two opposites sides, giving lines.These add up to .Solution 2There are ways to pick two points, but we've clearly overcounted all of the lines which pass through three points. In fact, each line which passes throughthree points will have been counted tim es, so we have to subtract foreach of these lines. Quick counting yields horizontal, vertical, and diagonallines, so the answer is distinct lines.Problem 14A sequence of three real numbers forms an arithmetic progression with a first term of . If is added to the second term and is added to the third term, the threeresulting numbers form a geometric progression. What is the sm allest possible value for the third term in the geom etric progression?SolutionLet be the common difference. Thenare the term s of the geom etric progression. Since the m iddle term is the geom etric m ean of the other two term s,. The sm allestpossible value occurs when , and the third term is .Problem 15Brenda and Sally run in opposite directions on a circular track, starting atdiametrically opposite points. They first m eet after Brenda has run meters. Theynext m eet after Sally has run meters past their first m eeting point. Each girl runs at a constant speed. What is the length of the track in m eters?SolutionSolution 1Call the length of the race track . When they m eet at the first m eeting point,Brenda has run m eters, while Sally has run meters. By the secondmeeting point, Sally has run meters, while Brenda has run meters.Since they run at a constant speed, we can set up a proportion:. Cross-m ultiplying, we get that .Solution 2The total distance the girls run between the start and the first m eeting is one half of the track length.The total distance they run between the two m eetings is the track lengt h.As the girls run at constant speeds, the interval between the m eetings is twice as long as the interval between the start and the first m eeting.Thus between the m eetings Brenda will run meters. Therefore thelength of the track is metersProblem 16The set of all real numbers for whichis defined is . What is the value of ?SolutionWe know that the dom ain of , where is a constant, is . So. By the definition of logarithms, we then have. Then and.Problem 17Let be a function with the following properties:, and, for any positive integer .What is the value of ?Solution. Problem 18Square has side length . A sem icircle with diameter is constructedinside the square, and the tangent to the semicircle from intersects side at .What is the length of ?SolutionSolution 1Let the point of tangency be . By the TwoTangent Theorem and . Thus . The Pythagorean Theorem on yieldsHence .Solution 2Clearly, . Thus, the sides of right triangle are in arithmeticprogression. Thus it is similar to the triangle and since ,.Problem 19Circles and are externally tangent to each other, and internally tangent tocircle . Circles and are congruent. Circle has radius and passes throughthe center of . What is the radius of circle ?SolutionSolution 1Note that since is the center of the larger circle of radius . Usingthe Pythagorean Theorem on ,Now using the Pythagorean Theorem on ,Substituting ,Solution 2We can apply Descartes' Circle Formula.The four circles have curvatures , and .We haveSimplifying, we getProblem 20Select numbers and between and independently and at random, and let betheir sum. Let and be the results when and , respectively, are roundedto the nearest integer. What is the probability that ?SolutionSolution 1Casework:1.. The probability that and is . Notice thatthe sum ranges from to with a symmetric distribution across, and we want . Thus the chance is .2.. The probability that and is , but now, which m akes autom atically. Hence the chanc e is.3.. This is the sam e as the previous case.4.. We recognize that this is equivalent to the first case.Our answer is .Solution 2Use areas to deal with this continuous probability problem. Set up a unit square with values of on x-axis and on y-axis.If then this will work because . Similarly ifthen this will work because in order for this to happen, and are eachgreater than making , and . Each of these triangles in theunit square has area of 1/8.The only case left is when . Then each of and must be 1 and 0, in anyorder. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.Then the area producing the desired result is 3/4. Since the area of the unit squareis 1, the probability is .Problem 21If , what is the value of ?SolutionThis is an infinite geom etric series, which sum s to. Using the formula.Problem 22Three m utually tangent spheres of radius rest on a horizontal plane. A sphere of radius rests on them. What is the distance from the plane to the top of the largersphere?SolutionThe height from the center of the bottom sphere to the plane is , and from the center of the top sphere to the tip is . We now need the vertical height of thecenters. If we connect the centers, we get a triangular pyramid with an equilateral triangle base. The distance from the vertex of the equilateral triangle to its centroidcan be found by s to be .By the Pythagorean Theorem, we have . Addingthe heights up, we get .Problem 23A polynomialhas real coefficients with and distinct com plex zeroes, with and real, , andWhich of the following quantities can be a nonzero number?SolutionWe have to evaluate the answer choices and use process of elimination:▪: We are given that , so . If one of the roots is zero,then .▪: By Vieta's formulas, we know that is the sum of all of the rootsof . Since that is real, , and , so .▪: All of the coefficients are real. For sake of contradiction suppose none ofare zero. Then for each complex root , its com plex conjugateis also a root. So the roots should pair up, but we have an odd num ber of im aginary roots! This gives us the contradiction, and therefore the product is equal to zero.▪: We are given that . Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of com plex conjugates is zero. Hence the RHS is zero.There is, however, no reason to believe that should be zero (in fact, thatquantity is , and there is no evidence that is a root of ).Problem 24A plane contains points and with . Let be the union of all disks ofradius in the plane that cover . What is the area of ?SolutionAs the red circles m ove about segment , they cover the area we are looking for.On the left side, the circle must m ove around pivoted on . On the right side, thecircle m ust m ove pivoted on However, at the top and bottom, the circle must lieon both A and B, giving us our upper and lower bounds.This egg-like shape is .The area of the region can be found by dividing it into several sectors, nam elyProblem 25For each integer , let denote the base-number . The productcan be expressed as , where and are positive integers and is as sm all as possible. What is the value of ?SolutionThis is an infinite geom etric series with common ratio and initial term, so.Alternatively, we could have used the algebraic m anipulation for repeatingdecim als,Telescoping,Som e factors cancel, (after all, isn't one of the answer choices)Since the only factor in the num erator that goes into is , is minimized.Therefore the answer is .。

2013 AMC 10B Problems

2013 AMC 10B ProblemsProblem 1 What is642531531642++++-++++? （A ）-1 （B ）365 （C ）127 （D ）2049 （E ）343 Problem 2Mr. Green measures his rectangular garden by walking two of the sides and finding that it is 15 steps by 20 steps. Each of Mr. Green's steps is 2 feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?（A ）600 （B ）800 （C ）1000 （D ）1200 （E ）1400 Problem 3On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and the low temperatures was 3º. In degrees, what was the low temperature in Lincoln that day?（A ）-13 （B ）-8 （C ）-5 （D ）-3 （E ）11Problem 4When counting from 3 to 201, 53 is the st 51 number counted. When counting backwards from 201 to 3, 53 is the th n number counted. What is ?（A ）146 （B ）147 （C ）148 （D ）149 （E ）150 Problem 5Positive integers and are each less than 6. What is the smallest possible value forab a -2 ?（A ）-20 （B ）-15 （C ）-10 （D ）0 （E ）2Problem 6The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?（A ）22 （B ）23.25 （C ）24.75 （D ）26.25 （E ）28 Problem 7Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?（A ）33 （B ）23 （C ）1 （D ）2 （E ）2Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?（A ）10 （B ）16 （C ）25 （D ）30 （E ）40Problem 9Three positive integers are each greater than 1, have a product of 27000, and are pairwise relatively prime. What is their sum?（A ）100 （B ）137 （C ）156 （D ）160 （E ）165Problem 10A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?（A ）10 （B ）15 （C ）20 （D ）25 （E ）30Problem 11Real numbers x and y satisfy the equation 3461022--=+y x y x . What is y x + ?（A ）1 （B ）2 （C ）3 （D ）6 （E ）8Problem 12Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length?（A ）52 （B ）94 （C ）21 （D ）95 （E ）54 Problem 13Jo and Blair take turns counting from 1 to one more than the last number said by the other person. Jo starts by saying "1", so Blair follows by saying "1,2" . Jo then says "1,2,3" , and so on. What is the 53rd number said?（A ）2 （B ）3 （C ）5 （D ）6 （E ）8Problem 14Define. Which of the following describes the set of points forwhich ? （A ）a finite set of points （B ）one line （C ）two parallel lines（D ）two intersecting lines （E ）three linesA wire is cut into two pieces, one of length and the other of length . The piece oflength is bent to form an equilateral triangle, and the piece of length is bent to form aregular hexagon. The triangle and the hexagon have equal area. What is a/b?（A ）1 （B ）26 （C ）3 （D ）2 （E ）223 Problem 16In triangle ABC , medians AD and C E intersect at P , PE =1.5, PD =2, and DE =2.5. What is the area of AEDC ?（A ）13 （B ）13.5 （C ）14 （D ）14.5 （E ）15Problem 17Alex has 75 red tokens and 75 blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?（A ）62 （B ）82 （C ）83 （D ）102 （E ）103Problem 18The number 2013 has the property that its units digit is the sum of its other digits, that is 2+0+1=3. How many integers less than 2013 but greater than 1000 share this property?（A ）33 （B ）34 （C ）45 （D ）46 （E ）58Problem 19The real numbers a b c ,,form an arithmetic sequence with 0≥≥≥c b a . The quadratic c bx ax ++2 has exactly one root. What is this root?（A ）347-- （B ）32-- （C ）-1 （D ）32+- （E ）347+- Problem 20The number 2013 is expressed in the form !!...!!!...!20132121n m b b b a a a =, where m a a a ≥≥≥...21 And n b b b ≥≥≥...21 are positive integers and 11b a + is as small as possible. What is 11b a - ?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is . What is the smallestpossible value of N?（A ）55 （B ）89 （C ）104 （D ）144 （E ）273Problem 22The regular octagon ABCDEFGH has its center at J . Each of the vertices and the center are to be associated with one of the digits 1 through 9, with each digit used once, in such a way that the sums of the numbers on the lines AJE, BJF , CJG , and DJH are all equal. In how many ways can this be done?（A ）384 （B ）576 （C ）1152 （D ）1680 （E ）3456Problem 23In triangle ABC , AB =13, BC =14, and CA =15. Distinct points D, E , and F lie on segments CA BC ,, and DE , respectively, such that ,,AC DE BC AD ⊥⊥ and BF AF ⊥. The length of segment DF can be written as n m /, where and are relatively primepositive integers. What is n m + ?（A ）18 （B ）21 （C ）24 （D ）27 （E ）30Problem 24A positive integer is nice if there is a positive integerwith exactly four positive divisors(including 1 and ) such that the sum of the four divisors is equal to . How manynumbers in the set {2010, 2011, 2012, …, 2019} are nice?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5Problem 25Bernardo chooses a three-digit positive integer N and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer S . For example, if N =749, Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains the sum S =13,689. For how many choices of N are the two rightmost digits of S , in order, the same as those of 2N ?（A ）5 （B ）10 （C ）15 （D ）20 （E ）252013 AMC 10B SolutionsProblem 1This expression is equivalent to 127129912=- Problem 2The following problem is from both the 2013 AMC 12B #2 and 2013 AMC 10B #2Since each step is 2 feet, his garden is 30 by 40 feet. Thus, the area of 30(40)=1200 square feet. Since he is expecting 1/2 of a pound per square foot, the totalamount of potatoes expected is 1200*1/2=600. Problem 3The following problem is from both the 2013 AMC 12B #1 and 2013 AMC 10B #3Let L be the low temperature. The high temperature is L +16. The average is32)16(=++L L . Solving for L , we get L =-5. Problem 4The following problem is from both the 2013 AMC 12B #3 and 2013 AMC 10B #4Note that is equal to the number of integers between 53 and 201, inclusive. Thus,149153201=+-=n . Problem 5Factoring the equation gives )2(b a - . From this we can see that to obtain the least possible value, b -2should be negative, and should be as small as possible. To doso, should be maximized. Because b -2 is negative, we should maximize the positivevalue of as well. The maximum values of both and are 5, so the answer is5(2-5)=-15. Problem 6The following problem is from both the 2013 AMC 12B #5 and 2013 AMC 10B #6The sum of the ages of the fifth graders is 33*11, while the sum of the ages of the parents is 55*33. Therefore, the total sum of all their ages must be 2178, and given33+55=88 people in total, their average age is 2178/88=99/4=24.75. .Problem 7If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the onlypossibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a 30—60—90 triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is 1, so the side opposite the thirty degree angle in the triangle isalso 1. From rules with 30—60—90 triangles, the area is 2/32/31=∙ Problem 8The following problem is from both the 2013 AMC 12B #4 and 2013 AMC 10B #8Let both Ray and Tom drive 40 miles. Ray's car would require 40/40=1 gallon of gas and Tom's car would require 40/10=4gallons of gas. They would have driven a total of40+40=80 miles, on 1+4=5 gallons of gas, for a combined rate ofProblem 9The prime factorization of 27000 is 333532∙∙. These three factors are pairwise relativelyprime, so the sum is 160532333=++ Problem 10Call the number of two point shots attempted and the number of three point shotsattempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, 54)3(4.0)2(5.0=+y x or 542.1=+y x . Because the team attempted 50% more two point shots then threes, y x 5.1=. Substituting y 5.1 for in the first equation gives542.15.1=+y y , so 20=y Problem 11The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11If we complete the square after bringing the and terms to the other side, we get0)3()5(22=++-y x . Squares of real numbers are nonnegative, so we need both 2)5(-x and 2)3(+y to be 0 which only happens when 5=x and 3-=y . Therefore,2=+y x Problem 12In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is 4/9.Problem 13The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13We notice that the number of numbers said is incremented by one each time; that is, Josays one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns," 1+2+3+4+5+6+7+8+9=45 numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, as 53-45=8. Since we're starting from 1each time, the 53rd number said will be 8. Problem 14and . Therefore, we have the equation2222yx x y xy y x -=-. Factoring out a -1 gives )(2222xy y x xy y x --=-. Factoring both sides further, )()(y x xy y x xy --=-. It follows that if 0,0==y x , or 0)(=-y x , both sides of the equation equal 0. By this, there are 3 lines (0,0==y x , ory x = ) so the answer is three lines. Problem 15Solution 1Using the formulas for area of a regular triangle )43(2s and regular hexagon )233(2s and plugging 3a and 6b into each equation, you find that 24336322b a =.Simplifying this, you get 26=b a . Solution 2The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle's area is 6 times the area of one of the little triangles. Therefore each side of the big triangle is 6 times the side of thesmall triangle. The desired ratio is 26663=. Problem 16Solution 1Let us use mass points: Assign B mass 1. Thus, because E is the midpoint of AB , A also has a mass of 1. Similarly, C has a mass of 1. D and E each have a mass of 2 because they are between B and C and A and B respectively. Note that the mass of D is twice the mass of A , so AP must be twice as long as PD . PD has length 2, so AP has length 4 and AD has length 6. Similarly, CP is twice PE and PE =1.5, so CP =3 and CE =4.5. Now note that triangle PED is a 3-4-5 right triangle with the right angle DPE . This means that the quadrilateral AEDC is a kite. The area of a kite is half the product of the diagonals, AD and CE . Recall that they are 6 and 4.5 respectively, so the area of AEDC is6*4.5/2=13.5. Solution 2Note that triangle DPE is a right triangle, and that the four angles that have point P are allright angles. Using the fact that the centroid (P ) divides each median in a 2:1 ratio, AP =4 and CP =3. Quadrilateral AEDC is now just four right triangles. The area is5.1325.1223345.14=∙+∙+∙+∙ Solution 3From the solution above, we can find the length of the diagonals to be 6 and 4.5. Now, since AEDC is a trapezoid, we use the area formula to find that the total area is5.1325.4*6= Problem 17The following problem is from both the 2013 AMC 12B #10 and 2013 AMC 10B #17 Solution 1We can approach this problem by assuming he goes to the red booth first. You start with 75R and 75B and at the end of the first booth, you will have 1R and 112B and 37S. We now move to the blue booth, and working through each booth until we have none left, wewill end up with: 1R, 2B and 103S. So, the answer is 103. Solution 2Let denote the number of visits to the first booth and denote the number of visits to thesecond booth. Then we can describe the quantities of his red and blue coins as follows: 752),(++-=y x y x R , 753),(+-=y x y x B . There are no legal exchanges when he has fewer than 2 red coins and fewer than 3 blue coins, namely when he has 1 red coin and 2 blue coins. We can then create a system of equations: 7521++-=y x , 7532+-=y x . Solving yields 59=x and 44=y . Since he gains one silver coin pervisit to each booth, he has 1035944=+=+y x silver coins in total. Problem 18First, note that the only integer 20132000<≤x is 2002. Now let's look at all numbers where 20001000<<x . Let the hundreds digit be 0. Then, the tens andunits digit can be 01, 12, 23, …, 89, which is 9 possibilities. We notice as the hundreds digit goes up by one, the number of possibilities goes down by one. Thus, the number ofProblem 19Solution 1It is given that 02=++c bx ax has 1 real root, so the discriminant is zero, or ac b 42=. Because a, b, c are in arithmetic progression, ,b c a b -=-or 2c a b +=. We need to find the unique root, or a b 2- (discriminant is 0). From ac b 42=, we have b c a b 22-=-.Ignoring the negatives, we have .1114224144124+===+==+++ca c c a c a c a c c a c cbc Fortunately, finding c a /is not very hard. Plug in 2c a b += to ac b 42=, we have ,16222ac c ac a =++ or 01422=+-c ac a , and dividing by 2c gives,01)(14)(2=+-c a c a so 347219214±=±=c a . But 1347<-, violating the assumptionthat c a ≥. Therefore,347+=ca . Plugging this in, we have 323211414-=+=+c a . Butwe need the negative of this, so the answer is . Solution 2Note that we can divide the polynomial by to make the leading coefficient 1 sincedividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation is must be of theform 2222)(r rx x r x +-=- where 0212≥≥-≥r r . We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the median. Thus, r r 412-=+ and 32±-=r . Since 21r >, we easily see that r has to be between 1 and 0. Thus, we can eliminate 32-- and areleft with 32+- as the answer. Solution 3Given that 02=++c bx ax has only 1 real root, we know that the discriminant must equal 0, or that ac b 42=. Because the discriminant equals 0, we have that the root of the quadratic is a b r 2/-=. We are also given that the coefficients of the quadratic are in arithmetic progression, where 0≥≥≥c b a . Letting the arbitrary difference equal variable , we have that d b a += and that d b c -=. Plugging those two equationsinto ac b 42=, we have )(4222d b b -= which yields 2243d b =. Isolating , we have 2/3b d =. Substituting that in for in d b a +=, we get )231(23+=+=b b b a . Once again, substituting that in for in a b r 2/-=, we have32321)1(223+-=+-=+-=b b r . The answer is Solution 4Let the double root be . Then by the arithmetic progression and Vieta's,ac a b a b c b b a -=-⇒-=-1 3201422122±-=⇒=++⇒--=+r r r r r rWe see 100≤≤⇒≤≤ab a b , and so we want 120≤-≤r . Note that since 1324)32(20≥+=---≤ and 1324)32(20≤-=+--≤, we can concludethat 32+-=r and the answer is .Problem 20The following problem is from both the 2013 AMC 12B #15 and 2013 AMC 10B #20The prime factorization of 2013 is 61·11·3. To have a factor of 61 in the numerator and to minimize11,a a must equal 61. Now we notice that there can be no prime which is not a factor of 2013 such that 611<<p b , because this prime will not be represented in the denominator, but will be represented in the numerator. The highest lessthan 61 is 59, so there must be a factor of 59 in the denominator. It follows that591=b (to minimize 1b as well), so the answer is 25961=- to express 2013 with )59,61(),(11=b a is !10!20!59!11!19!612013∙∙∙∙= Problem 21The following problem is from both the 2013 AMC 12B #14 and 2013 AMC 10B #21 Let the first two terms of the first sequence be 1x and 2x and the first two of the second sequence be 1y and 2y . Computing the seventh term, we see that 21218585y y x x +=+. Note that this means that 1x and 1y must have the same value modulo 8. To minimize, let one of them be 0; WLOG assume that 01=x . Thus, the smallest possible value of 1y is 8; since the sequences are non-decreasing 82≥y . To minimize, let 82=y .Thus, 10464408521=+=+y y . Problem 22First of all, note that J must be 1, 5, or 9 to preserve symmetry. We also notice that A+E=B+F=C+G=D+H.WLOG assume that J =1. Thus the pairs of vertices must be 9 and 2, 8 and 3, 7 and 4, and 6 and 5. There are 4!=24 ways to assign these to the vertices. Furthermore, there are 1624= ways to switch them (i.e. do 2 9 instead of 9 2).Thus, there are 16(24)=384 ways for each possible J value. There are 3 possible J valuesthat still preserve symmetry: 384(3)=1152. Solution 2As in solution 1, J must be 1, 5, or 9 giving us 3 choices. Additionally A+E=B+F=C+G=D+H . This means once we choose J there are 8 remaining choices. Going clockwise from A we count, 8 possibilities for A . Choosing A also determines E which leaves 6 choices for B , once B is chosen it also determines F leaving 4 choices for C . Once C is chosen it determines G leaving 2 choicesfor D . Choosing D determines H , exhausting the numbers. To get the answer we multiply2*4*6*8*3=1152. Problem 23The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, Solution 1Since 90=∠=∠ADB AFB , quadrilateral ABDF is cyclic. It follows that ABF ADE ∠=∠. In addition, since 90=∠=∠AED AFB , triangles ABF and ADE are similar. It follows that )5/3)(13(),5/4)(13(=BF AF . By Ptolemy , we have )5/3)(13)(12()5/4)(13)(5(13=+DF .Cancelling 13, the rest is easy. We obtain 215165/16=+⇒=DF . Solution 2Using the similar triangles in triangle ADC gives AE =48/5 and DE =36/5. Quadrilateral ABDF is cyclic, implying that ︒=∠+∠180DFA B . Therefore, EFA B ∠=∠, and triangles AEF and ADB are similar. Solving the resulting proportion gives EF =4. Therefore,DF=ED-EF =16/5. Problem 24A positive integer with only four positive divisors has its prime factorization in the form of b a *, where and are both prime positive integers or 3c where is a prime. One caneasily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of b a *. The four factors of this number would be 1, ,,b a and ab . The sum of these would be 1+++b a ab , which can be factored into the form )1)(1(++b a . Easily we can see that now we can take cases again.Case 1: Either or is 2.If this is true then we have to have that one of )1(+a or )1(+b is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either 132016- or 132010- is a prime. We see that in this case none of them work. Case 2: Both and are odd primes.This implies that both )1(+a and )1(+b are even which implies that in this case the number must be divisible by 4. This leaves only 2012 and 2016. 2012=4*503 so we have that a factor of 2 must go to both )1(+a and )1(+b . So we have that )1(+a and )1(+b equal the numbers (502+1)(3+1), but 502 is not an odd prime, so 2012 does not work. 2016=4*504, so we have (503+1)(3+1). 503 and 3 are both oddprimes, so 2016 is a solution. Thus the answer is 1. Problem 25The following problem is from both the 2013 AMC 12B #23 and 2013 AMC 10B #25,First, we can examine the units digits of the number base 5 and base 6 and eliminatesome possibilities.Say that )6(mod a N ≡ also that )5(mod b N ≡Substituting these equations into the question and setting the units digits of 2N and Sequal to each other, it can be seen thatb a =, and 5<b , so),6(mod a N ≡40),30(mod )5(mod ≤≤=⇒≡a a N a N 。

美国数学测评2020 AMC 12B

2020 AMC 12BProblem 1What is the value in simplest form of the following expression?下面表达式化简后的值为多少？Problem 2What is the value of the following expression?下面表达式的值是多少?Problem 3The ratio of to is , the ratio of to is , and the ratio of to is . What is the ratio of to ?w比x的比值为4:3，y比z的比值为3:2，z比x的比值为1:6，则w比y的比值是多少？The acute angles of a right triangle are and , where and both and are prime numbers. What is the least possible value of ?一个直角三角形的两个锐角的度数分别为a°和b°，其中a>b，且a和b均为质数，则b的最小可能值是多少？Problem 5Teams and are playing in a basketball league where each game results in a win for one teamand a loss for the other team. Team has won of its games and team has won of its games.Also, team has won more games and lost more games than team How many games hasteam played?A队和B队在一个篮球联盟内打比赛，每场比赛后，结果都是一个队伍赢，另一个队伍输。

2012 AMC 12A 详解

2012 AMC 12AProblem 1A bug crawls along a number line, starting at -2. It crawls to -6, then turns around and crawls to 5. How many units does the bug crawl altogether?SolutionCrawling from -2 to -6 takes it a distance of 4 units. Crawling from -6 to 5 takes it a distance of 11 units. Add 4 and 11 to getProblem 2Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?SolutionCagney can frost one in 20 seconds, and Lacey can frost one in 30 seconds. Working together, they can frost one in seconds. In 300 seconds (5 minutes), they can frost .Problem 3A box centimeters high, centimeters wide, and centimeters long can hold grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold grams of clay. What is ?SolutionThe first box has volume , and the second has volume. The second has a volume that is times greater, so it holds grams.Problem 4In a bag of marbles, of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?SolutionAssume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles.Thus, the answer is .Problem 5A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?SolutionSo let the number of blueberries be the number of raspberries be the number of grapes be and finally the number of cherries beObserve that since there are pieces of fruit,Since there are twice as many raspberries as blueberries,The fact that there are three times as many grapes as cherries implies,Because there are four times as many cherries as raspberries, we deduce the following:Note that we are looking for So, we try to rewrite all of the other variables in terms of The third equation gives us the value of in terms of already. Wedivide the fourth equation by to get that Finally, substituting this value of into the first equation provides us with the equation and substituting yields:Multiply this equation by to get:Problem 6The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?SolutionLet the three numbers be equal to , , and . We can now write three equations:Adding these equations together, we get thatandSubstituting the original equations into this one, we findTherefore, our numbers are 12, 7, and 5. The middle number isProblem 7Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?SolutionIf we let be the smallest sector angle and be the difference between consecutive sector angles, then we have the angles . Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.All sector angles are integers so must be a multiple of 2. Plug in even integers for starting from 2 to minimize We find this value to be 4 and the minimum value ofto beProblem 8An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?SolutionThe minimum and maximum can be achieved with the orders and .The difference between the two isProblem 9A year is a leap year if and only if the year number is divisible by 400 (such as 2000) or is divisible by 4 but not 100 (such as 2012). The 200th anniversary of the birth of novelist Charles Dickens was celebrated on February 7, 2012, a Tuesday. On what day of the week was Dickens born?SolutionEach year we go back is one day back, because . Each leap year we go back is two days back, since . A leap year is GENERALLYevery four years, so 200 years would have = leap years, but the problem points out that 1900 does not count as a leap year.This would mean a total of 150 regular years and 49 leap years, so= days back. Since , four days back from Tuesday would beProblem 10A triangle has area , one side of length , and the median to that side of length . Let be the acute angle formed by that side and the median. What is ?SolutionSolution 1is the side of length , and is the median of length . The altitude ofto is because the 0.5(altitude)(base)=Area of the triangle. is . To find , just use opposite over hypotenuse with the right triangle . Thisis equal to .Solution 2It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of in the above figure.Expressing the area in terms of , . Solving for gives . .Problem 11Alex, Mel, and Chelsea play a game that has rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round theprobability that Alex wins is , and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?SolutionIf is the probability Mel wins and is the probability Chelsea wins, and . From this we get and . For Alex to win three, Mel to win two, and Chelsea to win one, in that order, is . Multiply this by the number of permutations (orders they can win) which isProblem 12A square region is externally tangent to the circle with equationat the point on the side . Vertices and are on the circle with equation . What is the side length of this square?SolutionThe circles have radii of and . Draw the triangle shown in the figure above and write expressions in terms of (length of the side of the square) for the sides of the triangle. Because is the radius of the larger circle, which is equal to , we can write the Pythagorean Theorem.Use the quadratic formula.Problem 13Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?SolutionLet Paula work at a rate of , the two helpers work at a combined rate of , and the time it takes to eat lunch be , where and are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:Adding the second and third equations together gives us. Subtracting the first equation from this new onegives us , and solving for in terms of gives . We can plug this into the second equation:We can then subtract this from the third equation:Therefore . Plugging this into our third equation then gives usSince we let be in hours, the lunch break then takes minutes, which leads us to the correct answer of .Problem 14The following problem is from both the 2012 AMC 12A #14 and 2012 AMC 10A #18, so both problems redirect to this page.The closed curve in the figure is made up of 9 congruent circular arcs each of length, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?SolutionDraw the hexagon between the centers of the circles, and compute its area. Then add the areas of the three sectors outside thehexagon () and subtract the areas of the three sectors inside the hexagon () to get the area enclosed in the curved figure , which is .Problem 15The following problem is from both the 2012 AMC 12A #15 and 2012 AMC 10A #20, so both problems redirect to this page.A x square is partitioned into unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?SolutionFirst, the middle square is always black because it is rotated onto itself. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is case with all black squares. There are four cases with one white square and all work. There are six cases with two white squares, but only the with the white squares opposite from each other work. There are no cases with three white squares or four white squares. Then the total number of ways to color the corners is . In essence, the edges work the same way, so there are also ways to color them. The number of ways to fit the conditions over the number of ways to color the squares isProblem 16Circle has its center lying on circle . The two circles meet at and . Point in the exterior of lies on circle and , , and . What is the radius of circle ?SolutionSolution 1Let denote the radius of circle . Note that quadrilateral is cyclic. By Ptolemy's Theorem, we have and . Let t be the measure of angle . Since , the law of cosines on triangle gives us . Again since is cyclic, the measure of angle . We apply the law of cosines to triangle so that. Sincewe obtain . Butso that . .Solution 2Let us call the the radius of circle , and the radius of . Considerand . Both of these triangles have the same circumcircle (). From the Extended Law of Sines, we see that . Therefore,. We will now apply the Law of Cosines to andand get the equations,,respectively. Because , this is a system of two equations and two variables. Solving for gives . .Problem 17Let be a subset of with the property that no pair of distinct elements in has a sum divisible by . What is the largest possible size of ?SolutionOf the integers from to , there are six each of . We can create several rules to follow for the elements in subset . No element can be iff there is an element that is . No element can beiff there is an element that is . Ignoring those that are , we can get a subset with elements. Considering , there can be one element that is so because it will only be divisible by if paired with anotherelement that is . The final answer is .Problem 18Triangle has , , and . Let denote the intersection of the internal angle bisectors of . What is ?SolutionInscribe circle of radius inside triangle so that it meets at ,at , and at . Note that angle bisectors of triangle are concurrent at the center of circle . Let , and . Note that , and . Hence , , and. Subtracting the last 2 equations we have and adding this to the first equation we have .By Herons formula for the area of a triangle we have that the area of triangleis . On the other hand the area is given by. Then so that .Since the radius of circle is perpendicular to at , we have by the pythagorean theorem so that .Problem 19The following problem is from both the 2012 AMC 12A #19 and 2012 AMC 10A #23, so both problems redirect to this page.ProblemAdam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?SolutionNote that if is the number of friends each person has, then can be any integer from to , inclusive.Also note that the cases of and are the same, since a map showing a solution for can correspond one-to-one with a map of a solution for by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with when compared to configurations of . Thus, we have two cases to examine, and , and we count each of these combinations twice.For , if everyone has exactly one friend, that means there must be pairs of friends, with no other interconnections. The first person has choices for a friend. There are people left. The next person has choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are configurations with .For , there are two possibilities. The group of can be split into two groups of , with each group creating a "friendship triangle." The first person hasways to pick two friends from the other five, while the other three are forced together. Thus, there are triangular configurations.However, the group can also from a "friendship hexagon", with each person sitting on a vertex, and each side representing the two friends that person has. The firstperson may be seated anywhere on the hexagon WLOG. This person haschoices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are hexagonal configurations, and in total configurations for .As stated before, has configurations, and has configurations. This gives a total of configurations, which is optionProblem 20Consider the polynomialThe coefficient of is equal to . What is ?SolutionEvery term in the expansion of the product is formed by taking one term from each factor and multiplying them all together. Therefore, we pick a power of or a power of from each factor.Every number, including , has a unique representation by the sum of powers of two, and that representation can be found by converting a number to its binary form., meaning.Thus, the term was made by multiplying from thefactor, from the factor, and so on. The only numbers not used are , , and .Thus, from the factors, , , and were chosen as opposed to , and .Thus, the coefficient of the term is . So the answer is .Problem 21The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.ProblemLet , , and be positive integers with such thatand . What is ?SolutionAdd the two equations..Now, this can be rearranged:and factored:, , and are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that ., since is the biggest difference. It is impossible to determine by inspection whether or , or whether or .We want to solve for , so take the two cases and solve them each for an expression in terms of . Our two cases are or. Plug these values into one of the original equations to see if we can get an integer for ., after some algebra, simplifies to . 2021 is not divisible by 7, so is not an integer.The other case gives , which simplifies to . Thus, and the answer isProblem 22Distinct planes intersect the interior of a cube . Let be the unionof the faces of and let . The intersection of and consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of . What is the difference between the maximum and minimum possible values of ?SolutionProblem 23Let be the square one of whose diagonals has endpoints and. A point is chosen uniformly at random over all pairs of real numbers and such that and . Let be a translated copy of centered at . What is the probability that the square regiondetermined by contains exactly two points with integer coefficients in its interior?SolutionSolution 1We first notice that for the translated square to contain two points with integer coordinates (lattice points) in its interior, these two points must be adjacent. This can be shown by considering the diagonal of . The diagonal is, which is the length of the diagonal of a unit square. Becausesquare is not parallel to the axis, square cannot two points that are not adjacent.Because we have showed that the two lattice points contained in must be adjacent, let us consider the unit square with vertices and . Let us first consider only two vertices, and . We want to find the area of the region within that the point will create the translation of , such that it covers both and . By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.For to contain the point , must be inside square . Similarly, forto contain the point , must be inside a translated square with center at , which we will call . Therefore, the area we seek is Area.To calculate the area, we notice that Area Area by symmetry. Let . Let be the midpoint of , and along the line . Let be the intersection of and within , and be the intersection ofand outside . Therefore, the area we seek is Area. Because all have coordinate , they are collinear. Noting that the side length of and is (as shown above), we also see that, so . If follows thatand. Therefore, the area is Area.Because there are three other regions in the unit square that we need to count, the total area of within such that contains two adjacent lattice points is .By periodicity, this probability is the same if , where and . Therefore, the answer is .Note: the ranges of and in the problem are arbitrary as long as the maximum and minimum of the range are integers.Problem 24Let be the sequence of real numbers defined by, and in general,Rearranging the numbers in the sequence in decreasing order produces a new sequence . What is the sum of all integers , , such thatSolutionProblem 25Let where denotes the fractional part of . The numberis the smallest positive integer such that the equation has at least real solutions. What is ? Note:the fractional part of is a real number such that and is an integer.Solution2012 AMC 12A Answer Key1. E2. D3. D4. C5. D6. D7. C8. C9. A10. D11. B12. D13. D14. E15. A16. E17. B18. A19. B20. B21. E22. C23. C24. C25. C。

amc

2011 AMC 12AProblem 1A cell phone plan costs dollars each month, plus cents per text message sent, pluscents for each minute used over hours. In January Michelle sent text messages andtalked for hours. How much did she have to pay?SolutionProblem 2There are coins placed flat on a table according to the figure. What is the order of the coins from top to bottom?SolutionProblem 3A small bottle of shampoo can hold milliliters of shampoo, whereas a large bottle can holdmilliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?SolutionProblem 4At an elementary school, the students in third grade, fourth grade, and fifth grade run anaverage of , , and minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?SolutionProblem 5Last summer of the birds living on Town Lake were geese, were swans, were herons,and were ducks. What percent of the birds that were not swans were geese?Problem 6The players on a basketball team made some three-point shots, som e two-point shots, and som e one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number ofsuccessful two-point shots. The team's total score was points. How many free throws did they make?SolutionProblem 7A majority of the 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. 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 . What was the cost of a pencil in cents?SolutionProblem 8In the eight term sequence , , , , , , , , the value of is and the sum ofany three consecutive terms is . What is ?SolutionProblem 9At a twins and triplets convention, there were sets of twins and sets of triplets, all from different families. Each twin shook hands with all the twins except his/her siblings and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?SolutionProblem 10A pair of standard -sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?SolutionProblem 11Circles and each have radius 1. Circles and share one point of tangency. Circlehas a point of tangency with the midpoint of What is the area inside circle butoutside circle and circleSolutionProblem 12A power boat and a raft both left dock on a river and headed downstream. The raft driftedat the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock downriver, then immediately turned and traveledback upriver. It eventually met the raft on the river 9 hours after leaving dock How manyhours did it take the power boat to go from toSolutionProblem 13Triangle has side-lengths and The line throughthe incenter of parallel to intersects at and at What is theperimeter ofSolutionProblem 14Suppose and are single-digit positive integers chosen independently and at random. Whatis the probability that the point lies above the parabola ?2010 AMC 12BProblem 1Makarla attended two meetings during her -hour work day. The first meeting took minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?Problem 2A big is formed as shown. What is its area?Problem 3A ticket to a school play cost dollars, where is a whole number. A group of 9th graders buys tickets costing a total of $, and a group of 10th graders buys tickets costing a total of$. How many values for are possible?SolutionProblem 4A month with days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?Problem 5Lucky Larry's teacher asked him to substitute numbers for , , , , and in theexpression and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. Thenumber Larry sustitued for , , , and were , , , and , respectively. What numberdid Larry substitude for ?Problem 6At the beginning of the school year, of all students in Mr. Wells' math class answered"Yes" to the question "Do you love math", and answered "No." At the end of the schoolyear, answered "Yes" and answerws "No." Altogether, of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of ?SolutionProblem 7Shelby drives her scooter at a speed of miles per hour if it is not raining, and miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening,for a total of miles in minutes. How many minutes did she drive in the rain?SolutionProblem 8Every high school in the city of Euclid sent a team of students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carlaplaced th and th, respectively. How many schools are in the city?SolutionProblem 9Let be the smallest positive integer such that is divisible by , is a perfect cube, andis a perfect square. What is the number of digits of ?SolutionProblem 10The average of the numbers and is . What is ?SolutionProblem 11A palindrome between and is chosen at random. What is the probability that itis divisible by ?SolutionProblem 12For what value of doesSolutionProblem 13In , and . What is ?SolutionProblem 14Let , , , , and be postive integers with and let bethe largest of the sum , , and . What is the smallest possible value of ?2010 AMC 12AProblem 1What is ?SolutionProblem 2A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?SolutionProblem 3Rectangle , pictured below, shares of its area with square . Squareshares of its area with rectangle . What is ?SolutionProblem 4If , then which of the following must be positive?SolutionProblem 5Halfway through a 100-shot archery tournam ent, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next shots are bullseyes she will be guaranteed victory. What is the minimum value for ?Problem 6A , such as 83438, is a number that remains the same when its digits arereversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?Problem 7Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionProblem 8Triangle has . Let and be on and , respectively, suchthat . Let be the intersection of segments and , andsuppose that is equilateral. What is ?SolutionProblem 9A solid cube has side length inches. A -inch by -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?SolutionProblem 10The first four terms of an arithm etic sequence are , , , and . What is theterm of this sequence?SolutionProblem 11The solution of the equation can be expressed in the form . What is ?Problem 12In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statem ents.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?SolutionProblem 13For how many integer values of do the graphs of and not intersect?Problem 14Nondegenerate has integer side lengths, is an angle bisector, , and. What is the smallest possible value of the perimeter?。

2015年 AMC 12B真题含答案

2015AMC12BProblem1What is the value of?的值是多少？Problem2Marie does three equally time-consuming tasks in a row without taking breaks.She begins the first task at1:00PM and finishes the second task at2:40PM.When does she finish the third task?Marie连续做了3项耗时相同的任务，中间没有休息，她从下午1:00开始做第一个任务，在下午2:40完成第二个任务，她何时完成第三个任务？Problem3Isaac has written down one integer two times and another integer three times.The sum of the five numbers is100,and one of the numbers is28.What is the other number?Isaac把一个整数写了2遍，另一个整数写了3遍。

这5个数之和为100，其中一个数是28，另一个数是多少？David,Hikmet,Jack,Marta,Rand,and Todd were in a12-person race with6other people.Rand finished6places ahead of Hikmet.Marta finished1place behind Jack.David finished2places behind Hikmet.Jack finished2places behind Todd.Todd finished1place behind Rand.Marta finished in6th place.Who finished in8th place?David，Hikmet，Jack，Marta，Rand，Todd和其他6人参加了一场12人的比赛。

2008 AMC 12B Problems

2008 AMC 12B ProblemsProblem 1A basketball player made baskets during a game. Each basket was worth eitheror points. How many different numbers could represent the total points scoredby the player?Problem 2A block of calendar dates is shown. The order of the numbers in the secondrow is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?Problem 3A semipro baseball league has teams with players each. League rules statethat a player must be paid at least dollars, and that the total of all players'salaries for each team cannot exceed dollars. What is the maximum possiblle salary, in dollars, for a single player?On circle , points and are on the same side of diameter ,, and . What is the ratio of the area of the smallersector to the area of the circle?Problem 5A class collects dollars to buy flowers for a classmate who is in the hospital.Roses cost dollars each, and carnations cost dollars each. No other flowers areto be used. How many different bouquets could be purchased for exactlydollars?Problem 6Postman Pete has a pedometer to count his steps. The pedometer records up to steps, then flips over to on the next step. Pete plans to determine hismileage for a year. On January Pete sets the pedometer to . During theyear, the pedometer flips from to forty-four times. On Decemberthe pedometer reads . Pete takes steps per mile. Which of thefollowing is closest to the number of miles Pete walked during the year?For real numbers and , define . What is ?Problem 8Points and lie on . The length of is times the length of , andthe length of is times the length of . The length of is what fractionof the length of ?Problem 9Points and are on a circle of radius and . Point is the midpoint ofthe minor arc . What is the length of the line segment ?Problem 10Bricklayer Brenda would take hours to build a chimney alone, and bricklayerBrandon would take hours to build it alone. When they work together they talka lot, and their combined output is decreased by bricks per hour. Workingtogether, they build the chimney in hours. How many bricks are in the chimney?A cone-shaped mountain has its base on the ocean floor and has a height of 8000feet. The top of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?Problem 12For each positive integer , the mean of the first terms of a sequence is . What is the th term of the sequence?Problem 13Vertex of equilateral triangle is in the interior of unit square .Let be the region consisting of all points inside and outsidewhose distance from is between and . What is the area of ?Problem 14A circle has a radius of and a circumference of . What is?On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let be the region formed by the union ofthe square and all the triangles, and be the smallest convex polygon thatcontains . What is the area of the region that is inside but outside ?Problem 16A rectangular floor measures by feet, where and are positive integers with. An artist paints a rectangle on the floor with the sides of the rectangleparallel to the sides of the floor. The unpainted part of the floor forms a border of width foot around the painted rectangle and occupies half of the area of theentire floor. How many possibilities are there for the ordered pair ?Problem 17Let , and be three distinct points on the graph of such that lineis parallel to the -axis and is a right triangle with area . What isthe sum of the digits of the -coordinate of ?Problem 18A pyramid has a square base and vertex . The area of squareis , and the areas of and are and , respectively. Whatis the volume of the pyramid?A function is defined by for all complex numbers ,where and are complex numbers and . Suppose that andare both real. What is the smallest possible value of ?Problem 20Michael walks at the rate of feet per second on a long straight path. Trash pailsare located every feet along the path. A garbage truck travels at feet persecond in the same direction as Michael and stops for seconds at each pail. AsMichael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?Problem 21Two circles of radius 1 are to be constructed as follows. The center of circle ischosen uniformly and at random from the line segment joining and . The center of circle is chosen uniformly and at random, and independently ofthe first choice, from the line segment joining to . What is the probability that circles and intersect?Problem 22A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?Problem 23The sum of the base-logarithms of the divisors of is . What is ?Problem 24Let . Distinct points lie on the -axis, and distinct pointslie on the graph of . For every positive integer ,is an equilateral triangle. What is the least for which the length?Problem 25Let be a trapezoid with , , , , and. Bisectors of and meet at , and bisectors of and meetat . What is the area of hexagon ?。

2002AMC12B解答

A A/2 = minutes to mow hers, 2R 4R A/3 A = minutes to mow his. R 3R
A A < , Beth will ﬁnish ﬁrst. 4R 3R
18. (C) The area of the rectangular region is 2. Hence the probability that P is closer to (0, 0) than it is to (3, 1) is half the area of the trapezoid bounded by the lines y = 1, the x- and y - axes, and the perpendicular bisector of the segment joining (0, 0) and (3, 1). The perpendicular bisector goes through the point (3/2, 1/2), which is the center of the square whose vertices are (1, 0), (2, 0), (2, 1), and Hence, the line cuts the square into two quadrilaterals of equal area 1/2. Thus the area of the trapezoid is 3/2 and the probability is 3/4.
1 41 1 1 1 1 + + + = + 2 3 7 n 42 n is an integer precisely when it is equal to 1. This implies that n = 42, so the answer is (E).

2021年Fall AMC12B真题含答案

2021Fall AMC12B Problem1What is the value ofProblem2What is the area of the shaded figure shown below?At noon on a certain day,Minneapolis is degrees warmer than St.Louis.At the temperaturein Minneapolis has fallen by degrees while the temperature in St.Louis has risen by degrees,atwhich time the temperatures in the two cities differ by degrees.What is the product of all possiblevalues ofProblem4Let.Which of the following is equal toProblem5Call a fraction,not necessarily in the simplest form,special if and are positive integers whosesum is.How many distinct integers can be written as the sum of two,not necessarily different, special fractions?The largest prime factor of is because.What is the sum of the digits of thegreatest prime number that is a divisor of?Problem7Which of the following conditions is sufficient to guarantee that integers,,and satisfy theequationandandandandThe product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base.What is the measure,in degrees,of the vertex angle of this triangle?Problem9Triangle is equilateral with side length.Suppose that is the center of the inscribed circleof this triangle.What is the area of the circle passing through,,and?Problem10What is the sum of all possible values of between and such that the triangle in the coordinateplane whose vertices are is isosceles?Una rolls standard-sided dice simultaneously and calculates the product of the numbersobtained.What is the probability that the product is divisible byProblem12For a positive integer,let be the quotient obtained when the sum of all positive divisorsof is divided by For example,WhatisProblem13Let What is the value ofSuppose that,and are polynomials with real coefficients,having degrees,, and,respectively,and constant terms,,and,respectively.Let be the number of distinctcomplex numbers that satisfy the equation.What is the minimum possiblevalue of?Problem15Three identical square sheets of paper each with side length are stacked on top of each other.Themiddle sheet is rotated clockwise about its center and the top sheet is rotatedclockwise about its center,resulting in the-sided polygon shown in the figure below.Thearea of this polygon can be expressed in the form,where,,and are positive integers,and is not divisible by the square of any prime.What is?Suppose,,are positive integers suchthat and What is the sum of all possibledistinct values of?Problem17A bug starts at a vertex of a grid made of equilateral triangles of side length.At each step the bugmoves in one of the possible directions along the grid lines randomly and independently with equalprobability.What is the probability that after moves the bug never will have been more than unitaway from the starting position?Set,and for let be determined by the recurrenceThis sequence tends to a limit;call it.What is the least value of such thatProblem19Regular polygons with,,,and sides are inscribed in the same circle.No two of the polygonsshare a vertex,and no three of their sides intersect at a common point.At how many points inside the circle do two of their sides intersect?Problem20A cube is constructed from white unit cubes and blue unit cubes.How many different ways arethere to construct the cube using these smaller cubes?(Two constructions are consideredthe same if one can be rotated to match the other.)For real numbers,let where.For how many values of with doesProblem22Right triangle has side lengths,,and.A circle centered at is tangent to line at and passes through.A circle centered at istangent to line at and passes through.What is?What is the average number of pairs of consecutive integers in a randomly selected subsetof distinct integers chosen from the set?(For example theset has pairs of consecutive integers.)P roblem24Triangle has side lengths,and.The bisectorof intersects in point,and intersects the circumcircle of in point.The circumcircle of intersects the line in points and.What is?Problem25For a positive integer,let be the sum of the remainders when is dividedby,,,,,,,,and.Forexample,.How many two-digit positiveintegers satisfy。

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What isSolutionJosanna's test scores to date are , , , , and . Her goal is to raise her test average at least points with her next test. What is the minimum test score she would need to accomplish this goal?SolutionLeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy hadpaid dollars and Bernardo had paid dollars, where . How many dollars must LeRoy give to Bernardo so that they share the costs equally?SolutionIn multiplying two positive integers and , Ron reversed the digits of thetwo-digit number . His erroneous product was 161. What is the correct value of the product of and ?SolutionLet be the second smallest positive integer that is divisible by every positive integer less than . What is the sum of the digits of ?SolutionTwo tangents to a circle are drawn from a point . The points ofcontact and divide the circle into arcs with lengths in the ratio . What is the degree measure of ?SolutionLet and be two-digit positive integers with mean . What is the maximum value of the ratio ?SolutionKeiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track haswidth meters, and it takes her seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?SolutionTwo real numbers are selected independently and at random from theinterval . What is the probability that the product of those numbers is greater than zero?SolutionRectangle has and . Point is chosen on side so that . What is the degree measure of ?SolutionA frog located at , with both and integers, makes successive jumps of length and always lands on points with integer coordinates. Suppose that the frog starts at and ends at . What is the smallest possible number of jumps the frog makes?SolutionA dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?SolutionBrian writes down four integers whose sum is . The pairwise positive differences of these numbers are and . What is the sum of the possible values of ?SolutionA segment through the focus of a parabola with vertex is perpendicularto and intersects the parabola in points and . What is ?SolutionHow many positive two-digit integers are factors of ?SolutionRhombus has side length and . Region consists of all points inside of the rhombus that are closer to vertex than any of the other three vertices. What is the area of ?SolutionLet ,and for integers . What is the sum of the digitsof ?SolutionA pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?SolutionA lattice point in an -coordinate system is any point whereboth and are integers. The graph of passes through no lattice point with for all such that . What is the maximum possible value of ?SolutionTriangle has , and . The points ,and are the midpoints of , and respectively. Let be the intersection of the circumcircles of and . Whatis ?SolutionThe arithmetic mean of two distinct positive integers and is a two-digit integer. The geometric mean of and is obtained by reversing the digits of the arithmetic mean. What is ?SolutionLet be a triangle with sides , and . For ,if and , and are the points of tangency of the incircleof to the sides , and , respectively, then is a triangle with side lengths , and , if it exists. What is the perimeter of the last triangle in the sequence ?SolutionA bug travels in the coordinate plane, moving only along the lines that are parallel to the -axis or -axis. Let and . Consider all possiblepaths of the bug from to of length at most . How many points with integer coordinates lie on at least one of these paths?SolutionLet . What is the minimum perimeter among all the -sided polygons in the complex plane whose vertices are precisely the zeros of ?SolutionFor every and integers with odd, denote by the integer closest to . For every odd integer , let be the probability thatfor an integer randomly chosen from the interval . What is the minimum possible value of over the odd integers in theinterval ?。