2024 AMC 12A Problems/Problem 1
Contents1234ProblemSolution 1Video SolutionSee AlsoProblemCarlos took of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?Solution 1If Carlos took of the pie, is left.Therefore: must be remaining. After Maria takes of the remaining -Contributed by Awesome2.1, latex by quacker88Video Solutionhttps://youtu.be/qJF3G7_IDgc~IceMatrixSee Also2024 AMC 12A (Problems ? Answer Key ? Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2024))Preceded?by First ProblemFollowed?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 AMC 12 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s AmericanMathematics Competitions (http://amc.maa.org). Retrieved from \Copyright ? 2024 Art of Problem Solving2024 AMC 12A Problems/Problem 2Contents1234ProblemSolutionVideo SolutionSee AlsoProblemThe acronym AMC is shown in the rectangular grid below with grid lines spaced unit apart. In units, what is the sum of thelengths of the line segments that form the acronym AMCSolutionEach of the straight line segments have length and each of the slanted line segments have length using , pythag, trig, or just sense)There area a total of straight lines segments and slanted line segments. The sum is (this can be deducted ~quacker88 was You could have also just counted slanted line segments and realized that the only answer choice involving .Video Solutionhttps://youtu.be/qJF3G7_IDgc~IceMatrixSee Also2024 AMC 12A (Problems ? Answer Key ? Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2024))Preceded?by Problem 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 AMC 12 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s AmericanMathematics Competitions (http://amc.maa.org). 2024 AMC 10A Problems/Problem 4The following problem is from both the 2024 AMC 12A #3 and 2024 AMC 10A #4, so both problems redirect to this page.Contents1234ProblemSolutionVideo SolutionSee AlsoProblemA driver travels for hours at miles per hour, during which her car gets miles per gallon of gasoline. She is paid permile, and her only expense is gasoline at per gallon. What is her net rate of pay, in dollars per hour, after this expense?SolutionSince the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends $4 per hour on gas. If she gets$0.50 per mile, then she gets $30 per hour of driving. Subtracting the gas cost, her net rate of pay per hour is .Video Solutionhttps://youtu.be/WUcbVNy2uv0~IceMatrixSee Also2024 AMC 10A (Problems ? Answer Key ? Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2024))Preceded?by Problem 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 AMC 10 Problems and Solutions2024 AMC 12A (Problems ? Answer Key ? Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2024))Preceded?by Problem 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 AMC 12 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s AmericanMathematics Competitions (http://amc.maa.org). Retrieved from \Copyright ? 2024 Art of Problem Solving2024 AMC 10A Problems/Problem 6The following problem is from both the 2024 AMC 12A #4 and 2024 AMC 10A #6, so both problems redirect to this page.Contents1234ProblemSolutionVideo SolutionSee AlsoProblemHow many -digit positive integers (that is, integers between and , inclusive) having only even digits are divisible by SolutionThe ones digit, for all numbers divisible by 5, must be either or . However, from the restriction in the problem, it must be even,giving us exactly one choice () for this digit. For the middle two digits, we may choose any even integer from , meaning thatwe have total options. For the ?rst digit, we follow similar intuition but realize that it cannot be , hence giving us 4 possibilities.Therefore, using the multiplication rule, we get . ~ciceronii swrebbyVideo Solutionhttps://youtu.be/JEjib74EmiY~IceMatrixSee Also2024 AMC 10A (Problems ? Answer Key ? Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2024))Preceded?by Problem 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 AMC 10 Problems and Solutions2024 AMC 12A (Problems ? Answer Key ? Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2024))Preceded?by Problem 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 AMC 12 Problems and SolutionsThe problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s AmericanMathematics Competitions (http://amc.maa.org). Retrieved from \2024 AMC 10A Problems/Problem 7The following problem is from both the 2024 AMC 12A #5 and 2024 AMC 10A #7, so both problems redirect to this page.Contents1Problem2Solution2.1Solution 22.2Solution 33Video Solution4See AlsoProblemThe integers from to inclusive, can be arranged to form a -by- square in which the sum of the numbers in eachrow, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. Whatis the value of this common sum? SolutionWithout loss of generality, consider the ?ve rows in the square. Each row must have the same sum of numbers, meaning that thesum of all the numbers in the square divided by is the total value per row. The sum of the integers is , and the common sum is .Solution 2Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get as our answer. ~Baolan Solution 3Taking the average of the ?rst and last terms, and , we have that the mean of the set is . There are 5 values in each row,, or . ~Arctic_Bunny, edited by KINGLOGICcolumn or diagonal, so the value of the common sum is Video Solutionhttps://youtu.be/JEjib74EmiY~IceMatrixSee Also2024 AMC 10A (Problems ? Answer Key ? Resources (http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2024))Preceded?by Problem 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 AMC 10 Problems and Solutions
美国中学数学竞赛(真题加详解)2024 AMC-12A
