AMC8试题及答案.pdf

1、Copyright 2016 Art of Problem SolvingHow many square yards of carpet are required to cover a rectangular floor that is feet long and feetwide?(There are 3 feet in a yard.)First,we multiply to get that you need square feet of carpet you need to cover.Since thereare square feet in a square yard,you di

2、vide by to get square yards,so our answer is.Since there are feet in a yard,we divide by to get,and by to get.To find the area ofthe carpet,we then multiply these two values together to get.2015 AMC 8(Problems Answer Key Resources(http:/ First ProblemFollowedby Problem 21 2 3 4 5 6 7 8 9 10 11 12 13

3、 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(http:/www.maa.org)sAmerican Mathematics Competitions(http:/amc.maa.org).Placement:Easy GeometryRetrieved from http:/ AMC 8 Problems/Problem

4、1SolutionSolution 2See AlsoPoint is the center of the regular octagon,and is the midpoint of the side What fraction of the area of the octagon is shaded?1 Solution 12 Solution 23 Solution 34 See AlsoSince octagon is a regular octagon,it is split into equal parts,such as triangles,etc.These parts,sin

5、ce they are all equal,are of the octagon each.The shaded region consists of of these equal parts plus half of another,so the fraction of the octagonthat is shaded is 2015 AMC 8 Problems/Problem 2ContentsSolution 1Solution 2The octagon has been divided up into identical triangles(and thus they each h

6、ave equal area).Since theshaded region occupies out of the total triangles,the answer is.For starters what I find helpful is to divide the whole octagon up into triangles as shown here:Now it is just a matter of counting the larger triangles remember that and are notfull triangles and are only half

7、for these purposes.We count it up and we get a total of of the shapeshaded.We then simplify it to get our answer of.2015 AMC 8(Problems Answer Key Resources(http:/ Problem 1Followedby 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 SolutionsSol

8、ution 3See AlsoCopyright 2016 Art of Problem SolvingJack and Jill are going swimming at a pool that is one mile from their house.They leave homesimultaneously.Jill rides her bicycle to the pool at a constant speed of miles per hour.Jack walks tothe pool at a constant speed of miles per hour.How many

9、 minutes before Jack does Jill arrive?Using,we can set up an equation for when Jill arrives at the swimming pool:Solving for,we get that Jill gets to the pool in of an hour,which is minutes.Doing the same forJack,we get thatJack arrives at the pool in of an hour,which in turn is minutes.Thus,Jill ha

10、s to wait minutes for Jack to arrive at the pool.2015 AMC 8(Problems Answer Key Resources(http:/ Problem 2Followedby 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 SolutionsThe problems on this page are copyrighted by the Mathematical Associat

11、ion of America(http:/www.maa.org)sAmerican Mathematics Competitions(http:/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 3SolutionSee AlsoCopyright 2016 Art of Problem SolvingThe Centerville Middle School chess team consists of two boys and three girls.A photographer wants to takea pictur

12、e of the team to appear in the local newspaper.She decides to have them sit in a row with a boy ateach end and the three girls in the middle.How many such arrangements are possible?There are ways to order the boys on the end,and there are ways to order the girls in the middle.We get the answer to be

13、.2015 AMC 8(Problems Answer Key Resources(http:/ Problem 3Followedby 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 SolutionsThe problems on this page are copyrighted by the Mathematical Association of America(http:/www.maa.org)sAmerican Mathe

14、matics Competitions(http:/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 4SolutionSee AlsoCopyright 2016 Art of Problem SolvingBillys basketball team scored the following points over the course of the first 11 games of the season:If his team scores 40 in the 12th game,which of the followi

15、ng statistics will show an increase?When they score a on the next game,the range increases from to.Thismeans the increased.Because is less than the score of every game theyve played so far,the measures of center will neverrise.Only measures of spread,such as the,may increase.2015 AMC 8(Problems Answ

16、er Key Resources(http:/ Problem 4Followedby Problem 61 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(http:/www.maa.org)sAmerican Mathematics Competitions(http:

17、/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 5SolutionSolution 2Copyright 2016 Art of Problem SolvingIn,and.What is the area of?We know the semi-perimeter of is.Next,we use Herons Formula to findthat the area of the triangle is just.Splitting the isosceles triangle in half,we get a rig

18、ht triangle with hypotenuse and leg.Using thePythagorean Theorem,we know the height is.Now that we know the height,the area is.2015 AMC 8(Problems Answer Key Resources(http:/ Problem 5Followedby 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 S

19、olutionsThe problems on this page are copyrighted by the Mathematical Association of America(http:/www.maa.org)sAmerican Mathematics Competitions(http:/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 6Solution 1Solution 2See AlsoEach of two boxes contains three chips numbered,.A chip is dr

20、awn randomly from each box and thenumbers on the two chips are multiplied.What is the probability that their product is even?We can instead find the probability that their product is odd,and subtract this from.In order to get anodd product,we have to draw an odd number from each box.We have a probab

21、ility of drawing an odd numberfrom one box,so there is a probability of having an odd product.Thus,there is a probability of having an even product.You can also make this problem into a spinner problem.You have the first spinner with equally dividedsections,and You make a second spinner that is iden

22、tical to the first,with equal sections of,and.If the first spinner lands on,to be even,it must land on two.You write down the firstcombination of numbers.Next,if the spinner lands on,it can land on any number on the secondspinner.We now have the combinations of.Finally,if the first spinner endson,we

23、have Since there are possible combinations,and we have evens,the final answer is.2015 AMC 8(Problems Answer Key Resources(http:/ Problem 6Followedby 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 copyrigh

24、ted by the Mathematical Association of America(http:/www.maa.org)sAmerican Mathematics Competitions(http:/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 7SolutionSolution 2See AlsoCopyright 2016 Art of Problem SolvingWhat is the smallest whole number larger than the perimeter of any trian

25、gle with a side of length and aside of length?We know from the triangle inequality that the last side,fulfills.Adding to both sides of the inequality,we get,and because is the perimeter ofour triangle,is our answer.2015 AMC 8(Problems Answer Key Resources(http:/ Problem 7Followedby Problem 91 2 3 4

26、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(http:/www.maa.org)sAmerican Mathematics Competitions(http:/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 8S

27、olutionSee AlsoCopyright 2016 Art of Problem SolvingOn her first day of work,Janabel sold one widget.On day two,she sold three widgets.On day three,shesold five widgets,and on each succeeding day,she sold two more widgets than she had sold on the previousday.How many widgets in total had Janabel sol

28、d after working days?The sum of is The sum is just the sum of the first odd integers,which is 2015 AMC 8(Problems Answer Key Resources(http:/ Problem 8Followedby 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 SolutionsThe problems on this pag

29、e are copyrighted by the Mathematical Association of America(http:/www.maa.org)sAmerican Mathematics Competitions(http:/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 9Solution 1Solution 2See AlsoCopyright 2016 Art of Problem SolvingHow many integers between and have four distinct digits?

30、The question can be rephrased to How many four-digit positive integers have four distinct digits?,sincenumbers between and are four-digit integers.There are choices for the first number,sinceit cannot be,there are only choices left for the second number since it must differ from the first,choices fo

31、r the third number,since it must differ from the first two,and choices for the fourth number,since it must differ from all three.This means there are integersbetween and with four distinct digits.2015 AMC 8(Problems Answer Key Resources(http:/ Problem 9Followedby Problem 111 2 3 4 5 6 7 8 9 10 11 12

32、 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(http:/www.maa.org)sAmerican Mathematics Competitions(http:/amc.maa.org).Retrieved from http:/ AMC 8 Problems/Problem 10Solution 1See Also

33、Copyright 2016 Art of Problem SolvingIn the small country of Mathland,all automobile license plates have four symbols.The first must be a vowel(A,E,I,O,or U),the second and third must be two different letters among the 21 non-vowels,and thefourth must be a digit(0 through 9).If the symbols are chosen at random subject to these conditions,whatis the probability that the plate will read AMC8?There i