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, yo
2、u divide 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:/ However, all of our slipsare bigger than , so th
3、is is impossible. Cup has a sum of , but we are told that it already has a slip, leaving , which is too small for the slip. Cup is a little bit trickier, but stillmanageable. It must have a value of , so adding the slip leaves room for . This looksgood at first, as we do have slips smaller than that
4、, but upon closer inspection, we see that no slip fitsexactly, and the smallest sum of two slips is , which is too big, so this case is alsoimpossible. Cup has a sum of , but we are told it already has a slip, so we are left with , which is identical to the Cup C case, and thus also impossible. With
5、 all other choicesremoved, we are left with the answer: Cup 2015 AMC 8 (Problems Answer Key Resources(http:/ by Problem 22Followed by Problem 241 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 Mat
6、hematical Association of America (http:/www.maa.org)sAmerican Mathematics Competitions (http:/amc.maa.org). Retrieved from “http:/ AMC 8 Problems/Problem 23SolutionSee AlsoCopyright 2016 Art of Problem SolvingA baseball league consists of two four-team divisions. Each team plays every other team in
7、its division games. Each team plays every team in the other division games with and . Each teamplays a 76 game schedule. How many games does a team play within its own division?On one team they play games in their division and games in the other. This gives Since we start by trying . This doesnt wor
8、k because is not divisible by .Next , does not work because is not divisible by We try this does work giving and thus games in theirdivision., giving . Since , we have Since is , we must have equal to , so .This gives , as desired. The answer is .2015 AMC 8 (Problems Answer Key Resources(http:/ by P
9、roblem 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 (http:/www.maa.org)sAmerican Mathematics Competitions (http:/amc.maa.org). Retri
10、eved from “http:/ AMC 8 Problems/Problem 24Solution 1Solution 2See AlsoOne-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of thelargest square that can be fitted into the remaining space?1 Solution 12 Solution 23 Solution 34 See AlsoWe can draw a diagr
11、am as shown.Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit squarebetween the inscribed square, are similiar to the big triangles by Let the height of a big trianglebe then .2015 AMC 8 Problems/Problem 25ContentsSolution 1Thus , because by symmetry
12、, .This means the area of each triangle is This the area of thesquare is We draw a square as shown:We wish to find the area of the square. The area of the larger square is composed of the smaller square andthe four red triangles. The red triangles have base and height , so the combined area of the f
13、ourtriangles is . The area of the smaller square is . We add these to see that the area of thelarge square is .Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They bothhave one leg of length , and lets label the other legs for one of the triangles a
14、nd for the other.Note that . The area of each of the triangles is and , and there are of each. So now weneed to find .Solution 2Solution 3Copyright 2016 Art of Problem SolvingRemember that , so substituting this inwe find that the area of all of the triangles is . The area of the unit squares is , s
15、othe area of the square we need is 2015 AMC 8 (Problems Answer Key Resources(http:/ by Problem 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 (http:/www.maa.org)sAmerican Mathematics Competitions (http:/amc.maa.org). Thank you for reading the solutions of the 2015 AMC 8 Problems made by people on AoPS.Retrieved from “http:/ Also
