2015amc8试题与答案.pdf

Copyright © 2016 Art of Problem Solving How many square yards of carpet are required to cover a rectangular floor that is feet long and feet wide? (There are 3 feet in a yard.) First, we multiply to get that you need square feet of carpet you need to cover. Since there are square feet in a square yard, you 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 of the carpet, we then multiply these two values together to get . 2015 AMC 8 (Problems • Answer Key • Resources ( However, all of our slips are bigger than , so this 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 still manageable. It must have a value of , so adding the slip leaves room for . This looks good at first, as we do have slips smaller than that, but upon closer inspection, we see that no slip fits exactly, and the smallest sum of two slips is , which is too big, so this case is also impossible. 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 all other choices removed, we are left with the answer: Cup 2015 AMC 8 (Problems • Answer Key • Resources ( Preceded by Problem 22 Followed by Problem 24 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American Mathematics Competitions (http://amc.maa.org). Retrieved from “ title=2015_AMC_8_Problems/Problem_23&oldid=73520“ 2015 AMC 8 Problems/Problem 23 Solution See Also Copyright © 2016 Art of Problem Solving A baseball league consists of two four-team divisions. Each team plays every other team in its division games. Each team plays every team in the other division games with and . Each team plays 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 doesn't work because is not divisible by . Next , does not work because is not divisible by We try this does work giving and thus games in their division. , 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 ( Preceded by Problem 23 Followed by Problem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America (http://www.maa.org)'s American Mathematics Competitions (http://amc.maa.org). Retrieved from “ title=2015_AMC_8_Problems/Problem_24&oldid=79168“ 2015 AMC 8 Problems/Problem 24 Solution 1 Solution 2 See Also One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space? 1 Solution 1 2 Solution 2 3 Solution 3 4 See Also We can draw a diagram as shown. Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the big triangles by Let the height of a big triangle be then . 2015 AMC 8 Problems/Problem 25 Contents Solution 1 Thus , because by symmetry, . This means the area of each triangle is This the area of the square 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 and the four red triangles. The red triangles have base and height , so the combined area of the four triangles is . The area of the smaller square is . We add these to see that the area of the large square is . Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length , and let's label the other legs for one of the triangles and for the other. Note that . The area of each of the triangles is and , and there are of each. So now we need to find . Solution 2 Solution 3 Copyright © 2016 Art of Problem Solving Remember that , so substituting this in we find that the area of all of the triangles is . The area of the unit squares is , so the area of the square we need is 2015 AMC 8 (Problems • Answer Key • Resources ( Preceded by Problem 24 Followed by La。