A 16 B 31 C 48 D 62 E 93 2011 AMC 10B AMC 12B Problem 13 Write

A 16 b 31 c 48 d 62 e 93 2011 amc 10b amc 12b problem

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(A) 16 (B) 31 (C) 48 (D) 62 (E) 93 2011 AMC 10B, Problem #21— 2011 AMC 12B, Problem #13— “Write the largest pairwise difference using the variables.” Solution Answer (B): The largest pairwise difference is 9, so w z = 9 . Let n be either x or y . Because n is between w and z , 9 = w z = ( w n ) + ( n z ) . Therefore the positive differences w n and n z must sum to 9. The given pairwise differences that sum to 9 are 3 + 6 and 4 + 5 . The remaining pairwise difference must be x y = 1 . The second largest pairwise difference is 6, so either w y = 6 or x z = 6 . In the first case the set of four numbers may be expressed as { w, w 5 , w 6 , w 9 } . Hence 4 w 20 = 44 , so w = 16 . In the second case w x = 3 , and the four numbers may be expressed as { w, w 3 , w 4 , w 9 } . Therefore 4 w 16 = 44 , so w = 15 . The sum of the possible values for w is 16 + 15 = 31 . Note: The possible sets of four numbers are { 16 , 11 , 10 , 7 } and { 15 , 12 , 11 , 6 } . Difficulty: Hard CCSS-M: A-CED.3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. 21
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A pyramid has a square base with sides 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? (A) 5 2 7 (B) 7 4 3 (C) 2 2 27 (D) 2 9 (E) 3 9 2011 AMC 10B, Problem #22— 2011 AMC 12B, Problem #18— “What is the height of the cube?” Solution Answer (A): Let A be the apex of the pyramid, and let the base be the square BCDE . Then AB = AD = 1 and BD = 2 , so BAD is an isosceles right triangle. Let the cube have edge length x . The intersection of the cube with the plane of BAD is a rectangle with height x and width 2 x . It follows that 2 = BD = 2 x + 2 x , from which x = 2 1 . 2 x 2 x A D B Hence the cube has volume ( 2 1) 3 = ( 2) 3 3( 2) 2 + 3 2 1 = 5 2 7 . OR Let A be the apex of the pyramid, let O be the center of the base, let P be the midpoint of one base edge, and let the cube intersect AP at Q . Let a coordinate plane intersect the pyramid so that O is the origin, A on the positive y -axis, and P = ( 1 2 , 0 ) . Segment AP is an altitude of a lateral side of the pyramid, so AP = 3 2 , and it follows that A = ( 0 , 2 2 ) . Thus the equation of line AP is y = 2 2 2 x . If the side length of the cube is s , then Q = ( s 2 , s ) , so s = 2 2 2 · s 2 . Solving gives s = 2 1 , and the result follows that in the first solution. Difficulty: Hard CCSS-M: G-GMD.4. Identify the shapes of two-dimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects. 22
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What is the hundreds digit of 2011 2011 ? (A) 1 (B) 4 (C) 5 (D) 6 (E) 9 2011 AMC 10B, Problem #23— “Express as (2000+11) 2011 and find all the terms divisible by 1000.” Solution Answer (D): In the expansion of (2000 + 11) 2011 , all terms except 11 2011 are divisible by 1000, so the hundreds digit of 2011 2011 is equal to that of 11 2011 . Furthermore, in the expansion of (10 + 1) 2011 , all terms except 1 2011 , ( 2011 1 ) (10) ( 1 2010 ) , and ( 2011 2 ) (10) 2 ( 1 2009 ) are divisible by 1000. Thus the hundreds digit of 2011 2011 is equal to that of 1 + ( 2011 1 ) (10) ( 1 2010 ) + ( 2011 2 ) (10) 2 ( 1 2009 ) = 1 + 2011 · 10 + 2011 · 1005 · 100 = 1 + 2011 · 100510 .
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