2007 AMC 10B ws-25.pdf - Isabella\u2019s house has 3 bedrooms Each bedroom is 12 feet long 10 feet wide and 8 feet high Isabella must paint the walls of

2007 AMC 10B ws-25.pdf - Isabellau2019s house has 3...

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Isabella’s house has 3 bedrooms. Each bedroom is 12feet long, 10 feet wide, and 8 feet high. Isabella mustpaint the walls of all the bedrooms.Doorways andwindows, which will not be painted, occupy 60 squarefeet in each bedroom. How many square feet of wallsmust be painted? (A)678(B)768(C)786(D)867(E)876 2007 AMC 10 B, Problem #1“Paintonlythewalls,thatmeansnoceilingsandfloors.” Difficulty: Medium-easy NCTM Standard: Geometry Standard: analyze characteristics and properties of two- and three- dimensional geometric shapes and develop mathematical arguments about geometric relationships. Mathworld.com Classification: Geometry > Solid Geometry > Polyhedra > Cubes
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Define the operation ? by a ? b = ( a + b ) b . What is (3 ? 5) - (5 ? 3) ? (A) - 16 (B) - 8 (C) 0 (D) 8 (E) 16 2007 AMC 10 B, Problem #2 “Plug in 3, 5 for a, b.” Solution Answer (E): Since 3 ? 5 = (3 + 5)5 = 8 · 5 = 40 and 5 ? 3 = (5 + 3)3 = 8 · 3 = 24 , we have 3 ? 5 - 5 ? 3 = 40 - 24 = 16 . Difficulty: Easy NCTM Standard: Number and Operations Standard: Understand meanings of operations and how they relate to one another. Mathworld.com Classification: Recreational Mathematics > Cryptograms > Cryptarithmetic
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A college student drove his compact car 120 mileshome for the weekend and averaged 30 miles pergallon.On the return trip the student drove hisparents’ SUV and averaged only 20 miles per gallon.What was the average gas mileage, in miles per gallon,for the round trip? 2007 AMC 10 B, Problem #3“Solve for total distance and total number of gallonsused.” Difficulty: Medium NCTM Standard: Algebra Standard: Analyze change in various contexts. Mathworld.com Classification: Geometry > Distance
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The pointOis the center of the circle circumscribedabout4ABC, withBOC= 120andAOB=140, as shown.What is the degree measure ofABC? 120 140 A C O B (A) 35 (B) 40 (C) 45 (D) 50 (E) 60 2007 AMC 10 B, Problem #4 OA = OB = OC .” Solution Answer (D): Since OA = OB = OC , triangles AOB , BOC , and COA are all isosceles. Hence ABC = ABO + OBC = 180 - 140 2 + 180 - 120 2 = 50 . OR Since AOC = 360 - 140 - 120 = 100 , the Central Angle Theorem implies that ABC = 1 2 AOC = 50 . Difficulty: Medium NCTM Standard: Geometry Standard: Analyze characteristics and properties of two- and three- dimensional geometric shapes and develop mathematical arguments about geometric relationships. Mathworld.com Classification: Geometry > Plane Geometry > Circles Geometry > Plane Geometry > Triangles > Special Triangles > Other Triangles > Triangle
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