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...
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)8762007 AMC 10 B, Problem #1—“Paintonlythewalls,thatmeansnoceilingsandfloors.”Difficulty:Medium-easyNCTM 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
Define the operation?bya ? b= (a+b)b. What is(3?5)-(5?3)?(A)-16(B)-8(C)0(D)8(E)162007 AMC 10 B, Problem #2—“Plug in 3, 5 for a, b.”SolutionAnswer (E):Since3?5 = (3 + 5)5 = 8·5 = 40and5?3 = (5 + 3)3 =8·3 = 24, we have3?5-5?3 = 40-24 = 16.Difficulty:EasyNCTM Standard:Number and Operations Standard:Understand meanings of operations andhow they relate to one another.Mathworld.com Classification:Recreational Mathematics>Cryptograms>Cryptarithmetic
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:MediumNCTM Standard:Algebra Standard: Analyze change in various contexts.Mathworld.com Classification:Geometry>Distance
The pointOis the center of the circle circumscribedabout4ABC, with∠BOC= 120◦and∠AOB=140◦, as shown.What is the degree measure of∠ABC?120140ACOB(A)35(B)40(C)45(D)50(E)602007 AMC 10 B, Problem #4—“OA=OB=OC.”SolutionAnswer (D):SinceOA=OB=OC, trianglesAOB,BOC, andCOAare all isosceles. Hence∠ABC=∠ABO+∠OBC=180◦-140◦2+180◦-120◦2= 50◦.ORSince∠AOC= 360◦-140◦-120◦= 100◦,the Central Angle Theorem implies that∠ABC=12∠AOC= 50◦.Difficulty:MediumNCTM 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>CirclesGeometry>Plane Geometry>Triangles>Special Triangles>Other Triangles>Triangle