Academic Challenge 2009 Regional Mathematics Solution Set. #2 Ans. C. Let a be the side of the cube. Then its surface area equals 6a = 10, so

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Meghan Hodges
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1 Academic Challenge 009 Regional Mathematics Solution Set #1 Ans. C: x 4 = x 9 = -5 # Ans. C. Let a be the side of the cube. Then its surface area equals 6a = 10, so a = 10 / 6 and volume V = a = ( 10 / 6 ).15 (in ). 4 5 # Ans D: The series is a, ar, ar, ar, ar, ar,., with a = 5 and r =. Rather than finding them separately and adding, the sum of the first n numbers is n 0 ar ( 1) 5( 1) = = r 1 1 #4 Ans: C: 1 Ο sin (10 /14) = 45.6 #5 Ans A. The area of a triangle equals A = r s, where r is the radius of the inscribed circle and s is the semi-perimeter. The hypotenuse of the triangle is = 17, so the semi-perimeter equals ( )/ = 0. The area of the triangle equals A = ½ 8 15 = 60. Thus, the radius of the inscribed circle equals 60/0 =. #6 Ans A: Let x be the radius of the circle. A line going from the right angle to the tangent on the long side would have a length of x+ x = x (1+ ). This can act as the height of the triangle, and the base (long side), would be double this. This makes 1 the area of the triangle ( x(1+ ) ) ( x (1+ ) ), or x (1+ ). Since the area of the circle is π x, dividing x (1+ ) by π x gives us the ratio of 1 to #7 Ans. E: P(at least 1 boy) = 1 P(no boys)=1 1/8 = 7/8 #8 Ans. E. Since 5 π /< β < 7 π /, cos β < 0. Therefore, cos β + sin β = 1 implies that cos β = 1 sin β = 1 (5 / 1) = 1 / 1. Therefore, tan β = sin β / cos β = 5 / 1. #9 Ans B: We can split the hexagon into six equilateral triangles each with sides of 10 inches. Because all of the triangle angles are 60 degree angles, the height of each triangle is 5. The area of each triangle is therefore = 4., so the hexagon is 6*4.01, or 59.8 square inches #10 Ans C: log() / log(1.04) = 17.7

2 #11 Ans. B. Let x x = t, then ( t 1)( t + 4) < 0 4< t < 1 4< < 1 x < 0. #1 Ans A: If a = semi major axis length, b = semi minor axis length and c = focal length, then a = 1, c/a = 0.8, c = 0.8, and b = = 0.6. The area of the ellipse is π ab = 1.88 square inches. #1 Ans. B: Since ϕ ϕ = 1, ϕ 1 = ϕ 1. Therefore, ( ) ϕ = ( ϕ 1) = ϕ ϕ + 1= ϕ + 1 ϕ + 1= ϕ. #14 Ans A: ( ) x + + x x + x f( x) = x + x = x + x = = x + + x x + + x x + + x as x. Therefore, y = is the equation of the horizontal asymptote. #15 Ans B: amount = rate * time. (1/0) * t + (1/0) * t = 00, so t =,600 minutes #16 Ans. D: ( 75 m/ h) (580 f / m) (1 h/ min) (1min/ 60sec) = 110 f / sec #17 Ans. B. The derivative of the function f( x) equals x, so the slope of the tangent at x = 1 equals f (1) = 1 = 1. Therefore, the slope of the line is 1: y = x + b. Since it passes (, 5), the value of b can be found from the equation 5 = + b. Therefore b = and y = x + #18 Ans A: ± ± 16 ± 4i x = = = = 1 ± i *1 Ο #19 Ans A: htan(40 ) = 0 and h =.84 #0 Ans. D: x 14x + 49 y xy + x = ( x 7) ( y x) = x 7 y x. Since x > y and x < 7, x 7 y x = 7 x ( x y) = 7 x + y #1 Ans B: By looking at I, II, and III, we should notice that the three brothers must take turns sitting out for each of the first three. The set of two brothers sit out the fourth and the fifth time. Since Ed sits out the third time, he is one of the three brothers. Chris is in both the fourth and fifth time, so he is also one of the three brothers. Since Chris rides with his cousin Dave in the fifth ride, and his only other cousin in the third, he must be sitting out in the second and riding with his brother Bob in the fourth. This makes Bob the third brother, who sits out in the first ride. For the sake of completeness, the list is: 1 st : B out, C&E, A&D. nd : C out, B&A, D&E. rd : E out, A&C, B&D. 4 th : D out, B&C, E&A. 5 th : A out, B&E, C&D # Ans A: 500*5 1*500-*00-*100 = = 100 net

3 # Ans. D i 1 i ( 1 + i ) = = = = = = ( 1 + i) ( 1 + i) ( 1 + i) i( 1 + i) (1 + i) 1+ i 1 i 4 #4 Ans D: Let P be the center of the right circle, T be the point of tangency for AE, and x be the radius of the right circle. This means TP = x, DP = x, AP = 5x, and AC = 6x. Let s give angle TAP the name α, since we will use it multiple times. TP x sinα = = = 0., so α degrees (be careful with the rounding!). AP 5x EC tanα =. This means EC AC = AC tan α = 6x = 1.47x. The ratio of DP to EC is 1 to 1., and this is the same ratio for DF to EG. #5 Ans C: This is a permutation: P(7,4) = 840 x 8 ( x )( x + x + 4) #6 Ans. B lim = lim = lim( x + x + 4) = = 1 x sin( x ) x sin( x ) x x x t (since lim = lim = lim = 1). x sin( x ) x 0 sin( x ) t 0 sint #7 Ans B: Rearrange d = r * t to get t = d/r. Time up is ¼ / = 1/8 hours, time down is ¼ / 4 = 1/16 hours. Total time is /16 of an hour #8 Ans B: SA = π rl + ( πr ) = π 5 + π 4 = 8π #9 Ans. C: The radius of the cylinder is the leg of the right triangle with the hypotenuse 6 ft and the other leg ft: r = 6 = 7 = (ft). The lateral area equals the product of the circumference and the height: A = πr h = π 6 = 6π (ft ). ( 10 ) #0 Ans B: Let 1990 be t = 0. If Pt ( ) = (1.1) t, then P(19) gives us the 10 population. An alternate method is = e k, k = , then evaluate P = e. Be careful with rounding! #1 Ans E: We essentially have two triangles that will set up two tangent equations. h The left triangle will give us tan6 = x and the right one gives us = h tan74 0 x. Now solve for x and h: x tan6 = h, (0 x)tan74 = h, xtan6 = (0 x )tan74, x 0 tan74 = tan6 + tan74 =, plug this into h = x tan 6 to get h = * tan6 = (All angles were given in degree measure).. # Ans. C: Let us square sinα + cosα = 4 / : sin α + sinα cosα + cos α = 16 / 9. Therefore, 1+ sinα = 16/9, so sinα = 7 / 9. L

4 # Ans E: BA 1,1 = 0*1+ 1* 1= 1, BA1, = 0* + ( 1)* x = x, BA,1 = 1*1+ * 1= 1 BA, = 1* + * x = x +, so #4 Ans D: 1 x BA = + 1 x 1 1 #5 Ans. B: r = sec θ = = r cos θ r sin θ = 1 cos θ cos θ sin θ x y = 1, #6 Ans A: Solve 9.8t + 10.t = 1. Students with graphing calculator can use a graphing method. Otherwise, 9.8t + 10.t = 0, use quadratic formula to get 10. ± ( 9.8) 0.7 t =, t = -0.1 and 1.1, only 1.1 works. ( 9.8) #7 Ans E: (1/)[(/6)+(1/5)] = 4/15 #8 Ans. C: First, c can only be nonnegative since it is equal to the absolute value of a number. x = c x = ± c x = ± c. If c =, then the equation has exactly solutions: x = 6 or x = 0. If c >, then the equation has exactly two solutions: x = + c x =± ( + c ). If c = 0, then the equation has exactly two solutions: x =±. If 0 < c <, then the equation has four solutions: x = ± ( ± c ). The problem can also be solved by constructing the graph of the function f( x) = x. 4 #9 Ans E: V = πr, = 1000 π = r 6.05, SA = 4πr = 4 π(6.05) = #40 Ans D: y = () = Sectional Mathematics #1

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MLC Practice Final Exam

MLC Practice Final Exam Name: Section: Recitation/Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through. Show all your work on the standard response