2004 AMC 12B Solutions

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Solutions 2004 55 th AMC 12 B 21. (A) At Jenny’s fourth practice she made 1 2(48) = 24 free throws. At her thirdpractice she made 12, at her second practice she made 6, and at her first practiceshe made 3.2. (D) If d ≠ 0, the value of the expression can be increased by interchanging 0with the value of d. Therefore the maximum value must occur when d = 0. Ifa = 1, the value is c, which is 2 or 3. If b = 1, the value is c · a = 6. If c = 1,the value is a b , which is 2 3 = 8 or 3 2 = 9. Thus the maximum value is 9.3. (A) Note that 1296 = 6 4 = 2 4 3 4 , so x = y = 4 and x + y = 8.4. (B) There are 90 possible choices for x. Ten of these have a units digit of 7,and nine have a tens digit of 7. Because 77 has been counted twice, there are10 + 9 − 1 = 18 choices of x for which at least one digit is a 7. Therefore theprobability is 1890 = 1 5 .5. (A) Isabella received 10d/7 Canadian dollars at the border and spent 60 ofthem. Thus 10d/7 − 60 = d, from which it follows that d = 140, and the sum ofthe digits of d is 5.6. (A) Let downtown St. Paul, downtown Minneapolis, and the airport be locatedat S, M, and A, respectively. Then △MAS has a right angle at A, so by thePythagorean Theorem,MS = √ 10 2 + 8 2 = √ 164 ≈ √ 169 = 13.7. (B) The areas of the regions enclosed by the square and the circle are 10 2 = 100and π(10) 2 = 100π, respectively. One quarter of the second region is alsoincluded in the first, so the area of the union is100 + 100π − 25π = 100 + 75π.8. (D) If there are n rows in the display, the bottom row contains 2n − 1 cans.The total number of cans is therefore the sum of the arithmetic series1 + 3 + 5 + · · · + (2n − 1),which isThus n 2 = 100, so n = 10.n2 [(2n − 1) + 1] = n2 .
Solutions 2004 55 th AMC 12 B 39. (E) The rotation takes (−3, 2) into B = (2, 3), and the reflection takes B intoC = (3, 2).A(-3,2)(2,3)BC(3,2)10. (A) The area of the annulus is the difference between the areas of the two circles,which is πb 2 − πc 2 . Because the tangent XZ is perpendicular to the radius OZ,b 2 − c 2 = a 2 , so the area is πa 2 .11. (D) Each score of 100 is 24 points above the mean, so the five scores of 100represent a total of (5)(24) = 120 points above the mean. Those scores must bebalanced by scores totaling 120 points below the mean. Since no student scoredmore than 76 − 60 = 16 points below the mean, the number of other students inthe class must be an integer no less than 120/16. The smallest such integer is8, so the number of students in the class is at least 13. Note that the conditionsof the problem are met if 5 students score 100 and 8 score 61.ORIf there are k students in the class, the sum of their scores is 76k. If the fivescores of 100 are excluded, the sum of the remaining scores is 76k − 500. Sinceeach student scored at least 60, the sum is at least 60(k − 5). Thus76k − 500 ≥ 60(k − 5),so k ≥ 12.5. Since k must be an integer, k ≥ 13.12. (C) Let a k be the k th term of the sequence. For k ≥ 3,a k+1 = a k−2 + a k−1 − a k , so a k+1 − a k−1 = −(a k − a k−2 ).Because the sequence begins2001, 2002, 2003, 2000, 2005, 1998, . . . ,it follows that the odd-numbered terms and the even-numbered terms each formarithmetic progressions with common differences of 2 and −2, respectively. The2004 th term of the original sequence is the 1002 nd term of the sequence 2002,2000, 1998,. . ., and that term is 2002 + 1001(−2) = 0.
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