2004 AMC amc12a 真题 答案 详解

英文真题

2004 AMC 12A Problems Problem 1 Alicia earns 20 dollars per hour, of which 1.45% is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes? (A) 0.0029 (B) 0.029 (C) 0.29 (D) 2.9 (E) 29 Problem 2 On the AMC 12, each correct answer is worth 6 points, each incorrect answer is worth 0 points, and each problem left unanswered is worth 2.5 points. If Charlyn leaves 8 of the 25 problems unanswered, how many of the remaining problems must she answer correctly in order to score at least 100? (A) 11 (B) 13 (C) 14 (D) 16 (E) 17 Problem 3 For how many ordered pairs of positive integers (x, y) is x + 2y = 100? (A) 33 (B) 49 (C) 50 (D) 99 (E) 100 Problem 4 Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters? (A) 22 (B) 23 (C) 24 (D) 25 (E) 26 Problem 5 The graph of the line y = mx + b is shown. Which of the following is true? (A) mb < -1 (B) -1 < mb < 0 (C) mb = 0 (D) 0 < mb < 1 (E) mb > 1 Problem 6 Let U = 2 · 2004^2005, V = 2004^2005, W = 2003 · 2004^2004, X = 2 · 2004^2004, Y = 2004^2004 and Z = 2004^2003 . Which of the following is the largest? (A) U - V (B) V - W (C) W - X (D) X - Y (E) Y - Z Problem 7 A game is played with tokens according to the following rules. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players A, B and C start with 15, 14 and 13 tokens, respectively. How many rounds will there be in the game? (A) 36 (B) 37 (C) 38 (D) 39 (E) 40 Problem 8 In the overlapping triangles △ABC and △ABE sharing common side AB, ∠EAB and ∠ABC are right angles, AB = 4, BC = 6, AE = 8, and AC and BE intersect at D. What is the difference between the areas of △ADE and △BDC? (A) 2 (B) 4 (C) 5 (D) 8 (E) 9

真题文字详解(英文)

Problem 1 The following problem is from both the 2004 AMC 12A #1 and 2004 AMC 10A #3, so both problems redirect to this page. Alicia earns 20 dollars per hour, of which 1.45% is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes? (A) 0.0029 (B) 0.029 (C) 0.29 (D) 2.9 (E) 29 Solution 20 dollars is the same as 2000 cents, and 1.45% of 2000 is 0.0145 × 2000 = 29 cents. →(E) 29 Alternate Solution Since there can't be decimal values of cents, the answer must be →(E) 29 Problem 2 On the AMC 12, each correct answer is worth 6 points, each incorrect answer is worth 0 points, and each problem left unanswered is worth 2.5 points. If Charlyn leaves 8 of the 25 problems unanswered, how many of the remaining problems must she answer correctly in order to score at least 100? (A)11 (B)13 (C)14 (D)16 (E)17 Solution She gets 8 * 2.5 = 20 points for the problems she didn't answer. She must get [100 - 20] / 6 = 14 →(C) problems right to score at least 100. Problem 3 For how many ordered pairs of positive integers (x, y) is x + 2y = 100? (A)33 (B)49 (C)50 (D)99 (E)100 Solution 1 Every integer value of y leads to an integer solution for x. Since y must be positive, y ≥ 1 Also, y = (100 - x) / 2 Since x must be positive, y < 50 1 ≤ y < 50 This leaves 49 values for y, which mean there are 49 solutions →(B) Solution 2 If x and 2y must each be positive integers, then we can say that x is at least 1 and 2y is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer adds to 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because y must be an integer, 2y must be even; thus only [99 / 2] = 49 ways exist to distribute these ones. Problem 4 The following problem is from both the 2004 AMC 12A #4 and 2004 AMC 10A #6, so both problems redirect to this page. Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and granddaughters have no daughters? (A) 22 (B) 23 (C) 24 (D) 25 (E) 26 Solutions Solution 1 Since Bertha has 6 daughters, she has 30 - 6 = 24 granddaughters, of which none have daughters. Of Bertha's daughters, 24/6 = 4 have daughters, so 6 - 4 = 2 do not have daughters. Therefore, of Bertha's daughters and granddaughters, 24 + 2 = 26 do not have daughters. →(E) 26 Solution 2 (From Alcumus) Bertha has 30 - 6 = 24 granddaughters, none of whom have any daughters. The granddaughters are the children of 24/6 = 4 of Bertha's daughters, so the number of women having no daughters is 30 - 4 = [26] Solution 3 Draw a tree diagram and see that the answer can be found in the sum of 6 + 6 granddaughters, 5 + 5 daughters, and 4 more daughters. Adding them together gives the answer of →(E) 26 Problem 5 The graph of the line y = mx + b is shown. Which of the following is true?

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2004 AMC 12A #1 and 2004 AMC 10A #3, so both problems redirect to this page. Alicia earns 20 dollars per hour, of which 1.45% is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes? (A) 0.0029 (B) 0.029 (C) 0.29 (D) 2.9 (E) 29 Solution 20 dollars is the same as 2000 cents, and 1.45% of 2000 is 0.0145 × 2000 = 29 cents. → (E) 29 20美元相当于2000分，1.45%的2000是0.0145×2000=29分。→(E) 29 Alternate Solution Since there can't be decimal values of cents, the answer must be → (E) 29 由于不能有小数值，答案必须是→(E) 29 Problem 2 On the AMC 12, each correct answer is worth 6 points, each incorrect answer is worth 0 points, and each problem left unanswered is worth 2.5 points. If Charlyn leaves 8 of the 25 problems unanswered, how many of the remaining problems must she answer correctly in order to score at least 100? (A)11 (B)13 (C)14 (D)16 (E)17 Solution She gets 8 * 2.5 = 20 points for the problems she didn't answer. She must get (100 - 20)/6 = 14 → (C) problems right to score at least 100. 她未回答的问题得8*2.5=20分。她必须正确回答(100-20)/6=14个问题才能至少得100分。 Problem 3 For how many ordered pairs of positive integers (x, y) is x + 2y = 100? 正整数(x,y)的有序对有多少个使得x+2y=100? (A)33 (B)49 (C)50 (D)99 (E)100 Solution 1 Every integer value of y leads to an integer solution for x. Since y must be positive, y ≥ 1. y的每个整数值都导致x的整数解。由于y必须是正数，所以y≥1。 Also, y = (100 - x)/2 Since x must be positive, y < 50 另外，y=(100-x)/2由于x必须是正数，y<50 1 ≤ y < 50 This leaves 49 values for y, which means there are 49 solutions to the equation → (B) 1≤y<50这留下了49个y的值，这意味着方程有49个解→(B)有49个解 Solution 2 If x and 2y must each be positive integers, then we can say that x is at least 1 and 2y is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer adds up to 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because y must be an integer, 2y must be even; thus only [99/2] = 49 ways exist to distribute these ones. 如果x和2y必须是正整数，那么我们可以说x至少为1，2y至少为1。从那里，我们想要找出将其他98个（最小的正整数加起来等于100）分配到两个不同的箱子中的方法。98个相同的对象可以以99种方式（想想星星和棒子）分配到两个不同的箱子里，但这个99是一个过度计数。因为y必须是整数，2y必须是偶数；因此只有[99/2]=49种方法可以分配这些对象。 Problem 4 The following problem is from both the 2004 AMC 12A #4 and 2004 AMC 10A #6, so both problems redirect to this page.