2018 AMC amc12a 真题 答案 详解

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中英双语真题

2018 AMC 12A

Problem 1
A large urn contains 100 balls, of which 36% are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be 72%? (No red balls are to be removed.)
(A) 28 (B) 32 (C) 36 (D) 50 (E) 64

Problem 2
While exploring a cave, Carl comes across a collection of 5-pound rocks worth $14 each, 4-pound rocks worth $11 each, and 1-pound rocks worth $2 each. There are at least 20 of each size. He can carry at most 18 pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?
(A) 48 (B) 49 (C) 50 (D) 51 (E) 52

Problem 3
How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.)
(A) 3 (B) 6 (C) 12 (D) 18 (E) 24

英文真题

2018 AMC 12A Problems Problem 1 A large urn contains 100 balls, of which 36% are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be 72%? (No red balls are to be removed.) (A) 28 (B) 32 (C) 36 (D) 50 (E) 64 Problem 2 While exploring a cave, Carl comes across a collection of 5-pound rocks worth $14 each, 4-pound rocks worth $11 each, and 1-pound rocks worth $2 each. There are at least 20 of each size. He can carry at most 18 pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave? (A) 48 (B) 49 (C) 50 (D) 51 (E) 52 Problem 3 How many ways can a student schedule 3 mathematics courses - algebra, geometry, and number theory - in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.) (A) 3 (B) 6 (C) 12 (D) 18 (E) 24 Problem 4 Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least 6 miles away," Bob replied, "We are at most 5 miles away." Charlie then remarked, "Actually the nearest town is at most 4 miles away." It turned out that none of the three statements were true. Let d be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of d? (A) (0, 4) (B) (4, 5) (C) (4, 6) (D) (5, 6) (E) (5, ∞) Problem 5 What is the sum of all possible values of k for which the polynomials x^2 − 3x + 2 and x^2 − 5x + k have a root in common? (A) 3 (B) 4 (C) 5 (D) 6 (E) 10 Problem 6 For positive integers m and n such that m + 10 < n + 1, both the mean and the median of the set {m, m+4, m+10, n+1, n+2, 2n} are equal to n. What is m + n? (A) 20 (B) 21 (C) 22 (D) 23 (E) 24 Problem 7 For how many (not necessarily positive) integer values of n is the value of 4000 · (2/5)^n an integer? (A) 3 (B) 4 (C) 6 (D) 8 (E) 9 Problem 8 All of the triangles in the diagram below are similar to isosceles triangle ABC, in which AB = AC. Each of the 7 smallest triangles has area 1, and △ABC has area 40. What is the area of trapezoid DBCE? (A) 16 (B) 18 (C) 20 (D) 22 (E) 24 Problem 9 Which of the following describes the largest subset of values of y within the closed interval [0, π] for which

真题文字详解(英文)

Problem 1 A large urn contains 100 balls, of which 36% are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be 72%? (No red balls are to be removed.) (A) 28 (B) 32 (C) 36 (D) 50 (E) 64 Solution 1 There are 36 red balls; for these red balls to comprise 72% of the urn, there must be only 14 blue balls. Since there are currently 64 blue balls, this means we must remove (D) 50 Solution 2 There are 36 red balls and 64 blue balls. For the percentage of the red balls to double from 36% to 72% of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is (D) 50 Solution 3 There are 36 red balls out of the total 100 balls. We want to continuously remove blue balls until the percentage of red balls in the urn is 72%. Therefore, we want $$\frac{36}{100 - x} = \frac{72}{100}$$ Solving for x gives that we must remove (D) 50 blue balls. Problem 2 While exploring a cave, Carl comes across a collection of 5-pound rocks worth $14 each, 4-pound rocks worth $11 each, and 1-pound rocks worth $2 each. There are at least 20 of each size. He can carry at most 18 pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave? (A) 48 (B) 49 (C) 50 (D) 51 (E) 52 Solution 1 The value of 5-pound rocks is $14 ÷ 5 = $2.80 per pound, and the value of 4-pound rocks is $11 ÷ 4 = $2.75 per pound. Clearly, Carl should not carry more than three 1 -pound rocks. Otherwise, he can replace some 1 -pound rocks with some heavier rocks, preserving the weight but increasing the total value. We perform casework on the number of 1 -pound rocks Carl can carry: 1-Pound Rocks ($2 Each) | 4-Pound Rocks ($11 Each) | 5-Pound Rocks ($14 Each) | Total Value 0 2 2 $50 1 3 1 $49 2 4 0 $48 3 0 3 $48 Clearly, the maximum value of the rocks Carl can carry is (C) 50 dollars. Remark Note that an upper bound of the total value is $2.80 · 18 = $50.40, from which we can eliminate choices (D) and (E). Solution 2 Since each rock is worth 1 dollar less than 3 times its weight (in pounds), the answer is just 3 · 18 = 54 minus the minimum number of rocks we need to make 18 pounds. Note that we need at least 4 rocks (two 5-pound rocks and two 4-pound rocks) to make 18 pounds, so the answer is 54 – 4 = (C) 50. Problem 3 The following problem is from both the 2018 AMC 10A #4 and 2018 AMC 12A #3, so both problems redirect to this page. How many ways can a student schedule 3 mathematics courses — algebra, geometry, and number theory — in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.) (A) 3 (B) 6 (C) 12 (D) 18 (E) 24 Solution 1 We must place the classes into the periods such that no two classes are in the same period or in consecutive periods. Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive: Periods 1, 3, 5

真题文字详解(中英双语)

Problem 1 A large urn contains 100 balls, of which 36% are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be 72%? (No red balls are to be removed.) 一个大urn里有100个球，其中36%是红色的，其余的是蓝色的。需要移除多少个蓝色球，才能使红色球在urn中的百分比变为72%？（不移动红色球。） (A) 28 (B) 32 (C) 36 (D) 50 (E) 64 Solution 1 There are 36 red balls; for these red balls to comprise 72% of the urn, there must be only 14 blue balls. Since there are currently 64 blue balls, this means we must remove (D) 50 有36个红色球；为了使红色球占urn的72%，只有14个蓝色球。由于目前有64个蓝色球，这意味着我们必须移除(D) 50个。 Solution 2 There are 36 red balls and 64 blue balls. For the percentage of the red balls to double from 36% to 72% of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is (D) 50 有36个红色球和64个蓝色球。为了使红色球在urn中的比例从36%增加到72%，必须移除一半的总球数。因此，需要移除的蓝色球数量是(D) 50。 Solution 3 There are 36 red balls out of the total 100 balls. We want to continuously remove blue balls until the percentage of red balls in the urn is 72%. Therefore, we want $$\frac{36}{100 - x} = \frac{72}{100}$$ Solving for x gives that we must remove (D) 50 blue balls. 有36个红色球，总共100个球。我们希望不断移除蓝球，直到红色球在urn中的百分比为72%。因此，求x给出我们必须移除(D) 50个蓝色球。 Problem 2 While exploring a cave, Carl comes across a collection of 5-pound rocks worth $14 each, 4-pound rocks worth $11 each, and 1-pound rocks worth $2 each. There are at least 20 of each size. He can carry at most 18 pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave? 在探索洞穴时，卡尔发现了一堆重5磅的石头，每块价值$14美元，重4磅的石头，每块价值$11美元，以及重1磅的石头，每块价值$2美元。每种石头的数量至少为20块。他最多能携带18磅的重量。他能带出的石头的最大价值是多少美元？ (A) 48 (B) 49 (C) 50 (D) 51 (E) 52 Solution 1 The value of 5-pound rocks is $14 ÷ 5 = $2.80 per pound, and the value of 4-pound rocks is $11 ÷ 4 = $2.75 per pound. Clearly, Carl should not carry more than three 1-pound rocks. Otherwise, he can replace some 1-pound rocks with some heavier rocks, preserving the weight but increasing the total value. 5磅石头的价值是每磅$14÷5=$2.80美元，4磅石头的价值是每磅$11÷4=$2.75美元。显然，卡尔不应该携带超过三块1磅的石头。否则，他可以用一些更重的石头替换一些1磅的石头，同时增加总价值。 We perform casework on the number of 1-pound rocks Carl can carry: 我们对卡尔能携带的1磅石头的数量进行分类讨论： | 1-Pound Rocks ($2 Each) | 4-Pound Rocks ($11 Each) | 5-Pound Rocks ($14 Each) | Total Value | |--------------------------|---------------------------|---------------------------|-------------| | 0 | 2 | 2 | $50 | | 1 | 3 | 1 | $49 | | 2 | 4 | 0 | $48 | | 3 | 0 | 3 | $48 | Clearly, the maximum value of the rocks Carl can carry is (C) 50 dollars. 显然，卡尔能携带的石头的最大价值是(C) 50美元。 Remark Note that an upper bound of the total value is $2.80 · 18 = $50.40, from which we can eliminate choices (D) and (E). 注意，总价值的上限是$2.80·18=$50