2011 AMC amc12b 真题 答案 详解

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中英双语真题

2011 AMC12B 2011 AMC12B Problem 1 What is $$\frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}$$? 表达式的值是多少 $$\frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}$$ (A) -1 (B) (\frac{5}{36}) (C) (\frac{7}{12}) (D) (\frac{147}{60}) (E) (\frac{43}{3}) Problem 2 Josanna's test scores to date are 90, 80, 70, 60, and 85. Her goal is to raise her test average at least 3 points with her next test. What is the minimum test score she would need to accomplish this goal? Josanna目前的考试成绩是90、80、70、60和85，她的目标是靠下次考试将她的考试平均分提高至少3分，那么她下次至少需要考多少分才能完成这个目标？ (A) 80 (B) 82 (C) 85 (D) 90 (E) 95 Problem 3 LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid A dollars and Bernardo had paid B dollars, where A < B. How many dollars must LeRoy give to Bernardo so that they share the costs equally? LeRoy 和 Bernardo 一起出去旅游了一周，并且商定两人均摊费用，在这一周中，他们都支付了各种共同支出，例如汽油和租车费用。旅游结束后，发现 LeRoy 花费 A 美元，Bernardo 花费 B 美元，这里 A<B，那么 LeRoy 需要支付 Bernardo 多少美元才能保证他们均摊了费用？ (A) (\frac{A+B}{2}) (B) (\frac{A-B}{2}) (C) (\frac{B-A}{2}) (D) (B-A) (E) (A+B)

英文真题

2011 AMC 12B Problems Problem 1 What is \frac{2+4+6}{1+3+5}-\frac{1+3+5}{2+4+6}? (A) -1 (B) \frac{5}{36} (C) \frac{7}{12} (D) \frac{147}{60} (E) \frac{43}{3} Problem 2 Josanna's test scores to date are 90, 80, 70, 60, and 85. Her goal is to raise her test average at least 3 points with her next test. What is the minimum test score she would need to accomplish this goal? (A) 80 (B) 82 (C) 85 (D) 90 (E) 95 Problem 3 LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid A dollars and Bernardo had paid B dollars, where A < B. How many dollars must LeRoy give to Bernardo so that they share the costs equally? (A) \frac{A+B}{2} (B) \frac{A-B}{2} (C) \frac{B-A}{2} (D) B-A (E) A+B Problem 4 In multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His erroneous product was 161. What is the correct value of the product of a and b? (A) 116 (B) 161 (C) 204 (D) 214 (E) 224 Problem 5 Let N be the second smallest positive integer that is divisible by every positive integer less than 7. What is the sum of the digits of N? (A) 3 (B) 4 (C) 5 (D) 6 (E) 9 Problem 6 Two tangents to a circle are drawn from a point A. The points of contact B and C divide the circle into arcs with lengths in the ratio 2:3. What is the degree measure of ∠BAC? (A) 24 (B) 30 (C) 36 (D) 48 (E) 60 Problem 7 Let x and y be two-digit positive integers with mean 60. What is the maximum value of the ratio \frac{x}{y}? (A) 3 (B) \frac{33}{7} (C) \frac{39}{7} (D) 9 (E) \frac{99}{10} Problem 8 Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has width 6 meters, and it takes her 36 seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second? (A) \frac{\pi}{3} (B) \frac{2\pi}{3} (C) \pi (D) \frac{4\pi}{3} (E) \frac{5\pi}{3}

真题文字详解(英文)

Problem 1 What is \frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}?
(A) -1 (B) \frac{5}{36} (C) \frac{7}{12} (D) \frac{147}{60} (E) \frac{43}{3}

Solutions
Solution 1
Add up the numbers in each fraction to get \frac{12}{9} - \frac{9}{12}, which equals \frac{4}{3} - \frac{3}{4}. Doing the subtraction yields \frac{7}{12}.

Solution 2
Notice that the numerators and denominators of each expression are 3-term arithmetic series. The sum of an arithmetic series is the middle term multiplied with the number of terms. Since each of the arithmetic series have the same number of terms, we can replace the fractions with the middle terms, which gives us \frac{4}{3} - \frac{3}{4} = \frac{7}{12}

See also

Problem 2
Josanna's test scores to date are 90, 80, 70, 60, and 85. Her goal is to raise her test average at least 3 points with her next test. What is the minimum test score she would need to accomplish this goal?
(A) 80 (B) 82 (C) 85 (D) 90 (E) 95

Solution
Take the average of her current test scores, which is \frac{90 + 80 + 70 + 60 + 85}{5} = \frac{385}{5} = 77
This means that she wants her test average after the sixth test to be 80. Let x be the score that Josanna receives on her sixth test. Thus, our equation is
\frac{90 + 80 + 70 + 60 + 85 + x}{6} = 80
385 + x = 480
x = 95

(E)

See also

Problem 3

真题文字详解(中英双语)

Problem 1 What is \frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}?
(A) -1 (B) \frac{5}{36} (C) \frac{7}{12} (D) \frac{147}{60} (E) \frac{43}{3}

Solutions
Solution 1
Add up the numbers in each fraction to get \frac{12}{9} - \frac{9}{12}, which equals \frac{4}{3} - \frac{3}{4}. Doing the subtraction yields \frac{7}{12} (C).
将每个分数中的数字相加得到\frac{12}{9} - \frac{9}{12}，等于\frac{4}{3} - \frac{3}{4}。进行减法运算得到\frac{7}{12} (C)

Solution 2
Notice that the numerators and denominators of each expression are 3-term arithmetic series. The sum of an arithmetic series is the middle term multiplied with the number of terms. Since each of the arithmetic series have the same number of terms, we can replace the fractions with the middle terms, which gives us \frac{4}{3} - \frac{3}{4} = \frac{7}{12}
注意到每个表达式的分子和分母都是三项等差数列。等差数列的和是中间项乘以项数。由于每个等差数列的项数相同，我们可以用中间项来代替分数，得到\frac{4}{3} - \frac{3}{4} = \frac{7}{12}

See also

Problem 2
Josanna's test scores to date are 90, 80, 70, 60, and 85. Her goal is to raise her test average at least 3 points with her next test. What is the minimum test score she would need to accomplish this goal?
乔莎娜到目前为止的测试成绩是90、80、70、60和85。她的目标是下一次测试将她的平均分至少提高3分。她需要至少得到多少分才能实现这个目标？
(A) 80 (B) 82 (C) 85 (D) 90 (E) 95

Solution
Take the average of her current test scores, which is \frac{90 + 80 + 70 + 60 + 85}{5} = \frac{385}{5} = 77
取她当前测试成绩的平均值，即\frac{90 + 80 + 70 + 60 + 85}{5} = \frac{385}{5} = 77
This means that she wants her test average after the sixth test to be 80. Let x be the score that Josanna receives on her sixth test. Thus, our equation is
这意味着她希望第六次测试后的平均分是80。设x为乔莎娜在第六次测试中得到的分数。因此，我们的方程是