2013 AMC amc12b 真题 答案 详解

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中英双语真题

2013 AMC12B 2013 AMC12B Problem 1 On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and low temperatures was 3. In degrees, what was the low temperature in Lincoln that day? 内布拉斯加州林肯市在一月份的某天，高温比低温高16度，且高温和低温的平均值是3度，问林肯市那天的低温是多少度？ (A) -13 (B) -8 (C) -5 (D) -3 (E) 11 Problem 2 Mr. Green measures his rectangular garden by walking two of the sides and finds that it is 15 steps by 20 steps. Each of Mr. Green's steps is 2 feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden? 格林先生通过用自己的步长来测量他的矩形花园，发现大小是15步x20步，格林先生的每一步是2英尺长，他期望他的花园每平方英尺的土地可以产出半磅的土豆，问格林先生期望他的花园总共可以产出多少磅的土豆？ (A) 600 (B) 800 (C) 1000 (D) 1200 (E) 1400 Problem 3 When counting from 3 to 201, 53 is the 51st number counted. When counting backwards from 201 to 3, 53 is the nth number counted. What is n? 当从3数到201，数到的第51个数是53，当从201倒数到3，数到的第n个数是53，问n是多少？ (A) 146 (B) 147 (C) 148 (D) 149 (E) 150

英文真题

2013 AMC 12B Problems

Problem 1

On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and low temperatures was 3. In degrees, what was the low temperature in Lincoln that day?

(A) -13 (B) -8 (C) -5 (D) -3 (E) 11

Problem 2

Mr. Green measures his rectangular garden by walking two of the sides and finds that it is 15 steps by 20 steps. Each of Mr. Green's steps is 2 feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?

(A) 600 (B) 800 (C) 1000 (D) 1200 (E) 1400

Problem 3

When counting from 3 to 201, 53 is the 51st number counted. When counting backwards from 201 to 3, 53 is the nth number counted. What is n?

(A) 146 (B) 147 (C) 148 (D) 149 (E) 150

Problem 4

Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?

(A) 10 (B) 16 (C) 25 (D) 30 (E) 40

Problem 5

The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders?

(A) 22 (B) 23.25 (C) 24.75 (D) 26.25 (E) 28

Problem 6

Real numbers x and y satisfy the equation (x^2 + y^2 = 10x - 6y - 34). What is (x + y)?

(A) 1 (B) 2 (C) 3 (D) 6 (E) 8

Problem 7

Jo and Blair take turns counting from 1 to one more than the last number said by the other person. Jo starts by saying "1", so Blair follows by saying "1, 2". Jo then says "1, 2, 3", and so on. What is the 53rd number said?

(A) 2 (B) 3 (C) 5 (D) 6 (E) 8

Problem 8

Line l1 has equation (3x - 2y = 1) and goes through A = (-1, -2). Line l2 has equation (y = 1) and meets line l1 at point B. Line l3 has positive slope, goes through point A, and meets l2 at point C. The area of (\triangle ABC) is 3. What is the slope of l3?

(A) (\frac{2}{3}) (B) (\frac{3}{4}) (C) 1 (D) (\frac{4}{3}) (E) (\frac{3}{2})

Problem 9

What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides 12!?

真题文字详解(英文)

Problem 1 The following problem is from both the 2013 AMC 12B #1 and 2013 AMC 10B #3, so both problems redirect to this page. On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and low temperatures was 3. In degrees, what was the low temperature in Lincoln that day? (A)-13 (B)-8 (C)-5 (D)-3 (E)11 Solution Let L be the low temperature. The high temperature is L + 16. The average is \frac{L+(L+16)}{2}=3. Solving for L, we get L = (C)-5 Problem 2 The following problem is from both the 2013 AMC 12B #2 and 2013 AMC 10B #2, so both problems redirect to this page. Mr. Green measures his rectangular garden by walking two of the sides and finds that it is 15 steps by 20 steps. Each of Mr. Green's steps is 2 feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden? (A) 600 (B) 800 (C) 1000 (D) 1200 (E) 1400 Solution Since each step is 2 feet, his garden is 30 by 40 feet. Thus, the area of 30(40)=1200 square feet. Since he is expecting \frac{1}{2} of a pound per square foot, the total amount of potatoes expected is 1200\times\frac{1}{2}=\textbf{(A) }600 See also Problem 3 The following problem is from both the 2013 AMC 12B #3 and 2013 AMC 10B #4, so both problems redirect to this page. When counting from 3 to 201, 53 is the 51st number counted. When counting backwards from 201 to 3, 53 is the nth number counted. What is n? (A) 146 (B) 147 (C) 148 (D) 149 (E) 150 Solution Note that n is equal to the number of integers between 53 and 201, inclusive. Thus, n=201-53+1=(D) 149 See also Problem 4 The following problem is from both the 2013 AMC 12B #4 and 2013 AMC 10B #8, so both problems redirect to this page. Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? (A) 10 (B) 16 (C) 25 (D) 30 (E) 40 Solution 1 Let Ray and Tom drive 40 miles. Ray's car would require \frac{40}{40}=1 gallon of gas and Tom's car would require \frac{40}{10}=4 gallons of gas. They would have driven a total of 40+40=80 miles, on 1+4=5 gallons of gas, for a combined rate of \frac{80}{5}=\textbf{(B) }16 Solution 2 Taking the harmonic mean of the two rates, we get (\frac{40^{-1}+10^{-1}}{2})^{-1}=\frac{2}{\frac{1}{40}+\frac{1}{10}}=\frac{2}{\frac{5}{40}}=\frac{2}{\frac{1}{8}}=\textbf{(B) }16

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2013 AMC 12B #1 and 2013 AMC 10B #3, so both problems redirect to this page. On a particular January day, the high temperature in Lincoln, Nebraska, was 16 degrees higher than the low temperature, and the average of the high and low temperatures was 3. In degrees, what was the low temperature in Lincoln that day? (A)-13 (B)-8 (C)-5 (D)-3 (E)11 Solution Let L be the low temperature. The high temperature is L + 16. The average is \frac{L+(L+16)}{2}=3. Solving for L, we get L = (C)-5 Problem 2 The following problem is from both the 2013 AMC 12B #2 and 2013 AMC 10B #2, so both problems redirect to this page. Mr. Green measures his rectangular garden by walking two of the sides and finds that it is 15 steps by 20 steps. Each of Mr. Green's steps is 2 feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden? (A) 600 (B) 800 (C) 1000 (D) 1200 (E) 1400 Solution Since each step is 2 feet, his garden is 30 by 40 feet. Thus, the area of 30(40)=1200 square feet. Since he is expecting \frac{1}{2} of a pound per square foot, the total amount of potatoes expected is 1200\times\frac{1}{2}=\textbf{(A) }600 See also Problem 3 The following problem is from both the 2013 AMC 12B #3 and 2013 AMC 10B #4, so both problems redirect to this page. When counting from 3 to 201, 53 is the 51st number counted. When counting backwards from 201 to 3, 53 is the nth number counted. What is n? (A) 146 (B) 147 (C) 148 (D) 149 (E) 150 Solution Note that n is equal to the number of integers between 53 and 201, inclusive. Thus, n=201-53+1=(D) 149 See also Problem 4 The following problem is from both the 2013 AMC 12B #4 and 2013 AMC 10B #8, so both problems redirect to this page. Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? (A) 10 (B) 16 (C) 25 (D) 30 (E) 40 Solution 1 Let Ray and Tom drive 40 miles. Ray's car would require \frac{40}{40}=1 gallon of gas and Tom's car would require \frac{40}{10}=4 gallons of gas. They would have driven a total of 40+40=80 miles, on 1+4=5 gallons of gas, for a combined rate of \frac{80}{5}=\textbf{(B) }16 Solution 2 Taking the harmonic mean of the two rates, we get (\frac{40^{-1}+10^{-1}}{2})^{-1}=\frac{2}{\frac{1}{40}+\frac{1}{10}}=\frac{2}{\frac{5}{40}}=\frac{2}{\frac{1}{8}}=\textbf{(B) }16