2008 AMC amc12a 真题 答案 详解

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英文真题

2008 AMC 12A Problems
Problem 1
A bakery owner turns on his doughnut machine at 8:30 AM. At 11:10 AM the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?
(A) 1:50 PM (B) 3:00 PM (C) 3:30 PM (D) 4:30 PM (E) 5:50 PM

Problem 2
What is the reciprocal of ( \frac{1}{2} + \frac{2}{3} )?
(A) ( \frac{6}{7} ) (B) ( \frac{7}{6} ) (C) ( \frac{5}{3} ) (D) 3 (E) ( \frac{7}{2} )

Problem 3
Suppose that ( \frac{2}{3} ) of 10 bananas are worth as much as 8 oranges. How many oranges are worth as much as ( \frac{1}{2} ) of 5 bananas?
(A) 2 (B) ( \frac{5}{2} ) (C) 3 (D) ( \frac{7}{2} ) (E) 4

Problem 4
Which of the following is equal to the product
( \frac{8}{4} \cdot \frac{12}{8} \cdot \frac{16}{12} \cdots \frac{4n+4}{4n} \cdots \frac{2008}{2004} )?
(A) 251 (B) 502 (C) 1004 (D) 2008 (E) 4016

Problem 5
Suppose that
( \frac{2x}{3} - \frac{x}{6} )
is an integer. Which of the following statements must be true about ( x )?
(A) It is negative. (B) It is even, but not necessarily a multiple of 3.
(C) It is a multiple of 3, but not necessarily even.
(D) It is a multiple of 6, but not necessarily a multiple of 12.
(E) It is a multiple of 12.

Problem 6
Heather compares the price of a new computer at two different stores. Store A offers 15% off the sticker price followed by a $90 rebate, and store B offers 25% off the same sticker price with no rebate. Heather saves $15 by buying the computer at store A instead of store B. What is the sticker price of the computer, in dollars?
(A) 750 (B) 900 (C) 1000 (D) 1050 (E) 1500

Problem 7
While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing towards the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?
(A) 2 (B) 4 (C) 6 (D) 8 (E) 10

真题文字详解(英文)

Problem 1 The following problem is from both the 2008 AMC 12A #1 and 2008 AMC 10A #1, so both problems redirect to this page. A bakery owner turns on his doughnut machine at 8:30 AM. At 11:10 AM the machine has completed one third of the day's job. At what time will the doughnut machine complete the job? (A) 1:50 PM (B) 3:00 PM (C) 3:30 PM (D) 4:30 PM (E) 5:50 PM Solution The machine completes one-third of the job in $11:10 - 8:30 = 2:40$ hours. Thus, the entire job is completed in $3 \cdot (2:40) = 8:00$ hours. Since the machine was started at 8:30 AM, the job will be finished 8 hours later, at 4:30 PM. The answer is (D). Note: $2:40$ means 2 hours and 40 minutes. $3$ multiplied by this time interval is 8 hours.

Problem 2 What is the reciprocal of $\frac{1}{2} + \frac{2}{3}$? (A) $\frac{6}{7}$ (B) $\frac{7}{6}$ (C) $\frac{5}{3}$ (D) 3 (E) $\frac{7}{2}$ Solution 1 Here's a cheapsot: Obviously, $\frac{1}{2} + \frac{2}{3}$ is greater than 1. Therefore, its reciprocal is less than 1, and the answer must be $\boxed{\frac{6}{7}}$. Solution 2 $\left(\frac{1}{2} + \frac{2}{3}\right)^{-1} = \left(\frac{3}{6} + \frac{4}{6}\right)^{-1} = \left(\frac{7}{6}\right)^{-1} = \boxed{(A)\ \frac{6}{7}}$

Problem 3 The following problem is from both the 2008 AMC 12A #3 and 2008 AMC 10A #4, so both problems redirect to this page. Suppose that $\frac{2}{3}$ of 10 bananas are worth as much as 8 oranges. How many oranges are worth as much as $\frac{1}{2}$ of 5 bananas? (A) 2 (B) $\frac{5}{2}$ (C) 3 (D) $\frac{7}{2}$ (E) 4 Solution If $\frac{2}{3} \cdot 10$ bananas = 8 oranges then $\frac{1}{2} \cdot 5$ bananas = $\left(\frac{1}{2} \cdot 5\right.$ bananas)$\cdot \left(\frac{8\text{ oranges}}{\frac{2}{3} \cdot 10\text{ bananas}}\right) = 3$ oranges $\Rightarrow$ (C).

Problem 4 The following problem is from both the 2008 AMC 12A #4 and 2008 AMC 10A #5, so both problems redirect to this page. Which of the following is equal to the product $\frac{8}{4} \cdot \frac{12}{8} \cdot \frac{16}{12} \cdots \frac{4n+4}{4n} \cdots \frac{2008}{2004}$? (A) 251 (B) 502 (C) 1004 (D) 2008 (E) 4016 Solution 1 $\frac{8}{4} \cdot \frac{12}{8} \cdot \frac{16}{12} \cdots \frac{4n+4}{4n} \cdots \frac{2008}{2004} = \frac{1}{4} \cdot \left(\frac{8}{8} \cdot \frac{12}{12} \cdots \frac{4n}{4n} \cdots \frac{2004}{2004}\right) \cdot 2008 = \frac{2008}{4} = 502 \Rightarrow B.$ Solution 2 Notice that everything cancels out except for 2008 in the numerator and 4 in the denominator. Thus, the product is $\frac{2008}{4} = 502$ and the answer is (B).

Problem 5 The following problem is from both the 2008 AMC 12A #5 and 2008 AMC 10A #9, so both problems redirect to this page. Suppose that $\frac{2x}{3} - \frac{x}{6}$ is an integer. Which of

真题文字详解(中英双语)

Problem 1
The following problem is from both the 2008 AMC 12A #1 and 2008 AMC 10A #1, so both problems redirect to this page.
A bakery owner turns on his doughnut machine at 8:30 AM. At 11:10 AM the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?
(A) 1:50 PM (B) 3:00 PM (C) 3:30 PM (D) 4:30 PM (E) 5:50 PM

Solution
The machine completes one-third of the job in $11:10 - 8:30 = 2:40$ hours. Thus, the entire job is completed in $3 \cdot (2:40) = 8:00$ hours.
Since the machine was started at 8:30 AM, the job will be finished 8 hours later, at 4:30 PM. The answer is (D).
Note: $2:40$ means 2 hours and 40 minutes. 3 multiplied by this time interval is 8 hours.

Problem 2
What is the reciprocal of $\frac{1}{2} + \frac{2}{3}$?
$\left(\frac{1}{2} + \frac{2}{3}\right)$ 的倒数是什么？
(A) $\frac{6}{7}$ (B) $\frac{7}{6}$ (C) $\frac{5}{3}$ (D) $3$ (E) $\frac{7}{2}$

Solution 1
Here's a cheap shot: Obviously, $\frac{1}{2} + \frac{2}{3}$ is greater than 1. Therefore, its reciprocal is less than 1, and the answer must be $\boxed{\frac{6}{7}}$.

Solution 2
$\left(\frac{1}{2} + \frac{2}{3}\right)^{-1} = \left(\frac{3}{6} + \frac{4}{6}\right)^{-1} = \left(\frac{7}{6}\right)^{-1} = \boxed{(A)\ \frac{6}{7}}.$

Problem 3
The following problem is from both the 2008 AMC 12A #3 and 2008 AMC 10A #4, so both problems redirect to this page.
Suppose that $\frac{2}{3}$ of 10 bananas are worth as much as 8 oranges. How many oranges are worth as much as $\frac{1}{2}$ of 5 bananas?
(A) 2 (B) $\frac{5}{2}$ (C) 3 (D) $\frac{7}{2}$ (E) 4

Solution
If $\frac{2}{3} \cdot 10$ bananas $= 8$ oranges, then $\frac{1}{2} \cdot 5$ bananas $= \left(\frac{1}{2} \cdot 5\text{ bananas}\right) \cdot \left(\frac{8\text{ oranges}}{\frac{2}{3} \cdot 10\text{ bananas}}\right) = 3$ oranges $\Rightarrow$ (C).

Problem 4
The following problem is from both the 2008 AMC 12A #4 and 2008 AMC 10A #5, so both problems redirect to this page.
Which of the following is equal to the product
$\frac{8}{4} \cdot \frac{12}{8} \cdot \frac{16}{12} \cdots \frac{4n+4}{4n} \cdots \frac{2008}{2004}$?
(A) 251 (B) 502 (C) 1004 (D) 2008 (E) 4016

Solution 1
$\frac{8}{4} \cdot \frac{12}{8} \cdot \frac{16}{12} \cdots \frac{4n+4}{4n} \cdots \frac{2008}{2004} = \frac{1}{4} \cdot \left(\frac{8}{8} \cdot \frac{12}{12} \cdots \frac{4n}{4n} \cdots \frac{2008}{2004}\right) \cdot 2008 = \frac{2008}{4} = 502 \Rightarrow B.$

Solution 2
Notice that everything cancels out except for 2008 in the numerator and 4 in the denominator.
Thus, the product is $\frac{2008}{4} = 502$, and the answer is (B).

Problem 5