2001 AMC amc12 真题 答案 详解

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英文真题

2001 AMC 12 Problems

Problem 1
The sum of two numbers is ( S ). Suppose ( 3 ) is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?
(A) ( 2S + 3 )
(B) ( 3S + 2 )
(C) ( 3S + 6 )
(D) ( 2S + 6 )
(E) ( 2S + 12 )

Problem 2
Let ( P(n) ) and ( S(n) ) denote the product and the sum, respectively, of the digits of the integer ( n ). For example, ( P(23)=6 ) and ( S(23)=5 ). Suppose ( N ) is a two-digit number such that ( N = P(N)+S(N) ). What is the units digit of ( N )?
(A) ( 2 )
(B) ( 3 )
(C) ( 6 )
(D) ( 8 )
(E) ( 9 )

Problem 3
The state income tax where Kristin lives is levied at the rate of ( p \% ) of the first $28000$ of annual income plus ( (p+2)\% ) of any amount above $28000$. Kristin noticed that the state income tax she paid amounted to ( (p+0.25)\% ) of her annual income. What was her annual income?
(A) $28000$
(B) $32000$
(C) $35000$
(D) $42000$
(E) $56000$

Problem 4
The mean of three numbers is ( 10 ) more than the least of the numbers and ( 15 ) less than the greatest. The median of the three numbers is ( 5 ). What is their sum?
(A) ( 5 )
(B) ( 20 )
(C) ( 25 )
(D) ( 30 )
(E) ( 36 )

Problem 5
What is the product of all positive odd integers less than ( 10000 )?
(A) ( \frac{10000!}{(5000!)^2} )
(B) ( \frac{10000!}{25000} )
(C) ( \frac{9999!}{25000} )
(D) ( \frac{10000!}{25000\cdot5000!} )
(E) ( \frac{5000!}{25000} )

Problem 6
A telephone number has the form ABC-DEF-GHIJ, where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, ( A>B>C, D>E>F ), and ( G>H>I>J ). Furthermore, ( D, E ), and ( F ) are consecutive even digits; ( G, H, I ), and ( J ) are consecutive odd digits; and ( A+B+C=9 ). Find ( A ).
(A) ( 4 )
(B) ( 5 )
(C) ( 6 )
(D) ( 7 )
(E) ( 8 )

Problem 7
A charity sells ( 140 ) benefit tickets for a total of $2001. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?
(A) $782$
(B) $986$
(C) $1158$
(D) $1219$
(E) $1449$

Problem 8
Which of the cones listed below can be formed from a ( 252^\circ ) sector of a circle of radius ( 10 ) by aligning the two straight sides?

真题文字详解(英文)

Problem1
The sum of two numbers is ( S ). Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?
[ \begin{array}{ll} (\text{A}) & 2S+3 \ (\text{B}) & 3S+2 \ (\text{C}) & 3S+6 \ (\text{D}) & 2S+6 \ (\text{E}) & 2S+12 \ \end{array} ]

Solution
Suppose the two numbers are ( a ) and ( b ), with ( a+b=S ). Then the desired sum is
[ 2(a+3)+2(b+3)=2(a+b)+12=2S+12, ]
which is answer (E).

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2001 AMC 12 #1 and 2001 AMC 10 #3, so both problems redirect to this page. 以下问题既来自2001年AMC 12第1题，也来自2001年AMC 10第3题，因此这两个问题都指向这一页。

The sum of two numbers is S. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers? 两数的和是S。假设将3加到每个数上，然后将得到的结果数都乘以2。那么这两个最终数的和是多少？

(A) 2S + 3 (B) 3S + 2 (C) 3S + 6 (D) 2S + 6 (E) 2S + 12

Solution Suppose the two numbers are a and b, with a + b = S. Then the desired sum is 2(a + 3) + 2(b + 3) = 2(a + b) + 12 = 2S + 12, which is answer (E).

假设这两个数是a和b，即a + b = S。那么期望的和是2(a + 3) + 2(b + 3) = 2(a + b) + 12 = 2S + 12，这是答案(E)。

Problem 2 The following problem is from both the 2001 AMC 12 #2 and 2001 AMC 10 #6, so both problems redirect to this page. 下面的问题是来自2001年AMC 12第2题和2001年AMC 10第6题，因此这两个问题都重定向到此页面。

Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What is the units digit of N? 让P(n)和S(n)分别表示整数n的数字积和和。例如，P(23)=6和S(23)=5。假设N是一个两位数，使得N=P(N)+S(N)。N的个位数字是多少？

(A) 2 (B) 3 (C) 6 (D) 8 (E) 9

Solution 1 Denote a and b as the tens and units digit of N, respectively. Then N = 10a + b. It follows that 10a + b = ab + a + b, which implies that 9a = ab. Since a ≠ 0, b = 9. So the units digit of N is (E) 9.

让a和b分别表示N的十位和个位数字。那么N=10a+b。由此可得10a+b=ab+a+b，这意味着9a=ab。由于a≠0，所以b=9。因此N的个位数是(E) 9。

Problem 3 The following problem is from both the 2001 AMC 12 #3 and 2001 AMC 10 #9, so both problems redirect to this page. 以下问题来自2001年AMC 12第3题和2001年AMC 10第9题，因此这两个问题都重定向到此页面。

The state income tax where Kristin lives is levied at the rate of p% of the first $28000 of annual income plus (p + 2)% of any amount above $28000. Kristin noticed that the state income tax she paid amounted to (p + 0.25)% of her annual income. What was her annual income? Kristin居住的州所得税税率为年收入的前$28000部分的p%，加上超过$28000部分的(p+2)%。Kristin注意到她缴纳的州所得税占她年收入的比例为(p+0.25)%。她的年收入是多少？