2002 AMC amc12a 真题 答案 详解

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英文真题

2002 AMC 12A Problems
Problem 1
Compute the sum of all the roots of $(2x+3)(x-4)+(2x+3)(x-6)=0$
(A) $\frac{7}{2}$ (B) $4$ (C) $5$ (D) $7$ (E) $13$

Problem 2
Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
(A) $15$ (B) $34$ (C) $43$ (D) $51$ (E) $138$

Problem 3
According to the standard convention for exponentiation,
$$2^{2^{2^2}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.$$
If the order in which the exponentiations are performed is changed, how many other values are possible?
(A) $0$ (B) $1$ (C) $2$ (D) $3$ (E) $4$

Problem 4
Find the degree measure of an angle whose complement is 25% of its supplement.
(A) $48$ (B) $60$ (C) $75$ (D) $120$ (E) $150$

Problem 5
Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
(A) $\pi$ (B) $1.5\pi$ (C) $2\pi$ (D) $3\pi$ (E) $3.5\pi$

Problem 6
For how many positive integers $m$ does there exist at least one positive integer $n$ such that $m \cdot n \leq m + n$?
(A) $4$ (B) $6$ (C) $9$ (D) $12$ (E) infinitely many

Problem 7
A $45^\circ$ arc of circle A is equal in length to a $30^\circ$ arc of circle B. What is the ratio of circle A's area and circle B's area?
(A) $\frac{4}{9}$ (B) $\frac{2}{3}$ (C) $\frac{5}{6}$ (D) $\frac{3}{2}$ (E) $\frac{9}{4}$

Problem 8
Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct?

真题文字详解(英文)

Problem 1 The following problem is from both the 2002 AMC 12A #1 and 2002 AMC 10A #10, so both problems redirect to this page. Compute the sum of all the roots of $(2x+3)(x-4)+(2x+3)(x-6)=0$
$\textbf{(A)}\ \frac{7}{2}\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 13$

Solution 1 We expand to get $2x^2 -8x + 3x - 12 + 2x^2 - 12x + 3x - 18 = 0$ which is $4x^2 -14x -30=0$ after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is $\frac{14}{4}=\textbf{(A)}\ 7/2$.

Solution 2 We see a $(2x+3)$ term in both addends so we can factor the equation to $(2x+3)((x-4)+(x-6))=(2x+3)(2x-10)=0$. The two roots become obvious: they are $-\frac{3}{2}$ and $\frac{10}{2}$. Adding these two gives $\textbf{(A)}\ 7/2$.

-starwarmonkey

Problem 2 The following problem is from both the 2002 AMC 12A #2 and 2002 AMC 10A #6, so both problems redirect to this page. Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 51\qquad\textbf{(E)}\ 138$

Solution We work backwards; the number that Cindy started with is $3(43)+9=138$. Now, the correct result is $\frac{138-3}{9}=\frac{135}{9}=15$. Our answer is $\textbf{(A)}\ 15$.

Solution 2 Let the number be $x$. We transform the problem into an equation: $\frac{x-9}{3}=43$. Solve for $x$ gives us $x=138$. Therefore, the correct result is $\frac{138-3}{9}=\frac{135}{9}=\textbf{(A)}\ 15$.

Problem 3 The following problem is from both the 2002 AMC 12A #3 and 2002 AMC 10A #3, so both problems redirect to this page. According to the standard convention for exponentiation, $2^{2^{2^2}}=2^{(2^{(2^2)})}=2^{16}=65536$.

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2002 AMC 12A #1 and 2002 AMC 10A #10, so both problems redirect to this page. 计算所有根的和 $(2x+3)(x-4)+(2x+3)(x-6)=0$ (A) $7/2$ (B) 4 (C) 5 (D) 7 (E) 13 Solution 1 We expand to get $2x^2 -8x + 3x -12 + 2x^2 -12x + 3x -18 = 0$ which is $4x^2 -14x -30=0$ after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is $\frac{14}{4}=\textbf{(A)}\ 7/2$. 我们展开得到 $2x^2 -8x + 3x -12 + 2x^2 -12x + 3x -18 = 0$, 合并同类项后是 $4x^2 -14x -30 = 0$. 使用韦达定理的二次部分，我们找到根的和是 $\frac{14}{4}=\textbf{(A)}\ 7/2$. Solution 2 We see a $(2x+3)$ term in both addends so we can factor the equation to $(2x+3)((x-4)+(x-6))=(2x+3)(2x-10)=0$. The two roots become obvious: they are $-\frac{3}{2}$ and $\frac{10}{2}$. Adding these two gives $\textbf{(A)}\ 7/2$. 我们看到两个加数中都有一个 $(2x+3)$ 项，所以我们可以将方程分解为 $(2x+3)((x-4)+(x-6))=(2x+3)(2x-10)=0$. 这两个根就变得明显了：它们是 $-\frac{3}{2}$ 和 $\frac{10}{2}$. 将这两个相加得到 $\textbf{(A)}\ 7/2$. starwarmonkey Problem 2 The following problem is from both the 2002 AMC 12A #2 and 2002 AMC 10A #6, so both problems redirect to this page. 辛迪被她的老师要求从一个数中减去3，然后将结果除以9。但她却减去了9，然后将结果除以3，得到了43。如果她正确地解题，答案会是多少？ (A) 15 (B) 34 (C) 43 (D) 51 (E) 138 Solution