2006 AMC amc12a 真题 答案 详解

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英文真题

2006 AMC 12A Problems Problem 1 Sandwiches at Joe's Fast Food cost $3$ dollars each and sodas cost $2$ dollars each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?
(A) $31$ (B) $32$ (C) $33$ (D) $34$ (E) $35$

Problem 2 Define $x \otimes y = x^3 - y$. What is $h \otimes (h \otimes h)$?
(A) $-h$ (B) $0$ (C) $h$ (D) $2h$ (E) $h^3$

Problem 3 The ratio of Mary's age to Alice's age is $3:5$. Alice is $30$ years old. How old is Mary?
(A) $15$ (B) $18$ (C) $20$ (D) $24$ (E) $50$

Problem 4 A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
(A) $17$ (B) $19$ (C) $21$ (D) $22$ (E) $23$

Problem 5 Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?
(A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) $5$

Problem 6 The $8\times 18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?
(A) $6$ (B) $7$ (C) $8$ (D) $9$ (E) $10$

Problem 7 Mary is $20\%$ older than Sally, and Sally is $40\%$ younger than Danielle. The sum of their ages is $23.2$ years. How old will Mary be on her next birthday?
(A) $7$ (B) $8$ (C) $9$ (D) $10$ (E) $11$

Problem 8 How many sets of two or more consecutive positive integers have a sum of $15$?
(A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) $5$

真题文字详解(英文)

Problem 1 The following problem is from both the 2006 AMC 12A #1 and 2006 AMC 10A #1, so both problems redirect to this page. Sandwiches at Joe's Fast Food cost $3 each and sodas cost $2 each. How many dollars will it cost to purchase 5 sandwiches and 8 sodas? (A) 31 (B) 32 (C) 33 (D) 34 (E) 35 Solution The 5 sandwiches cost $5 \cdot 3 = 15$ dollars. The 8 sodas cost $8 \cdot 2 = 16$ dollars. In total, the purchase costs $15 + 16 = (A) 31$ dollars. See also Problem 2 The following problem is from both the 2006 AMC 12A #2 and 2006 AMC 10A #2, so both problems redirect to this page. Define $x \otimes y = x^3 - y$. What is $h \otimes (h \otimes h)$? (A) $-h$ (B) 0 (C) $h$ (D) $2h$ (E) $h^3$ Solution 1 By the definition of $\otimes$, we have $h \otimes h = h^3 - h$. Then, $h \otimes (h \otimes h) = h \otimes (h^3 - h) = h^3 - (h^3 - h) = (C) h$. See also Problem 3 The following problem is from both the 2006 AMC 12A #3 and 2006 AMC 10A #3, so both problems redirect to this page. The ratio of Mary's age to Alice's age is $3:5$. Alice is 30 years old. How old is Mary? (A) 15 (B) 18 (C) 20 (D) 24 (E) 50 Solution 1 Let $m$ be Mary's age. Then $\frac{m}{30} = \frac{3}{5}$ Solving for $m$, we obtain $m = (B) 18$. Solution 2 We can see this is a combined ratio of $8,(5+3)$. We can equalize by doing $30 \div 5 = 6$, and $6 \cdot 3 = (B) 18$. With the common ratio of $8$ and difference ratio of $6$, we see $6 \cdot 8 = 30 + 18$. Therefore, we can see our answer is correct. See also Problem 4 The following problem is from both the 2006 AMC 12A #4 and 2006 AMC 10A #4, so both problems redirect to this page. A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display? (A) 17 (B) 19 (C) 21 (D) 22 (E) 23 Solution 1 From the greedy algorithm, we have $9$ in the hours section and $59$ in the minutes section. $9 + 5 + 9 = (E) 23$ Solution 2 (matrix) With a matrix, we can see $1+2\quad 9\quad 6\quad 3\ 1+1\quad 8\quad 5\quad 2\ 1+0\quad 7\quad 4\quad 1$ The largest single digit sum we can get is $9$. For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9 + 5 = 14$, and finally $14 + 9 = (E) 23$ Solution 3 We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$. Therefore, the maximum value of the digits for the hours is when the watch displays $09$, which gives us $0 + 9 = 9$. Next, we look at the value of the minutes section, which varies from $00-59$. Let this value be a number $ab$. We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$. Adding these up, we get $9 + 5 + 9 = (E) 23$ See also Problem 5

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2006 AMC 12A #1 and 2006 AMC 10A #1, so both problems redirect to this page. Sandwiches at Joe's Fast Food cost $3 each and sodas cost $2 each. How many dollars will it cost to purchase 5 sandwiches and 8 sodas?
(A) 31 (B) 32 (C) 33 (D) 34 (E) 35

Solution The 5 sandwiches cost $5 \cdot 3 = 15$ dollars. The 8 sodas cost $8 \cdot 2 = 16$ dollars. In total, the purchase costs $15 + 16 =$ (A) 31 dollars.

See also Problem 2 The following problem is from both the 2006 AMC 12A #2 and 2006 AMC 10A #2, so both problems redirect to this page. Define $x \otimes y = x^3 - y$. What is $h \otimes (h \otimes h)$?
(A) −h (B) 0 (C) h (D) 2h (E) h³

Solution 1 By the definition of $\otimes$, we have $h \otimes h = h^3 - h.$ Then, $h \otimes (h \otimes h) = h \otimes (h^3 - h) = h^3 - (h^3 - h) = $ (C) h.

See also Problem 3 The following problem is from both the 2006 AMC 12A #3 and 2006 AMC 10A #3, so both problems redirect to this page. The ratio of Mary's age to Alice's age is $3:5$. Alice is 30 years old. How old is Mary?
(A) 15 (B) 18 (C) 20 (D) 24 (E) 50

Solution 1 Let $m$ be Mary's age. Then $\frac{m}{30} = \frac{3}{5}$ Solving for $m$, we obtain $m = $ (B) 18.

Solution 2 We can see this is a combined ratio of $8,(5+3).$ We can equalize by doing $30 \div 5 = 6,$ and $6 \cdot 3 =$ (B) 18. With the common ratio of $8$ and difference ratio of $6$, we see $6 \cdot 8 = 30 + 18$. Therefore, we can see our answer is correct.

See also Problem 4 The following problem is from both the 2006 AMC 12A #4 and 2006 AMC 10A #4, so both problems redirect to this page. A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
(A) 17 (B) 19 (C) 21 (D) 22 (E) 23

Solution 1 From the greedy algorithm, we have $9$ in the hours section and $59$ in the minutes section. $9 + 5 + 9 =$ (E) 23

Solution 2 (matrix)
With a matrix, we can see
$\begin{bmatrix} 1+2 & 9 & 6 & 3\ 1+1 & 8 & 5 & 2\ 1+0 & 7 & 4 & 1 \end{bmatrix}$ The largest single digit sum we can get is $9$. For the minutes digits, we can combine the largest $2$ digits, which are $9, 5 \Rightarrow 9 + 5 = 14$, and finally $14 + 9 =$ (E) 23

Solution 3 We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$. Therefore, the maximum value of the digits for the hours is when the watch displays $09$, which gives us $0 + 9 = 9$. Next, we look at the value of the minutes section, which varies from $00-59$. Let this value be a number $ab$. We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$. Adding these up, we get $9 + 5 + 9 =$ (E) 23