2020 AMC amc12b 真题 答案 详解

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中英双语真题

2020 AMC 12B

Problem 1
What is the value in simplest form of the following expression?
$$\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}$$

下面表达式的值为多少？
$$\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}$$
(A) 5 (B) $4+\sqrt{7}+\sqrt{10}$ (C) 10 (D) 15 (E) $4+3\sqrt{3}+2\sqrt{5}+\sqrt{7}$

Problem 2
What is the value of the following expression?
$$\frac{100^2 - 7^2}{70^2 - 11^2}\cdot\frac{(70-11)(70+11)}{(100-7)(100+7)}$$

下面表达式的值是多少？
$$\frac{100^2 - 7^2}{70^2 - 11^2}\cdot\frac{(70-11)(70+11)}{(100-7)(100+7)}$$
(A) 1 (B) $\frac{9951}{9950}$ (C) $\frac{4780}{4779}$ (D) $\frac{108}{107}$ (E) $\frac{81}{80}$

Problem 3
The ratio of w to x is 4:3, the ratio of y to z is 3:2, and the ratio of z to x is 1:6. What is the ratio of w to y?
w比x的比值是4:3，y比z的比值是3:2，z比x的比值是1:6，则w比y的比值是多少？
(A) 4:3 (B) 3:2 (C) 8:3 (D) 4:1 (E) 16:3

英文真题

2020 AMC 12B Problems

Problem 1
What is the value in simplest form of the following expression?
$$\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}$$
(A) $5$ (B) $4+\sqrt{7}+\sqrt{10}$ (C) $10$ (D) $15$ (E) $4+3\sqrt{3}+2\sqrt{5}+\sqrt{7}$

Problem 2
What is the value of the following expression?
$$\frac{100^2 - 7^2}{70^2 - 11^2}\cdot\frac{(70-11)(70+11)}{(100-7)(100+7)}$$
(A) $1$ (B) $\frac{9951}{9950}$ (C) $\frac{4780}{4779}$ (D) $\frac{108}{107}$ (E) $\frac{81}{80}$

Problem 3
The ratio of $w$ to $x$ is $4:3$, the ratio of $y$ to $z$ is $3:2$, and the ratio of $z$ to $x$ is $1:6$. What is the ratio of $w$ to $y$?
(A) $4:3$ (B) $3:2$ (C) $8:3$ (D) $4:1$ (E) $16:3$

Problem 4
The acute angles of a right triangle are $a^\circ$ and $b^\circ$, where $a > b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$?
(A) $2$ (B) $3$ (C) $5$ (D) $7$ (E) $11$

Problem 5
Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\frac{2}{3}$ of its games and team $B$ has won $\frac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A$. How many games has team $A$ played?
(A) $21$ (B) $27$ (C) $42$ (D) $48$ (E) $63$

Problem 6
For all integers $n \geq 9$, the value of
$$\frac{(n+2)!-(n+1)!}{n!}$$
is always which of the following?
(A) a multiple of $4$ (B) a multiple of $10$ (C) a prime number (D) a perfect square (E) a perfect cube

Problem 7
Two nonhorizontal, nonvertical lines in the $xy$-coordinate plane intersect to form a $45^\circ$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
(A) $\frac{1}{6}$ (B) $\frac{2}{3}$ (C) $\frac{3}{2}$ (D) $3$ (E) $6$

Problem 8
How many ordered pairs of integers $(x,y)$ satisfy the equation
$x^{2020} + y^2 = 2y?$
(A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) infinitely many

Problem 9
A three-quarter sector of a circle of radius $4$ inches together with its interior can be rolled up to form the lateral surface of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?

真题文字详解(英文)

Problem 1 What is the value in simplest form of the following expression? \sqrt{1}+\sqrt{1+3}+\sqrt{1+3+5}+\sqrt{1+3+5+7} (A) 5 (B) 4+\sqrt{7}+\sqrt{10} (C) 10 (D) 15 (E) 4+3\sqrt{3}+2\sqrt{5}+\sqrt{7}

Solution We have \sqrt{1}+\sqrt{1+3}+\sqrt{1+3+5}+\sqrt{1+3+5+7}=\sqrt{1}+\sqrt{4}+\sqrt{9}+\sqrt{16}=1+2+3+4= (C) 10.

Note: This comes from the fact that the sum of the first n odds is n^2 Problem 2 What is the value of the following expression? \frac{100^2-7^2}{70^2-11^2}\cdot\frac{(70-11)(70+11)}{(100-7)(100+7)} (A) 1 (B)\frac{9951}{9950} (C)\frac{4780}{4779} (D)\frac{108}{107} (E)\frac{81}{80}

Solution Using difference of squares to factor the left term, we get \frac{100^2-7^2}{70^2-11^2}\cdot\frac{(70-11)(70+11)}{(100-7)(100+7)}=\frac{(100-7)(100+7)}{(70-11)(70+11)}\cdot\frac{(100-7)(100+7)}{(70-11)(70+11)}. Cancelling all the terms, we get (A) 1 as the answer.

Problem 3 The following problem is from both the 2020 AMC 10 B #3 and 2020 AMC 12 B #3, so both problems redirect to this page. The ratio of w to x is 4 : 3, the ratio of y to z is 3 : 2, and the ratio of z to x is 1 : 6. What is the ratio of w to y? (A) 4 : 3 (B) 3 : 2 (C) 8 : 3 (D) 4 : 1 (E) 16 : 3 Solution 1 (One Sentence) We have \frac{w}{y}=\frac{w}{x}\cdot\frac{x}{z}\cdot\frac{z}{y}=\frac{4}{3}\cdot\frac{6}{2}\cdot\frac{2}{3}=\frac{16}{3}, from which w:y=(E) 16:3.

Solution 2 We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two. z:x=1:6=2:12 and since y:z is 3:2 we can link them together to get y:x=3:2:12. Finally, since x:w is 3:4=12:16 we can link this again to get y:z:x:w=3:2:12:16 so w:y=(E) 16:3.

Solution 3 We have the equations \frac{w}{x}=\frac{4}{3},\frac{y}{z}=\frac{3}{2},and\frac{z}{x}=\frac{1}{6}. Clearing denominators, we have 3w=4x,2y=3z,and6z=x. Since we want \frac{w}{y},we look to find y in terms of x since we know the relationship between x and w. We begin by multiplying both sides of 2y=3z by two, obtaining 4y=6z. We then substitute that into 6z=x to get 4y=x. Now, to be able to substitute this into out first equation, we need to have 4x on the RHS. Multiplying both sides by 4, we have 16y=4x. Substituting this into our first equation, we have 3w=16y or \frac{w}{y}=\frac{16}{3},so our answer is(E) 16:3.

Solution 4 WLOG, let w = 4 and x = 3. Since the ratio of z to x is 1 : 6, we can substitute in the value of x to get \frac{z}{3}=\frac{1}{6}\Rightarrow z=\frac{1}{2}. The ratio of y to z is 3 :

真题文字详解(中英双语)

Problem 1 What is the value in simplest form of the following expression?
$$\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}$$
(A) 5 (B) $4+\sqrt{7}+\sqrt{10}$ (C) 10 (D) 15 (E) $4+3\sqrt{3}+2\sqrt{5}+\sqrt{7}$

Solution We have
$$\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16} = 1 + 2 + 3 + 4 = \text{(C) } 10.$$
Note: This comes from the fact that the sum of the first n odds is $n^2$.

Problem 2 What is the value of the following expression?
$$\frac{100^2 - 7^2}{70^2 - 11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}$$
(A) 1 (B) $\frac{9951}{9950}$ (C) $\frac{4780}{4779}$ (D) $\frac{108}{107}$ (E) $\frac{81}{80}$

Solution Using difference of squares to factor the left term, we get
$$\frac{100^2 - 7^2}{70^2 - 11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.$$
Canceling all the terms, we get (A) 1 as the answer.

Problem 3 The following problem is from both the 2020 AMC 10B #3 and 2020 AMC 12B #3, so both problems redirect to this page.
The ratio of w to x is 4 : 3, the ratio of y to z is 3 : 2, and the ratio of z to x is 1 : 6. What is the ratio of w to y?
(A) 4 : 3 (B) 3 : 2 (C) 8 : 3 (D) 4 : 1 (E) 16 : 3

Solution 1 (One Sentence)
We have
$$\frac{w}{y} = \frac{w}{x} \cdot \frac{x}{z} \cdot \frac{z}{y} = \frac{4}{3} \cdot \frac{6}{2} \cdot \frac{2}{3} = \frac{16}{3},$$
from which $w:y = \text{(E) } 16:3.$

Solution 2
We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two.
$z:x = 1:6 = 2:12$ and since $y:z = 3:2$ we can link them together to get $y:x = 3:2:12.$
Finally, since $x:w = 3:4 = 12:16$ we can link this again to get $y:z:x:w = 3:2:12:16$ so $w:y = \text{(E) } 16:3.$

Solution 3
We have the equations
$$\frac{w}{x} = \frac{4}{3}, \frac{y}{z} = \frac{3}{2}, \frac{z}{x} = \frac{1}{6}.$$
Clearing denominators, we have $3w = 4x$, $2y = 3z$, and $6z = x$. Since we want $\frac{w}{y}$, we look to find $y$ in terms of $x$ since we know the relationship between $x$ and $w$. We begin by multiplying both sides of $2y = 3z$ by two, obtaining $4y = 6z$. We then substitute that into $6z = x$ to get $4y = x$. Now, to be able to substitute this into out first equation, we need to have $4x$ on the RHS. Multiplying both sides by 4, we have $16y =