2021 AMC FALL amc12a 真题 答案 详解

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中英双语真题

2021Fall AMC12A 2021Fall AMC12A Problem 1 What is the value of \frac{(2112-2021)^{2}}{169} ? 表达式\frac{(2112-2021)^{2}}{169} 的值是多少? (A)7 (B)21 (C)49 (D)64 (E)91 Problem 2 Menkara has a 4 x 6 index card. If she shortens the length of one side of this card by 1 inch, the card would have area 18 square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by 1 inch? Menkara 有一张 4 英寸 ×6 英寸的索引卡片。如果她将这张卡片一边的长度缩短 1 英寸，则该卡片的面积变为 18 平方英寸。如果她将另一边的长度缩短 1 英寸，那么卡片的面积会是多少平方英寸? (A)16 (B)17 (C)18 (D)19 (E)20

英文真题

2021 Fall AMC 12A Problems Problem 1 What is the value of \frac{(2112-2021)^{2}}{169}? (A) 7 (B) 21 (C) 49 (D) 64 (E) 91 Problem 2 Menkara has a $4\times6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch? (A) 16 (B) 17 (C) 18 (D) 19 (E) 20 Problem 3 Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\frac12$-mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A? (A) $2\frac34$ (B) $3\frac34$ (C) $4\frac12$ (D) $5\frac12$ (E) $6\frac34$ Problem 4 The six-digit number $20210A$ is prime for only one digit $A$. What is $A$? (A) 1 (B) 3 (C) 5 (D) 7 (E) 9 Problem 5 Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$st pole along this road is exactly one mile ($5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride? (A) 6 (B) 8 (C) 10 (D) 11 (E) 15 Problem 6 As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE=110^{\circ}$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$? (A) 160 (B) 164 (C) 166 (D) 170 (E) 174

真题文字详解(英文)

Problem1 What is the value of \frac{(2112-2021)^{2}}{169}?
(A) 7 (B) 21 (C) 49 (D) 64 (E) 91

Solution 1 (Laws of Exponents) We have
[ \frac{(2112-2021)^{2}}{169} = \frac{91^{2}}{13^{2}} = \left(\frac{91}{13}\right)^{2} = 7^{2} = \boxed{\text{(C) }49}. ]

~MRENTHUSIASM

Solution 2 (Difference of Squares) We have
[ \frac{(2112-2021)^{2}}{169} = \frac{91^{2}}{13^{2}} = \frac{(10^2 - 3^2)^{2}}{13^{2}} = \frac{((10+3)(10-3))^{2}}{13^{2}} = \frac{(13 \cdot 7)^{2}}{13^{2}} = \frac{13^{2} \cdot 7^{2}}{13^{2}} = 7^{2} = \boxed{\text{(C) }49}. ]

Solution 3 (Estimate) We know that (2112 - 2021 = 91.). Approximate this as (100) as it is pretty close to it. Also, approximate (169) to (170). We then have
[ \frac{(2112-2021)^{2}}{169} \approx \frac{100^{2}}{170} \approx \frac{1000}{17} \approx 58. ] Now check the answer choices. The two closest answers are (49) and (64). As the numerator is actually bigger than it should be, it should be the smaller answer, or (\boxed{\text{(C) }49}).

for AMC 10: https://youtu.be/o98vGHAUYjM for AMC 12: https://youtu.be/jY-17W6dA3c

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2021 Fall AMC 10A #1 and 2021 Fall AMC 12A #1, so both problems redirect to this page. 以下问题来自2021年秋季AMC 10A第1题和2021年秋季AMC 12A第1题，因此这两个问题都重定向到该页面。

What is the value of (\frac{(2112-2021)^{2}}{169})?

((2112 - 2021)^{2} \div 169) 的值是多少？

(A) 7 (B) 21 (C) 49 (D) 64 (E) 91

Solution 1 (Laws of Exponents)

解法 1（指数法则）

We have (\frac{(2112-2021)^{2}}{169} = \frac{91^{2}}{13^{2}} = \left( \frac{91}{13} \right)^{2} = 7^{2} = (C) 49.)

~MRENTHUSIASM 我们有 ~MRENTHUSIASM

Solution 2 (Difference of Squares)

解法 2（平方差公式）

We have (\frac{(2112-2021)^{2}}{169} = \frac{91^{2}}{13^{2}} = \frac{(10^2 - 3^2)^{2}}{13^{2}} = \frac{((10+3)(10-3))^{2}}{13^{2}} = \frac{(13 \cdot 7)^{2}}{13^{2}} = \frac{13^{2} \cdot 7^{2}}{13^{2}} = 7^{2} = (C) 49.)

Solution 3 (Estimate) 解法 3（估算）

We know that (2112 - 2021 = 91.) Approximate this as (100,) as it is pretty close to it. Also, approximate (169) to (170.) We then have (\frac{(2112-2021)^{2}}{169} \approx \frac{100^{2}}{170} \approx \frac{1000}{17} \approx 58.)

Now check the answer choices. The two closest answers are (49) and (64.) As the numerator is actually bigger than it should be, it should be the smaller answer, or (C) 49.

我们知道 (2112 - 2021 = 91.) 将其近似为 (100,) 因为它非常接近。同时，将 (169) 近似为 (170.) 然后我们有 (\frac{(2112-2021)^{2}}{169} \approx \frac{100^{2}}{170} \approx \frac{1000}{17} \approx 58.)

现在检查答案选项。最接近的两个答案是 (49) 和 (64。\由于分子实际上比它应该的大，所以应该是较小的答案，即 (C) 49。

Problem 2 The following problem is from both the 2021 Fall AMC 10A #2 and 2021 Fall AMC 12A #2, so both problems redirect to this page. 以下问题来自2021年秋季AMC 10A第2题和2021年秋季AMC 12A第2题，因此这两个问题都重定向到该页面。