2021 AMC FALL amc10a 真题 答案 详解

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中英双语真题

2021Fall AMC10A 2021Fall AMC10A 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 x 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 Problem 3 What is the maximum number of balls of clay of radius 2 that can completely fit inside a cube of side length 6 assuming the balls can be reshaped but not compressed before they are packed in the cube? 在边长为 6 的立方体里，最多可以放入多少个半径为 2 的粘土球，这里假定球在装入立方体之前可以重塑但不能压缩? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7

英文真题

2021 Fall AMC 10A 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 What is the maximum number of balls of clay of radius (2) that can completely fit inside a cube of side length (6) assuming the balls can be reshaped but not compressed before they are packed in the cube?
((A)) (3) ((B)) (4) ((C)) (5) ((D)) (6) ((E)) (7)

Problem 4 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 5 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 6 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 7 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)?

真题文字详解(英文)

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. 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}}{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}}{169}=\frac{(10^2-3^2)^{2}}{13^{2}}=\frac{((10+3)(10-3))^{2}}{13^{2}}=\frac{(13\cdot7)^{2}}{13^{2}}=\frac{13^{2}\cdot7^{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}\approx58. 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 $\boxed{\text{(C) }49}$.

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. 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
Solution We construct the following table:
| Scenario | Length | Width | Area | |---|---|---|---| | Initial | 4 | 6 | 24 | | Menkara shortens one side. | 3 | 6 | 18 | | Menkara shortens other side instead. | 4 | 5 | 20 |

Therefore, the answer is $\boxed{\text{(E) }20}.$

Problem 3 What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7
Solution 1 (Inequality) The volume of the cube is $V_{cube}=6^3=216$, and the volume of a clay ball is $V_{ball}=\frac43\pi\cdot2^3=\frac{32}3\pi$.

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 #1和2021年秋季AMC 12A #1，因此这两个问题都重定向到此页面。

What is the value of (\frac{(2112-2021)^{2}}{169})? ((2112 - 2021)^{2} / 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}=\boxed{\text{(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\cdot7)^{2}}{13^{2}}=\frac{13^{2}\cdot7^{2}}{13^{2}}=7^{2}=\boxed{\text{(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}\approx58.)

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}).

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}\approx58.)

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 (\boxed{\text{(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 #2和2021年秋季AMC 12A #2，因此这两个问题都重定向到此页面。

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? Menkara有一张(4\times6)索引卡。如果她缩短这张卡的一边的长度为1英寸，卡片面积将是18平方英寸。如果她将另一边缩短1英寸，卡片的面积将是多少平方英寸？

(A) 16 (B) 17 (C) 18 (D) 19 (E) 20