2015 AMC amc12a 真题 答案 详解

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中英双语真题

2015 AMC12A

Problem 1 What is the value of $(2^0 - 1 + 5^2 - 0)^{-1} \times 5$?
下列算式的值是多少 $(2^0 - 1 + 5^2 - 0)^{-1} \times 5$?
(A) $-125$ (B) $-120$ (C) $\frac{1}{5}$ (D) $\frac{5}{24}$ (E) $25$

Problem 2 Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
一个三角形的其中两条边长度是 20 和 15，下面哪个数字不可能成为三角形的周长？
(A) $52$ (B) $57$ (C) $62$ (D) $67$ (E) $72$

Problem 3 Mr. Patrick teaches math to 15 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 80. After he graded Payton's test, the class average became 81. What was Payton's score on the test?
Patrick 先生给 15 个学生上数学课，他正批改考试试卷，发现当他把除了 Payton 以外的所有其他学生的试卷都改好后，班级的平均分是 80，当他批改好 Payton 的试卷后，班级的平均分变成了 81 分，问 Payton 这次考试得分是多少？
(A) $81$ (B) $85$ (C) $91$ (D) $94$ (E) $95$

Problem 4 The sum of two positive numbers is 5 times their difference. What is the ratio of the larger number to the smaller?
两个正数之和是他们之差的 5 倍，那么大的数比小的数的比值是多少？
(A) $\frac{5}{4}$ (B) $\frac{3}{2}$ (C) $\frac{9}{5}$ (D) $2$ (E) $\frac{5}{2}$

英文真题

2015 AMC 12A Problems

Problem 1
What is the value of ((2^0 - 1 + 5^2 - 0)^{-1} \times 5)? (A) -125 (B) -120 (C) ( \frac{1}{5} ) (D) ( \frac{5}{24} ) (E) 25

Problem 2
Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
(A) 52 (B) 57 (C) 62 (D) 67 (E) 72

Problem 3
Mr. Patrick teaches math to 15 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 80. After he graded Payton's test, the class average became 81. What was Payton's score on the test?
(A) 81 (B) 85 (C) 91 (D) 94 (E) 95

Problem 4
The sum of two positive numbers is 5 times their difference. What is the ratio of the larger number to the smaller?
(A) ( \frac{5}{4} ) (B) ( \frac{3}{2} ) (C) ( \frac{9}{5} ) (D) 2 (E) ( \frac{5}{2} )

Problem 5
Amelia needs to estimate the quantity ( \frac{a}{b} - c ), where ( a ), ( b ), and ( c ) are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of ( \frac{a}{b} - c )?
(A) She rounds all three numbers up.
(B) She rounds ( a ) and ( b ) up, and she rounds ( c ) down.
(C) She rounds ( a ) and ( c ) up, and she rounds ( b ) down.
(D) She rounds ( a ) up, and she rounds ( b ) and ( c ) down.
(E) She rounds ( c ) up, and she rounds ( a ) and ( b ) down.

Problem 6
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be ( 2:1 )?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 8

Problem 7
Two right circular cylinders have the same volume. The radius of the second cylinder is 10% more than the radius of the first. What is the relationship between the heights of the two cylinders?
(A) The second height is 10% less than the first.
(B) The first height is 10% more than the second.
(C) The second height is 21% less than the first.
(D) The first height is 21% more than the second.
(E) The second height is 80% of the first.

Problem 8
The ratio of the length to the width of a rectangle is ( 4:3 ). If the rectangle has diagonal of length ( d ), then the area may be expressed as ( kd^2 ) for some constant ( k ). What is ( k )?
(A) ( \frac{2}{7} ) (B) ( \frac{3}{7} ) (C) ( \frac{12}{25} ) (D) ( \frac{16}{25} ) (E) ( \frac{3}{4} )

Problem 9
A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?
(A) ( \frac{1}{10} ) (B) ( \frac{1}{6} ) (C) ( \frac{1}{5} ) (D) ( \frac{1}{3} ) (E) ( \frac{1}{2} )

Problem 10
Integers ( x ) and ( y ) with ( x > y > 0 ) satisfy ( x + y + xy = 80 ). What is ( x

真题文字详解(英文)

Problem 1 The following problem is from both the 2015 AMC 12A #7 and 2015 AMC 10A #1, so both problems redirect to this page. What is the value of $(2^0 - 1 + 5^2 - 0)^{-1} \times 5$?
(A) $-125$ (B) $-120$ (C) $\frac{1}{5}$ (D) $\frac{1}{24}$ (E) $25$

Solution
$(2^0 - 1 + 5^2 - 0)^{-1} \times 5 = (1 - 1 + 25 - 0)^{-1} \times 5 = 25^{-1} \times 5 = \frac{1}{25} \times 5 = \left(\text{(C)}\ \frac{1}{5}\right)$

Problem 2 Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
(A) $52$ (B) $57$ (C) $62$ (D) $67$ (E) $72$

Solution
Letting $x$ be the third side, by the triangle inequality, $20 - 15 < x < 20 + 15$, or $5 < x < 35$. Therefore the perimeter must be greater than $40$ but less than $70$. $72$ is not in this range, so $\text{(E) } 72$ is our answer.

Problem 3 The following problem is from both the 2015 AMC 12A #3 and 2015 AMC 10A #5, so both problems redirect to this page.
Mr. Patrick teaches math to 15 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$. After he graded Payton's test, the test average became $81$. What was Payton's score on the test?
(A) $81$ (B) $85$ (C) $91$ (D) $94$ (E) $95$

Solution
If the average of the first $14$ people's scores was $80$, then the sum of all of their tests is $14 \cdot 80 = 1120$. When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$. So, Payton's score must be $1215 - 1120 = \textbf{(E) } 95$.

Alternate Solution
The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first $14$ students each scored $80$. If Payton also scored $80$, the average would still be $80$. In order to increase the overall average to $81$, we need to add one more point to all of the scores, including Payton's. This means we need to add a total of $15$ points, so Payton needs $80 + 15 = \textbf{(E) } 95$.

Problem 4 The following problem is from both the 2015 AMC 12A #4 and 2015 AMC 10A #6, so both problems redirect to this page.
The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
(A) $\frac{5}{4}$ (B) $\frac{3}{2}$ (C) $\frac{9}{5}$ (D) $2$ (E) $\frac{5}{2}$

Solution 1
Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac{3}{2}$ so the answer is $\boxed{\text{(B) }\frac{3}{2}}$

Solution 2
Without loss of generality, let the two numbers be $3$ and $2$, as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{\text{(B) }\frac{3}{2}}$

Problem 5 Sreeshma needs to estimate the quantity $\frac{a}{b} - c$, where $a$, $b$, and $c$ are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will

真题文字详解(中英双语)

Problem 1

The following problem is from both the 2015 AMC 12A #4 and 2015 AMC 10A #4, so both problems redirect to this page.

What is the value of $(2^0 - 1 + 5^2 - 0)^{-1} \times 5$?

$(2^0 - 1 + 5^2 - 0)^{-1} \times 5$ 的值是多少？

(A) $-125$ (B) $120$ (C) $\frac{1}{5}$ (D) $\frac{5}{24}$ (E) $25$

Solution

$(2^0 - 1 + 5^2 - 0)^{-1} \times 5 = (1 - 1 + 25 - 0)^{-1} \times 5 = 25^{-1} \times 5 = \frac{1}{25} \times 5 = \boxed{\text{(C)}\ \frac{1}{5}}$

Problem 2

Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?

(A) 52 (B) 57 (C) 62 (D) 67 (E) 72

Solution

Let ( x ) be the third side, then by the triangle inequality $20 - 15 < x < 20 + 15$, or $5 < x < 35$. Therefore, the perimeter must be greater than $40$ but less than $70$. $72$ is not in this range, so $\boxed{\text{(E)}\ 72}$ is our answer.

Problem 3

The following problem is from both the 2015 AMC 12A #3 and 2015 AMC 10A #5, so both problems redirect to this page.

Mr. Patrick teaches math to 15 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 80. After he graded Payton's test, the test average became 81. What was Payton's score on the test?

(A) 81 (B) 85 (C) 91 (D) 94 (E) 95

Solution

If the average of the first 14 peoples' scores was 80, then the sum of all of their tests is $14 \cdot 80 = 1120$. When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$. So, Payton's score must be $1215 - 1120 = \boxed{\text{(E)}\ 95}$.

Alternate Solution

The average of a set of numbers is the value we get if we evenly distribute the total across all entries. Assume that the first 14 students each scored 80. If Payton also scored an 80, the average would still be 80. In order to increase the overall average to 81, we need to add one more point to all of the scores, including Payton's. This means we need to add a total of 15 more points, so Payton needs $80 + 15 = \boxed{\text{(E)}\ 95}$.

Problem 4

The following problem is from both the 2015 AMC 12A #4 and 2015 AMC 10A #6, so both problems redirect to this page.

The sum of two positive numbers is 5 times their difference. What is the ratio of the larger number to the smaller number?

(A) $\frac{5}{4}$ (B) $\frac{3}{2}$ (C) $\frac{9}{5}$ (D) $\frac{5}{2}$ (E) 2

Solution 1

Let ( a ) be the bigger number and ( b ) be the smaller. $$a + b = 5(a - b)$$ Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives $$\frac{a}{b} = \frac{3}{2} \text{so the answer is } \boxed{\text{(B)}\ \frac{3}{2}}$$

Solution 2

Without loss of generality, let the two numbers be 3 and 2 as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{\text{(B)}\ \frac{3}{2}}$.

Problem 5

Sreeshma needs to estimate the quantity $\frac{a}{b} - c$, where ( a ), ( b ), and ( c ) are large