美国数学竞赛2013 Amc 8试题与详解 665l1a

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way? (C) 3

(D) 4

(E) 5

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(B) 2

Solution

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2013 AMC 8, Problem #1—

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(A) 1

Answer (A): The smallest multiple of 6 that is at least 23 is 24, so Danica must buy 1 additional car.

A sign at the fish market says, “50% off, today only: half-pound packages for just $3 per package.” What is the regular price for a full pound of fish, in dollars? (C) 10

(D) 12

(E) 15

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(B) 9

Solution

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2013 AMC 8, Problem #2—

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(A) 6

Answer (D): A half-pound package costs $3 at the sale price, so it would cost $6 at the regular price. A whole pound would cost $12 at the regular price.

What is the value of 4 · (−1 + 2 − 3 + 4 − 5 + 6 − 7 + · · · + 1000)? (B) 0

(C) 1

(D) 500

(E) 2000

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(A) − 10

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Solution

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2013 AMC 8, Problem #3—

Answer (E): Inside the parentheses are 500 pairs of numbers, each with a sum of 1. Therefore the expression equals 4 · 500 = 2000.

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill? (B) $128

(C) $140

(D) $144

(E) $160

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(A) $120

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Solution

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2013 AMC 8, Problem #4—

Answer (C): Judi’s share of the bill was 7($2.50) = $17.50, so the total bill was 8($17.50) = $140.

Hammie is in the 6th grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds? (B) median, by 20

(C) average, by 5

(D) average, by 15

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(A) median, by 60 (E) average, by 20

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Solution

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2013 AMC 8, Problem #5—

Answer (E): The total weight is 130 pounds, so the average is 26 pounds. The median is 6 pounds, so the average is greater by 20 pounds.

The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, 30 = 6 × 5. What is the missing number in the top row?

6

5

30

600

(C) 4

(D) 5

(E) 6

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(B) 3

Solution

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2013 AMC 8, Problem #6—

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(A) 2

Answer (C): The product of the two numbers in the second row is 600, so the missing number in that row is 600 30 = 20. The product of 5 with the missing number in the top row is 20, so the missing number in the top row is 20 5 = 4.

Trey and his mom stopped at a railroad crossing to let a train . As the train began to , Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train? (B) 80

(C) 100

(D) 120

(E) 140

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(A) 60

Solution

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2013 AMC 8, Problem #7—

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Answer (C): Because Trey counted 6 cars in 10 seconds, close to 6 · 6 = 36 cars ed in 10 · 6 = 60 seconds or 1 minute. In 2 minutes 45 seconds, 3 about 2 · 36 + 45 60 (36) = 72 + 4 (36) = 72 + 27 = 99 cars ed, so there were approximately 100 cars in the train. OR Two minutes and 45 seconds is 2(60) + 45 = 165 seconds. Let N be the number of cars, and set up a proportion: N 6 = 10 165

Solving gives 10N = 6(165) = 990, so N = 99. Approximately 100 train cars ed.

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads? 1 8

(B)

1 4

(C)

3 8

(D)

1 2

(E)

3 4

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(A)

Solution

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2013 AMC 8, Problem #8—

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Answer (C): List the 8 possible equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Only HHH, HHT, THH have at least 2 consecutive heads, so the probability of at least 2 consecutive heads is 38 .

The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer? (B) 10th

(C) 11th

(D) 12th

(E) 13th

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(A) 9th

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Solution

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2013 AMC 8, Problem #9—

Answer (C):

Jump 1 2 3 .. . 10 11

Distance (meters) 1 = 20 2 = 21 4 = 22 .. . 512 = 29 1024 = 210

Because 1024 meters is greater than 1 kilometer, he first exceeds 1 kilometer on the 11th jump.

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594? (B) 165

(C) 330

(D) 625

(E) 660

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(A) 110

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Solution

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2013 AMC 8, Problem #10—

Answer (C): Because the prime factorizations of 180 and 594 are 22 · 32 ·5 and 2 · 33 · 11, respectively, the least common multiple of 180 and 594 is 22 · 33 · 5 · 11, and their greatest common factor is 2 · 32 . The ratio of their least common 2 3 multiple to their greatest common factor is 2 ·32·3·5·11 = 2 · 3 · 5 · 11 = 330. 2

Ted’s grandfather used his treill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treill. How many minutes less? (B) 2

(C) 3

(D) 4

(E) 5

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(A) 1

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Solution

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2013 AMC 8, Problem #11—

Answer (D): Because time equals distance divided by rate, Grandfather was on the treill for 25 hours or 24 minutes on Monday. Similarly, he walked for 23 hours, or 40 minutes, on Wednesday and 24 hours, or 30 minutes, on Friday. The total time Grandfather spent on the treill was 24 + 40 + 30 = 94 minutes. If he had walked the entire 6 miles at 4 miles per hour, he would have spent 64 hours, or 90 minutes, on the treill, so he would have saved 4 minutes.

At the 2013 Winnebago County Fair a vendor is offering a “fair special” on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the “fair special” to buy three pairs of sandals. What percentage of the $150 regular price did he save? (B) 30

(C) 33

(D) 40

(E) 45

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(A) 25

Solution

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2013 AMC 8, Problem #12—

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Answer (B): Javier spent $50 on the first pair, $30 on the second pair, and $25 on the third pair, for a total of $105. This is a savings of $45 off the $150 list price, which is 30% off.

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one? (B) 46

(C) 47

(D) 48

(E) 49

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(A) 45

Solution

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2013 AMC 8, Problem #13—

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Answer (A): Switching a digit from the units column to the tens column increases the sum by 9 times the value of that digit. For example, switching a 7 from the units column to the tens column increases the sum by 70−7 = 63 = 9·7. Similarly, switching a digit from the tens column to the units column decreases the sum by 9 times the value of that digit. Therefore reversing two digits changes the sum by an amount that must be a multiple of 9. Among the given choices, only 45 is a possible difference. (Note: Other multiples of 9 are also possible.)

Abe holds 1 green and 1 red jelly bean in his hand. Bea holds 1 green, 1 yellow, and 2 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match? (B)

1 3

(C)

3 8

(D)

1 2

(E)

2 3

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1 4

Solution

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2013 AMC 8, Problem #14—

Answer (C): Denote Abe’s jelly beans by g and r. Denote Bea’s jelly beans by G, Y , R1 , and R2 . There are 8 equally likely pairings: (g, G), (g, Y ), (g, R1 ), (g, R2 ), (r, G), (r, Y ), (r, R1 ), and (r, R2 ). Only (g, G), (r, R1 ), and (r, R2 ) match, so the probability that the colors match is 38 .

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(A)

If 3p + 34 = 90, 2r + 44 = 76, and 53 + 6s = 1421, what is the product of p, r, and s? (B) 40

(C) 50

(D) 70

(E) 90

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(A) 27

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Solution

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2013 AMC 8, Problem #15—

Answer (B): From the first equation, 3p +81 = 90, so 3p = 9, and p = 2. From the second equation, 2r = 32, so r = 5. From the third equation, 6s +125 = 1421, so 6s = 1296, and s = 4. The product of p, r, and s is 2 · 5 · 4 = 40.

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of 8th-graders to 6th-graders is 5 : 3, and the the ratio of 8th-graders to 7th-graders is 8 : 5. What is the smallest number of students that could be participating in the project? (B) 40

(C) 55

(D) 79

(E) 89

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(A) 16

Solution

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2013 AMC 8, Problem #16—

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Answer (E): The number of 8th -graders must be a multiple of both 5 and 8, so it must be at least 40. If there are 40 8th -graders, then there are 35 (40) = 24 6th -graders and 58 (40) = 25 7th -graders, for a total of 40 + 24 + 25 = 89 students.

The sum of six consecutive positive integers is 2013. What is the largest of these six integers? (B) 338

(C) 340

(D) 345

(E) 350

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(A) 335

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Solution

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2013 AMC 8, Problem #17—

= 335.5, so 2013 = Answer (B): The average of the six integers is 2013 6 333 + 334 + 335 + 336 + 337 + 338. The largest of the six integers is 338.

Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain? (B) 280

(C) 320

(D) 340

(E) 600

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(A) 204

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Solution

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2013 AMC 8, Problem #18—

Answer (B): The fort, including the inside, occupies a volume of 12×10×5 = 600 cubic feet. The inside of the fort is 12 − 2 = 10 feet long, 10 − 2 = 8 feet wide, and 5 − 1 = 4 feet high, so it occupies a volume of 10 × 8 × 4 = 320 cubic feet. Therefore the walls and floor occupy 600 − 320 = 280 cubic feet, so the fort contains 280 blocks.

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don’t show theirs to anyone. Cassie says, “I didn’t get the lowest score in our class,” and Bridget adds, “I didn’t get the highest score.” What is the ranking of the three girls from highest to lowest? (B) Hannah, Bridget, Cassie (D) Cassie, Hannah, Bridget

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(A) Hannah, Cassie, Bridget (C) Cassie, Bridget, Hannah (E) Bridget, Cassie, Hannah

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Solution

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2013 AMC 8, Problem #19—

Answer (D): Cassie says, “I didn’t get the lowest score,” so her score is higher than Hannah’s score. Bridget says, “I didn’t get the highest score,” so her score is lower than Hannah’s score. Therefore the order, from highest to lowest, must be Cassie, Hannah, Bridget.

A 1 × 2 rectangle is inscribed in a semicircle with longer side on the diameter. What is the area of the semicircle? (B)

2π 3

(C) π

(D)

4π 3

(E)

5π 3

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π 2

Solution

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2013 AMC 8, Problem #20—

Answer (C):

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(A)

√

2 1

1

1

By√the Pythagorean Theorem, the radius of the semicircle is π( 2)2 = π. 2

√

2, so its area is

Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take? (C) 9

(D) 12

(E) 18

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(B) 6

2013 AMC 8, Problem #21— Solution

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Answer (E): There are 3 ways she can bike from home to the southwest corner of the park, EEN, ENE, or NEE. There are 6 ways to bike from the northeast corner of the park to school, EENN, ENEN, ENNE, NEEN, NENE, or NNEE. So there are 6 · 3 = 18 routes.

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(A) 3

OR

Using a Pascal’s Triangle approach starting from the house to the school, count the routes to each intermediate point with the following diagram, moving only north or east at each corner.

Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?

(B) 1952

(C) 1980

(D) 2013

(E) 3932

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(A) 1920

2013 AMC 8, Problem #22—

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Solution

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Answer (E): A grid 60 toothpicks long and 32 toothpicks high needs 61 columns of 32 toothpicks and 33 rows of 60 toothpicks. Therefore a total of (61 × 32) + (33 × 60) = 3932 toothpicks are needed.

Angle ABC of 4ABC is a right angle. The sides of 4ABC are the diameters of semicircles as shown. The area of the semicircle on AB equals 8π, and the arc of the semicircle on AC has length 8.5π. What is the radius of the semicircle on BC?

A

B

(B) 7.5

(C) 8

(D) 8.5

(E) 9

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(A) 7

C

2013 AMC 8, Problem #23—

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Solution

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Answer (B): The circle with diameter AB has twice the area of the corresponding semicircle; thus the area of the circle is 16π and its radius is 4. Consequently AB = 8. The circle with diameter AC has circumference 17π, so AC = 17. AC is the hypotenuse of the right triangle. By the Pythagorean Theorem, 172 = 82 + (BC)2 . Therefore BC = 15, and the radius is 7.5.

Squares ABCD, EFGH, and GHIJ are equal in area. Points C and D are the midpoints of sides IH ad HE, respectively. What is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares? A

D

F 7 24

(C)

1 3

J

G (D)

3 8

(E)

5 12

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I

2013 AMC 8, Problem #24— Solution Answer (C):

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1 4

C

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(A)

H

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E

B

Let the length of the side of each square be 1 and extend side AD to Y as shown. The total area of the three squares is 3. The unshaded area is area (EDY F ) + area (AY J) = 1( 12 ) + 12 · 32 · 2 = 12 + 32 = 2, so the shaded area is 1 and the desired ratio is 13 . OR Label point X as shown. Intuitively, rotating XIJ 180◦ about X takes it to XDA so the shaded area is the same as the area of square ABCD and the desired ratio is 13 . More precisely, segments AX, XD, and DA are parallel to segments JX, XI, and IJ, respectively. Also, DA = IJ, so ADX is congruent to J I X .

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are R1 = 100 inches, R2 = 60 inches, and R3 = 80 inches, respectively. The ball always remains in with the track and does not slip. What is the distance the center of the ball travels over the course from A to B? R2 A R3

R1

(B) 240π

(C) 260π

(D) 280π

(E) 500π

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(A) 238π

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Solution

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2013 AMC 8, Problem #25—

Answer (A): The diameter of the ball is 4 inches, so its radius is 2 inches. The center of the ball rolls through semicircles of radii R1 − 2 = 100 − 2 = 98 inches, R2 + 2 = 60 + 2 = 62 inches, and R3 − 2 = 80 − 2 = 78 inches, respectively. The length of the path is then π(98 + 62 + 78) = 238π inches.