美国数学竞赛2013 Amc 8试题与详解

2013 AMC 8, Problem #1—

Solution

Difficulty, Percent correct Easy 100-80% Med Easy 80-60% Medium 60-40% Med Hard 40-20% Hard 20-0%

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. Shenow has 23 model cars. What is the smallest number of additional cars she must buy inorder to be able to arrange all her cars this way?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

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

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Solution


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.

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

(A) 6 (B) 9 (C) 10 (D) 12 (E) 15

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Solution


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

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

(A) − 10 (B) 0 (C) 1 (D) 500 (E) 2000

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Solution


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

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

(A) $120 (B) $128 (C) $140 (D) $144 (E) $160

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Solution


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.

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

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Solution


30

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

5number in the top row is 20, so the missing number in the top row is 20 = 4.

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?

600

30

6 5

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

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Solution


Answer (C): Because Trey counted 6 cars in 10 seconds, close to 6 · 6 = 36cabarsout 2

passed36

i+n

4510 ·(36)

6==7260+ 3seconds(36) =

or721+m27inu=te9.9 cInars

2pmiassednut

,esso45there

secondswere

,· 60 4

approximately 100 cars in the train.

OR

Two minutes and 45 seconds is 2(60)+ 45 = 165 seconds. Let N be the numberof cars, and set up a proportion:

6

10=

N

165

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

Trey and his mom stopped at a railroad crossing to let a train pass. As the train beganto pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45seconds to clear the crossing at a constant speed. Which of the following was the mostlikely number of cars in the train?

(A) 60 (B) 80 (C) 100 (D) 120 (E) 140

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Solution


8

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 consec-utive heads, so the probability of at least 2 consecutive heads is 3 .

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

(A) 18 (B) 1

4 (C) 38 (D) 1

2 (E) 34

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Solution


Answer (C):

Jump123

Distance (meters)1 = 20

2 = 21

4 = 22

......

10 512 = 29

11 1024 = 210

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

The Incredible Hulk can double the distance he jumps with each succeeding jump. Ifhis first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, andso on, then on which jump will he first be able to jump more than 1 kilometer?

(A) 9th (B) 10th (C) 11th (D) 12th (E) 13th

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Solution


What is the ratio of the least common multiple of 180 and 594 to the greatest commonfactor of 180 and 594?

(A) 110 (B) 165 (C) 330 (D) 625 (E) 660

2An

3s3w11,er

resp(C):

ectivelBecaus

y, tehtehleeastprime f

commoacto

nrizmautlitoinplseooff180180

andand

594594

arise222

2 ·333

2 ·55an11,d

· · · · ·and their greatest common factor is 2 · 32.

2

The ratio of their least common

multiple to their greatest common factor is2·32

3

3·25·11

· = 2 · 3 · 5 · 11 = 330.kaoguti.com


Solution


5

Answer (D): Because time equals distance divided by rate, Grandfather wason the treadmill for 2 hours or 24 minutes on Monday. Similarly, he walked for 2

3

4hours, or 40 minutes, on Wednesday and 2 hours, or 30 minutes, on Friday. Thetotal time Grandfather spent on the treadmill was 24 + 40 + 30 = 94 minutes.If he had walked the entire 6 miles at 4 miles per hour, he would have spent 6

4hours, or 90 minutes, on the treadmill, so he would have saved 4 minutes.

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

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

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Solution


At the 2013 Winnebago County Fair a vendor is offering a “fair special” on sandals. If youbuy 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 buythree pairs of sandals. What percentage of the $150 regular price did he save?

(A) 25 (B) 30 (C) 33 (D) 40 (E) 45

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 $150list price, which is 30% off.ka

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Solution


Answer (A): Switching a digit from the units column to the tens columnincreases the sum by 9 times the value of that digit. For example, switching a 7from 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 decreasesthe sum by 9 times the value of that digit. Therefore reversing two digits changesthe 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.)

When Clara totaled her scores, she inadvertently reversed the units digit and the tensdigit of one score. By which of the following might her incorrect sum have differed fromthe correct one?

(A) 45 (B) 46 (C) 47 (D) 48 (E) 49

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Solution


8

Answer (C): Denote Abe’s jelly beans by g and r. Denote Bea’s jelly beansby 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 3 .

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

(A) 14 (B) 1

3 (C) 38 (D) 1

2 (E) 23

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Solution


Answer (B): From the first equation, 3p+81 = 90, so 3p = 9, and p = 2. Fromthe 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.

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

(A) 27 (B) 40 (C) 50 (D) 70 (E) 90

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Solution


5

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 3 (40) = 24

86th-graders and 5 (40) = 25 7th-graders, for a total of 40+24+25 = 89 students.

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?

(A) 16 (B) 40 (C) 55 (D) 79 (E) 89

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Solution


Answer (B): The average of the six integers is 20136 = 335.5, so 2013 =

333 + 334 + 335 + 336 + 337 + 338. The largest of the six integers is 338.

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

(A) 335 (B) 338 (C) 340 (D) 345 (E) 350

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Solution


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 feetwide, and 5− 1 = 4 feet high, so it occupies a volume of 10× 8× 4 = 320 cubicfeet. Therefore the walls and floor occupy 600 − 320 = 280 cubic feet, so thefort contains 280 blocks.

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

(A) 204 (B) 280 (C) 320 (D) 340 (E) 600

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Solution


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

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannahshows 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 getthe highest score.” What is the ranking of the three girls from highest to lowest?

(B) Hannah, Bridget, Cassie(D) Cassie, Hannah, Bridget

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

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Solution


Answer (C):

√2

1

1

1

By the Pythagorean Theorem, the radius of the semicircle is√2, so its area is

π(√2)2

2 = π.

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

(A) π2 (B) 2π

3 (C) π (D) 4π3 (E) 5π

3

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Solution


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

OR

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

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. Onschool days she bikes on streets to the southwest corner of City Park, then takes adiagonal path through the park to the northeast corner, and then bikes on streets toschool. If her route is as short as possible, how many different routes can she take?

(A) 3 (B) 6 (C) 9 (D) 12 (E) 18

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Solution


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

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

(A) 1920 (B) 1952 (C) 1980 (D) 2013 (E) 3932

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Solution


Answer (B): The circle with diameter AB has twice the area of the cor-responding 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 PythagoreanTheorem, 172 = 82 + (BC)2. Therefore BC = 15, and the radius is 7.5.

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

A

B C

(A) 7 (B) 7.5 (C) 8 (D) 8.5 (E) 9

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Solution


Answer (C):

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 )

2+ area (AY J) = 1( 1 ) + 12 · 3

2 · 2 = 12 + 3

2 = 2, so the shaded area is 1 and thedesired ratio is 1

3 .

OR

3

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 thedesired ratio is 1 . More precisely, segments AX, XD, and DA are parallel to

tosegme

JnItXs.JX, XI, and IJ , respectively. Also, DA = IJ , so�ADX is congruent

�

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

A B

CDE

F G

H I

J

(A) 14 (B) 7

24 (C) 13 (D) 3

8 (E) 512

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Solution


Answer (A): The diameter of the ball is 4 inches, so its radius is 2 inches. Thecenter 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. Thelength of the path is then π(98 + 62 + 78) = 238π inches.

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

R1

R2

R3

A

(A) 238π (B) 240π (C) 260π (D) 280π (E) 500π

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美国数学竞赛2013 Amc 8试题与详解