AMC 8 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 1 Hard

From City A to City B, one can travel by bus, ship, or train. The bus runs 3 times a day, the ship 2 times a day, and the train 6 times a day. In total, how many different ways are there that travel from City A to City B in one day?

  • A.

    7

  • B.

    9

  • C.

    11

  • D.

    13

  • E.

    15

Answer:C

Since one can travel directly from City A to City B by bus, ship, or train, this problem can be considered in three cases.

Taking the bus gives 3 ways, taking the ship gives 2 ways, and taking the train gives 6 ways.

Therefore, the total number of ways to travel from City A to City B is 3 + 2 + 6 = 11 \quad \text{(ways)}.

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Problem 2 Medium

Four students form a study group. A leader and a deputy leader are to be chosen from among the four. In total, there are            different ways to make this selection.

  • A.

    6

  • B.

    8

  • C.

    10

  • D.

    12

  • E.

    14

Answer:D

4\times 3=12

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Problem 3 Medium

4 colors are available (not necessary to use all) to paint each part of the figure below. If the colors in adjacent parts cannot be the same, how many ways of painting are there?

  • A.

    144

  • B.

    288

  • C.

    72

  • D.

    36

  • E.

    6

Answer:C

Case analysis:

First fill the middle part, which has 4 possibilities.

(1) AC has the same color:

4 \times 3 \times 1 \times 2 \times 2 = 48

(2) AC has different colors:

4 \times 3 \times 2 \times 1 \times 1 = 24

Thus,

48 + 24 = 72~\text{ways}.

Therefore, the answer is: \text{C}.

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Problem 4 Medium

In a certain school’s ballroom dance troupe, there are 43 members in total. Among them, 15 can dance Latin, 13 can dance Tango, and 5 can dance both. The number of people who can dance neither of these two dances is            .

  • A.

    18

  • B.

    20

  • C.

    22

  • D.

    24

  • E.

    26

Answer:B

43-\left( 15+13-5 \right)=20

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Problem 5 Medium

Among 40 students solving three math problems, 25 solved the first problem correctly, 28 solved the second problem correctly, and 31 solved the third problem correctly. How many students solved all three problems correctly, at least.

  • A.

    3

  • B.

    4

  • C.

    5

  • D.

    6

  • E.

    7

Answer:B

The number of students who solved both the second and the third problems is 28 + 31 - 40 = 19 \quad (\text{students}),

and the number of students who solved all three problems is 40 - (25 + 28 + 31 - 13 - 16 - 19) = 4 \quad (\text{students}).

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Problem 6 Medium

There are 9 different books: 3 math books, 4 Chinese books, and 2 English books. These books are to be arranged in a row on a shelf, with the math books kept together and the Chinese books kept together. Then there are            possible arrangements in total.

Answer:3456

Since the books of the same subject must be placed together, we can “bundle” them. First, bundle the math books and the Chinese books, making two bundles. Together with the two English books, we have four items to arrange. After arranging these, we then order the books within each bundle.

Therefore, the total number of arrangements is _{4}P_{4} \times _{4}P_{4} \times _{3}P_{3} = 3456 \quad (\text{ways}).

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Problem 7 Medium

Five people line up for a photo. Person A does not want to stand at either end. How many different possible arrangements are there?

  • A.

    72

  • B.

    48

  • C.

    36

  • D.

    24

  • E.

    12

Answer:A

3\times4\times3\times2\times1=72

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Problem 8 Medium

It’s primary school graduation, and June’s group is taking a photo. There are 4 boys and two girls, June and Ming. The two girls must not stand at either end and must stand next to each other. How many different possible arrangements are there?

  • A.

    24

  • B.

    48

  • C.

    96

  • D.

    144

  • E.

    184

Answer:D

_{4}P_{4} \times 2\times _{3}C_{1}

=24\times 6

=144

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Problem 9 Medium

There are 3 boys and 2 girls standing in a line. The two girls are not allowed to stand next to each other. Then there are            different possible arrangements.

Answer:72

For 5 people in a line, normally there are _{5}P_{5} = 120 arrangements.

If the two girls are adjacent, there are 2 \times _{4}P_{4} = 48 arrangements.

Therefore, the total number of arrangements is 120 - 48 = 72.

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Problem 10 Medium

Class 3 of Grade 4 is holding a Children’s Day celebration. The entire program consists of 2 dances, 2 songs, and 3 skits. If programs of the same type must be performed consecutively, then there are            different possible performance orders.

  • A.

    122

  • B.

    144

  • C.

    155

  • D.

    166

  • E.

    177

Answer:B

Since programs of the same type must be performed consecutively, we can apply the “bundling method.” First, arrange the three types of programs (dance, song, skit). Then, within each type, arrange the specific programs.

Therefore, the total number of performance orders is _{3}P_{3} \times _{2}P_{2} \times _{2}P_{2} \times _{3}P_{3} = 144 \quad (\text{orders}).

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Problem 11 Medium

Pick 3 from 8 children to take part in interview. How many different combination(s) is / are there ?

  • A.

    36

  • B.

    42

  • C.

    48

  • D.

    56

  • E.

    72

Answer:D

8\times 7\times 6\div 6=56．

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Problem 12 Medium

3 of the 5 children are selected to participate in a competition, and Chris and Debbie are 2 of the 5 children. If at least one people between Chris and Dibbie is selected, how many ways of selecting participants are there?

  • A.

    17

  • B.

    2021

  • C.

    9

  • D.

    5

  • E.

    1

Answer:C

From 5 children, choosing 3 can be done in _{5}C_{5} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 ways.

The case where neither Chris nor Debbie is chosen means choosing all 3 from the remaining 3 children, which is _{3}C_{3} = 1 way.

Therefore, the number of ways in which at least one of the two is chosen is 10 - 1 = 9.

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Problem 13 Easy

Brandon needs to choose three courses to study from Physics, Chemistry, Biology, Politics, History, and Geography. He has already chosen Physics. How many different ways are there for him to choose the other two courses?

  • A.

    10

  • B.

    15

  • C.

    20

  • D.

    30

Answer:A

There are _{5}C_{2} = 1 0 ways.

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Problem 14 Medium

Connect the diagonals of square ABCD, and color each of the four vertices either red or yellow. A triangle whose vertices are all the same color is called a monochromatic triangle. How many coloring methods in which there is at least one monochromatic triangle?

  • A.

    12

  • B.

    17

  • C.

    15

  • D.

    22

  • E.

    10

Answer:E

Each vertex can be colored in two ways, so there are 2 \times 2 \times 2 \times 2 = 16 coloring methods in total.

For there to be a monochromatic triangle, the case of “two vertices red and two vertices yellow” must be excluded. This case has _{4}C_{2} = 6 methods. Therefore, the number of coloring methods that yield at least one monochromatic triangle is 16 - 6 = 10.

Equivalently, to have a monochromatic triangle, we must exclude the case where “the two diagonals are colored differently.” This case has 2 \times 3 = 6 methods. Hence, the number of coloring methods with at least one monochromatic triangle is also 16 - 6 = 10.

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Problem 15 Medium

From the 8 points shown in the figure, choosing any 3 as vertices, how many triangles can be formed?

  • A.

    40

  • B.

    42

  • C.

    44

  • D.

    46

  • E.

    52

Answer:D

_{8}C_{3}-_{5}C_{3}=46

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Problem 16 Medium

There are three cards with numbers 8, 6, and 10. If any two cards are drawn, the difference that is most likely to occur is            .

Answer:2

If 8 and 6 are drawn, their difference is 8 - 6 = 2

If 8 and 10 are drawn, their difference is 10 - 8 = 2

If 6 and 10 are drawn, their difference is 10 - 6 = 4

Since the difference of 2 occurs more frequently, the most likely difference is 2.

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Problem 17 Medium

There are two dice, one large and one small. Each die has six faces numbered from 1 to 6. When the two dice are thrown simultaneously, what is the probability that the product of the two numbers is 12?

  • A.

    \frac{5}{36}

  • B.

    \frac{1}{9}

  • C.

    \frac{1}{18}

  • D.

    \frac{5}{18}

  • E.

    \frac{2}{9}

Answer:B

List all possibilities:

 

1

2

3

4

5

6

1

1

2

3

4

5

6

2

2

4

6

8

10

12

3

3

6

9

12

15

18

4

4

8

12

16

20

24

5

5

10

15

20

25

30

6

6

12

18

24

30

36

The probability that the product of the two numbers is 12 is \frac{4}{36} = \frac{1}{9}.

Therefore, the answer is \frac{1}{9}.

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Problem 18 Medium

In the sixth grade of a certain primary school, there are 6 classes, each with 40 students. Two classes are randomly selected from the 6 classes to participate in a live entertainment event hosted by a TV station. During the event, there is one lottery in which 4 lucky audience members are chosen. What is the probability that Bunny, a sixth-grade student, becomes one of the lucky winners?

  • A.

    \frac{1}{60}

  • B.

    \frac{1}{20}

  • C.

    \frac{1}{30}

  • D.

    \frac{1}{15}

  • E.

    \frac{1}{10}

Answer:A

The probability that Bunny’s class is selected to participate in the event is \frac{_{5}C_{1}}{_{6}C_{2}} = \frac{5}{15} = \frac{1}{3}.

If Bunny participates in the event, then the probability that he becomes a lucky winner is \frac{4}{40 \times 2} = \frac{1}{20}.

 

Therefore, the probability that Bunny becomes a lucky winner is \frac{1}{3} \times \frac{1}{20} = \frac{1}{60}.

Alternatively, the probability that Bunny’s class is selected can also be written as \frac{5}{6 \times 5 \div 2} = \frac{1}{3}.

Then, combining with the lottery probability, we again obtain \frac{1}{3} \times \frac{1}{20} = \frac{1}{60}.

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Problem 19 Medium

A school offers 3 labor-technology courses and 4 art courses. A student selects 3 courses from them. The probability that the student selects at least one course from each category is            .

  • A.

    \frac{6}{7}

  • B.

    \frac{5}{7}

  • C.

    \frac{4}{7}

  • D.

    \frac{3}{7}

  • E.

    \frac{1}{7}

Answer:A

\frac{_{7}C_{3}-_{3}C_{3}-_{4}C_{3}}{_{7}C_{3}}=\frac{6}{7}．

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Problem 20 Medium

Weier is playing a coin-tossing game: If a coin is tossed 5 times, what is the probability of getting heads no more than twice?

  • A.

    \frac{1}{2}

  • B.

    \frac{1}{8}

  • C.

    \frac{5}{32}

  • D.

    \frac{1}{32}

  • E.

    \frac{1}{4}

Answer:A

The probability of getting 0 heads is \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{32}, the probability of getting 1 head is 5 \times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{5}{32}, the probability of getting 2 heads is _{5}C_{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{5}{16}.

Therefore, the probability of getting no more than 2 heads is \frac{1}{32}+\frac{5}{32}+\frac{5}{16}=\frac{1}{2}.

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Problem 21 Medium

Using the digits 1, 2, 3, 4 to form three-digit numbers without repeating digits, if one such number is chosen at random, what is the probability that the selected number is a multiple of 3?

  • A.

    \frac{1}{4}

  • B.

    \frac{1}{3}

  • C.

    \frac{1}{2}

  • D.

    \frac{3}{4}

  • E.

    \frac{1}{3}

Answer:C

There are 4 possible combinations to form three-digit numbers. Among them, the combinations (1,2,3) and (2,3,4) produce multiples of 3, giving a total of 2 such combinations.

Therefore, the probability is \frac{2}{4} = \frac{1}{2}.

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Problem 22 Medium

Given the four digits 2, 0, 1, 7, each digit may be used at most once. How many natural numbers less than 2017 can be formed?

  • A.

    33

  • B.

    35

  • C.

    37

  • D.

    41

  • E.

    45

Answer:C

The natural numbers that satisfy the condition can be divided into 4 categories:

One-digit numbers: 0, 1, 2, 7, a total of 4.

Two-digit numbers: The first digit cannot be 0, and digits cannot be repeated, so there are 3 \times 3 = 9 numbers.

Three-digit numbers: The first digit cannot be 0, and digits cannot be repeated, so there are 3 \times 3 \times 2 = 18 numbers.

Four-digit numbers: The first digit cannot be 0, and digits cannot be repeated, so there are 3 \times 2 \times 1 = 6 numbers.

Therefore, in total, the number of natural numbers less than 2017 that can be formed is 4 + 9 + 18 + 6 = 37.

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Problem 23 Medium

From 3 ones, 2 twos, and 1 three, if 3 digits are selected, how many different three-digit numbers can be formed?

  • A.

    15

  • B.

    16

  • C.

    17

  • D.

    18

  • E.

    19

Answer:E

Case 1:

3 digits all the same: 111, only 1.

Case 2:

2 of three digits are the same:

322, 232, 223, 311, 131, 113, 122, 212, 221, 211, 121, 112. There are 3+3+6=12 numbers.

Case 3:

All three digits are different: there are _{3}P_{3}=3\times 2\times 1=6 numbers.

Totally, 1+12+6=19 numbers.

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Problem 24 Medium

Among the positive integers less than 1000, how many are there with no repeating digit?

  • A.

    648

  • B.

    738

  • C.

    758

  • D.

    828

  • E.

    670

Answer:B

Three-digit numbers: 648

Two-digit numbers: 81

One-digit numbers: 9

In total: 738.

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Problem 25 Hard

Using the digits 1, 2, 3, 4, 5 to form a five-digit number (each digit used at most once), such that the difference between any two adjacent digits is at least 2. How many such five-digit numbers are there?

  • A.

    10

  • B.

    12

  • C.

    14

  • D.

    16

  • E.

    18

Answer:C

Starting with 1: 2 numbers — 13524, 14253;

so starting with 5, there are also 2 numbers.

Starting with 2: 3 numbers — 24135, 24153, 25314;

so starting with 4, there are also 3 numbers.

Starting with 3: 4 numbers — 31425, 31524, 35241, 35142.

Starting with 4: 3 numbers — 41352, 42513, 42531.

Starting with 5: 2 numbers — 52413, 53142.

Therefore, the total number of such five-digit numbers is 14.

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Problem 26 Medium

There are 5 red balls and 3 white balls in the box. If two balls are drawn at random, what is the probability of getting one red and one white?

  • A.

    \frac{13}{28}

  • B.

    \frac{15}{28}

  • C.

    \frac{15}{56}

  • D.

    \frac{5}{14}

  • E.

    \frac{5}{7}

Answer:B

\frac{5\times 3}{8\times 7\div 2}=\frac{15}{28}．

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Problem 27 Medium

Box A contains 3 white ping-pong balls numbered 1, 2, 3. Box B contains 3 yellow ping-pong balls numbered 4, 5, 6. One ball is randomly drawn from each box. What is probability that the sum of the two numbers is greater than 6?

  • A.

    \frac{1}{4}

  • B.

    \frac{1}{2}

  • C.

    \frac{2}{3}

  • D.

    \frac{3}{4}

  • E.

    \frac{1}{3}

Answer:C

The possible sums of the numbers are: 1+4=5,\; 1+5=6,\; 1+6=7,\; 2+4=6,\; 2+5=7,\; 2+6=8,\; 3+4=7,\; 3+5=8,\; 3+6=9.

The probability that the sum is greater than 6 is \frac{6}{9} = \frac{2}{3}.

Therefore, the answer is \frac{2}{3}.

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Problem 28 Medium

A bag contains 5 red balls, 6 white balls, and 3 black balls. In order for the probability of drawing a black ball to be \tfrac{2}{3}, how many additional black balls must be put into the bag?

  • A.

    15

  • B.

    16

  • C.

    17

  • D.

    18

  • E.

    19

Answer:E

Suppose x additional black balls need to be put into the bag. Then \frac{3+x}{5+6+3+x} = \frac{2}{3}.

Solving gives x = 19.

Therefore, 19 more black balls should be added to the bag.

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Problem 29 Medium

Vendors often run lottery games at the school gate. One vendor has a black bag containing 50 balls of different colors: 1 red, 2 yellow, 10 green, and the rest white. After mixing them thoroughly, the rule is: each draw costs 2 dollars for 1 ball.

Drawing a red ball wins a prize worth 8 dollars.

Drawing a yellow ball wins a prize worth 5 dollars.

Drawing a green ball wins a prize worth 2 dollars.

Drawing a white ball wins no prize.

If you spend 4 dollars to draw 2 balls at the same time, what is the probability of obtaining a prize worth 10 dollars?

  • A.

    \frac{18}{1225}

  • B.

    \frac{11}{1225}

  • C.

    \frac{121}{225}

  • D.

    \frac{11}{245}

  • E.

    \frac{2}{245}

Answer:B

There are two cases for winning a prize worth 10 dollars:

Case 1: Drawing one 8-dollar prize and one 2-dollar prize, i.e., one red ball and one green ball. The probability of drawing a red ball first and then a green ball, or a green ball first and then a red ball, is the same, giving \frac{1 \times 10}{50 \times 49} \times 2 = \frac{2}{245}.

Case 2: Drawing two yellow balls. The probability of drawing the first yellow ball is \frac{2}{50}, and the probability of drawing the second yellow ball is \frac{1}{49}. Thus, \frac{2}{50} \times \frac{1}{49} = \frac{1}{1225}.

Therefore, the total probability is \frac{2}{245} + \frac{1}{1225} = \frac{11}{1225}.

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Problem 30 Medium

A bag contains 5 balls that are identical in size and shape: 3 white balls and 2 red balls. Two balls are drawn with replacement. What is the probability that the two balls are of different colors?

  • A.

    \frac{2}{5}

  • B.

    \frac{11}{25}

  • C.

    \frac{12}{25}

  • D.

    \frac{13}{25}

  • E.

    \frac{3}{5}

Answer:C

The probability that both balls are white is \frac{3 \times 3}{5 \times 5} = \frac{9}{25}, and the probability that both balls are red is \frac{2 \times 2}{5 \times 5} = \frac{4}{25}.

So the probability that the two balls are the same color is \frac{9}{25} + \frac{4}{25} = \frac{13}{25}.

Therefore, the probability that the two balls are of different colors is 1 - \frac{13}{25} = \frac{12}{25}.

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Related Paper
AMC 8 Daily Practice Round 5 AMC 8 Daily Practice Round 4 AMC 8 Daily Practice Round 3 AMC 8 Daily Practice Round 2
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