AMC 8 Daily Practice Round 9

Complete problem set with solutions and individual problem pages

Problem 1 Easy

A bakery owner received a picnic order requiring 125 sandwiches. If he can make one sandwich in 24 seconds, how many minutes will it take him to complete the order?

  • A.

    45

  • B.

    48

  • C.

    50

  • D.

    86

  • E.

    90

Answer:C

125 \times 24 \div 60 = 50

The answer is \text{C}.

Link Problem
Problem 2 Easy

Andy is drawing on a grid. He has drawn a small boat on a one-inch grid. What is the area of the boat in square inches?

  • A.

    20.5

  • B.

    21.5

  • C.

    22

  • D.

    22.5

  • E.

    23.5

Answer:D

Square: 2 \times 2 = 4

Small trapezoid: \frac 12 \times (4 + 5) \times 1 =4.5

Big trapezoid: \frac 12 \times (5 + 9) \times 2 =14

In total: 4 + \frac 92 + 14 = 22.5

The answer is \text{D}.

Link Problem
Problem 3 Easy

A train took H hours to travel D kilometers from Applewood to Aurora, arriving 2 hours late. What speed should it have traveled at to arrive on time?

  • A.

    H+2

  • B.

    H-2

  • C.

    \frac{D}{H}+2

  • D.

    \frac{D}{H+2}

  • E.

    \frac{D}{H-2}

Answer:E

The required time is now H - 2 hours, while the distance remains D kilometers.

∴ The required speed is \frac{D}{H-2}.

The answer is \text{E}.

Link Problem
Problem 4 Easy

Dino bought a giant jelly and plans to share it with his friends. He is going to make three cuts. What is the maximum number of pieces he can get?

  • A.

    5

  • B.

    6

  • C.

    7

  • D.

    8

  • E.

    9

Answer:D

By making two vertical cuts from top to bottom, the jelly is divided into 4 pieces. Then, by making a horizontal cut through the middle, the jelly can be further divided into 8 pieces.

The answer is \text{D}.

Link Problem
Problem 5 Easy

Bill is choosing his courses for next year. He has to choosc two different courses, one from Foundational courses and one from Elective courses.

Foundational courses

Elective courses
Game AIGame AI
Database systemBiology
AlgebraItalian
GeometryHistory

How many different pairs of combinations are possible?

  • A.

    7

  • B.

    8

  • C.

    12

  • D.

    15

  • E.

    16

Answer:D

It is worth noting that Game AI courses are both in Foundational courses and in elective courses.

C_{3}^{1}\times C_{4}^{1}+1\times3=3\times4+3=15.

The answer is \text{D}.

Link Problem
Problem 6 Easy

A trapezoid has a top base of length 5, a bottom base of length 10, and two legs of lengths 3 and 4, respectively. What is the area of the trapezoid?

  • A.

    18

  • B.

    22.5

  • C.

    26.25

  • D.

    30

  • E.

    32

Answer:A

As shown in the diagram, draw AE \parallel BC through point A, intersecting DC at point F. Draw AF \perp DC at point F.

Since quadrilateral ABCE is a parallelogram, we have:   AE = BC = 4

Also,   DE = DC - AB = 10 - 5 = 5

Given that AD = 3, applying the Pythagorean theorem in right triangle \triangle ADE, we find:   AF = \frac{3 \times 4}{5} = 2.4

Thus, the area of trapezoid ABCD is:   S_{ABCD} = \frac{(5+10) \times 2.4}{2} = 18

Link Problem
Problem 7 Easy

Emily's teacher asks her to plot all the ordered pairs (r, s) of positive integers, where r represents the radius of a circle and s represents the area of the same circle. What should her graph look like?

  • A.

  • B.

  • C.

  • D.

  • E.

Answer:D

S=\pi r^2

r and s present a quadratic function relationship.

The answer is \text{D}.

Link Problem
Problem 8 Easy

Given that {x^{2}-6x+1=0}, and x \ne 0. Find the value of {x^{4}+\dfrac{1}{x^{4}}}.

  • A.

    1131

  • B.

    1154

  • C.

    1156

  • D.

    1197

  • E.

    1290

Answer:B

{{x}^{2}}-6x+1=0

x-6+\frac{1}{x}=0(∵x\ne 0

x+\frac{1}{x}=6

{{\left( x+\frac{1}{x} \right)}^{2}}={{6}^{2}}

{{x}^{2}}+\frac{1}{{{x}^{2}}}=36-2

~~~~~~~~~~~~~~~=34

{{\left( x+\frac{1}{x} \right)}^{4}}={{6}^{4}}

{{x}^{4}}+4{{x}^{2}}+6+\frac{4}{{{x}^{2}}}+\frac{1}{{{x}^{4}}}=1296

\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+4\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=1290

{{x}^{4}}+\frac{1}{{{x}^{4}}}=1290-136

~~~~~~~~~~~~~~~=1154

The answer is \text{B}.

Link Problem
Problem 9 Easy

The school's basketball team currently has 5 boys and 6 girls. The coach wants to randomly select 3 people from the group to participate in the schools' 3v3 basketball league. How many ways can the 3 selected students be arranged such that at least one girl is included?

  • A.

    45

  • B.

    105

  • C.

    145

  • D.

    155

  • E.

    159

Answer:D

Number of choices of choosing 3 students from 11 students is:

P_1=\frac{11\times10\times9}{3\times2\times1}

Number of choices of choosing 3 students from 5 boys is:

P_2=\frac{5\times4\times3}{3\times2\times1}

P=P_1-P_2=\frac{11\times10\times9}{3\times2\times1}-\frac{5\times4\times3}{3\times2\times1}=155.

The answer is \text{D}.

Link Problem
Problem 10 Hard

What is the value of \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots +\frac{99}{100!} (Factorial can be used in the final result)?

  • A.

    \frac{1}{100!}

  • B.

    \frac{1}{99!}

  • C.

    \frac {1}{100}

  • D.

    1-\frac{1}{100!}

  • E.

    1

Answer:D

\therefore \frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+\cdots +\frac{100-1}{100!}

=\left( \frac{2}{2!}+\frac{3}{3!}+\frac{4}{4!}+\cdots +\frac{100}{100!} \right)-\left( \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{100!} \right)

=\left( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{99!} \right)-\left( \frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{100!} \right)

=1-\frac{1}{100!};

\therefore The answer is \text{D}.

Link Problem
Problem 11 Easy

A sequence of 6 lights (either on or off) is used to represent numbers. The diagram provided shows the representations for the numbers 1 to 5. Based on this pattern, what number does the sequence ●○○●●○ represent?

  • A.

    23

  • B.

    24

  • C.

    25

  • D.

    26

  • E.

    27

Answer:C

The given sequence can be converted into a binary number, where a lit light is represented as 1 and an unlit light as 0.

Thus, the sequence ●○○●●○ corresponds to the binary number: 011001_2

Converting this to decimal:

0 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 16 + 8 + 1 = 25

Thus, the number represented by the sequence is 25.

The answer is \text{C}.

Link Problem
Problem 12 Medium

There are several boxes of cards, with each box containing the same number of cards.

- If only one box is distributed, and each child receives 8 cards, there will be a shortage of 5 cards.

- If all the boxes are distributed, each child receives 56 cards, with 8 cards remaining.

How many cards are there in each box?

  • A.

    6

  • B.

    8

  • C.

    43

  • D.

    48

  • E.

    344

Answer:C

Let there be x children. Then, the number of cards in each box is 8x - 5, and the total number of cards is 56x + 8.

Since \frac{56x + 8}{8x - 5} = \frac{43}{8x - 5} + 7

it follows that 8x - 5 must be either 1 or 43.

Solving for x, we get x = 6 (since \frac{3}{4} is discarded as it is not an integer).

Thus, the number of cards in each box is 43.

The answer is \text{C}.

Link Problem
Problem 13 Easy

As shown in the figure, the cross-sections of 7 cylindrical chopsticks are all circles with radius r. What is the total length of the string needed to wrap around all 7 chopsticks?

  • A.

    12r + 2\pi r

  • B.

    16r + 2\pi r

  • C.

    6r + 12\pi r

  • D.

    18r

  • E.

    18 \pi r

Answer:A

As shown in the figure, let M and N be the centers of the two circles, and let BC be their external common tangent with A as the point of tangency.

Since BC = MN = 2r, the same reasoning applies to each pair of adjacent circles, forming rectangles and circular sectors.

Since there are 6 such tangents, their total length is:   6BC = 6 \times 2r = 12r

The central angle corresponding to each arc is:   6 \angle AMB = 360^\circ \quad \Rightarrow \quad \angle AMB = 60^\circ

Thus, the arc length of each circular segment is:   AB = \frac{60^\circ}{360^\circ} \times 2\pi r = \frac{1}{3} \pi r

The total string length consists of the 6 straight segments and 6 arc segments:   6BC + 6AB = 12r + 2\pi r

The answer is \text{A}.

Link Problem
Problem 14 Medium

As shown in the figure, in triangle \triangle ABC, the sides AB and AC are divided into five equal parts by four segments parallel to BC. If the area of \triangle ABC is S, what is the sum of the areas of the shaded regions labeled ② and ④?

  • A.

    \frac{S}{3}

  • B.

    \frac{2S}{5}

  • C.

    \frac{3S}{7}

  • D.

    \frac{5S}{8}

  • E.

    \frac{3S}{5}

Answer:B

From the graph, {{S}_{②}}+{{S}_{④}}=\frac{2S}{5}.

The answer is \text{B}.

Link Problem
Problem 15 Easy

A switch is connected to three independent light bulbs, meaning each bulb operates independently of the others. When the switch is pressed, each bulb has a 50\% probability of turning on. What is the probability that at least two bulbs will light up with a single press of the switch?

  • A.

    \frac13

  • B.

    \frac38

  • C.

    \frac12

  • D.

    \frac23

  • E.

    \frac56

Answer:C

The probability of two bulbs being on and one bulb being off is:   P_1 = \binom{3}{2} \left( \frac{1}{2} \right)^2 \times \frac{1}{2} = \frac{3}{8}

The probability of all three bulbs being on is:   P_2 = \left( \frac{1}{2} \right)^3 = \frac{1}{8}

Thus, the total probability is:   P = P_1 + P_2 = \frac{3}{8} + \frac{1}{8} = \frac{1}{2}

Link Problem
Problem 16 Medium

Among the 100 natural numbers from 1 to 100, how many numbers are not multiples of 2, not multiples of 3, and not multiples of 5?

  • A.

    24

  • B.

    25

  • C.

    26

  • D.

    27

  • E.

    28

Answer:C

Among the 100 natural numbers from 1 to 100:

- Multiples of 2: 50

- Multiples of 3: 33

- Multiples of 5: 20

- Multiples of both 2 and 3: 16

- Multiples of both 2 and 5: 10

- Multiples of both 3 and 5: 6

- Multiples of 2, 3, and 5: 3

Using the principle of Inclusion-Exclusion, the total count of numbers that are multiples of 2, 3, or 5 is:

50 + 33 + 20 - 16 - 10 - 6 + 3 = 74

Thus, the count of numbers not being multiples of 2, 3, or 5 is:

100 - 74 = 26

The answer is \text{C}.

Link Problem
Problem 17 Medium

An opaque bag contains two balls, one red and one black, which are identical in size and shape. A ball is randomly drawn from the bag three times, with replacement. If drawing a red ball earns 2 points and drawing a black ball earns 1 point, what is the probability that the total score after three draws is exactly 5 points?

  • A.

    \frac{1}{4}

  • B.

    \frac{1}{2}

  • C.

    \frac{3}{8}

  • D.

    \frac{5}{8}

  • E.

    \frac{3}{4}

Answer:C

The possible total scores after 3 draws are 3, 4, 5, and 6, resulting in 4 different cases.

- The probability of scoring 3 points (drawing three black balls) is:     \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}

- The probability of scoring 6 points (drawing three red balls) is:     \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}

- Scoring 4 points requires 1 red ball and 2 black balls.

- Scoring 5 points requires 2 red balls and 1 black ball.

Since there are only two types of balls, each ball has an equal probability of being drawn (\frac{1}{2} per draw). The probabilities of scoring 4 and 5 must be the same.

Thus, the probability of scoring 5 points is:   \frac{1 - \frac{1}{8} - \frac{1}{8}}{2} = \frac{3}{8}

The answer is \text{C}.

Link Problem
Problem 18 Hard

A total of 25 unit cubes, each with an edge length of 1, are stacked together to form a solid shape. What is the minimum possible surface area of the resulting geometric figure?

  • A.

    50

  • B.

    54

  • C.

    56

  • D.

    70

  • E.

    75

Answer:B

The surface area is minimized when the unit cubes overlap as much as possible.

Consider a cube composed of 27 unit cubes (3 \times 3 \times 3), which has the smallest possible surface area when fully assembled.

To reduce the total count to 25 unit cubes, we need to remove 2 cubes in a way that does not increase the surface area. This can be achieved by removing:

1. Two cubes from two different corners, or

2. Two adjacent cubes from the same corner.

In both cases, the surface area remains unchanged at 54.

Thus, the minimum possible surface area of the resulting shape is 54.

The answer is \text{B}.

Link Problem
Problem 19 Hard

A pile of fruit has its water content decrease from 97\% to 96\% after being stored for a long period. By what percentage has the total weight of the fruit decreased?

  • A.

    1\%

  • B.

    10\%

  • C.

    15\%

  • D.

    20\%

  • E.

    25\%

Answer:E

Let the original total weight be 100a. Then, the weight of the non-water part is 100a \times (1 - 97\%) = 3a.

Now, with the water content at 96\%, the weight of the non-water part is 4\%, so the non-water weight becomes:

\frac{3a}{4\%} = 75a

Thus, the total weight decreased by:

\frac{100a - 75a}{100a} = 25\%

The answer is \text{E}.

Link Problem
Problem 20 Medium

Given that the \text{L.C.M.} of 2 consecutive even numbers is 10 times of the larger number. What is the square difference between these two integers?

  • A.

    12

  • B.

    28

  • C.

    60

  • D.

    76

  • E.

    84

Answer:E

Let a be half of the larger integer.

2\times a\times (a-1)=2\times a\times 10

~~~~~~~~~~~~~~~~~a-1=10

~~~~~~~~~~~~~~~~~~~~~~~~a=11

∴ These two integers are 20 and 22.

{{22}^{2}}-{{20}^{2}}

=84

The answer is \text{E}.

Link Problem
Problem 21 Medium

Alice has four grandchildren of different ages, all under 20, and the product of their ages is 2025. What is the sum of their ages?

  • A.

    30

  • B.

    34

  • C.

    28

  • D.

    29

  • E.

    32

Answer:E

2025=3^4 \times 5^2

The ages of four children are 3, 5, 9, 15, and 3\times 5 \times 9 \times 15=2025.

So the sum of their ages is 32.

Link Problem
Problem 22 Medium

As shown in the diagram, if a, b, c are distinct nonzero digits, and the two-digit numbers \overline{ab} and \overline{bc}, formed in the counterclockwise direction, are both multiples of 7, then what is the sum of the largest and smallest possible three-digit numbers \overline{abc} that can be formed?

  • A.

    991

  • B.

    1056

  • C.

    1126

  • D.

    1133

  • E.

    1198

Answer:C

Since the two-digit numbers \overline{ab} and \overline{bc} are both multiples of 7, they must be one of the following:   14, 21, 28, 35, 42, 49, 56, 63, 70, 84, 91, 98.

From these, the valid three-digit numbers \overline{abc} that can be formed are:   142, 214, 284, 356, 421, 428, 491, 498, 563, 635, 149, 842, 849, 914, 984.

The largest three-digit number is 984, and the smallest is 142. Their sum is:   984 + 142 = 1126.

The answer is \text{C}.

Link Problem
Problem 23 Medium

As shown in the figure, a rectangle OABC with length 4 and width 3 rolls continuously along the positive x-axis for 2025 times. The point A sequentially lands at positions A_1, A_2, A_3, \dots, A_{2025}. What are the coordinates of A_{2025}?

  • A.

    (7080, 0)

  • B.

    (7084, 0)

  • C.

    (7088, 3)

  • D.

    (7091,3)

  • E.

    (7094,0)

Answer:D

From the given conditions, we obtain the following sequence of coordinates for point A:   A_1(7,3), \quad A_2(10,0), \quad A_3(10,0), \quad A_4(14,4), \quad A_5(21,3), \quad A_6(24,0), \quad A_7(24,0), \quad A_8(28,4), \dots

Since every four rolls form a cycle, we analyze the displacement per cycle. Each full cycle results in a horizontal shift of:   3 + 4 + 3 + 4 = 14

Thus, after one full cycle, point A moves 14 units to the right.

The additional positions within a cycle are as follows:

- After one extra roll, A moves 7 units further, and the y-coordinate is 3.

- After two extra rolls, A moves 10 units further, and the y-coordinate is 0.

- After three extra rolls, the coordinates are the same as after two extra rolls.

Since:   2025 \div 4 = 506 \text{ remainder } 1

Point A completes 506 full cycles and moves one extra step beyond that.

Thus, the final coordinates are:   \left(14 \times 506 + 7, 3\right) = (7091,3)

The answer is \text{D}.

Link Problem
Problem 24 Easy

Among five pairs of distinct shoes, four shoes are randomly selected. How many ways can at least one matching pair be formed?

  • A.

    130

  • B.

    178

  • C.

    194

  • D.

    205

  • E.

    210

Answer:A

The total number of ways to choose any 4 shoes from 10 shoes is:   C(10,4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

The number of ways to select 4 shoes without forming any complete pair is:

- Choose 4 different pairs out of the 5 available:     C(5,4) = \frac{5!}{4!(5-4)!} = 5

- For each selected pair, choose one shoe (2 choices per pair):     2^4 = 16

- Total ways to pick 4 shoes without forming a pair:     5 \times 16 = 80

Thus, the number of ways to select 4 shoes with at least one matching pair is:   210 - 80 = 130

The answer is \text{A}.

Link Problem
Problem 25 Medium

How many four-digit numbers \overline{abcd} can be formed using the digits 1, 2, 3, 4 (digits may be repeated) such that they satisfy the condition:   a + c = b + d?

  • A.

    36

  • B.

    40

  • C.

    44

  • D.

    48

  • E.

    52

Answer:C

We classify the four-digit numbers into four cases:

① All four digits are the same. Possible numbers: 1111, 2222, 3333, 4444

Total: 4 numbers.

② Two different digits appear twice each. For example, numbers like 1221, 1122, 2112, 2211.

We choose two different digits from \{1,2,3,4\}, which can be done in \binom{4}{2} = 6 ways.

Each choice allows 4 different digit arrangements, so this case contributes:      6 \times 4 = 24 \text{ numbers.}

③ Three different digits appear, with one digit repeated. Examples include 1232, 3212, 2123, 2321.

The valid digit sum conditions are 1+3 = 2+2 and 2+4 = 3+3, giving 2 valid digit sets.

Each set allows 4 different digit arrangements, so this case contributes: 2 \times 4 = 8 \text{ numbers.}

④ All four digits are different. Valid numbers include 1243, 1342, 4213, 4312, 2134, 2431, 3124, 3421, totaling:      8 \text{ numbers.}

Adding all cases together:   4 + 24 + 8 + 8 = 44.

The answer is \text{C}.

Link Problem
Table of Contents
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
  • 16
  • 17
  • 18
  • 19
  • 20
  • 21
  • 22
  • 23
  • 24
  • 25