2006 AMC 8 Exam Problems

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Used with permission of the Mathematical Association of America.

Tiempo restante:

40:00

1.

Mindy made three purchases for $1.98\$ 1.98 dollars, $5.04\$ 5.04 dollars, and $9.89\$ 9.89 dollars. What was her total, to the nearest dollar?

$10

$15

$16

$17

$18

Answer: D
Solution(s):

Since all the values are already close to a whole number, we can just add the sums of the rounded numbers.

These prices round to $2,\$ 2, $5,\$ 5, and $10,\$ 10, which add to $17.\$ 17.

Thus, D is the correct answer.

2.

On the AMC 88 contest Billy answers 1313 questions correctly, answers 77 questions incorrectly and doesn't answer the last 5.5. What is his score?

11

66

1313

1919

2626

Answer: C
Solution(s):

Since the AMC 88 only awards 11 point for each correct question, Billy will get 1313 points.

Thus, C is the correct answer.

3.

Elisa swims laps in the pool. When she first started, she completed 1010 laps in 2525 minutes. Now, she can finish 1212 laps in 2424 minutes. By how many minutes has she improved her lap time?

12\dfrac{1}{2}

34\dfrac{3}{4}

11

22

33

Answer: A
Solution(s):

Initially, Elisa swam 11 lap in 2510=52 \dfrac{25}{10} = \dfrac{5}{2} minutes. Now, she swims 11 lap in 2412=2 \dfrac{24}{12} = 2 minutes.

Therefore, she improved her lap time by 522=12\dfrac{5}{2} - 2 = \dfrac{1}{2} minutes.

Thus, A is the correct answer.

4.

Initially, a spinner points west. Chenille moves it clockwise 214 2 \frac{1}{4} revolutions and then counterclockwise 334 3 \frac{3}{4} revolutions. In what direction does the spinner point after the two moves?

North

East

South

West

Northwest

Answer: B
Solution(s):

If the spinner goes 2142 \frac{1}{4} revolutions clockwise and 3343 \frac{3}{4} revolutions counterclockwise, it goes 1121 \frac{1}{2} revolutions counterclockwise.

This means that the spinner will be pointing east.

Thus, B is the correct answer.

5.

Points A,B,CA, B, C and DD are midpoints of the sides of the larger square. If the larger square has area 60,60, what is the area of the smaller square?

1515

2020

2424

3030

4040

Answer: D
Solution(s):

Note that we can fold all the triangles in to perfectly cover the smaller square. This means that area of the smaller square is half the area of the larger square.

This makes the area of the smaller square 60÷2=30.60 \div 2 = 30.

Thus, D is the correct answer.

6.

The letter T is formed by placing two 2×42 \times 4 inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?

1212

1616

2020

2222

2424

Answer: C
Solution(s):

If we found the total perimeter of the two rectangles separately, we would have gotten 2(2(2+4))=226=24. 2(2(2 + 4)) = 2 \cdot 2 \cdot 6 = 24.

In the letter T, we can see that their intersection removes a piece of length 22 from each of the rectangles. Therefore, the perimeter of the T is 2422=20. 24 - 2 \cdot 2 = 20.

Thus, C is the correct answer.

7.

Circle XX has a radius of π.\pi. Circle YY has a circumference of 8π.8 \pi. Circle ZZ has an area of 9π.9 \pi. List the circles in order from smallest to largest radius.

X,Y,ZX, Y, Z

Z,X,YZ, X, Y

Y,X,ZY, X, Z

Z,Y,XZ, Y, X

X,Z,YX, Z, Y

Answer: B
Solution(s):

Recall that C=2πrC = 2 \pi r and A=πr2.A = \pi r^2. Using these formulas we get that the radius of YY is 8π÷(2π)=4. 8 \pi \div (2 \pi) = 4. We also get that the radius of ZZ is 9π÷π=3. \sqrt{9 \pi \div \pi} = 3. As π\pi is greater than 33 and less than 4,4, the correct order is Z,X,Y.Z,X,Y.

Thus, the answer is B.

8.

The table shows some of the results of a survey by radio station KAMC. What percentage of the males surveyed listen to the station?

3939

4848

5252

5555

7575

Answer: E
Solution(s):

The total number of males surveyed is the total number surveyed by the total number of women surveyed: 20096=104. 200 - 96 = 104. The percentage of males that listen to the station is 100%100 \% minus the percent that don't listen to the station: 100%10026104% 100 \% - 100 \cdot \dfrac{26}{104} \% =100%25%= 100 \% - 25 \% =75%.= 75 \%.

Thus, E is the correct answer.

9.

What is the product of 32×43×54××20062005?\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\cdots\times\dfrac{2006}{2005}?

11

10021002

10031003

20052005

20062006

Answer: C
Solution(s):

Note that the numerator of every fraction cancels with the denominator of the following fraction. This leaves two numbers: 20062=1003. \dfrac{2006}{2} = 1003.

Thus, C is the correct answer.

10.

Jorge's teacher asks him to plot all the ordered pairs (w,l)(w, l) of positive integers for which ww is the width and ll is the length of a rectangle with area 12.12. What should his graph look like?

Answer: A
Solution(s):

We know that wl=12,wl = 12, so w=12l.w = \dfrac{12}{l}.

This shows that ww and ll are inversely proportional, which can only be represented by a non-linear graph.

Thus, A is the correct answer.

11.

How many two-digit numbers have digits whose sum is a perfect square?

1313

1616

1717

1818

1919

Answer: C
Solution(s):

There is 11 number whose digit sum is 1:10.1: 10.

There are 44 numbers whose digit sum is 4:13,22,31,4: 13, 22, 31, and 40.40.

There are 99 numbers whose digit sum is 9:18,27,36,45,54,9: 18, 27, 36, 45, 54, 63,72,81,63, 72, 81, and 90.90.

There are 33 numbers whose digit sum is 16:79,88,16: 79, 88, and 97.97.

Therefore, there are 1717 numbers that satisfy the problem statement.

Thus, C is the correct answer.

12.

Antonette gets 70%70 \% on a 1010-problem test, 80%80 \% on a 2020-problem test and 90%90 \% on a 3030-problem test. If the three tests are combined into one 6060-problem test, which percent is closest to her overall score?

4040

7777

8080

8383

8787

Answer: D
Solution(s):

Antonette got .710=7.7 \cdot 10 = 7 questions write on the first test. Similarly, she got .820=16.8 \cdot 20 = 16 and .930=27.9 \cdot 30 = 27 problems right on her second and third tests respectively.

Adding these up yields a total of 5050 correct questions. Her score on the 6060-problem test would have been a 1005060=100.8383%. 100 \cdot \dfrac{50}{60} = 100 \cdot .8\overline{3} \approx 83 \%.

Thus, D is the correct answer.

13.

Cassie leaves Escanaba at 8:308:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 1212 miles per hour. Brian leaves Marquette at 9:009:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 1616 miles per hour. They both bike on the same 6262-mile route between Escanaba and Marquette. At what time in the morning do they meet?

10:0010:00

10:1510:15

10:3010:30

11:0011:00

11:3011:30

Answer: D
Solution(s):

By the time Brian starts biking, Marquette has already traveled 1212=6\dfrac{1}{2} \cdot 12 = 6 miles. This means that Brian and Marquette must then travel a total of 626=5662 - 6 = 56 miles.

Combined, the two bike at 12+16=2812 + 16 = 28 miles per hour. This means that they can travel 5656 miles in 56÷28=256 \div 28 = 2 hours.

This means that they meet at 11:00.11:00.

Thus, D is the correct answer.

14.

The students in Mrs. Reed’s English class are reading the same 760760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 2020 seconds, Bob reads a page in 4545 seconds and Chandra reads a page in 3030 seconds.

If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?

7,6007,600

11,40011,400

12,50012,500

15,20015,200

22,80022,800

Answer: B
Solution(s):

Bob will take 76045760 \cdot 45 seconds to read the book, and Chandra will take 76030760 \cdot 30 seconds.

The difference between the time they spent reading is 7604576030 760 \cdot 45 - 760 \cdot 30 =760(4530)= 760(45 - 30) =76015= 760 \cdot 15 =11,400= 11,400 seconds.

Thus, B is the correct answer.

15.

The students in Mrs. Reed’s English class are reading the same 760760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 2020 seconds, Bob reads a page in 4545 seconds and Chandra reads a page in 3030 seconds.

Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 11 to a certain page and Bob will read from the next page through page 760,760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?

425425

444444

456456

484484

506506

Answer: C
Solution(s):

Let xx be the number of pages that Chandra will read. Then Bob will read 760x760 - x pages.

For them to read for the same amount of time, 30x=45(760x)30x=4576045x75x=457605x=3760x=3152x=456. \begin{align*} 30x &= 45(760 - x) \\ 30x &= 45 \cdot 760 - 45x \\ 75x &= 45 \cdot 760 \\ 5x &= 3 \cdot 760 \\ x &= 3 \cdot 152 \\ x &= 456. \end{align*}

Thus, C is the correct answer.

16.

The students in Mrs. Reed’s English class are reading the same 760760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 2020 seconds, Bob reads a page in 4545 seconds and Chandra reads a page in 3030 seconds.

Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?

64006400

66006600

68006800

70007000

72007200

Answer: E
Solution(s):

If all 33 individuals read the same amount of time, then the number of pages Bob, Chandra, and Alice will read will be in the ratio 4:6:94:6:9 respectively.

This means that they will read 160,240,160, 240, and 360360 pages respectively. Since they all read for the same amount of time, we can just calculate how long it takes for Bob to read his portion.

Bob will take 45160=720045 \cdot 160 = 7200 seconds to read his portion.

Thus, E is the correct answer.

17.

Jeff rotates spinners P,P, QQ and RR and adds the resulting numbers. What is the probability that his sum is an odd number?

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

Answer: B
Solution(s):

Note that adding an even number to a number does not affects is parity. Therefore, whatever the second spinner lands on will not impact whether the sum is odd.

The only way for a number on PP and a number on RR to add to an odd number is if PP lands on 2.2. Otherwise, it would be the sum of two odd numbers, which is even.

PP lands on 22 with a 13\dfrac{1}{3} probability.

Thus, B is the correct answer.

18.

A cube with 33-inch edges is made using 2727 cubes with 11-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?

19\dfrac{1}{9}

14\dfrac{1}{4}

49\dfrac{4}{9}

59\dfrac{5}{9}

1927\dfrac{19}{27}

Answer: D
Solution(s):

Since each face has the same black and white surface area, we can analyze what fraction of one side is white.

On one side, there are 99 unit squares. 44 of them are black. This means that 59\dfrac{5}{9} of each face is white.

Thus, D is the correct answer.

19.

Triangle ABCABC is an isosceles triangle with AB=BC.\overline{AB}=\overline{BC}. Point DD is the midpoint of both BC\overline{BC} and AE,\overline{AE}, and CE\overline{CE} is 1111 units long. Triangle ABDABD is congruent to triangle ECD.ECD. What is the length of BD?\overline{BD}?

44

4.54.5

55

5.55.5

66

Answer: D
Solution(s):

By the congruency condition, we know that AB=EC=11.AB = EC = 11.

Also from the isosceles condition, we know that BC=AB=11.BC = AB = 11.

Since DD is the midpoint of BC,\overline{BC}, we know that BD=BC÷2=5.5BD = BC \div 2 = 5.5

Thus, D is the correct answer.

20.

A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 44 games, Ines won 33 games, Janet won 33 games, Kendra won 22 games and Lara won 22 games, how many games did Monica win?

00

11

22

33

44

Answer: C
Solution(s):

In every match, there was exactly one winner. There are 65÷2=156 \cdot 5 \div 2 = 15 games and therefore 1515 wins.

There are already 4+3+3+2+2=13 4 + 3 + 3 + 2 + 2 = 13 wins accounted for, so Monica won 1513=215 - 13 = 2 games.

Thus, C is the correct answer.

21.

An aquarium has a rectangular base that measures 100100 cm by 4040 cm and has a height of 5050 cm. The aquarium is filled with water to a depth of 3737 cm. A rock with volume 1000 cm31000\text{ cm}^3 is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?

0.250.25

0.50.5

11

1.251.25

2.52.5

Answer: A
Solution(s):

The original volume of the water was 1004037=168,000 cm3. 100 \cdot 40 \cdot 37 = 168,000 \text{ cm}^3.

The volume after the rock was added is 168,000+1,000=169,000 cm3. 168,000 + 1,000 = 169,000 \text{ cm}^3.

This means the new height is 169,000÷100÷40=37.25 cm. 169,000 \div 100 \div 40 = 37.25 \text{ cm}.

This means that water level rose by .25.25 cm.

Thus, A is the correct answer.

22.

Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?

1616

2424

2525

2626

3535

Answer: D
Solution(s):

If the lower cells contain a,a, b,b, and c,c, the middle row will have a+ba + b and b+c.b + c.

This means that the top row will have a+2b+c.a + 2b + c. To minimize this, we can let b=1,b = 1, a=2,a = 2, and c=3.c = 3. This yields a top element of 7.7.

To maximize, this we can let b=9,b = 9, a=8,a = 8, and c=7.c = 7. This yields a top element of 33.33. The desired difference is 337=26.33 - 7 = 26.

Thus, D is the correct answer.

23.

A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?

00

11

22

33

55

Answer: A
Solution(s):

The positive integers that leave a remainder of 44 when divided by 66 are 4,10,16,22,28,34,40,. 4, 10, 16, 22, 28, 34, 40, \cdots. The positive integers that leave a remainder of 33 when divided by 55 are 3,8,13,18,23,28,33,. 3, 8, 13, 18, 23, 28, 33, \cdots.

From this, we can see that the smallest number of coins that work is 28.28. This leaves a remainder of 00 when divided by 7.7.

Thus, A is the correct answer.

24.

In the multiplication problem below A,A, B,B, C,C, DD are different digits. What is A+B?A+B? ABA×CDCDCD \begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}

11

22

33

44

99

Answer: A
Solution(s):

Note that CDCD=101CD. CDCD = 101 \cdot CD. This forces ABA=101.ABA = 101. Then A=1,A = 1, B=0,B = 0, and A+B=1.A + B = 1.

Thus, A is the correct answer.

25.

Barry wrote 66 different numbers, one on each side of 33 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?

1313

1414

1515

1616

1717

Answer: B
Solution(s):

Note that if we add an odd number to 59,59, we will get an even number. To make the other two sums even numbers, the primes on the back of 4444 and 3838 must both be even.

They also have to be different, however. Since there is only one even prime, the number on the back of 5959 must be an even prime, namely 2.2.

This means that the sum of the front and back of all the cards is 59+2=61.59 + 2 = 61. This makes the other two primes 6144=1761 - 44 = 17 and 6138=23.61- 38 = 23.

The average of the primes is therefore 2+17+233=423=14. \dfrac{2 + 17 + 23}{3} = \dfrac{42}{3} = 14.

Thus, B is the correct answer.