2009 AMC 8 Exam Solutions

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1.

Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 33 apples, keeping 44 apples for herself. How many apples did Bridget buy?

33

44

77

1111

1414

Solution(s):

We can work backwards, starting with the 44 apples that Bridget kept for herself. Adding the 33 apples that she gave Cassie, she now has 77 apples.

Finally, we multiply this value by 22 since she gave half of her initial apples to Ann. 72=14,7 \cdot 2 = 14, so Bridget started off with 1414 apples.

Thus, E is the correct answer.

2.

On average, for every 44 sports cars sold at the local dealership, 77 sedans are sold. The dealership predicts that it will sell 2828 sports cars next month. How many sedans does it expect to sell?

77

3232

3535

4949

112112

Solution(s):

We can set up the following proportion: 47=28x. \dfrac{4}{7} = \dfrac{28}{x}. Cross multiplying, we get 4x=728 4x = 7 \cdot 28 x=49 x = 49

Thus, D is the correct answer.

3.

The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?

55

5.55.5

66

6.56.5

77

Solution(s):

From the graph, we can see that Suzanna rides 33 miles in 1515 minutes. This means that in 3030 minutes, she will have ridden 66 miles.

Thus, C is the correct answer.

4.

The five pieces shown below can be arranged to form four of the five figures below. Which figure cannot be formed?

Solution(s):

Note that option B does not have any segments in it that are 55 blocks long. This means that it is impossible to arrange the 55 block long piece to fit within the figure.

Thus, B is the correct answer.

5.

A sequence of numbers starts with 1,1, 2,2, and 3.3. The fourth number of the sequence is the sum of the previous three numbers in the sequence: 1+2+3=6.1 + 2 + 3 = 6. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?

1111

2020

3737

6868

9999

Solution(s):

We can create the following list to find all the numbers up till the eight number aia_i denotes the ii-th number in the sequence.

a1=1,a_1 = 1, a2=2,a_2 = 2, and a3=3a_3 = 3 from the problem statement.

a4=1+2+3=6a_4 = 1 + 2 + 3 = 6

a5=2+3+6=11 a_5 = 2 + 3 + 6 = 11

a6=3+6+11=20 a_6 = 3 + 6 + 11 = 20

a7=6+11+20=37 a_7 = 6 + 11 + 20 = 37

a8=11+20+37=68 a_8 = 11 + 20 + 37 = 68

Thus, D is the correct answer.

6.

Steve's empty swimming pool will hold 24,00024,000 gallons of water when full. It will be filled by 44 hoses, each of which supplies 2.52.5 gallons of water per minute. How many hours will it take to fill Steve's pool?

4040

4242

4444

4646

4848

Solution(s):

The 44 hoses together fill the pool with 2.54=102.5 \cdot 4 = 10 gallons of water per minute.

To fill 24,00024,000 gallons, it will take the hoses 24,000÷10=2,40024,000 \div 10 = 2,400 minutes to fill the pool.

2,4002,400 minutes is the same as 2,400÷60=402,400 \div 60 = 40 hours.

Thus, A is the correct answer.

7.

The triangular plot of ACDACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?ACD?

22

33

4.54.5

66

99

Solution(s):

The base of ADCADC is CD,CD, which is 3.3. The altitude is 33 as well.

Therefore, the area of ADCADC is 1233=4.5.\dfrac{1}{2} \cdot 3 \cdot 3 = 4.5.

Thus, C is the correct answer.

8.

The length of a rectangle is increased by 10%10\% percent and the width is decreased by 10%10\% percent. What percent of the old area is the new area?

9090

9999

100100

101101

110110

Solution(s):

Let ll be the old width of the rectangle and ww be the old width. The new length is then 1.1l,1.1l, and the new width is .9w..9w. This means that the new area is 1.1×.9lw=.99lw.1.1 \times .9 lw = .99 lw.

This shows that the new area is 99%99\% of the original area.

Thus, B is the correct answer.

9.

Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?

2121

2323

2525

2727

2929

Solution(s):

Notice that every polygon in the middle of the chain (every shape except for the triangle and octagon) contribute all but 11 of their sides to the overall polygon. The triangle and octagon contribute all but 11 of their sides.

Therefore, the total number of sides is (31)+(42)+(52)+ (3 - 1) + (4 - 2) + (5 - 2) + (62)+(72)+(81) (6 - 2) + (7 - 2) + (8 - 1) =2+2+3+4+5+7 = 2 + 2 + 3 + 4 + 5 + 7 =23. = 23.

Thus, B is the correct answer.

10.

On a checkerboard composed of 6464 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

116\dfrac{1}{16}

716\dfrac{7}{16}

12\dfrac{1}{2}

916\dfrac{9}{16}

4964\dfrac{49}{64}

Solution(s):

There are (82)2=62=36(8 - 2)^2 = 6^2 = 36 squares on the interior.

This means that the probability of choosing one of these squares is 3664=916. \dfrac{36}{64} = \dfrac{9}{16}.

Thus, D is the correct answer.

11.

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of 1.431.43 dollars. Some of the 3030 sixth graders each bought a pencil, and they paid a total of 1.951.95 dollars. How many more sixth graders than seventh graders bought a pencil?

11

22

33

44

55

Solution(s):

The number of seventh graders that bought a pencil is 143143 divided by the price of a pencil. Similarly, the number of sixth graders that bought a pencil is 195195 divided by the price of a pencil.

This means that the price of a pencil divides both 143143 and 195.195. Prime factorizing, we get 143=1113 143 = 11 \cdot 13 and 195=3513. 195 = 3 \cdot 5 \cdot 13. The only numbers that divide both 143143 and 195195 are 11 and 13.13.

If 11 cent was the price of the pencil, that means 195195 sixth graders bought pencils, which is not possible. Therefore, the price of a pencil is 1313 cents.

This means that 143÷13=11143 \div 13 = 11 seventh graders bought a pencil, and 195÷13=15195 \div 13 = 15 sixth graders bought a pencil. Therefore, 44 more sixth graders than seventh graders bought pencils.

Thus, D is the correct answer.

12.

The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

79\dfrac{7}{9}

56\dfrac{5}{6}

Solution(s):

We can find the sum of the two numbers in every possible outcome.

\begin{gather*} 1 + 2 = 3 \\ 1 + 4 = 5 \\ 1 + 6 = 7 \\ 3 + 2 = 5 \\ 3 + 4 = 7 \\ 3 + 6 = 9 \\ 5 + 2 = 7 \\ 5 + 4 = 9 \\ 5 + 6 = 11. \end{gather*}

There are only 22 outcomes where the sum is not prime (the two instances when the sum is 99). Therefore, the probability that the sum is prime is 79.\dfrac{7}{9}.

Thus, D is the correct answer.

13.

A three-digit integer contains one of each of the digits 1,1, 3,3, and 5.5. What is the probability that the integer is divisible by 5?5?

16\dfrac{1}{6}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

56\dfrac{5}{6}

Solution(s):

Then number is equally likely to end in a 1,1, 3,3, or 5.5. It is only divisible if the last digit is a 5,5, which happens with a 13\dfrac{1}{3} probability.

Thus, B is the correct answer.

14.

Austin and Temple are 5050 miles apart along Interstate 35.35. Bonnie drove from Austin to her daughter's house in Temple, averaging 6060 miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged 4040 miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?

4646

4848

5050

5252

5454

Solution(s):

The trip from Austin to Temple took 50÷60=5650 \div 60 = \dfrac{5}{6} hours. The trip from Temple to Austin took 50÷40=5450 \div 40 = \dfrac{5}{4} hours. This means that the total time for the round trip was 56+54=2512\dfrac{5}{6} + \dfrac{5}{4} = \dfrac{25}{12} hours.

The total distance of the round trip was 250=1002 \cdot 50 = 100 miles. Therefore, the average speed for the round trip was 100÷2512=48100 \div \dfrac{25}{12} = 48 miles per hour.

Thus, B is the correct answer.

15.

A recipe that makes 55 servings of hot chocolate requires 22 squares of chocolate, 14\dfrac{1}{4} cup sugar, 11 cup water and 44 cups milk. Jordan has 55 squares of chocolate, 22 cups of sugar, lots of water, and 77 cups of milk. If he maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate he can make?

5185 \dfrac{1}{8}

6146 \dfrac{1}{4}

7127 \dfrac{1}{2}

8348 \dfrac{3}{4}

9789 \dfrac{7}{8}

Solution(s):

We need to find which of the ingredients is the limiting factor.

With 55 squares of chocolate, Jordan can only make 52\dfrac{5}{2} times what the recipe requires.

With 22 cups of sugar, Jordan can only make 214=8\dfrac{2}{\dfrac{1}{4}} = 8 times the required amount.

With 77 cups of milk, Jordan can only make 74\dfrac{7}{4} times what is needed.

Therefore, the number of servings is limited by the amount of milk. The amount of servings Jordan can make is 574=354=834.5 \cdot \dfrac{7}{4} = \dfrac{35}{4} = 8 \dfrac{3}{4}.

Thus, D is the correct answer.

16.

How many 33-digit positive integers have digits whose product equals 24?24?

1212

1515

1818

2121

2424

Solution(s):

The only triples of integers less than 1010 that multiply to 2424 are (1,3,8),(1,4,6),(2,2,6), (1, 3, 8), (1, 4, 6), (2, 2, 6), and (2,3,4).\text{and } (2, 3, 4).

The triples with 33 distinct numbers can be rearranged to form 66 distinct 33-digit positive integers. The other triple can be arranged to form 33 distinct 33-digit positive integers.

This leaves a total of 36+3=213 \cdot 6 + 3 = 21 integers.

Thus, D is the correct answer.

17.

The positive integers xx and yy are the two smallest positive integers for which the product of 360360 and xx is a square and the product of 360360 and yy is a cube. What is the sum of xx and y?y?

8080

8585

115115

165165

610610

Solution(s):

For a number to be a perfect square, every exponent in the prime factorization must be even. For it to be a cube, the exponents must be divisible by 3.3.

We can factor 360360 to get 360=23325. 360 = 2^3 \cdot 3^2 \cdot 5. For 360x360x to be a perfect square and xx to be minimized, xx must have one factor of 22 and one factor of 5.5. Therefore, we can let x=10.x = 10.

For 360y360y to be a cube, yy must have one factor of 33 and two factors of 5.5. Therefore, we can let y=75,y = 75, suggesting x+y=85.x + y = 85.

Thus, B is the correct answer.

18.

The diagram represents a 77-foot-by-77-foot floor that is tiled with 11-square-foot light tiles and dark tiles. Notice that the corners have dark tiles. If a 1515-foot-by-1515-foot floor is to be tiled in the same manner, how many dark tiles will be needed?

4949

5757

6464

9696

126126

Solution(s):

Looking at the example given, note that there are 424^2 dark tiles. This is because there are 44 rows that contain 44 dark tiles each.

For a 1515-foot-by-1515-foot floor, there are going to be 88 rows with 88 dark tiles in each row. This will give us 82=648^2 = 64 dark tiles.

Thus, C is the correct answer.

19.

Two angles of an isosceles triangle measure 7070^\circ and x.x^\circ. What is the sum of the three possible values of x?x?

9595

125125

140140

165165

180180

Solution(s):

All the following possibilities are shown below.

In the first scenario, we get x=70x = 70 by the properties of the isosceles triangle.

In the second scenario, we get that 702+x=180, 70 \cdot 2 + x = 180, from which we get that x=40.x = 40.

From the third scenario, we get that 2x+70=180, 2x + 70 = 180, from which we get that x=55.x = 55.

The sum of these values yields 70+40+55=165. 70 + 40 + 55 = 165.

Thus, D is the correct answer.

20.

How many non-congruent triangles have vertices at three of the eight points in the array shown below?

55

66

77

88

99

Solution(s):

We can label to points to find all the unique triangles.

All the possible distinct triangles are AEF,AEG,AEH,AFG,AFH, AEF, AEG, AEH, AFG, AFH,AGH,AFC, and AFD. AGH, AFC, \text{ and } AFD.

Any other triangle will be congruent to one of the 88 listed above.

Thus, D is the correct answer.

21.

Andy and Bethany have a rectangular array of numbers with 4040 rows and 7575 columns. Andy adds the numbers in each row. The average of his 4040 sums is A.A. Bethany adds the numbers in each column. The average of her 7575 sums is B.B. What is the value of AB?\dfrac{A}{B}?

64225\dfrac{64}{225}

815\dfrac{8}{15}

11

158\dfrac{15}{8}

22564\dfrac{225}{64}

Solution(s):

Each person includes every number in the array in their sums exactly once. This means that before they divide in the average, their sums are the exact same. This means that 40A=75B.40A = 75B. Therefore, AB=7540=158.\dfrac{A}{B} = \dfrac{75}{40} = \dfrac{15}{8}.

Thus, D is the correct answer.

22.

How many whole numbers between 11 and 10001000 do not contain the digit 1?1?

512512

648648

720720

728728

800800

Solution(s):

We can case on the number of digits.

There are 88 one digit numbers excluding 1.1.

There are 89=728 \cdot 9 = 72 two digit numbers that lack the digit 1.1.

There are 899=6488 \cdot 9 \cdot 9 = 648 three digit numbers that do not include 1.1.

This yields a total of 8+72+648=728 8 + 72 + 648 = 728 numbers that do not contain the digit 1.1.

Thus, D is the correct answer.

23.

On the last day of school, Mrs. Wonderful gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought 400400 jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

2626

2828

3030

3232

3434

Solution(s):

Let bb be the number of boys in the class and gg be the number of girls. From the problem, we get that b=g+2.b = g + 2.

If each boy gets bb jelly beans, then Mrs. Wonderful will give out a total of b2b^2 jelly beans to all the boys. Similarly, she will give out g2g^2 jelly beans to all the girls.

Therefore, b2+g2=4006(g+2)2+g2=3942g2+4g+4=394g2+2g195=0(g+15)(g13)=0. \begin{gather*} b^2 + g^2 = 400 - 6 \\ (g + 2)^2 + g^2 = 394 \\ 2g^2 + 4g + 4 = 394 \\ g^2 + 2g - 195 = 0 \\ (g + 15)(g - 13) = 0. \end{gather*} Since gg cannot be negative, we get that g=13.g = 13. This means that b=15,b = 15, so b+g=28.b + g = 28.

Thus, B is the correct answer.

24.

The letters A,A, B,B, CC and DD represent digits.
If AB+CADA \begin{array}{cc} &A &B \\ + &C &A \\ \hline &D &A \end{array} and ABCAA, \begin{array}{cc} &A &B \\ - &C &A \\ \hline &&A \end{array}, what digit does DD represent?

55

66

77

88

99

Solution(s):

Since A+B=A,A + B = A, we get that B=0.B = 0. Using this and the second equation, we also get that A=5.A = 5. This follows from carrying over a 1010 to B,B, resulting in 10A=A.10 - A = A. Therefore, AB=50.AB = 50.

Since 50C5=5,50 - C5 = 5, we also know that C=4.C = 4. Then D=C+5=9.D = C + 5 = 9.

Thus, E is the correct answer.

25.

A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is 12\dfrac{1}{2} foot from the top face. The second cut is 13\dfrac{1}{3} foot below the first cut, and the third cut is 117\dfrac{1}{17} foot below the second cut. From the top to the bottom the pieces are labeled A,A, B,B, C,C, and D.D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?

66

77

41951\dfrac{419}{51}

15817\dfrac{158}{17}

1111

Solution(s):

We can look at this shape from all 66 directions to add up the surface area from each view to get the total surface area.

Looking at it from the ends gives us the face of A,A, which has an area of 121=12 ft2.\dfrac{1}{2} \cdot 1 = \dfrac{1}{2} \text{ ft}^2.

Each side has the same area as if we were to stack up all the pieces into a unit cube and take the area of one side. This would yield 1 ft2.1 \text{ ft}^2.

The top and bottom each have 44 unit squares, for a total of 4 ft24 \text{ ft}^2 from each view.

Adding this all up gives us a total surface area of 122+12+42=11 ft2. \dfrac{1}{2} \cdot 2 + 1 \cdot 2 + 4 \cdot 2 = 11 \text{ ft}^2.

Thus, E is the correct answer.