## AMC 8 2009

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 1 point for each correct answer. There is no penalty for wrong answers.
- No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted.
- Figures are not necessarily drawn to scale.
- You will have
**40 minutes**working time to complete the test.

Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 14$

$\textbf{E}$

Before she gave Cassie 3 apples, sha had $3+4=7$ apples, which is half of the total number of apples she bought. So the total number is 14.

On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 112$

$\textbf{D}$

The number of sedan sold is $\dfrac74$ of the sports sold. Since 28 sports will be sold next month, the number of sedan should be $28\cdot\dfrac74=49$.

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?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$

$\textbf{C}$

We can see that it takes her 10 minutes to ride 2 miles. So she can ride 6 miles in 30 minutes.

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

$\textbf{B}$

The answer is B because the longest piece can not fit into the figure.

A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. 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?

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ 99$

$\textbf{D}$

List them out, adding the three previous numbers to get the next number, \[1,2,3,6,11,20,37,68\dots\]

Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?

$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 48$

$\textbf{A}$

Each of the four hoses fills $24000/4 = 6000$ gallons of water. At the rate of 2.5 gallons of water per minute, it will take $6000/2.5 = 2400$ minutes, or $40$ hours.

The triangular plot of ACD 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?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4.5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 9$

$\textbf{C}$

Let $CD$ be the base of $\triangle ACD$, and $AB$ be the height. The area of the triangle is $\dfrac12 \times 3 \times 3 = 4.5$.

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

$\textbf{(A)}\ 90\qquad\textbf{(B)}\ 99\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 101\qquad\textbf{(E)}\ 110$

$\textbf{B}$

The length is 1.1 times as before while the width is 0.9. So the percentage is $1.1\times0.9=0.99=99\%$.

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?

$\textbf{(A)} \ 21\qquad\textbf{(B)}\ 23\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 29$

$\textbf{B}$

There are 6 shapes from the triangle to the octagon. The total number of sides is $3+4+5+6+7+8=33$. However, since there are 5 interfaces with 2 adjacent sides each, the number of sides of the final polygon is $33-5\times2=23$.

On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

$\textbf{(A)}\ \dfrac{1}{16}\qquad\textbf{(B)}\ \dfrac{7}{16}\qquad\textbf{(C)}\ \dfrac{1}2\qquad\textbf{(D)}\ \dfrac{9}{16}\qquad\textbf{(E)}\ \dfrac{49}{64}$

$\textbf{D}$

Since there are 28 unit squares at the edge of the board, the probability that the chosen unit square is not at the edge is $\dfrac{64-28}{64}=\dfrac9{16}$.

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $\$1.43$. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $\$1.95$. How many more sixth graders than seventh graders bought a pencil?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

$\textbf{S}$

Since the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$. They can be factored into $143=11\times13$ and $195=3\times5\times13$. The common factor cannot be $1$, or there would be more than $30$ sixth graders. So the pencil costs $13$ cents. The difference in costs that the sixth and seventh graders paid is $195-143=52$ cents, which means $52/13 =4$ more sixth graders bought a pencil.

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?

$\textbf{(A)}\ \dfrac{1}{2}\qquad\textbf{(B)}\ \dfrac{2}{3}\qquad\textbf{(C)}\ \dfrac{3}{4}\qquad\textbf{(D)}\ \dfrac{7}{9}\qquad\textbf{(E)}\ \dfrac{5}{6}$

$\textbf{D}$

The possible sums are

Only $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $\dfrac79$.

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

$\textbf{(A)}\ \dfrac{1}{6}\qquad\textbf{(B)}\ \dfrac{1}{3}\qquad\textbf{(C)}\ \dfrac{1}{2}\qquad\textbf{(D)}\ \dfrac{2}{3}\qquad\textbf{(E)}\ \dfrac{5}{6}$

$\textbf{B}$

The number is divisible by 5 if and only if the number ends in $5$. If we randomly arrange the three digits, the probability of the last digit being $5$ is $\dfrac13$.

Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?

$\textbf{(A)}\ 46\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 50\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 54$

$\textbf{B}$

The way to Temple took $\dfrac{50}{60}=\dfrac56$ hours, and the way back took $\dfrac{50}{40}=\dfrac54$ for a total of $\dfrac56 + \dfrac54 = \dfrac{25}{12}$ hours. The trip is $50\times2=100$ miles. The average speed is $\dfrac{100}{25/12} =48$ miles per hour.

A recipe that makes 5 servings of hot chocolate requires 2 squares of chocolate, $1/4$ cup sugar, 1 cup water and 4 cups milk. Jordan has 5 squares of chocolate, 2 cups of sugar, lots of water and 7 cups of milk. If he maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate he can make?

$\textbf{(A)}\ 5\dfrac{1}8\qquad\textbf{(B)}\ 6\dfrac{1}4\qquad\textbf{(C)}\ 7\dfrac{1}2\qquad\textbf{(D)}\ 8\dfrac{3}4\qquad\textbf{(E)}\ 9\dfrac{7}8$

$\textbf{D}$

Comparing to the recipe, Jordan has 2.5 times as much chocolate, 4 times as much sugar, and $\dfrac74$ tmmes as much milk. Limited by the amount of milk, Jordan can make at most $5\times\dfrac74=8\dfrac34$ servings of hot chocolate.

How many 3-digit positive integers have digits whose product equals $24$?

$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ 24$

$\textbf{D}$

The three-digit group can be $(1,3,8)$, $(1,4,6)$ and $(2,3,4)$ with distinct digits, or $(2,2,6)$. For the three groups with distinct digits, there are $3\times3!=18$ ways to arrange the three digits. For the group $(2,2,6)$, we have $\dbinom{3}{1}=3$ ways to arrange the position of digit 6. So the total number of possible 3-digit positive integers is $18+3=21$.

The positive integers $x$ and $y$ are the two smallest positive integers for which the product of $360$ and $x$ is a square and the product of $360$ and $y$ is a cube. What is the sum of $x$ and $y$?

$\textbf{(A)}\ 80\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 115\qquad\textbf{(D)}\ 165\qquad\textbf{(E)}\ 610$

$\textbf{B}$

The prime factorization of $360=2^3 \times 3^2 \times 5$. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is the minimum possible value of $x$. Similarly, $y$ can be found by making all the exponents divisible by 3, so the minimum possible value of $y$ is $3 \times 5^2=75$. Thus, our answer is $x+y=10+75=85$.

The diagram represents a 7-foot-by-7-foot floor that is tiled with 1-square-foot black tiles and white tiles. Notice that the corners have white tiles. If a 15-foot-by-15-foot floor is to be tiled in the same manner, how many white tiles will be needed?

$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 126$

$\textbf{C}$

The number of white tiles for a 7-foot-by-7-foot floor is $4\times4=16$. 4 is the result of dividing 7 by 2 and rounding it up. Similarly, The number of white tiles for a 15-foot-by-15-foot floor is $8\times8=64$.

Two angles of an isosceles triangle measure $70^\circ$ and $x^\circ$. What is the sum of the three possible values of $x$?

$\textbf{(A)}\ 95\qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 165\qquad\textbf{(E)}\ 180$

$\textbf{D}$

There are 3 cases: where $x^\circ$ is a base angle with the $70^\circ$ as the other base angle, where $x^\circ$ is a base angle with $70^\circ$ as the vertex angle, and where $x^\circ$ is the vertex angle with $70^\circ$ as a base angle.

Case 1: $x^\circ$ is a base angle with $70^\circ$ as the other base angle. Here, $x=70$, since base angles are congruent.

Case 2: $x^\circ$ is a base angle with $70^\circ$ as the vertex angle. Here, the 2 base angles are both $x^\circ$, so we can use the equation $2x+70=180$, which simplifies to $x=55$.

Case 3: $x^\circ$ is the vertex angle with $70^\circ$ as a base angle. Here, both base angles are $70^\circ$, since base angles are congruent. Thus, we can use the equation $x+140=180$, which simplifies to $x=40$.

Adding up all the cases, we get the sum of the three possible values $70+55+40=165$.

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

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

$\textbf{D}$

Assuming the base of the triangle is on the bottom four points because a congruent triangle can be made by reflecting the base on the top four points.

For a triangle with a base of length $1$, there are $3$ triangles (Be careful! You need to eliminate all congruent triangles).

For a triangle with a base of length $2$, there are $3$ triangles.

For length $3$, there are $2$.

In total, the number of non-congruent triangles is $3+3+2=8$.

Andy and Bethany have a rectangular array of numbers greater than 0 with $40$ rows and $75$ columns. Andy adds the numbers in each row. The average of his $40$ sums is $A$. Bethany adds the numbers in each column. The average of her $75$ sums is $B$. What is the value of $\dfrac{A}{B}$?

$\textbf{(A)}\ \dfrac{64}{225}\qquad\textbf{(B)}\ \dfrac{8}{15}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \dfrac{15}{8}\qquad\textbf{(E)}\ \dfrac{225}{64}$

$\textbf{D}$

The sum of all numbers in the array is $40A$, and also $75B$. So we have $$40A=75B\rightarrow \dfrac{A}{B}=\dfrac{15}{8}$$

How many whole numbers between 1 and 1000 do not contain the digit 1?

$\textbf{(A)}\ 512\qquad\textbf{(B)}\ 648\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 800$

$\textbf{D}$

We consider the 3 cases, where the number has 1,2 or 3 digits.

Case 1 : The number has a $1$ digit. Well, 1-9 all work except for $1$, so $9-1=8$ numbers, for numbers that has 1 digit.

Case 2 : The number has $2$ digits. Since the tens digit can not be 0 or 1, we have 8 choices for the tens digit. For the units digit, we have 9 choices because the units digit can not be 1. This gives us a total of $8\times9=72$ two-digit numbers.

Case 3 : The number has $3$ digits. Using similar logic to Case 2, we have $8\times9\times9=648$ choices for the third number.

By adding the cases up, we have $648+72+8=728$ numbers in total.

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 $400$ 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?

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

$\textbf{B}$

Let the number of girls be $x$. Then the number of boys is $x+2$. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. So we have $$x^2+(x+2)^2 = 394\rightarrow x=13$$ Hence, the number of students is $13+(13+2)=28$.

The letters $A$, $B$, $C$ and $D$ represent digits. If and , what digit does $D$ represent?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

$\textbf{E}$

Think about the units digit first. Since $B+A=A$, $B$ must be $0$. Next, because $B-A=A\implies0-A=A,$ we get $A=5$ ( $A$ can not be 0 because $A$ also works as the tens digit) as the “0” mentioned above is actually 10 in this case.

Now think about the tens digit, we have $A+C=D$ and $A-1-C=0$. From the second equation we get $C=A-1=4$. Then from the first equation we have $D=A+C=9$.

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 $1/2$ foot from the top face. The second cut is $1/3$ foot below the first cut, and the third cut is $1/17$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and 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?

$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\dfrac{419}{51}\qquad\textbf{(D)}\:\dfrac{158}{17}\qquad\textbf{(E)}\:11$

$\textbf{E}$

The areas of the tops of $A$, $B$, $C$, and $D$ in the figure formed has a sum $1+1+1+1 = 4$, as well as the bottoms. Thus, the total so far is $8$. Now, one of the sides has an area of 1, since it combines all of the heights of $A$, $B$, $C$, and $D$, which is $1$. The other side is also the same. Thus the total area now is $10$. From the front, the surface area is 1/2, because if you looked at it straight from the front it would look exactly like a side of $A$, with a surface area of half. From the back, it is the same thing. Thus, the total surface area is $10+\dfrac{1}{2}+\dfrac{1}{2}= 11$.