## AMC 8 2011

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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.

Margie bought $3$ apples at a cost of $50$ cents per apple. She paid with a 5-dollar bill. How much change did Margie receive?

$\textbf{(A) }\ \$1.50 \qquad \textbf{(B) }\ \$2.00 \qquad \textbf{(C) }\ \$2.50 \qquad \textbf{(D) }\ \$3.00 \qquad \textbf{(E) }\ \$3.50$

$\textbf{E}$

Since the price of an apple is 50 cents, three apples cost $3 \times \$0.50 = \$1.50.$ The change Margie receives is $\$5.00 - \$1.50 =\$3.50$.

Karl's rectangular vegetable garden is $20$ feet by $45$ feet, and Makenna's is $25$ feet by $40$ feet. Which of the following statements are true?

$\textbf{(A) }\text{Karl's garden is larger by 100 square feet.}\newline$

$\textbf{(B) }\text{Karl's garden is larger by 25 square feet.}\newline$

$\textbf{(C) }\text{The gardens are the same size.}\newline$

$\textbf{(D) }\text{Makenna's garden is larger by 25 square feet.}\newline$

$\textbf{(E) }\text{Makenna's garden is larger by 100 square feet.}$

$\textbf{E}$

The area of Karl's garden is $20\times45=900$ square feet. The area of Makenna's garden is $25\times40=1000$ square feet. So Makenna's garden is larger by 100 square feet.

Extend the square pattern of 8 black and 17 white square tiles by attaching a border of black tiles around the square. What is the ratio of black tiles to white tiles in the extended pattern?

$\textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17$

$\textbf{D}$

The extended pattern is made up of $7\times7=49$ tiles. Since 17 of them are white, the number of black tiles is $49-17=32$. So the ratio of black tiles to white tiles is $32:17$.

Here is a list of the numbers of fish that Tyler caught in nine outings last summer:\[2,0,1,3,0,3,3,1,2.\]Which statement about the mean, median, and mode is true?

$\textbf{(A) }\text{median} < \text{mean} < \text{mode} \qquad\newline$

$\textbf{(B) }\text{mean} < \text{mode} < \text{median}\newline$

$\textbf{(C) }\text{mean} < \text{median} < \text{mode} \qquad \newline$

$\textbf{(D) }\text{median} < \text{mode} < \text{mean}\newline$

$\textbf{(E) }\text{mode} < \text{median} < \text{mean}$

$\textbf{C}$

We can rearrange the list in increasing order: $$0,0,1,1,2,2,3,3,3$$ The mean is $\dfrac{0+0+1+1+2+2+3+3+3}{9} = \dfrac{15}{9},$ the median is $2,$ and the mode is $3.$ So the answer is $\text{mean} < \text{median} < \text{mode}$.

What time was it $2011$ minutes after midnight on January 1, 2011?

$\textbf{(A) }\text{January 1 at 9:31 PM}\newline$

$\textbf{(B) }\text{January 1 at 11:51 PM}\newline$

$\textbf{(C) }\text{January 2 at 3:11 AM}\newline$

$\textbf{(D) }\text{January 2 at 9:31 AM}\newline$

$\textbf{(E) }\text{January 2 at 6:01 PM}$

$\textbf{D}$

By converting $2011\ \text{minutes}=33\ \text{hours and } 31\ \text{minutes}=1\ \text{day}+ 9\ \text{hours and } 31\ \text{minutes}$, we get the answer $\text{January 2 at 9:31 AM}$.

In a town of 351 adults, every adult owns a car, motorcycle, or both. If 331 adults own cars and 45 adults own motorcycles, how many of the car owners do not own a motorcycle?

$\textbf{(A) }20 \qquad\textbf{(B) }25 \qquad\textbf{(C) }45 \qquad\textbf{(D) }306 \qquad\textbf{(E) }351$

$\textbf{D}$

The number of adults who own both cars and motorcycles is $331+45-351=25$. So the number of adult who own cars but not motorcycles is $331-25=306$.

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded. What percent of the total area is partially bolded?

$\textbf{(A)}12\dfrac{1}{2}\qquad\textbf{(B)}20\qquad\textbf{(C)}25\qquad\textbf{(D)}33\dfrac{1}{3}\qquad\textbf{(E)}37\dfrac{1}{2}$

$\textbf{C}$

The ratios of bolded area in each figure are $\dfrac14$, $\dfrac18$, $\dfrac38$, and $\dfrac14$. So the percentage of the total bolded area is the average $\dfrac14\times\left(\dfrac14+\dfrac18+\dfrac38+\dfrac14\right)=\dfrac14=25\%$.

Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?

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

$\textbf{B}$

By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $5$.

Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?

$\textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5$

$\textbf{E}$

It takes her 7 hours to ride 35 miles. So the average speed is $35/7=5\ \text{mph}$.

The taxi fare in Gotham City is $\$2.40$ for the first $\dfrac12$ mile and additional mileage charged at the rate $\$0.20$ for each additional 0.1 mile. You plan to give the driver a $\$2$ tip. How many miles can you ride for $\$10$?

$\textbf{(A) } 3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$

$\textbf{C}$

First we subtract the $\$2$ tip and $\$2.40$ for the first $\dfrac12$ mile. We have $\$10-\$2-\$2.40=\$5.6$ for the additional mileage of $\$2.0$ per mile. Hence, the additional mileage is $5.6/2=2.8$ miles. So the total distance is $2.8+0.5=3.3$ miles.

The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

$\textbf{A}$

The differences of study time each day are 10, -10, 20, 30, and -20 minutes. So the average is $\dfrac{10-10+20+30-20}{5}=6$ minutes.

Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?

$\textbf{(A) } \dfrac14 \qquad\textbf{(B) } \dfrac13 \qquad\textbf{(C) } \dfrac12 \qquad\textbf{(D) } \dfrac23 \qquad\textbf{(E) } \dfrac34$

$\textbf{B}$

When the position of Angie is fixed, Carlos can be on the left side, on the right side, or on the opposite of Angie. Hence, the probability that Angie and Carlos are seated opposite each other is $\dfrac13$.

Two congruent squares, $ABCD$ and $PQRS$, have side length $15$. They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What percent of the area of rectangle $AQRD$ is shaded?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25$

$\textbf{C}$

The length of $\overline{BP}=\overline{AB}+\overline{PQ}-\overline{AQ}=15+15-25=5$. So the ratio of the area of rectangle $PBCS$ to $AQRD$ is $\overline{BP}:\overline{AQ}=5:25=20\%$.

There are $270$ students at Colfax Middle School, where the ratio of boys to girls is $5 : 4$. There are $180$ students at Winthrop Middle School, where the ratio of boys to girls is $4 : 5$. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?

$\textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45}$

$\textbf{C}$

The total number of students in the two middle schools is $270+180=450$, while the number of girls is $270\times\dfrac49+180\times\dfrac59=220$. So the ratio is $\dfrac{220}{450}=\dfrac{22}{45}$.

How many digits are in the product $4^5 \cdot 5^{10}$?

$\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12$

$\textbf{D}$

\[4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.\]

That is one $1$ followed by ten $0$'s, which is $11$ digits.

Let $A$ be the area of the triangle with sides of length $25, 25$, and $30$. Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$. What is the relationship between $A$ and $B$?

$\textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad\textbf{(D) } A = \dfrac43B \qquad \textbf{(E) }A = \dfrac{16}9B$

$\textbf{C}$

Using Heron's formula, we can calculate the area of the two triangles. The formula states that\[A = \sqrt{s(s - a)(s - b)(s - c)}\]where $s$ is the semiperimeter of a triangle with side lengths $a$, $b$, and $c$.

For the 25-25-30 triangle,\[s = \frac{25 + 25 + 30}{2} = 40\]Therefore,\[A = \sqrt{40 \times 15 \times 15 \times 10} = 300\] For the 25-25-40 triangle,\[s = \frac{25 + 25 + 40}{2} = 45\]Therefore,\[B = \sqrt{45 \times 20 \times 20 \times 5} = 300\] Hence,\[A = B\]

Let $w$, $x$, $y$, and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$, then what does $2w + 3x + 5y + 7z$ equal?

$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$

$\textbf{A}$

The prime factorization of $588$ is $588=2^2\times3\times7^2$. We can see $w=2, x=1,$ and $z=2.$ We can not see the factor 5 because $y=0$, $5^0=1$. Hence,\[2w+3x+5y+7z=4+3+0+14=21\]

A fair 6-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?

$\textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56$

$\textbf{D}$

When the first number is given, the probability that the second number is the same is $\dfrac16$. So the probability that the second number is not the same is $1-\dfrac16=\dfrac56$. By symmetry, the probability that the first number is greater is equal to the probability that the second number is greater. Hence, the probability that the first number is greater is $\dfrac12\times\dfrac56=\dfrac5{12}$. Furthermore, the probability that the first number is no less than the second number is $\dfrac5{12}+\dfrac16=\dfrac7{12}$.

How many rectangles are in this figure?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

$\textbf{D}$

The figure can be divided into $7$ indivisible sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There is none with only three sections. The number of rectangles with four sections is $3.$ So the total number of rectangles is $3+5+3=11$.

Quadrilateral $ABCD$ is a trapezoid, $AD = 15$, $AB = 50$, $BC = 20$, and the altitude is $12$. What is the area of the trapezoid?

$\textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800$

$\textbf{D}$

If we draw altitudes from $A$ and $B$ to $CD,$ the trapezoid will be divided into two right triangles and a rectangle. We can find the values of $a$ and $b$ with the Pythagorean theorem. \[a=\sqrt{15^2-12^2}=\sqrt{81}=9\] \[b=\sqrt{20^2-12^2}=\sqrt{256}=16\] So $CD=a+XY+b=9+50+16=75$. The area of the trapezoid is \[12\cdot \frac{(50+75)}{2}=750\]

Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$

$\textbf{C}$

If at least half the guesses are too low, then Norb's age must be greater than $36.\newline$

If two of the guesses are off by one, then his age is between two guesses whose difference is $2.$ It could be $31,37,$ or $48$. But because his age is greater than $36$, it can only be $37$ or $48.\newline$

Lastly, Norb's age is a prime number. So the answer must be $37$.

What is the $\textbf{tens}$ digit of $7^{2011}$?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$

$\textbf{D}$

Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic.\[7^1\equiv 07 \pmod{100}\]\[7^2\equiv 49 \pmod{100}\]\[7^3\equiv 43 \pmod{100}\]\[7^4\equiv 01 \pmod{100}\] We find that the last two digits will recur when the power is increased by 4. We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$. Hence, we get \[7^{2011}\equiv 7^3\equiv 43\pmod{100}.\]From the above, we can conclude that the last two digits of $7^{2011}$ are 43. The tens digit is $4$.

How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$

$\textbf{D}$

Since the integer is a multiple of $5,$ the last digit must be either $5$ or $0.$ We can separate this into two cases.$\newline$

Case 1: The last digit is $5.$ The leading digit can be $1,2,3,$ or $4.$ Because the second digit can be $0$ but not the leading digit, there are also $4$ choices. The third digit can not be the leading digit or the second digit, so there are $3$ choices. The number of integers is this case is $4\times4\times3\times1=48.\newline$

Case 2: The last digit is $0.$ Because $5$ is the largest digit, one of the remaining three digits must be $5.$ There are $3$ ways to choose which digit should be $5.$ The remaining digits can be $1,2,3,$ or $4,$ but since they have to be different there are $4\times3$ ways to choose. The number of integers in this case is $1\times3\times4\times3=36.\newline$

Therefore, the answer is $48+36=84$.

In how many ways can 10001 be written as the sum of two primes?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

$\textbf{A}$

For the sum of two numbers to be odd, one must be odd and the other must be even. The only even prime is 2. So the other number must be 9999. However, 9999 is not a prime because it is a multiple. So 10001 can not be written as the sum of two primes.

A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?

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

$\textbf{A}$

The side length of the small square is $\sqrt2$. And the side length of the large square is 2. So the area of the shaded area of the circle is $$\pi\times1^2-\left(\sqrt2\right)^2=\pi-2$$ The area between the two squares is $$2^2-\left(\sqrt2\right)^2=2$$ Hence, the ratio is $$\dfrac{\pi-2}{2}=\dfrac{1.14}{2}\approx\dfrac12$$