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2011 AMC 8 真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2017年7月25日下午1:29

2011 AMC 8 真题

答案详细解析请参考文末

Problem 1

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

Problem 2

Karl's rectangular vegetable garden is feet by feet, and Makenna's is feet by feet. Whose garden is larger in area?

Problem 3

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? 2011 AMC 8 Problems

Problem 4

Here is a list of the numbers of fish that Tyler caught in nine outings last summer: Which statement about the mean, median, and mode is true? 2011 AMC 8 Problems

Problem 5

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

Problem 6

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?

Problem 7

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? 2011 AMC 8 Problems

Problem 8

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?

Problem 9

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? 2011 AMC 8 Problems

Problem 10

The taxi fare in Gotham City is $2.40 for the first 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?

Problem 11

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? 2011 AMC 8 Problems

Problem 12

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?

Problem 13

Two congruent squares, and , have side length . They overlap to form the by rectangle shown. What percent of the area of rectangle is shaded? 2011 AMC 8 Problems

Problem 14

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

Problem 15

How many digits are in the product ?

Problem 16

Let be the area of the triangle with sides of length , and . Let be the area of the triangle with sides of length and . What is the relationship between and ? 2011 AMC 8 Problems

Problem 17

Let , , , and be whole numbers. If , then what does equal?

Problem 18

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?

Problem 19

How many rectangles are in this figure? 2011 AMC 8 Problems

Problem 20

Quadrilateral is a trapezoid, , , , and the altitude is . What is the area of the trapezoid? 2011 AMC 8 Problems

Problem 21

Students guess that Norb's age is , and . 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?

Problem 22

What is the tens digit of ?

Problem 23

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?

Problem 24

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

Problem 25

A circle with radius 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? 2011 AMC 8 Problems $50$ cents is equivalent to $textdollar 0.50.$ Then the three apples cost $3 times textdollar 0.50 = textdollar 1.50.$ The change Margie receives is $textdollar 5.00 - textdollar 1.50 = boxed{textbf{(E)} textdollar 3.50}$ The area of a rectangle is given by the formula length times width.

Karl's garden is $20 times 45 = 900$ square feet and Makenna's garden is $25 times 40 = 1000$ square feet. Since $1000 > 900,$ Makenna's garden is larger by $1000-900=100$ square feet. $Rightarrow boxed{ textbf{(E)} text{Makenna's garden is larger by 100 square feet.} }$ One way of approaching this is drawing the next circle of boxes around the current square. $[asy] filldraw((-1,-1)--(6,-1)--(6,6)--(-1,6)--cycle,mediumgray,black); filldraw((0,0)--(5,0)--(5,5)--(0,5)--cycle,white,black); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle,mediumgray,black); filldraw((2,2)--(3,2)--(3,3)--(2,3)--cycle,white,black); draw((4,-1)--(4,6)); draw((3,-1)--(3,6)); draw((2,-1)--(2,6)); draw((1,-1)--(1,6)); draw((-1,4)--(6,4)); draw((-1,3)--(6,3)); draw((-1,2)--(6,2)); draw((-1,1)--(6,1)); draw((0,-1)--(0,6)); draw((-1,5)--(6,5)); draw((-1,0)--(6,0)); draw((5,-1)--(5,6)); [/asy]$ We can now count the number of black and white tiles; 32 black tiles and 17 white tiles. This means the answer is $boxed{textbf{(D) }32:17}$ .
First, put the numbers in increasing order. $[0,0,1,1,2,2,3,3,3]$ The mean is $frac{0+0+1+1+2+2+3+3+3}{9} = frac{15}{9},$ the median is $2,$ and the mode is $3.$ Because, $frac{15}{9} < 2 < 3,$ the answer is $boxed{textbf{(C)} text{mean} < text{median} < text{mode}}$ There are $60$ minutes in an hour. $2011/60=33text{r}31,$ or $33$ hours and $31$ minutes. There are $24$ hours in a day, so the time is $9$ hours and $31$ minutes after midnight on January 2, 2011. $Rightarrowboxed{textbf{(D)}text{ January 2 at 9:31AM}}$ There are $351$ total adults, and $45$ own a motorcycle. The number of adults that don't own a motorcycle is $351 - 45 = 306$ . Since everyone owns a car or motorcycle and one who doesn't own a motorcycle owns a car, the answer is $boxed{textbf{(D)} 306}$ .
Assume that the area of each square is $1$ . Then, the area of the bolded region in the top left square is $frac{1}{4}$ .

The area of the top right bolded region is $frac{1}{8}$ . The area of the bottom left bolded region is $frac{3}{8}$ . And the area of the bottom right bolded region is $frac{1}{4}$ . Add the four fractions: $frac{1}{4} + frac{1}{8} + frac{3}{8} + frac{1}{4} = 1$ . The four squares together have an area of $4$ , so the percentage bolded is $frac{1}{4} cdot 100 = boxed{textbf{(C)} 25}$ .

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 $boxed{textbf{(B)} 5}$ . You can also approach this problem by noticing the sums will make odd multiples from $3$ to $11$ inclusive.
We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is $boxed{textbf{(E)} 5}$ Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation: $begin{align*} 10 &= 2.4 + 0.2 times frac{x-0.5}{0.1} + 2 5.6 &= 2(x-0.5) 2.8 &= x-0.5 x &= boxed{textbf{(C)} 3.3}end{align*}$ Average the differences between each day. We get $10, -10,text{ } 20,text{ } 30,-20$ . We find the average of this list to get $boxed{textbf{(A)} 6}$ If we designate Angie to be on a certain side, then all placements of the other people can be considered unique. There are then $3!=6$ total seating arrangements.

If Carlos is across from Angie, there are only $2!=2$ ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is $frac26=boxed{textbf{(B)} frac13}$ The overlap length is $5$ , so the shaded area is $5 cdot 15 =75$ . The area of the whole shape is $25 cdot 15 = 375$ . The fraction $dfrac{75}{375}$ reduces to $dfrac{1}{5}$ or 20%. Therefore, the answer is $boxed{ textbf{(C)} text{20} }$ At Colfax Middle School, there are $frac49 times 270 = 120$ girls.

At Winthrop Middle School, there are $frac59 times 180 = 100$ girls. The ratio of girls to the total number of students is $frac{120+100}{270+180} = frac{220}{450} = boxed{textbf{(C)} frac{22}{45}}$ $[4^5 cdot 5^{10} = 2^{10} cdot 5^{10} = 10^{10}.]$ That is one $1$ followed by ten $0$ 's, which is $boxed{textbf{(D)} 11}$ digits. 25-25-30We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent.

Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem, we have $[15^2 + x^2 =25^2]$ $[x^2 = 25^2 - 15^2]$ $[x^2 = (25 + 15)(25-15)]$ $[x^2= 40cdot 10]$ $[x^2= 400]$ $[x = sqrt{400}]$ $[x= 20]$ Thus we have two 15-20-25 right triangles.25-25-40We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20.

From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles. Let the area of a 15-20-25 right triangle be $x$ . $[a = 2x]$ $[b = 2x]$ $[boxed{textbf{(C) } A = B}]$ The prime factorization of $588$ is $2^2cdot3cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$ $[2w+3x+5y+7z=4+3+0+14=boxed{textbf{(A)} 21}]$ There are $6cdot6=36$ ways to roll the two dice, and $6$ of them result in two of the same number.

Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $30/2=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is $[frac{15+6}{36}=frac{21}{36}=boxed{textbf{(D)} frac{7}{12}}]$ The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$ $3+5+3=boxed{textbf{(D)} 11}$ $[asy] unitsize(1.5mm); defaultpen(linewidth(.9pt)+fontsize(10pt)); dotfactor=3; pair A,B,C,D,X,Y; A=(9,12); B=(59,12); C=(75,0); D=(0,0); X=(9,0); Y=(59,0); draw(A--B--C--D--cycle); draw(A--X); draw(B--Y); pair[] ps={A,B,C,D,X,Y}; dot(ps); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$X$",X,SE); label("$Y$",Y,S); label("$a$",D--X,S); label("$b$",Y--C,S); label("$15$",D--A,NW); label("$50$",B--A,N); label("$20$",B--C,NE); label("$12$",X--A,E); label("$12$",Y--B,W); [/asy]$ If you draw altitudes from $A$ and $B$ to $CD,$ the trapezoid will be divided into two right triangles and a rectangle. You can find the values of $a$ and $b$ with the Pythagorean theorem. $[a=sqrt{15^2-12^2}=sqrt{81}=9]$ $[b=sqrt{20^2-12^2}=sqrt{256}=16]$ $ABYX$ is a rectangle so $XY=AB=50.$ $[CD=a+XY+b=9+50+16=75]$ The area of the trapezoid is $[12cdot frac{(50+75)}{2} = 6(125) = boxed{textbf{(D)} 750}]$ If at least half the guesses are too low, then his age must be greater than $36.$ If two of the guesses are off by one, then his age is in between two guesses whose difference is $2.$ It could $31,37,$ or $48,$ but because it is greater than $36$ it can only be $37$ or $48.$ Lastly, his age is a prime number so the answer must be $boxed{textbf{(C)} 37}$ The first couple powers of $7$ are $7, 49, 343, 2401, 16807.$ As you can see, the last two digits cycle after every 4 powers. $7^{1} (text{mod }100) equiv 7^{5} (text{mod }100) equiv 7^{2009} (text{mod }100).$ From there, we go two more powers. The last two digits are $43$ so the tens digit is $boxed{textbf{(D)} 4}$ We can separate this into two cases.

If an integer is a multiple of $5,$ the last digit must be either $0$ or $5.$ 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 cannot be the leading digit or the second digit, so there are $3$ choices. The number of integers is this case is $4cdot4cdot3cdot1=48.$ 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 $4cdot3$ ways to choose. The number of integers in this case is $1cdot3cdot4cdot3=36.$

Therefore, the answer is $48+36=boxed{textbf{(D)} 84}$ For the sum of two numbers to be odd, one must be odd and the other must be even, because All odd numbers are of the form $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. $[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1]$ and $m+n$ is an integer because $m$ and $n$ are both integers.

The only even prime number is $2,$ so our only combination could be $2$ and $9999.$ However, $9999$ is clearly divisible by $3$ so the number of ways $10001$ can be written as the sum of two primes is $boxed{textbf{(A)} 0}$ The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: $2*2*1/2=2.$

The circle's shaded area is the area of the smaller square subtracted from the area of the circle: $pi - 2.$ If you draw the diagonals of the smaller square, you will see that the larger square is split $4$ congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: $2.$ Approximating $pi$ to $3.14,$ the ratio of the circle's shaded area to the area between the two squares is about $[frac{pi-2}{2} approx frac{3.14-2}{2} = frac{1.14}{2} approx boxed{textbf{(A)} frac12}]$