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2007 AMC 8 真题及详解

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月3日下午7:01

2007 AMC 8 真题

答案详细解析请参考文末

Problem 1

Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of? $10$ ?hours per week helping around the house for? $6$ ?weeks. For the first? $5$ ?weeks she helps around the house for? $8$ ,? $11$ ,? $7$ ,? $12$ ?and? $10$ ?hours. How many hours must she work for the final week to earn the tickets?

$mathrm{(A)} 9 qquadmathrm{(B)} 10 qquadmathrm{(C)} 11 qquadmathrm{(D)} 12 qquadmathrm{(E)} 13$

Problem 2

$650$ ?students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?

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

Problem 3

What is the sum of the two smallest prime factors of? $250$ ?

$mathrm{(A)} 2 qquadmathrm{(B)} 5 qquadmathrm{(C)} 7 qquadmathrm{(D)} 10 qquadmathrm{(E)} 12$

Problem 4

A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

$mathrm{(A)} 12 qquadmathrm{(B)} 15 qquadmathrm{(C)} 18 qquadmathrm{(D)} 30 qquadmathrm{(E)} 36$

Problem 5

Chandler wants to buy a? $textdollar 500$ ?mountain bike. For his birthday, his grandparents send him? $textdollar 50$ , his aunt sends him? $textdollar 35$ ?and his cousin gives him? $textdollar 15$ . He earns? $textdollar 16$ ?per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?

$mathrm{(A)} 24 qquadmathrm{(B)} 25 qquadmathrm{(C)} 26 qquadmathrm{(D)} 27 qquadmathrm{(E)} 28$

Problem 6

The average cost of a long-distance call in the USA in? $1985$ ?was? $41$ ?cents per minute, and the average cost of a long-distance call in the USA in? $2005$ ?was? $7$ ?cents per minute. Find the approximate percent decrease in the cost per minute of a long- distance call.

$mathrm{(A)} 7 qquadmathrm{(B)} 17 qquadmathrm{(C)} 34 qquadmathrm{(D)} 41 qquadmathrm{(E)} 80$

Problem 7

The average age of? $5$ ?people in a room is? $30$ ?years. An? $18$ -year-old person leaves the room. What is the average age of the four remaining people?

$mathrm{(A)} 25 qquadmathrm{(B)} 26 qquadmathrm{(C)} 29 qquadmathrm{(D)} 33 qquadmathrm{(E)} 36$

Problem 8

In trapezoid? $ABCD$ ,? $AD$ ?is perpendicular to? $DC$ ,? $AD$ ?=? $AB$ ?=? $3$ , and? $DC$ ?=? $6$ . In addition,? $E$ ?is on? $DC$ , and? $BE$ ?is parallel to? $AD$ . Find the area of? $triangle BEC$ .

$text{(A)} 3 qquad text{(B)} 4.5 qquad text{(C)} 6 qquad text{(D)} 9 qquad text{(E)} 18$

Problem 9

To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square?

$mathrm{(A)} 1 qquad mathrm{(B)} 2 qquad mathrm{(C)} 3 qquad mathrm{(D)} 4 qquad mathrm{(E)}$ ?cannot be determined

Problem 10

For any positive integer? $n$ , define? $boxed{n}$ ?to be the sum of the positive factors of? $n$ . For example,? $boxed{6} = 1 + 2 + 3 + 6 = 12$ . Find? $boxed{boxed{11}}$ ?.

$mathrm{(A)} 13 qquad mathrm{(B)} 20 qquad mathrm{(C)} 24 qquad mathrm{(D)} 28 qquad mathrm{(E)} 30$

Problem 11

Tiles? $I, II, III$ ?and? $IV$ ?are translated so one tile coincides with each of the rectangles? $A, B, C$ ?and? $D$ . In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle? $C$ ?

$mathrm{(A)} I qquad mathrm{(B)} II qquad mathrm{(C)} III qquad mathrm{(D)} IV qquad mathrm{(E)}$ ?cannot be determined

Problem 12

A unit hexagram is composed of a regular hexagon of side length? $1$ ?and its? $6$ ?equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?

$mathrm{(A)} 1:1 qquad mathrm{(B)} 6:5 qquad mathrm{(C)} 3:2 qquad mathrm{(D)} 2:1 qquad mathrm{(E)} 3:1$

Problem 13

Sets? $A$ ?and? $B$ , shown in the Venn diagram, have the same number of elements. Their union has? $2007$ ?elements and their intersection has? $1001$ ?elements. Find the number of elements in? $A$ .

$mathrm{(A)} 503 qquad mathrm{(B)} 1006 qquad mathrm{(C)} 1504 qquad mathrm{(D)} 1507 qquad mathrm{(E)} 1510$

Problem 14

The base of isosceles? $triangle ABC$ ?is? $24$ ?and its area is? $60$ . What is the length of one of the congruent sides?

$mathrm{(A)} 5 qquad mathrm{(B)} 8 qquad mathrm{(C)} 13 qquad mathrm{(D)} 14 qquad mathrm{(E)} 18$

Problem 15

Let? $a, b$ ?and? $c$ ?be numbers with? $0 < a < b < c$ . Which of the following is impossible?

$mathrm{(A)} a + c < b qquad mathrm{(B)} a cdot b < c qquad mathrm{(C)} a + b < c qquad mathrm{(D)} a cdot c < b qquad mathrm{(E)}frac{c} = a$

Problem 16

Amanda draws five circles with radii? $1, 2, 3, 4$ ?and? $5$ . Then for each circle she plots the point? $(C,A)$ , where? $C$ ?is its circumference and? $A$ ?is its area. Which of the following could be her graph?

Problem 17

A mixture of? $30$ ?liters of paint is? $25%$ ?red tint,? $30%$ ?yellow tint and? $45%$ ?water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?

$mathrm{(A)} 25 qquad mathrm{(B)} 35 qquad mathrm{(C)} 40 qquad mathrm{(D)} 45 qquad mathrm{(E)} 50$

Problem 18

The product of the two? $99$ -digit numbers

$303,!030,!303,!...,!030,!303$ ?and? $505,!050,!505,!...,!050,!505$

has thousands digit? $A$ ?and units digit? $B$ . What is the sum of? $A$ ?and? $B$ ?

$mathrm{(A)} 3 qquad mathrm{(B)} 5 qquad mathrm{(C)} 6 qquad mathrm{(D)} 8 qquad mathrm{(E)} 10$

Problem 19

Pick two consecutive positive integers whose sum is less than? $100$ . Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

$mathrm{(A)} 2 qquad mathrm{(B)} 64 qquad mathrm{(C)} 79 qquad mathrm{(D)} 96 qquad mathrm{(E)} 131$

Problem 20

Before district play, the Unicorns had won? $45%$ ?of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

$mathrm{(A)} 48 qquad mathrm{(B)} 50 qquad mathrm{(C)} 52 qquad mathrm{(D)} 54 qquad mathrm{(E)} 60$

Problem 21

Two cards are dealt from a deck of four red cards labeled? $A, B, C, D$ ?and four green cards labeled? $A, B, C, D$ . A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

$mathrm{(A)} frac{2}{7} qquad mathrm{(B)} frac{3}{8} qquad mathrm{(C)} frac{1}{2} qquad mathrm{(D)} frac{4}{7} qquad mathrm{(E)} frac{5}{8}$

Problem 22

A lemming sits at a corner of a square with side length? $10$ ?meters. The lemming runs? $6.2$ ?meters along a diagonal toward the opposite corner. It stops, makes a? $90$ ?degree right turn and runs? $2$ ?more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?

$mathrm{(A)} 2 qquad mathrm{(B)} 4.5 qquad mathrm{(C)} 5 qquad mathrm{(D)} 6.2 qquad mathrm{(E)} 7$

Problem 23

What is the area of the shaded part shown in the? $5$ ?x? $5$ ?grid?

$mathrm{(A)} 4 qquadmathrm{(B)} 6 qquadmathrm{(C)} 8 qquadmathrm{(D)} 10 qquadmathrm{(E)} 12$

Problem 24

A bag contains four pieces of paper, each labeled with one of the digits "1, 2, 3" or "4", with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

$mathrm{(A)} frac{1}{4} qquad mathrm{(B)} frac{1}{3} qquad mathrm{(C)} frac{1}{2} qquad mathrm{(D)} frac{2}{3} qquad mathrm{(E)} frac{3}{4}$

Problem 25

On the dart board shown in the Figure, the outer circle has radius? $6$ ?and the inner circle has a radius of 3. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?

$mathrm{(A)} frac{17}{36} qquad mathrm{(B)} frac{35}{72} qquad mathrm{(C)} frac{1}{2} qquad mathrm{(D)} frac{37}{72} qquad mathrm{(E)} frac{19}{36}$

2007AMC8详细解析

Let? $x$ ?be the number of hours she must work for the final week. We are looking for the average, so $[frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10]$ Solving gives: $[frac{48 + x}{6} = 10]$ $[48 + x = 60]$ $[x = 12]$ So, the answer is? $boxed{textbf{(D)} 12}$ Use average deviation:
The average is 10 hour per day. If work 8 hours then it is 2 hours short; if work 11 hours then there is 1 hour surplus, the last day need to cancel out the collective deviation from the previous 5 days.

So we got

$[-2+1-3+2+0=-2]$

The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so? $boxed{textbf{(D)} 12}$ .
The answer is? $dfrac{text{number of students who preferred spaghetti}}{text{number of students who preferred manicotti}}$ So, $frac{250}{100}$ Simplify, $frac{5}{2}$
The answer is? $boxed{textbf{(E)} dfrac{5}{2}}$
The prime factorization of? $250$ ?is? $2 cdot 5^3$ . The smallest two are? $2$ ?and? $5$ .? $2+5 = boxed{text{(C) }7}$ .
Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows.So, Georgie has a total of? $6 cdot 5$ ?ways he can enter the house by one window and leave by a different window.Therefore, we have? $boxed{textbf{(D)} 30}$ ?ways.
Let? $x$ ?be the number of weeks.Thus, we have the equation? $50 + 35 + 15 + 16x = 500$ .Simplify, $100 + 16x = 500$ $16x = 400$
$x = 25$

The answer is? $boxed{textbf{(B)} 25}$
The percent decrease is (the amount of decrease)/(original amount)the amount of decrease is? $41 - 7 = 34$ so the percent decrease is? $frac{34}{41}$ ?which is about? $boxed{textbf{(E)} 80%}$
Let? $x$ ?be the average of the remaining? $4$ ?people.The equation we get is? $frac{4x + 18}{5} = 30$ Simplify, $4x + 18 = 150$ $4x = 132$
$x = 33$

Therefore, the answer is? $boxed{textbf{(D)} 33}$

Since an? $18$ ?year old left from a group of people averaging? $30$ , The remaining people must total? $30 - 18 = 12$ ?years older than? $30$ . Therefore, the average is? $dfrac{12}{4} = 3$ ?years over? $30$ . Giving us? $boxed{textbf{(D)} 33}$

The total ages would be? $30*5=150$ . Then, if one? $18$ ?year old leaves, we subtract? $18$ ?from? $150$ ?and get? $132$ . Then, we divide? $132$ ?by? $4$ ?to get the new average,? $boxed{textbf{(D)} 33}$
We know that? $ABED$ ?is a square with side length? $3$ . We subtract? $DC$ ?and? $DE$ ?to get the length of? $EC$ . $EC = DC - DE = 6 - 3 = 3$ We are trying to find the area of? $triangle BEC$ .So,? $frac{1}{2} cdot 3 cdot 3 = boxed{textbf{(B)} 4.5}$
The number in the first row, last column must be a? $3$ ?due to the fact if a? $3$ ?was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a? $1$ . Therefore the number in the lower right-hand square is? $boxed{textbf{(B)} 2}$ .
First we find? $boxed{11}$ . $boxed{11} = 1 + 11 = 12$ Then we find? $boxed{12}$ . $boxed{boxed{11}} = boxed{12} = 1 + 2 + 3 + 4 + 6 + 12 = boxed{textbf{(D)} 28}$
We first notice that tile III has a? $0$ ?on the bottom and a? $5$ ?on the right side. Since no other tile has a? $0$ ?or a? $5$ , Tile III must be in rectangle? $D$ . Tile III also has a? $1$ ?on the left, so Tile IV must be in Rectangle? $C$ .The answer is? $boxed{textbf{(D)}}$
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is? $boxed{textbf{(A) }1:1}$
Let? $x$ ?be the number of elements in? $A$ ?and? $B$ .Since the union is the sum of all elements in? $A$ ?and? $B$ ,and? $A$ ?and? $B$ ?have the same number of elements then, $2x-1001 = 2007$ $2x = 3008$
$x = 1504$ .

The answer is? $boxed{textbf{(C)} 1504}$
The area of a triangle is shown by? $frac{1}{2}bh$ . We set the base equal to? $24$ , and the area equal to? $60$ , and we get the height, or altitude, of the triangle to be? $5$ . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem,? $a^2+b^2=c^2$ , we can solve for one of the legs of the triangle (it will be the the hypotenuse,? $c$ ).? $a = 12$ ,? $b = 5$ ,? $c = 13$ . The answer is? $boxed{textbf{(C)} 13}$
According to the given rules, every number needs to be positive. Since? $c$ ?is always greater than? $b$ , adding a positive number ( $a$ ) to? $c$ ?will always make it greater than? $b$ .Therefore, the answer is? $boxed{textbf{(A)} a+c<b}$
he circumference of a circle is obtained by simply multiplying the radius by? $2pi$ . So, the C-coordinate (in this case, it is the x-coordinate) will increase at a steady rate. The area, however, is obtained by squaring the radius and multiplying it by? $pi$ . Since squares do not increase in an evenly spaced arithmetic sequence, the increase in the A-coordinates ( aka the y- coordinates) will be much more significant. The answer is? $boxed{textbf{(A)}},$
Since? $30%$ ?of the original? $30$ ?liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are? $9+5=14$ ?liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This gives? $40%$ ?of yellow tint in the new mixture, which is? $boxed{textbf{(C) 40}}$ .
We can first make a small example to find out? $A$ ?and? $B$ . So, $303times505=153015$ The ones digit plus thousands digit is? $5+3=8$ .Note that the ones and thousands digits are, added together,? $8$ . (and so on...) So the answer is? $boxed{textbf{(D)} 8}$ ?This is a direct multlipication way.
Let the smaller of the two numbers be? $x$ . Then, the problem states that? $(x+1)+x<100$ .? $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$ .? $2x+1$ ?is obviously odd, so only answer choices C and E need to be considered. $2x+1=131$ ?refutes the fact that? $2x+1<100$ , so the answer is? $boxed{mathrm{(C)} 79}$
At the beginning of the problem, the Unicorns had played? $y$ ?games and they had won? $x$ ?of these games. From the information given in the problem, we can say that? $frac{x}{y}=0.45.$ ?Next, the Unicorns win 6 more games and lose 2 more, for a total of? $6+2=8$ ?games played during district play. We are told that they end the season having won half of their games, or? $0.5$ ?of their games. We can write another equation:? $frac{x+6}{y+8}=0.5.$ ?This gives us a system of equations:? $frac{x}{y}=0.45$ ?and? $frac{x+6}{y+8}=0.5.$ ?We first multiply both sides of the first equation by? $y$ ?to get? $x=0.45y.$ ?Then, we multiply both sides of the second equation by? $(y+8)$ ?to get? $x+6=0.5(y+8).$ ?Applying the Distributive Property gives yields? $x+6=0.5y+4.$ ?Now we substitute? $0.45y$ ?for? $x$ ?to get? $0.45y+6=0.5y+4.$ ?Solving gives us? $y=40.$ ?Since the problem asks for the total number of games, we add on the last 8 games to get the solution? $boxed{textbf{(A)} 48}$ .Simplifying 45% to? $frac{9}{20}$ , we see that the numbers of games are a multiple of 20. After that the Unicorns played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is? $boxed{48}$ , which is 20(2)+8.-harsha12345First we simplify 45% to? $frac{9}{20}$ . After they won 6 more games and lost 2 more games the number of games they won is? $9x+6$ , and the total number of games is? $20x+8$ . Turning it into a fraction we get? $frac{9x+6}{20x+8}=frac{1}{2}$ , so solving for? $x$ ?we get? $x=2.$ ?Plugging in 2 for? $x$ ?we get? $20(2)+8=boxed{48}$ .-harsha12345
There are 4 ways of choosing a winning pair of the same letter, and? $2 left( dbinom{4}{2} right) = 12$ ?ways to choose a pair of the same color.There's a total of? $dbinom{8}{2} = 28$ ?ways to choose a pair, so the probability is? $dfrac{4+12}{28} = boxed{textbf{(D)} frac{4}{7}}$ .
The shortest segments would be perpendicular to the square. The lemming went? $x$ ?meters horizontally and? $y$ ?meters vertically. No matter how much it went, the lemming would have been? $x$ ?and? $y$ ?meters from the sides and? $10-x$ ?and? $10-y$ ?meters from the remaining two. To find the average, add the lengths of the four segments and divide by four:? $frac {cancel{x}+10-cancel{x}+cancel{y}+10-cancel{y}}{4} = 5$ ? $boxed{textbf{(C)} 5}$ .
The area of the square around the pinwheel is 25. The area of the pinwheel is equal to? $text{the square } - text{ the white space.}$ ?Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is? $25-(15+4)$ ?which is? $boxed{textbf{(B) 6}}$
The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4). The number of ways to choose three digits out of four is 4. Therefore, the probability is? $boxed{textbf{(C)} frac{1}{2}}$
To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. Before we do that, we also have to figure out some relative areas. You could either find the absolute areas (as is done below for completeness), or apply some proportional reasoning and symmetry to realize the relative areas (which is really all that matters).To find the areas of the sections, notice that the three smaller sections trisect a circle with radius? $3$ . The area of this entire circle is? $9pi$ . The area of each smaller section then must be? $frac{9pi}{3}$ ?or? $3pi$ . The larger sections trisect a "ring" which is the difference of two circles, one with radius? $3$ , the other radius? $6$ . So, the area of the ring (annulus) is? $36pi - 9pi$ ?or? $27pi$ . The area of each larger section must be? $frac{27pi}{3}$ ?or? $9pi$ . Note that the area of the whole circle is? $36pi$ .One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is? $3pi + (2 cdot 9pi) = 21pi$ . Since the area of the whole circle is? $36pi$ , the probability of getting an odd is? $frac{21pi}{36pi} = frac{21}{36} = frac{7}{12}$ .Since the remaining sections contain even numbers (that is, 2), the probability of throwing an even is the complement, or? $1 - frac{7}{12} = frac{5}{12}$ .Now, the two cases: You could either get an odd then an even, or an even then an odd.
Case 1: Odd then even Multiply the probabilities to get? $frac{7}{12} cdot frac{5}{12} = frac{35}{144}$ .

Case 2: Even then odd Multiply the probabilities to get? $frac{5}{12} cdot frac{7}{12} = frac{35}{144}$ . Notice that this is the same.

Thus, the total probability of an odd sum is? $frac{35}{144} cdot frac{2}{1}$ ?=? $boxed{B = frac{35}{72}}$ .