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2003AMC10B真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月5日下午12:30

2003AMC10B真题

真题及解析

Problem 1

Which of the following is the same as

$[frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?]$

$textbf{(A)} -1 qquadtextbf{(B)} -frac{2}{3} qquadtextbf{(C)} frac{2}{3} qquadtextbf{(D)} 1 qquadtextbf{(E)} frac{14}{3}$

Problem 2

Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $$$$ $1$ more than a pink pill, and Al's pills cost a total of $$$$ $546$ for the two weeks. How much does one green pill cost?

$textbf{(A) }$ $7 qquadtextbf{(B) }$ $14 qquadtextbf{(C) }$ $19qquadtextbf{(D) }$ $20qquadtextbf{(E) }$ $39$

Problem 3

The sum of $5$ consecutive even integers is $4$ less than the sum of the first $8$ consecutive odd counting numbers. What is the smallest of the even integers?

$textbf{(A) } 6 qquadtextbf{(B) } 8 qquadtextbf{(C) } 10 qquadtextbf{(D) } 12 qquadtextbf{(E) } 14$

Problem 4

Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $$$$ $1$ each, begonias $$$$ $1.50$ each, cannas $$$$ $2$ each, dahlias $$$$ $2.50$ each, and Easter lilies $$$$ $3$ each. What is the least possible cost, in dollars, for her garden?

$[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((6,0)--(0,0)--(0,1)--(6,1)); draw((0,1)--(0,6)--(4,6)--(4,1)); draw((4,6)--(11,6)--(11,3)--(4,3)); draw((11,3)--(11,0)--(6,0)--(6,3)); label("1",(0,0.5),W); label("5",(0,3.5),W); label("3",(11,1.5),E); label("3",(11,4.5),E); label("4",(2,6),N); label("7",(7.5,6),N); label("6",(3,0),S); label("5",(8.5,0),S);[/asy]$

$textbf{(A) } 108 qquadtextbf{(B) } 115 qquadtextbf{(C) } 132 qquadtextbf{(D) } 144 qquadtextbf{(E) } 156$

Problem 5

Moe uses a mower to cut his rectangular $90$ -foot by $150$ -foot lawn. The swath he cuts is $28$ inches wide, but he overlaps each cut by $4$ inches to make sure that no grass is missed. He walks at the rate of $5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn?

$textbf{(A) } 0.75 qquadtextbf{(B) } 0.8 qquadtextbf{(C) } 1.35 qquadtextbf{(D) } 1.5 qquadtextbf{(E) } 3$

Problem 6

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4 : 3$ . The horizontal length of a " $27$ -inch" television screen is closest, in inches, to which of the following?

$[asy] import math; unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)--(4,3)); fill((0,0)--(4,0)--(4,3)--cycle,mediumgray); label(rotate(aTan(3.0/4.0))*"Diagonal",(2,1.5),NW); label(rotate(90)*"Height",(4,1.5),E); label("Length",(2,0),S);[/asy]$ $textbf{(A) } 20 qquadtextbf{(B) } 20.5 qquadtextbf{(C) } 21 qquadtextbf{(D) } 21.5 qquadtextbf{(E) } 22$

Problem 7

The symbolism $lfloor x rfloor$ denotes the largest integer not exceeding $x$ . For example, $lfloor 3 rfloor = 3,$ and $lfloor 9/2 rfloor = 4$ . Compute $[lfloor sqrt{1} rfloor + lfloor sqrt{2} rfloor + lfloor sqrt{3} rfloor + cdots + lfloor sqrt{16} rfloor.]$

$textbf{(A) } 35 qquadtextbf{(B) } 38 qquadtextbf{(C) } 40 qquadtextbf{(D) } 42 qquadtextbf{(E) } 136$

Problem 8

The second and fourth terms of a geometric sequence are $2$ and $6$ . Which of the following is a possible first term?

$textbf{(A) } -sqrt{3} qquadtextbf{(B) } -frac{2sqrt{3}}{3} qquadtextbf{(C) } -frac{sqrt{3}}{3} qquadtextbf{(D) } sqrt{3} qquadtextbf{(E) } 3$

Problem 9

Find the value of $x$ that satisfies the equation $[25^{-2} = frac{5^{48/x}}{5^{26/x} cdot 25^{17/x}}.]$

$textbf{(A) } 2 qquadtextbf{(B) } 3 qquadtextbf{(C) } 5 qquadtextbf{(D) } 6 qquadtextbf{(E) } 9$

Problem 10

Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of the three letters followed by three digits. By how many times is the number of possible license plates increased?

$textbf{(A) } frac{26}{10} qquadtextbf{(B) } frac{26^2}{10^2} qquadtextbf{(C) } frac{26^2}{10} qquadtextbf{(D) } frac{26^3}{10^3} qquadtextbf{(E) } frac{26^3}{10^2}$

Problem 11

A line with slope $3$ intersects a line with slope $5$ at point $(10,15)$ . What is the distance between the $x$ -intercepts of these two lines?

$textbf{(A) } 2 qquadtextbf{(B) } 5 qquadtextbf{(C) } 7 qquadtextbf{(D) } 12 qquadtextbf{(E) } 20$

Problem 12

Al, Betty, and Clare split $$$$ $1000$ among them to be invested in different ways. Each begins with a different amount. At the end of one year, they have a total of $$$$ $1500$ . Betty and Clare have both doubled their money, whereas Al has managed to lose $$$$ $100$ . What was Al's original portion?

$textbf{(A) }$ $250 qquadtextbf{(B) }$ $350 qquadtextbf{(C) }$ $400qquadtextbf{(D) }$ $450qquadtextbf{(E) }$ $500$

Problem 13

Let $clubsuit(x)$ denote the sum of the digits of the positive integer $x$ . For example, $clubsuit(8)=8$ and $clubsuit(123)=1+2+3=6$ . For how many two-digit values of $x$ is $clubsuit(clubsuit(x))=3$ ?

$textbf{(A) } 3 qquadtextbf{(B) } 4 qquadtextbf{(C) } 6 qquadtextbf{(D) } 9 qquadtextbf{(E) } 10$

Problem 14

Given that $3^8cdot5^2=a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$ .

$textbf{(A) } 25 qquadtextbf{(B) } 34 qquadtextbf{(C) } 351 qquadtextbf{(D) } 407 qquadtextbf{(E) } 900$

Problem 15

There are $100$ players in a single tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest $28$ players are given a bye, and the remaining $72$ players are paired off to play. After each round, the remaining players play in the next round. The tournament continues until only one player remains unbeaten. The total number of matches played is

$textbf{(A) } text{a prime number} qquadtextbf{(B) } text{divisible by 2} qquadtextbf{(C) } text{divisible by 5} qquadtextbf{(D) } text{divisible by 7} qquadtextbf{(E) } text{divisible by 11}$

Problem 16

A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year $2003$ ?

$textbf{(A) } 4 qquadtextbf{(B) } 5 qquadtextbf{(C) } 6 qquadtextbf{(D) } 7 qquadtextbf{(E) } 8$

Problem 17

An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies $75%$ of the volume of the frozen ice cream. What is the ratio of the cone's height to its radius? (Note: a cone with radius $r$ and height $h$ has volume $pi r^2 h / 3$ and a sphere with radius $r$ has volume $4 pi r^3 / 3$ .)

$textbf{(A) } 2:1 qquadtextbf{(B) } 3:1 qquadtextbf{(C) } 4:1 qquadtextbf{(D) } 16:3 qquadtextbf{(E) } 6:1$

Problem 18

What is the largest integer that is a divisor of $[(n+1)(n+3)(n+5)(n+7)(n+9)]$ for all positive even integers $n$ ?

$textbf{(A) } 3 qquadtextbf{(B) } 5 qquadtextbf{(C) } 11 qquadtextbf{(D) } 15 qquadtextbf{(E) } 165$

Problem 19

Three semicircles of radius $1$ are constructed on diameter $overline{AB}$ of a semicircle of radius $2$ . The centers of the small semicircles divide $overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

$[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("$A$",A,W); label("$B$",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S);[/asy]$

$textbf{(A) } pi - sqrt{3} qquadtextbf{(B) } pi - sqrt{2} qquadtextbf{(C) } frac{pi + sqrt{2}}{2} qquadtextbf{(D) } frac{pi +sqrt{3}}{2} qquadtextbf{(E) } frac{7}{6}pi - frac{sqrt{3}}{2}$

Problem 20

In rectangle $ABCD, AB=5$ and $BC=3$ . Points $F$ and $G$ are on $overline{CD}$ so that $DF=1$ and $GC=2$ . Lines $AF$ and $BG$ intersect at $E$ . Find the area of $triangle AEB$ .

$[asy] unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair A=(0,0), B=(5,0), C=(5,3), D=(0,3), F=(1,3), G=(3,3); pair E=extension(A,F,B,G); draw(A--B--C--D--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,SE); label("$G$",G,SW); label("$B$",B,SE); label("1",midpoint(D--F),N); label("2",midpoint(G--C),N); label("3",midpoint(B--C),E); label("3",midpoint(A--D),W); label("5",midpoint(A--B),S); [/asy]$ $textbf{(A) } 10 qquadtextbf{(B) } frac{21}{2} qquadtextbf{(C) } 12 qquadtextbf{(D) } frac{25}{2} qquadtextbf{(E) } 15$

Problem 21

A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?

$textbf{(A) } frac{1}{8} qquadtextbf{(B) } frac{5}{32} qquadtextbf{(C) } frac{9}{32} qquadtextbf{(D) } frac{3}{8} qquadtextbf{(E) } frac{7}{16}$

Problem 22

A clock chimes once at $30$ minutes past each hour and chimes on the hour according to the hour. For example, at $1 text{PM}$ there is one chime and at noon and midnight there are twelve chimes. Starting at $11:15 text{AM}$ on $text{February 26, 2003},$ on what date will the $2003text{rd}$ chime occur?

$textbf{(A) } text{March 8} qquadtextbf{(B) } text{March 9} qquadtextbf{(C) } text{March 10} qquadtextbf{(D) } text{March 20} qquadtextbf{(E) } text{March 21}$

Problem 23

A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$ ?

$[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); [/asy]$ $textbf{(A) } 1 - frac{sqrt{2}}{2} qquadtextbf{(B) } frac{sqrt{2}}{4} qquadtextbf{(C) } sqrt{2} - 1 qquadtextbf{(D) } frac{1}{2} qquadtextbf{(E) } frac{1+sqrt{2}}{4}$

Problem 24

The first four terms in an arithmetic sequence are $x + y, x - y, xy,$ and $x/y,$ in that order. What is the ﬁfth term?

$textbf{(A)} -frac{15}{8}qquadtextbf{(B)} -frac{6}{5}qquadtextbf{(C)} 0qquadtextbf{(D)} frac{27}{20}qquadtextbf{(E)} frac{123}{40}$

Problem 25

How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?

$textbf{(A) } 27 qquadtextbf{(B) } 30 qquadtextbf{(C) } 33 qquadtextbf{(D) } 81 qquadtextbf{(E) } 90$

2003AMC10B详细解析

The numbers in the numerator and denominator can be grouped like this: $begin{align*} 2+(-4+6)+(-8+10)+(-12+14)&=2*4\ 3+(-6+9)+(-12+15)+(-18+21)&=3*4\ frac{2*4}{3*4}&=frac{2}{3} Rightarrow text {(C)} end{align*}$ Alternatively, notice that each term in the numerator is $frac{2}{3}$ of a term in the denominator, so the quotient has to be $frac{2}{3}$ .
Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $boxed{textbf{(D) }textdollar 20}$ .
It is a well-known fact that the sum of the first $n$ odd numbers is $n^2$ . This makes the sum of the first $8$ odd numbers equal to $64$ .Let $n$ be equal to the smallest of the $5$ even integers. Then $(n+2)$ is the next highest, $(n+4)$ even higher, and so on.This sets up the equation $n+(n+2)+(n+4)+(n+6)+(n+8)+4=64$ Now we solve: $[5n+24=64]$ $[5n=40]$ $[n=8]$ Thus, the smallest integer is $boxed{textbf{(B) } 8}$ .
The areas of the five regions from greatest to least are $21,20,15,6$ and $4$ .If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1cdot21+1.50cdot20+2cdot15+2.50cdot6+3cdot4$ , which simplifies to $$$$ $108$ . Therefore the answer is $boxed{textbf{(A) } 108}$ .
Since the swath Moe actually mows is $24$ inches, or $2$ feet wide, he mows $10000$ square feet in one hour. His lawn has an area of $13500$ , so it will take Moe $1.35$ hours to finish mowing the lawn. Thus the answer is $boxed{textbf{(C) } 1.35}$ .
If you divide the television screen into two right triangles, the legs are in the ratio of $4 : 3$ , and we can let one leg be $4x$ and the other be $3x$ . Then we can use the Pythagorean Theorem. $begin{align*}(4x)^2+(3x)^2&=27^2\ 16x^2+9x^2&=729\ 25x^2&=729\ x^2&=frac{729}{25}\ x&=frac{27}{5}\ x&=5.4end{align*}$ The horizontal length is $5.4times4=21.6$ , which is closest to $boxed{textbf{(D) } 21.5}$ .One can realize that the diagonal, vertical, and horizontal lengths all make up a $3,4,5$ triangle. Therefore, the horizontal length, being the $4$ in the $4 : 3$ ratio, is simply $frac{4}{5}$ times the hypotenuse. $frac{4}{5}cdot27=21.6 approx boxed{textbf{(D) } 21.5}$ .
The first three values in the sum are equal to $1,$ the next five equal to $2,$ the next seven equal to $3,$ and the last one equal to $4.$ For example, since $2^2=4$ any square root of a number less than $4$ must be less than $2.$ Sum them all together to get $[3cdot1 + 5cdot2 + 7cdot3 + 1cdot4 = 3+10+21+4 = boxed{textbf{(B) } 38}]$
Let the first term be $a$ and the common difference be $r$ . Therefore, $[ar=2 (1) qquad text{and} qquad ar^3=6 (2)]$ Dividing $(2)$ by $(1)$ eliminates the $a$ , yielding $r^2=3$ , so $r=pmsqrt{3}$ .Now, since $ar=2$ , $a=frac{2}{r}$ , so $a=frac{2}{pmsqrt{3}}=pmfrac{2sqrt{3}}{3}$ .We therefore see that $boxed{textbf{(B)} -frac{2sqrt{3}}{3}}$ is a possible first term.
Manipulate the powers of $5$ in order to get a clean expression. $[frac{5^{frac{48}{x}}}{5^{frac{26}{x}} cdot 25^{frac{17}{x}}} = frac{5^{frac{48}{x}}}{5^{frac{26}{x}} cdot 5^{frac{34}{x}}} = 5^{frac{48}{x}-(frac{26}{x}+frac{34}{x})} = 5^{-frac{12}{x}}]$ $[25^{-2} = (5^2)^{-2} = 5^{-4}]$ $[5^{-4} = 5^{-frac{12}{x}}]$ If two numbers are equal, and their bases are equal, then their exponents are equal as well. Set the two exponents equal to each other. $begin{align*}-4&=frac{-12}{x}\ -4x&=-12\ x&=boxed{textbf{(B) } 3}end{align*}$
There are $26$ letters and $10$ digits. There were $26 cdot 10^4$ old license plates. There are $26^3 cdot 10^3$ new license plates. The number of license plates increased by $[frac{26^3 cdot 10^3}{26 cdot 10^4} = boxed{textbf{(C) } frac{26^2}{10}}]$
Using the point-slope form, the equation of each line is $[y-15=3(x-10) longrightarrow y=3x-15]$ $[y-15=5(x-10) longrightarrow y=5x-35]$ Substitute in $y=0$ to find the $x$ -intercepts. $[0=3x-15longrightarrow x=5]$ $[0=5x-35longrightarrow x=7]$ The difference between them is $7-5=boxed{textbf{(A) } 2}$ .The $x$ -intercepts of a line is where the line's $y$ -coordinate is $=0$ . The slope is defined as $frac{text{change in } ytext{-coordinate}}{text{change in } xtext{-coordinate}}$ . Therefore, the line with slope $3$ can be expressed as $[frac{text{change in }y{-coordinate}}{text{change in }xtext{-coordinate}}=3]$ From $(10,15)$ to the $x$ -intercept, the line must move $-15$ units to the $x$ -axis. Therefore, $[frac{text{change in }y{-coordinate}}{text{change in }xtext{-coordinate}}=frac{-15}{x_1}=3implies x_1=-5]$ .Therefore, the $x$ -intercept of the line with slope $3$ is $(10-5,0)=(5,0)$ .Note that some sources may state "the $x$ -intercept is $5$ " instead of "the $x$ -intercept is $(5,0)$ .
If an idea functions great for one part of the problem but does not find the solution to the problem, we want to use the idea again. Using the idea for the line with slope $5$ , we find $[frac{text{change in }y{-coordinate}}{text{change in }xtext{-coordinate}}=frac{-15}{x_2}=5implies x_2=-3]$ . Therefore, the $x$ -intercept of the line with slope $5$ is $(10-3,0)=(7,0)$ . The distance between $boxed{(5,0)--(7,0)=2implies A}$ .
For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let $a, b,$ and $c$ represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have $a-100, 2b,$ and $2c$ . From this, we can write two equations, marked by (1) and (2). $[a+b+c=1000]$ $[(1). text{ }2a+2b+2c=2000]$ $[a-100+2b+2c=1500]$ $[(2). text{ }a+2b+2c=1600]$ (Equations (1) and (2) are derived from each equation above them.)Since all we need to find is $a,$ subtract the second equation from the first equation to get $a=400.$ Al's original portion was $boxed{textbf{(C)} textdollar 400}$ .
Let $a$ and $b$ be the digits of $clubsuit(x)$ , $[clubsuit(clubsuit(x)) = a + b = 3]$ Clearly $clubsuit(x)$ can only be $3, 12, 21,$ or $30$ and only $3$ and $12$ are possible to have two digits sum to.If $clubsuit(x)$ sums to $3$ , there are 3 different solutions : $12, 21, text{or } 30$ If $clubsuit(x)$ sums to $12$ , there are 7 different solutions: $39, 48, 57, 66,75, 84, text{or } 93$ The total number of solutions is $3 + 7 =10 Rightarrow text (E)$
$[3^8cdot5^2 = (3^4)^2cdot5^2 = (3^4cdot5)^2 = 405^2]$ $405$ is not a perfect power, so the smallest possible value of $a+b$ is $405+2=boxed{textbf{(D)} 407}$ .
Notice that $99$ players need to be eliminated for there to be declared a winner. Notice also that every match eliminates exactly one person. Therefore, $99$ matches are needed to eliminate $99$ people and therefore declare a winner. The rest of the information is irrelevant. Therefore, the total number of matches is $text{divisible by 11}, boxed{text{E}}$ .28 people receive byes, so in the first round there are $36$ matches played. In the second round there are $28 + 36 = 64$ people, so there are 32 matches. In the subsequent rounds, there are $16, 8, 4, 2, 1$ matches played, for a total of $36 + 32 + 16 + 8 + 4 + 2 + 1 = 99$ matches. Divisible by 11 $Rightarrow$ $boxed{(E)}$ .
Let $m$ be the number of main courses the restaurant serves, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2mtimes mtimes3=6m^2$ . Since the customer wants to eat a different dinner in all $365$ days of $2003$ , we must have $begin{align*} 6m^2 &geq 365\ m^2 &geq 60.83ldots.end{align*}$ The smallest integer value that satisfies this is $boxed{textbf{(E)} 8}$ .Let $m$ denote the number of main courses needed to meet the requirement. Then the number of dinners available is $3cdot m cdot 2m = 6m^2$ . Thus $m^2$ must be at least $365/6 approx 61$ . Since $7^2 = 49<61<64 = 8^2$ , $boxed{8}$ main courses is enough, but 7 is not. The smallest integer value that satisfies this is $boxed{textbf{(E)} 8}$ .
Let $r$ be the radius of both the cone and the sphere of ice cream, and let $h$ be the height of the cone. The cone can hold a maximum of $pi r^2h/3$ melted ice cream. The volume of the frozen ice cream is $4pi r^3/3.$ The volume of the melted ice cream is $75 %$ of that, which is $pi r^3.$ Since the melted ice cream fills the cone exactly, $begin{align*}pi r^3&=pi r^2h/3\ r&=h/3\ 3&=h/rend{align*}$ Then the ratio of the cone's height to its radius is $h:r=boxed{textbf{(B)} 3:1}$
For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 cdot 5 = 15$ , so ${boxed{textbf{(D)15}}}$ is the correct answer.
$[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("$A$",A,W); label("$B$",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S); draw((1,0)--(0.5,0.866)); draw((0,0)--(0.5,0.866)); draw((-1,0)--(-0.5,0.866)); draw((0,0)--(-0.5,0.866));[/asy]$ By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$ . This gives the area of the white region as $frac{5}{6}pi+frac{2cdotsqrt3}{4}=frac{5}{6}pi+frac{sqrt3}{2}$ . The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2pi-left(frac{5}{6}pi+frac{sqrt3}{2}right)=frac{7}{6}pi-frac{sqrt3}{2}$ .Thus the answer is $boxed{textbf{(E)} frac{7}{6}pi-frac{sqrt3}{2}}$ .
$triangle EFG sim triangle EAB$ because $FG parallel AB.$ The ratio of $triangle EFG$ to $triangle EAB$ is $2:5$ since $AB=5$ and $FG=2$ from subtraction. If we let $h$ be the height of $triangle EAB,$ $[frac{2}{5} = frac{h-3}{h}]$ $[2h = 5h-15]$ $[3h = 15]$ $[h = 5]$ The height is $5$ so the area of $triangle EAB$ is $frac{1}{2}(5)(5) = boxed{textbf{(D)} frac{25}{2}}$ .We can look at this diagram as if it were a coordinate plane with point $A$ being $(0,0)$ . This means that the equation of the line $AE$ is $y=3x$ and the equation of the line $EB$ is $y=frac{-3}{2}x+frac{15}{2}$ . From this we can set of the follow equation to find the $x$ coordinate of point $E$ : $[3x=frac{-3}{2}x+frac{15}{2}]$ $[6x=-3x+15]$ $[9x=15]$ $[x=frac{5}{3}]$ We can plug this into one of our original equations to find that the $y$ coordinate is $5$ , meaning the area of $triangle EAB$ is $frac{1}{2}(5)(5) = boxed{textbf{(D)} frac{25}{2}}$ At points $A$ and $B$ , segment $AE$ is 5 units from segment $BE$ . At points $F$ and $G$ , the segments are 2 units from each other. This means that collectively, the two lines closed the distance between them by 3 units over a height of 3 units. Therefore, to close the next two units of distance, they will have to travel a height of 2 units.Then calculate the area of trapezoid $FGBA$ and triangle $EGF$ separately and add them. The area of the trapezoid is $frac {2+5}{2}cdot 3 = frac {21}{2}$ and the area of the triangle is $frac{1}{2}cdot 2 cdot 2 = 2$ . $frac{21}{2}+2=boxed{textbf{(D)} frac{25}{2}}$
We can divide the case of all beads in the bag being red after three replacements into three cases.The first case is in which the two green beads are the first two beads to be chosen. The probability for this is $[frac{2}{4} times frac{1}{4} times frac{4}{4} = frac{1}{8}]$ The second case is in which the green beads are chosen first and third. The probability for this is $[frac{2}{4} times frac{3}{4} times frac{1}{4} = frac{3}{32}]$ The third case is in which the green beads are chosen second and third. The probability for this is $[frac{2}{4} times frac{2}{4} times frac{1}{4} = frac{1}{16}]$ Add all these cases together $[frac{1}{8}+frac{3}{32}+frac{1}{16} = frac{4}{32}+frac{3}{32}+frac{2}{32} = boxed{textbf{(C) } frac{9}{32}}]$
First, find how many chimes will have already happened before midnight (the beginning of the day) of $text{February 27, 2003}.$ $13$ half-hours have passed, and the number of chimes according to the hour is $1+2+3+cdots+12.$ The total number of chimes is $13+78=91$ Every day, there will be $24$ half-hours and $2(1+2+3+cdots+12)$ chimes according to the arrow, resulting in $24+156=180$ total chimes.On $text{February 26},$ the number of chimes that still need to occur is $2003-91=1912.$ $1912 div 180=10 text{R}112.$ Rounding up, it is $11$ days past $text{February 26},$ which is $boxed{textbf{(B) } text{March 9}}$
Here is an easy way to look at this, where $p$ is the perimeter, and $a$ is the apothem:Area of Octagon: $frac{ap}{2}=1$ .Area of Rectangle: $frac{p}{8}times 2a=frac{ap}{4}$ .You can see from this that the octagon's area is twice as large as the rectangle's area is $boxed{textbf{(D)} frac{1}{2}}$ . $[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--E--F--B--C--G--H--D); draw(A--E--F--B--A,blue); draw(A--F--E--B--A,red); [/asy]$ Here is a less complicated way than that of the user above. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF( In red), you can see that two of the triangles (In blue) share the same base and height with half the rectangle. Therefore, the rectangle's area is the same as 4 of the 8 triangles, and is $boxed{textbf{(D)} frac{1}{2}}$ the area of the octagon. $[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--D--E--H--G--B--C--F); [/asy]$ Drawing lines $AD$ , $BG$ , $CF$ , and $EH$ , we can see that the octagon is comprised of $1$ square, $4$ rectangles, and $4$ triangles. The triangles each are $45-45-90$ triangles, and since their diagonal is length $x$ , each of their sides is $frac{sqrt{2}}{2}x$ . The area of the entire figure is, likewise, $x^2$ (the square) $+4x^2frac{sqrt{2}}{2}$ (the 4 rectangles) $+2cdot(frac{sqrt{2}}{2}x)^2$ (the triangles), which simplifies to $2x^2 + 2sqrt{2}x^2$ . The area of $ABEF$ is just $x(x+frac{2sqrt{2}}{2}x)$ , or $x^2$ + $x^2sqrt{2}$ , which we can see is the area of $frac{ABCDEFGH}{2} = boxed{textbf{(D)} frac{1}{2}}$ the area of the octagon.
The difference between consecutive terms is $(x-y)-(x+y)=-2y.$ Therefore we can also express the third and fourth terms as $x-3y$ and $x-5y.$ Then we can set them equal to $xy$ and $frac{x}{y}$ because they are the same thing. $begin{align*} xy&=x-3y\ xy-x&=-3y\ x(y-1)&=-3y\ x&=frac{-3y}{y-1} end{align*}$ Substitute into our other equation. $[frac{x}{y}=x-5y]$ $[frac{-3}{y-1}=frac{-3y}{y-1}-5y]$ $[-3=-3y-5y(y-1)]$ $[0=5y^2-2y-3]$ $[0=(5y+3)(y-1)]$ $[y=-frac35, 1]$ But $y$ cannot be $1$ because then every term would be equal to $x.$ Therefore $y=-frac35.$ Substituting the value for $y$ into any of the equations, we get $x=-frac98.$ Finally, $[frac{x}{y}-2y=frac{9cdot 5}{8cdot 3}+frac{6}{5}=boxed{textbf{(E)} frac{123}{40}}]$
To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are $23$ , the sum of the digits is $2+3 = 5$ (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three. $5+1 = 6$ , $5+4 = 9$ , and so on.However since the largest four-digit number ending with $23$ is $9923$ , the maximum sum is $5+18 = 23$ .Using that process we can fairly quickly compile a list of the sum of the first two digits of the number. $[{1, 4, 7, 10, 13, 16}]$ Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers $xy$ in separate cases.
$[I. x+y = 1, {10} = 1]$ $[II. x+y = 4, {13, 22, 31, 40} = 4]$ $[III. x+y = 7, {16, 25, 34, 43, 52, 61, 70} = 7]$ $[IV. x+y = 10, {19, 28, 37, 46, 55, 64, 73, 82, 91} = 9]$ $[V. x+y = 13, {49, 58, 67, 76, 85, 94} = 6]$ $[VI. x+y = 16, {79, 88, 97} = 3]$

And finally, we add the number of elements in each set.

$[1+4+7+9+6+3 = boxed{textbf{(B)} 30}]$

A number divisible by $3$ has all its digits add to a multiple of $3.$ The last two digits are $2$ and $3$ and add up to $5 equiv 2 (text{mod} 3).$ Therefore the first two digits must add up to $1 (text{mod} 3).$ $4$ digits (including $0$ ) are $0 (text{mod} 3),$ $3$ are $1 (text{mod} 3),$ and $3$ are $2 (text{mod} 3).$ The following combinations are equivalent to $1 (text{mod} 3)$ :

$[0 (text{mod} 3)+1 (text{mod} 3) equiv 1 (text{mod} 3)+0 (text{mod} 3) equiv 2 (text{mod} 3) +2 (text{mod} 3)]$

Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.

$[3cdot3 + 3cdot4 + 3cdot3 = 9 + 12 + 9 = boxed{textbf{(B)} 30}]$

We have the following: $n equiv 0 pmod{3}$ and $n equiv 23pmod{100}$ . Then $n = 3a = 100b+23$ for some integers $a$ and $b$ . Taking mod 3 gives: $0 equiv b+2 pmod 3 implies bequiv 1pmod 3$ so $b=3c+1$ for some integer $c$ . But $N = 100b+23 = 100(3c+1)+23$ so $n = 300c+123$ . Bounding this gives us: $999 < 300c+123 < 10000$ so $876 < 300c < 9877$ . Dividing by $300$ gives $2.92 < c < 32.9222$ so $3le c le 32$ . This gives $32-3+1 = boxed{textbf{(B)} 30 }$

The number is in the form $xy23$

Notice that the number is divisible by $3$ if the sum of the digits of $xy23$ is divisible by $3$

Our first number that is divisible by $3$ is $1023$ , next is $1323$ .. Notice $xy$ goes from $10implies 97$

Hence, there are $frac{97-10}{3}+1=30$ distinct four digit numbers.

Following the form $xy23$ in Solution 4, notice that $x+y equiv 1 pmod 3$ to satisfy our condition. Choose a value of $x$ from 1 to 9. For $x=1, 4, 7$ , there are exactly 4 values of $y: 0, 3, 6, 9$ . For the remaining 6 digits, there are 3 choices for y. So our answer is $(3)(4)+(6)(3)=30$ .

There are $9 cdot 10 = 90$ possible values for the first two digits. One-third of them yield a multiple of $3$ , so the answer is $frac{90}{3} = boxed{textbf{(B)} 30}$