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2008AMC10B真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月4日下午5:16

2008AMC10B真题

答案解析请参考文末

Problem 1

A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?

$mathrm{(A)} 2qquadmathrm{(B)} 3qquadmathrm{(C)} 4qquadmathrm{(D)} 5qquadmathrm{(E)} 6$

Problem 2

A $4times 4$ block of calendar dates has the numbers $1$ through $4$ in the first row, $8$ though $11$ in the second, $15$ though $18$ in the third, and $22$ through $25$ in the fourth. The order of the numbers in the second and the fourth rows are reversed. The numbers on each diagonal are added. What will be the positive difference between the diagonal sums?

$mathrm{(A)} 2qquadmathrm{(B)} 4qquadmathrm{(C)} 6qquadmathrm{(D)} 8qquadmathrm{(E)} 10$

Problem 3

Assume that $x$ is a positive real number. Which is equivalent to $sqrt[3]{xsqrt{x}}$ ?

$mathrm{(A)} x^{1/6}qquadmathrm{(B)} x^{1/4}qquadmathrm{(C)} x^{3/8}qquadmathrm{(D)} x^{1/2}qquadmathrm{(E)} x$

Problem 4

A semipro baseball league has teams with $21$ players each. League rules state that a player must be paid at least $textdollar 15,000$ and that the total of all players' salaries for each team cannot exceed $textdollar 700,000.$ What is the maximum possible salary, in dollars, for a single player?

$mathrm{(A)} 270,000qquadmathrm{(B)} 385,000qquadmathrm{(C)} 400,000qquadmathrm{(D)} 430,000qquadmathrm{(E)} 700,000$

Problem 5

For real numbers $a$ and $b$ , define $a textdollar b$ $=(a-b)^2$ . What is $(x-y)^2textdollar(y-x)^2$ ?

$mathrm{(A)} 0qquadmathrm{(B)} x^2+y^2qquadmathrm{(C)} 2x^2qquadmathrm{(D)} 2y^2qquadmathrm{(E)} 4xy$

Problem 6

Points $B$ and $C$ lie on $AD$ . The length of $AB$ is $4$ times the length of $BD$ , and the length of $AC$ is $9$ times the length of $CD$ . The length of $BC$ is what fraction of the length of $AD$ ?

$mathrm{(A)} 1/36qquadmathrm{(B)} 1/13qquadmathrm{(C)} 1/10qquadmathrm{(D)} 5/36qquadmathrm{(E)} 1/5$

Problem 7

An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?

$mathrm{(A)} 10qquadmathrm{(B)} 25qquadmathrm{(C)} 100qquadmathrm{(D)} 250qquadmathrm{(E)} 1000$

Problem 8

A class collects $textdollar50$ to buy flowers for a classmate who is in the hospital. Roses cost $textdollar3$ each, and carnations cost $textdollar2$ each. No other flowers are to be used. How many different bouquets could be purchased for exactly $textdollar50$ ?

$mathrm{(A)} 1 qquad mathrm{(B)} 7 qquad mathrm{(C)} 9 qquad mathrm{(D)} 16 qquad mathrm{(E)} 17$

Problem 9

A quadratic equation $ax^2 - 2ax + b = 0$ has two real solutions. What is the average of these two solutions?

$mathrm{(A)} 1qquadmathrm{(B)} 2qquadmathrm{(C)} frac baqquadmathrm{(D)} frac{2b}aqquadmathrm{(E)} sqrt{2b-a}$

Problem 10

Points $A$ and $B$ are on a circle of radius $5$ and $AB=6$ . Point $C$ is the midpoint of the minor arc $AB$ . What is the length of the line segment $AC$ ?

$mathrm{(A)} sqrt{10}qquadmathrm{(B)} frac{7}{2}qquadmathrm{(C)} sqrt{14}qquadmathrm{(D)} sqrt{15}qquadmathrm{(E)} 4$

Problem 11

Suppose that $(u_n)$ is a sequence of real numbers satisfying $u_{n+2}=2u_{n+1}+u_n$ ,

and that $u_3=9$ and $u_6=128$ . What is $u_5$ ?

$mathrm{(A)} 40qquadmathrm{(B)} 53qquadmathrm{(C)} 68qquadmathrm{(D)} 88qquadmathrm{(E)} 104$

Problem 12

Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year?

$mathrm{(A)} 2500qquadmathrm{(B)} 3000qquadmathrm{(C)} 3500qquadmathrm{(D)} 4000qquadmathrm{(E)} 4500$

Problem 13

For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the 2008th term of the sequence?

$mathrm{(A)} {{{2008}}} qquad mathrm{(B)} {{{4015}}} qquad mathrm{(C)} {{{4016}}} qquad mathrm{(D)} {{{4,030,056}}} qquad mathrm{(E)} {{{4,032,064}}}$

Problem 14

Triangle $OAB$ has $O=(0,0)$ , $B=(5,0)$ , and $A$ in the first quadrant. In addition, $angle ABO=90^circ$ and $angle AOB=30^circ$ . Suppose that $OA$ is rotated $90^circ$ counterclockwise about $O$ . What are the coordinates of the image of $A$ ?

$mathrm{(A)} left( - frac {10}{3}sqrt {3},5right) qquad mathrm{(B)} left( - frac {5}{3}sqrt {3},5right) qquad mathrm{(C)} left(sqrt {3},5right) qquad mathrm{(D)} left(frac {5}{3}sqrt {3},5right) qquad mathrm{(E)} left(frac {10}{3}sqrt {3},5right)$

Problem 15

How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$ ?

$mathrm{(A)} 6qquadmathrm{(B)} 7qquadmathrm{(C)} 8qquadmathrm{(D)} 9qquadmathrm{(E)} 10$

Problem 16

Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is $0$ .)

$mathrm{(A)} {{{frac{3} {8}}}} qquad mathrm{(B)} {{{frac{1} {2}}}} qquad mathrm{(C)} {{{frac{43} {72}}}} qquad mathrm{(D)} {{{frac{5} {8}}}} qquad mathrm{(E)} {{{frac{2} {3}}}}$

Problem 17

A poll shows that $70%$ of all voters approve of the mayor's work. On three separate occasions a pollster selects a voter at random. What is the probability that on exactly one of these three occasions the voter approves of the mayor's work?

$mathrm{(A)} {{{0.063}}} qquad mathrm{(B)} {{{0.189}}} qquad mathrm{(C)} {{{0.233}}} qquad mathrm{(D)} {{{0.333}}} qquad mathrm{(E)} {{{0.441}}}$

Problem 18

Bricklayer Brenda would take nine hours to build a chimney alone, and Bricklayer Brandon would take $10$ hours to build it alone. When they work together, they talk a lot, and their combined output decreases by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?

$mathrm{(A)} 500qquadmathrm{(B)} 900qquadmathrm{(C)} 950qquadmathrm{(D)} 1000qquadmathrm{(E)} 1900$

Problem 19

A cylindrical tank with radius $4$ feet and height $9$ feet is lying on its side. The tank is filled with water to a depth of $2$ feet. What is the volume of water, in cubic feet?

$mathrm{(A)} 24pi - 36 sqrt {2} qquad mathrm{(B)} 24pi - 24 sqrt {3} qquad mathrm{(C)} 36pi - 36 sqrt {3} qquad mathrm{(D)} 36pi - 24 sqrt {2} qquad mathrm{(E)} 48pi - 36 sqrt {3}$

Problem 20

The faces of a cubical die are marked with the numbers $1$ , $2$ , $2$ , $3$ , $3$ , and $4$ . The faces of another die are marked with the numbers $1$ , $3$ , $4$ , $5$ , $6$ , and $8$ . What is the probability that the sum of the top two numbers will be $5$ , $7$ , or $9$ ?

$mathrm{(A)} 5/18qquadmathrm{(B)} 7/18qquadmathrm{(C)} 11/18qquadmathrm{(D)} 3/4qquadmathrm{(E)} 8/9$

Problem 21

Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$ . Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?

$mathrm{(A)} 240qquadmathrm{(B)} 360qquadmathrm{(C)} 480qquadmathrm{(D)} 540qquadmathrm{(E)} 720$

Problem 22

Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?

$mathrm{(A)} 1/12qquadmathrm{(B)} 1/10qquadmathrm{(C)} 1/6qquadmathrm{(D)} 1/3qquadmathrm{(E)} 1/2$

Problem 23

A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers with $b > a$ . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width 1 foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair $(a, b)$ ?

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

Problem 24

Quadrilateral $ABCD$ has $AB = BC = CD$ , angle $ABC = 70$ and angle $BCD = 170$ . What is the measure of angle $BAD$ ?

$mathrm{(A)} 75qquadmathrm{(B)} 80qquadmathrm{(C)} 85qquadmathrm{(D)} 90qquadmathrm{(E)} 95$

Problem 25

Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck travels at $10$ feet per second in the same direction as Michael and stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?

$mathrm{(A)} 4qquadmathrm{(B)} 5qquadmathrm{(C)} 6qquadmathrm{(D)} 7qquadmathrm{(E)} 8$

2008AMC10B详细解析

The number of points could have been 10, 11, 12, 13, 14, or 15. Thus, the answer is $boxed{mathrm{(E)} 6}$ .
After reversing the numbers on the second and fourth rows, the block will look like this: $begin{tabular}[t]{|c|c|c|c|} multicolumn{4}{c}{}\hline 1&2&3&4\hline 11&10&9&8\hline 15&16&17&18\hline 25&24&23&22\hline end{tabular}$ The difference between the two diagonal sums is: $(4+9+16+25)-(1+10+17+22)=3-1-1+3=4 Rightarrow B$ .
$sqrt[3]{xsqrt{x}}=sqrt[3]{sqrt{x^3}}=sqrt[6]{x^3}=x^{3/6}=x^{1/2} mathrm{(D)}$
The maximum occurs when 20 players get the minimum wage and the total of all players' salaries is 700000. That is when one player gets $700000-15000*20=400000Rightarrow boxed{mathrm{(C)}}$ .
Since $(-a)^2 = a^2$ , it follows that $(x-y)^2 = (y-x)^2$ , and $[(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0 mathrm{(A)}.]$
Let $CD = 1$ . Then $AB = 4(BC + 1)$ and $AB + BC = 9cdot1$ . From this system of equations we obtain $BC = 1$ . Adding $CD$ to both sides of the second equation, we obtain $AD = AB + BC + CD = 9 + 1 = 10$ . Thus, $frac{BC}{AD} = frac{1}{10} impliestext{(C)}$
(C) The area of the large triangle is $frac{10^2sqrt3}{4}$ , while the area each small triangle is $frac{1^2sqrt3}{4}$ . Dividing these two quantities, we get 100, therefore $boxed{100}$ small triangles can fit in the large one. $[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]$ The number of triangles is $1+3+dots+19 = boxed{100}$ .Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have: $19+17+15...+3+1 = frac{19+1}{2}cdot 10 = boxed{100}$ A fourth solution is to notice that the small triangles are similar to the large triangle as they are both equilateral. Therefore, the ratio of their areas is the square of the ratios of their side lengths. Hence the ratio of their areas is $(1/10)^2=1/100$ , so the answer is $boxed{100}$ .
The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $rgeq 0$ . Then we have $50-3cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r geq 0$ , and as $r$ must be an integer, this solves to $rleq 8$ . Hence there are $boxed{9->C}$ possible values of $r$ , and each gives us one solution.
Dividing both sides by $a$ , we get $x^2 - 2x + b/a = 0$ . By Vieta's formulas, the sum of the roots is $2$ , therefore their average is $1Rightarrow boxed{A}$ .We know that for an equation $ax^2 + bx + c = 0$ , the sum of the roots is $-b/a$ . This means that the sum of the roots for $ax^2 - 2ax + b = 0$ is $2a/a$ , or 2. The average is the sum of the two roots divided by two, so the average is $2/2 = 1$ .
Let the center of the circle be $O$ , and let $D$ be the intersection of $overline{AB}$ and $overline{OC}$ (then $D$ is the midpoint of $overline{AB}$ ). $OA=OB=5$ , since they are both radii.By the Pythagorean Theorem, $OD = sqrt{OA^2 - DA^2} = 4$ , and by subtraction, $CD=OC - OD = 1$ .Using the Pythagorean Theorem again, $AC= sqrt{AD^2 + CD^2} = sqrt{3^2+1^2}=sqrt{10} Longrightarrow textbf{(A)}$ . $[asy] pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9); path p = CR((0,0),5); pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C); D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D",D,W)),A)); MP("5",(A+O)/2); MP("3",(A+D)/2,SW); [/asy]$
Plugging in $n=4$ , we get $128=2u_5+u_4.$ Plugging in $n=3$ , we get $u_5=2u_4+9.$ This is simply a system of two equations with two unknowns. Substituting gives $128=5u_4+18 Longrightarrow u_4=22$ , and $u_5=frac{128-22}{2}=53 longleftarrow textbf{(B)}$ .
Every time the pedometer flips from $99999$ to $00000$ Pete has walked $100000$ steps.So, if the pedometer flipped $44$ timesPete walked $44*100000+50000=4450000$ steps.Dividing by $1800$ gives $2472.overline{2}$ This is closest to answer $boxed{A}$ .
Since the mean of the first $n$ terms is $n$ , the sum of the first $n$ terms is $n^2$ . Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$ . Hence, the 2008th term is $2008^2-2007^2=(2008+2007)(2008-2007)=4015Rightarrow boxed{text{(B)}}$ Note that $n^2$ is the sum of the first n odd numbers.From inspection, we see that the sum of the sequence is basically $n^2$ . We also notice that $n^2$ Is the sum of the first $n$ ODD integers. Because $4015$ is the only odd integer, $boxed{B}$ is the answer.
Since $angle ABO=90^circ$ , and $angle AOB=30^circ$ , we know that this triangle is one of the Special Right Triangles.We also know that $A$ is $(5,0)$ , so $A$ lies on the x-axis. Therefore, $OA = 5$ .Then, since we know that this is a Special Right Triangle(30-60-90 triangle), we can use the proportion $[frac{5}{sqrt 3}=frac{x}{1}]$ to find $AB$ .We find that $[AB=frac{5sqrt 3}{3}]$ That means that the coordinates of $A$ are $left(5,frac{5sqrt 3}3right)$ .Rotate this triangle $90^circ$ counterclockwise around $O$ , and you will find that $A$ will end up in the second quadrant with the coordinates $boxed{ left( -frac{5sqrt 3}3, 5right) text{or B.}}$ .As $angle ABO=90^circ$ and $A$ in the first quadrant, we know that the $x$ coordinate of $A$ is $5$ . We now need to pick a positive $y$ coordinate for $A$ so that we'll have $angle AOB=30^circ$ .By the Pythagorean theorem we have $AO^2 = AB^2 + BO^2 = AB^2 + 25$ .By the definition of sine, we have $frac{AB}{AO} = sin AOB = sin 30^circ = frac 12$ , hence $AO=2cdot AB$ .
Substituting into the previous equation, we get $AB^2 = frac{25}3$ , hence $AB=frac{5sqrt 3}3$ .

This means that the coordinates of $A$ are $left(5,frac{5sqrt 3}3right)$ .

After we rotate $A$ $90^circ$ counterclockwise about $O$ , it will get into the second quadrant and have the coordinates $boxed{ left( -frac{5sqrt 3}3, 5right) }$ . So the answer is $boxed{text{B}}$ .
By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$ This means that $a^2=2b+1$ .We know that $a,b>0$ , and that $b<100$ .We also know that a must be odd, since the rightside is odd.So $a=3,5,7,9,11,13$ , and the answer is $boxed{A}$ .
We consider 3 cases based on the outcome of the coin:Case 1, 0 heads: The probability of this occurring on the coin flip is $frac{1} {4}$ . The probability that 0 rolls of a die will result in an odd sum is $0$ .Case 2, 1 head: The probability of this case occurring is $frac{1} {2}$ . The probability that one die results as an odd number is $frac{1} {2}$ .Case 3, 2 heads: The probability of this occurring is $frac{1} {4}$ . The probability that 2 dice result in an odd sum is $frac{1} {2}$ , because regardless of what we throw on the first die, we have $frac{1} {2}$ probability that the second die will have the opposite parity.Thus, the probability of having an odd sum rolled is $frac{1} {4} cdot 0 + frac{1} {2} cdot frac{1} {2} + frac{1} {4} cdot frac{1} {2}=frac{3} {8}Rightarrow boxed{A}$
The pollster could select responses in 3 different ways: YNN, NYN, and NNY, where Y stands for a voter who approved of the work, and N stands for a person who didn't approve of the work. The probability of each of these is $(0.7)(0.3)^2=0.063.$ Thus, the answer is $3 cdot 0.063=0.189Rightarrow boxed{B}$
Let $x$ be the number of bricks in the chimney. Using $d=rt$ , we get $x = (x/9 + x/10 - 10)cdot(5)$ . Solving for $x$ , we get $boxed{900} Rightarrow B$ .
Any vertical cross-section of the tank parallel with its base looks as follows: $[asy] unitsize(0.8cm); defaultpen(0.8); pair s=(0,0), bottom=(0,-4), mid=(0,-2); pair x[]=intersectionpoints( (-10,-2)--(10,-2), circle(s,4) ); fill( arc(s,x[0],x[1]) -- cycle, lightgray ); draw( circle(s,4) ); dot(s); draw( s -- bottom ); label( "$2$", (mid+bottom)/2, E ); draw ( s -- x[0] -- x[1] -- s ); label( "$4$", (s+x[0])/2, NW ); label( "$4$", (s+x[1])/2, NE ); label( "$A$", s, N ); label( "$B$", x[0], W ); label( "$C$", x[1], E ); label( "$D$", mid, NW ); label( "$E$", bottom, S ); [/asy]$ The volume of water can be computed as the height of the tank times the area of the shaded part.Let $theta$ be the size of the smaller angle $DAC$ . We then have $costheta = frac{AD}{AC}=frac 12$ , hence $theta=60^circ$ .Thus the angle $CAB$ has size $360^circ - 2cdot 60^circ = 120^circ$ . Hence the non-shaded part consists of $frac{120^circ}{360^circ} = frac 13$ of the circle, minus the area of the triangle $ABC$ .Using the Pythagorean theorem we can compute that $CD=sqrt{AC^2-AD^2}=sqrt{16-4}=2sqrt 3$ . Thus $BC=4sqrt 3$ , and the area of the triangle $ABC$ is $frac {2 cdot 4sqrt 3} 2 = 4sqrt 3$ .The area of the shaded part is then $frac{4^2pi}3 - 4sqrt 3$ , and the volume of water is $9cdotleft(frac{4^2pi}3 - 4sqrt 3right) = boxed{48pi - 36sqrt 3}$ . The answer is $text{E}$ .
The easiest way is to write a table of all possible outcomes, do the sums, and count good outcomes.
```
     1  3  4  5  6  8
   ------------------
1 |  2  4  5  6  7  9
2 |  3  5  6  7  8 10
2 |  3  5  6  7  8 10
3 |  4  6  7  8  9 11
3 |  4  6  7  8  9 11
4 |  5  7  8  9 10 12
```
We see that out of $36$ possible outcomes $4$ give the sum of $5$ , $6$ the sum of $7$ , and $4$ the sum of $9$ , hence the resulting probability is $frac{4+6+4}{36}=frac{14}{36}=boxed{frac{7}{18}}$ .

Each die is equally likely to roll odd or even, so the probability of an odd sum is $frac{1}{2}$ .

So we can find the probability of rolling $3$ or $11$ instead and just subtract that from $frac{1}{2}$ , which seems easier. Without writing out a table, we can see that there are two ways to make $3$ , and two ways to make $11$ , for a probability of $frac{4}{36}$ .

$frac{1}{2}-frac{4}{36}=frac{14}{36}=boxed{frac{7}{18}}$ .
For the first man, there are $10$ possible seats. For each subsequent man, there are $4$ , $3$ , $2$ , and $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10cdot 4cdot 3cdot 2cdot 1cdot 2 = boxed{(text{C}) 480}$ .
There are two ways to arrange the red beads, where $R$ represents a red bead and $B$ represents a blank space. $1. R B R B R B$ $2. R B B R B R$ In the first, there are three ways to place a bead in the first free space, two for the second free space, and one for the third, so there are $6$ arrangements. In the second, a white bead must be placed in the third free space, so there are two possibilities for the third space, two for the second, and one for the first. That makes $4$ arrangements. There are $6+4=10$ arrangements in total. The two cases above can be reversed, so we double $10$ to $20$ arrangements. Also, in each case, there are three ways to place the first red bead, two for the second, and one for the third, so we multiply by $6$ to get $20cdot 6 = 120$ arrangements. There are $6! = 720$ total arrangements so the answer is $120/720 = boxed{1/6}$ .If you list the case for the red balls and blanks, these are four scenarios: $1. R - - R - R$ $2. R - R - - R$ $3. R - R - R -$ $4. - R - R - R$
In the cases one and two, the white balls must go in the blank surrounded on either side by the red balls. Like this: $1. R - - R W R$ $2. R W R - - R$ However, now with only one white ball and one blue ball left, there are two ways to order them in both cases. So for a total of $2 + 2 = 4$ possibilities.

In the last two cases, the two white balls and the blue ball can go anywhere, in those three blanks, becuase they are separated by a red ball. So there are $3$ ways for each case. This adds up to a total of $6$ possibilities for the last two cases.

Adding them up, we get $10$ total orderings.

There are $frac{6!}{3! * 2! * 1!} = 60$ total orderings.

So the answer is $frac{10}{60} = boxed{1/6}$ .

There are 3 cases:

Case 1

There is a block of RWRWR. The B can go in any slot between two letters. 6 ways.

Case 2

There is a RBR. Then, there is a WRW, and there are 2 orderings.

Case 3

There is a WBW. Then, there is a RWR, and there are 2 orderings.

There are 60 ways total, and 10 valid ones, giving $boxed{frac{1}{6}}$ as the answer.
Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$ . With this information we can make the equation: $begin{eqnarray*} ab &=& 2left((a-2)(b-2)right) \ ab &=& 2ab - 4a - 4b + 8 \ ab - 4a - 4b + 8 &=& 0 end{eqnarray*}$ Applying Simon's Favorite Factoring Trick, we get $begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \ (a-4)(b-4) &=& 8 end{eqnarray*}$ Since $b > a$ , then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$ , or $(a-4) = 2$ and $(b-4) = 4$ . This gives 2 possibilities: (5,12) or (6,8), So the answer is $boxed{textbf{(B) 2}}$
Draw the angle bisectors of the angles $ABC$ and $BCD$ . These two bisectors obviously intersect. Let their intersection be $P$ . We will now prove that $P$ lies on the segment $AD$ .Note that the triangles $ABP$ and $CBP$ are congruent, as they share the side $BP$ , and we have $AB=BC$ and $angle ABP = angle CBP$ .Also note that for similar reasons the triangles $CBP$ and $CDP$ are congruent.Now we can compute their inner angles. $BP$ is the bisector of the angle $ABC$ , hence $angle ABP = angle CBP = 35^circ$ , and thus also $angle CDP = 35^circ$ . (Faster Solution picks up here) $CP$ is the bisector of the angle $BCD$ , hence $angle BCP = angle DCP = 85^circ$ , and thus also $angle BAP = 85^circ$ .It follows that $angle APB = angle BPC = angle CPD = 180^circ - 35^circ - 85^circ = 60^circ$ . Thus the angle $APD$ has $180^circ$ , and hence $P$ does indeed lie on $AD$ . Then obviously $angle BAD = angle BAP = boxed{ 85^circ }$ . $[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$P$",P,W); label("$35^circ$",B + dir(180-17.5)); label("$35^circ$",B + dir(180-35-17.5)); label("$85^circ$",C + .5*dir(120+42.5)); label("$85^circ$",C + .5*dir(120+85+42.5)); [/asy]$ Faster Solution: Because we now know three angles, we can subtract to get $360 - 35 - 85 - 85 - 35 - 35$ , or $boxed{85}$ .Draw the diagonals $overline{BD}$ and $overline{AC}$ , and suppose that they intersect at $E$ . Then, $triangle ABC$ and $triangle BCD$ are both isosceles, so by angle-chasing, we find that $angle BAC = 55^{circ}$ , $angle CBD = 5^{circ}$ , and $angle BEA = 180 - angle EBA - angle BAE = 60^{circ}$ . Draw $E'$ such that $EE'B = 60^{circ}$ and so that $E'$ is on $overline{AE}$ , and draw $E''$ such that $angle EE''C = 60^{circ}$ and $E''$ is on $overline{DE}$ . It follows that $triangle BEE'$ and $triangle CEE''$ are both equilateral. Also, it is easy to see that $triangle BEC cong triangle DE''C$ and $triangle BCE cong triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$ . Thus, $AE = AE' + E'E = EE'' + DE'' = DE$ , so $triangle ADE$ is isosceles. Since $angle AED = 120^{circ}$ , then $angle DAC = frac{180 - 120}{2} = 30^{circ}$ , and $angle BAD = 30 + 55 = 85^{circ}$ . $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label("$A$",(-0.096,0.005),NE*lsf); dot((1,0),ds); label("$B$",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label("$C$",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label("$D$",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label("$E$",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label("$E'$",(0.1,0.23),NE*lsf); label("$60^circ$",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label("$E''$",(0.423,0.957),NE*lsf); label("$60^circ$",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Again, draw the diagonals $overline{BD}$ and $overline{AC}$ , and suppose that they intersect at $E$ . We find by angle chasing the same way as in solution 2 that $mangle ABE = 65^circ$ and $mangle DCE = 115^circ$ . Applying the Law of Sines to $triangle AEB$ and $triangle EDC$ , it follows that $DE = frac{2sin 115^circ}{sin angle DEC} = frac{2sin 65^circ}{sin angle AEB} = EA$ , so $triangle AED$ is isosceles. We finish as we did in solution 2.
$[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$E$",P,W); [/asy]$

Start off with the same diagram as solution 1. Now draw $overline{CA}$ which creates isosceles $triangle CAB$ . We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is $boxed{85}.$
Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$ . Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds. Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds. Meetings occur whenever $D(t)=0$ . We have $D(0)=200$ .The truck always moves for $20$ seconds, then stands still for $30$ . During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$ , hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$ . During the remaining $30$ seconds $D(t)$ decreases by $150$ .From this observation it is obvious that after four full cycles, i.e. at $t=200$ , we will have $D(t)=0$ for the first time.During the fifth cycle, $D(t)$ will first grow from $0$ to $100$ , then fall from $100$ to $-50$ . Hence Michael overtakes the truck while it is standing at the pail.During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$ , then fall from $50$ to $-100$ . Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$ , then fall from $0$ to $-150$ . Hence the truck meets Michael at the moment when it arrives to the next pail.Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $boxed{5 Longrightarrow B}$ meetings.The movement of Michael and the truck is plotted below: Michael in blue, the truck in red. $[asy] import graph; size(400,300,IgnoreAspect); real[] xt = new real[21]; real[] yt = new real[21]; for (int i=0; i<11; ++i) xt[2*i]=50*i; for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20; for (int i=0; i<11; ++i) yt[2*i]=200*(i+1); for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2); real[] xm={0,500}; real[] ym={0,2500}; draw(graph(xt,yt),red); draw(graph(xm,ym),blue); xaxis("$mathrm{time}$",Bottom,LeftTicks); yaxis("$mathrm{position}$",Left,LeftTicks); [/asy]$