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2001AMC10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4:23

2001AMC10真题

答案解析请参考文末

Problem 1

The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$ . What is the mean?

$mathrm{(A)} 4 qquadmathrm{(B)} 6 qquadmathrm{(C)} 7 qquadmathrm{(D)} 10 qquadmathrm{(E)} 11$

Problem 2

A number $x$ is $2$ more than the product of its reciprocal and its additive inverse. In which interval does the number lie?

$mathrm{(A)} -4leq xleq -2 qquadmathrm{(B)} -2<xleq 0 qquadmathrm{(C)} 0<xleq 2$

$mathrm{(D)} 2<xleq 4 qquadmathrm{(E)} 4<xleq 6$

Problem 3

The sum of two numbers is $S$ . Suppose $3$ is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

$mathrm{(A)} 2S+3 qquadmathrm{(B)} 3S+2 qquadmathrm{(C)} 3S+6 qquadmathrm{(D)} 2S+6 qquadmathrm{(E)} 2S+12$

Problem 4

What is the maximum number for the possible points of intersection of a circle and a triangle?

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

Problem 5

How many of the twelve pentominoes pictured below have at least one line of symmetry?

$[asy] unitsize(5mm); defaultpen(linewidth(1pt)); draw(shift(2,0)*unitsquare); draw(shift(2,1)*unitsquare); draw(shift(2,2)*unitsquare); draw(shift(1,2)*unitsquare); draw(shift(0,2)*unitsquare); draw(shift(2,4)*unitsquare); draw(shift(2,5)*unitsquare); draw(shift(2,6)*unitsquare); draw(shift(1,5)*unitsquare); draw(shift(0,5)*unitsquare); draw(shift(4,8)*unitsquare); draw(shift(3,8)*unitsquare); draw(shift(2,8)*unitsquare); draw(shift(1,8)*unitsquare); draw(shift(0,8)*unitsquare); draw(shift(6,8)*unitsquare); draw(shift(7,8)*unitsquare); draw(shift(8,8)*unitsquare); draw(shift(9,8)*unitsquare); draw(shift(9,9)*unitsquare); draw(shift(6,5)*unitsquare); draw(shift(7,5)*unitsquare); draw(shift(8,5)*unitsquare); draw(shift(7,6)*unitsquare); draw(shift(7,4)*unitsquare); draw(shift(6,1)*unitsquare); draw(shift(7,1)*unitsquare); draw(shift(8,1)*unitsquare); draw(shift(6,0)*unitsquare); draw(shift(7,2)*unitsquare); draw(shift(11,8)*unitsquare); draw(shift(12,8)*unitsquare); draw(shift(13,8)*unitsquare); draw(shift(14,8)*unitsquare); draw(shift(13,9)*unitsquare); draw(shift(11,5)*unitsquare); draw(shift(12,5)*unitsquare); draw(shift(13,5)*unitsquare); draw(shift(11,6)*unitsquare); draw(shift(13,4)*unitsquare); draw(shift(11,1)*unitsquare); draw(shift(12,1)*unitsquare); draw(shift(13,1)*unitsquare); draw(shift(13,2)*unitsquare); draw(shift(14,2)*unitsquare); draw(shift(16,8)*unitsquare); draw(shift(17,8)*unitsquare); draw(shift(18,8)*unitsquare); draw(shift(17,9)*unitsquare); draw(shift(18,9)*unitsquare); draw(shift(16,5)*unitsquare); draw(shift(17,6)*unitsquare); draw(shift(18,5)*unitsquare); draw(shift(16,6)*unitsquare); draw(shift(18,6)*unitsquare); draw(shift(16,0)*unitsquare); draw(shift(17,0)*unitsquare); draw(shift(17,1)*unitsquare); draw(shift(18,1)*unitsquare); draw(shift(18,2)*unitsquare);[/asy]$

$mathrm{(A)} 3 qquadmathrm{(B)} 4 qquadmathrm{(C)} 5 qquadmathrm{(D)} 6 qquadmathrm{(E)} 7$

Problem 6

Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$ ?

$mathrm{(A)} 2 qquadmathrm{(B)} 3 qquadmathrm{(C)} 6 qquadmathrm{(D)} 8 qquadmathrm{(E)} 9$

Problem 7

When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?

$mathrm{(A)} 0.0002 qquadmathrm{(B)} 0.002 qquadmathrm{(C)} 0.02 qquadmathrm{(D)} 0.2 qquadmathrm{(E)} 2$

Problem 8

Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?

$mathrm{(A)} 42 qquadmathrm{(B)} 84 qquadmathrm{(C)} 126 qquadmathrm{(D)} 178 qquadmathrm{(E)} 252$

Problem 9

The state income tax where Kristin lives is levied at the rate of $p%$ of the first $textdollar 28000$ of annual income plus $(p + 2)%$ of any amount above $textdollar 28000$ . Kristin noticed that the state income tax she paid amounted to $(p + 0.25)%$ of her annual income. What was her annual income?

$textbf{(A) } textdollar 28,000qquadtextbf{(B) } textdollar 32,000qquadtextbf{(C) }textdollar 35,000qquadtextbf{(D) } textdollar 42,000qquadtextbf{(E) } textdollar 56,000$

Problem 10

If $x$ , $y$ , and $z$ are positive with $xy = 24$ , $xz = 48$ , and $yz = 72$ , then $x + y + z$ is

$textbf{(A) }18qquadtextbf{(B) }19qquadtextbf{(C) }20qquadtextbf{(D) }22qquadtextbf{(E) }24$

Problem 11

Consider the dark square in an array of unit squares, part of which is shown. The ﬁrst ring of squares around this center square contains $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100^text{th}$ ring is

$[asy] unitsize(3mm); defaultpen(linewidth(1pt)); fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray); fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black); for(real i=0; i<=9; ++i) { draw((i,0)--(i,9)); draw((0,i)--(9,i)); }[/asy]$

$textbf{(A)} 396 qquad textbf{(B)} 404 qquad textbf{(C)} 800 qquad textbf{(D)} 10,!000 qquad textbf{(E)} 10,!404$

Problem 12

Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$ . Which of the following is not necessarily a divisor of $n$ ?

$textbf{(A)} 6 qquad textbf{(B)} 14 qquad textbf{(C)} 21 qquad textbf{(D)} 28 qquad textbf{(E)} 42$

Problem 13

A telephone number has the form $ABC - DEF - GHIJ$ , where each letter represents a different digit. The digits in each part of the numbers are in decreasing order; that is, $A > B > C$ , $D > E > F$ , and $G > H > I > J$ . Furthermore, $D$ , $E$ , and $F$ are consecutive even digits; $G$ , $H$ , $I$ , and $J$ are consecutive odd digits; and $A + B + C = 9$ . Find $A$ .

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

Problem 14

A charity sells $140$ benefit tickets for a total of $textdollar2001$ . Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?

$text{(A) } textdollar 782 qquad text{(B) } textdollar 986 qquad text{(C) } textdollar 1158 qquad text{(D) } textdollar 1219 qquad text{(E) } textdollar 1449$

Problem 15

A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes?

$textbf{(A)} 9 qquad textbf{(B)} 10 qquad textbf{(C)} 12 qquad textbf{(D)} 15 qquad textbf{(E)} 25$

Problem 16

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$ . What is their sum?

$textbf{(A)} 5 qquad textbf{(B)} 20 qquad textbf{(C)} 25 qquad textbf{(D)} 30 qquad textbf{(E)} 36$

Problem 17

Which of the cones listed below can be formed from a $252^circ$ sector of a circle of radius $10$ by aligning the two straight sides?

$[asy]import graph;unitsize(1.5cm);defaultpen(fontsize(8pt));draw(Arc((0,0),1,-72,180),linewidth(.8pt));draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt));label("$10$",(-0.5,0),S);draw(Arc((0,0),0.1,-72,180));label("$252^{circ}$",(0.05,0.05),NE);[/asy]$

$textbf{(A)} text{A cone with slant height of 10 and radius 6}$ $textbf{(B)} text{A cone with height of 10 and radius 6}$ $textbf{(C)} text{A cone with slant height of 10 and radius 7}$ $textbf{(D)} text{A cone with height of 10 and radius 7}$ $textbf{(E)} text{A cone with slant height of 10 and radius 8}$

Problem 18

The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to

$[asy] unitsize(3mm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; for(int i=0; i<3; ++i) { for(int j=0; j<3; ++j) { draw(shift(3*i,3*j)*p); } }[/asy]$

$textbf{(A)} 50 qquad textbf{(B)} 52 qquad textbf{(C)} 54 qquad textbf{(D)} 56 qquad textbf{(E)} 58 qquad$

Problem 19

Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?

$textbf{(A)} 6 qquad textbf{(B)} 9 qquad textbf{(C)} 12 qquad textbf{(D)} 15 qquad textbf{(E)} 18$

Problem 20

A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$ . What is the length of each side of the octagon?

$textbf{(A)} frac{1}{3}(2000) qquad textbf{(B)} {2000(sqrt{2}-1)} qquad textbf{(C)} {2000(2-sqrt{2})} qquad textbf{(D)} {1000} qquad textbf{(E)} {1000sqrt{2}}$

Problem 21

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$ , and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

$textbf{(A)} frac{8}3qquadtextbf{(B)} frac{30}{11}qquadtextbf{(C)} 3qquadtextbf{(D)} frac{25}{8}qquadtextbf{(E)} frac{7}{2}$

Problem 22

In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ , $w$ , $x$ , $y$ , and $z$ . Find $y + z$ .

$textbf{(A)} 43 qquad textbf{(B)} 44 qquad textbf{(C)} 45 qquad textbf{(D)} 46 qquad textbf{(E)} 47$

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]$

Problem 23

A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?

$textbf{(A)} frac{3}{10}qquadtextbf{(B)} frac{2}{5}qquadtextbf{(C)} frac{1}{2}qquadtextbf{(D)} frac{3}{5}qquadtextbf{(E)} frac{7}{10}$

Problem 24

In trapezoid $ABCD$ , $overline{AB}$ and $overline{CD}$ are perpendicular to $overline{AD}$ , with $AB+CD=BC$ , $AB<CD$ , and $AD=7$ . What is $ABcdot CD$ ?

$textbf{(A)} 12 qquad textbf{(B)} 12.25 qquad textbf{(C)} 12.5 qquad textbf{(D)} 12.75 qquad textbf{(E)} 13$

Problem 25

How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$ ?

$textbf{(A)} 768 qquad textbf{(B)} 801 qquad textbf{(C)} 934 qquad textbf{(D)} 1067 qquad textbf{(E)} 1167$

2001AMC10详细解析

The median of the list is $10$ , and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$ .We substitute the median for $10$ and the equation becomes $n+6=10$ .Subtract both sides by 6 and we get $n=4$ . $n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$ .The mean of those numbers is $frac{9n+63}{9}$ which is $n+7$ .Substitute $n$ for $4$ and $4+7=boxed{textbf{(E) }11}$ .
We can write our equation as $x= left(frac{1}{x} right) cdot (-x) +2 = -1+2 = 1$ . Therefore, $boxed{textbf{(C) }0 < xle 2}$ .
Suppose the two numbers are $a$ and $b$ , with $a+b=S$ . Then the desired sum is $2(a+3)+2(b+3)=2(a+b)+12=2S +12$ , which is answer $boxed{text{(E)}}$ .
We can draw a circle and a triangle, such that each side is tangent to the circle. This means that each side would intersect the circle at one point.You would then have $3$ points, but what if the circle was bigger? Then, each side would intersect the circle at 2 points.Therefore, $2 times 3 = boxed{textbf{(E) }6}$ .We know that the maximum amount of points that a circle and a line segment can intersect is $2$ . Therefore, because there are $3$ line segments in a triangle, the maximum amount of points of intersection is $boxed{textbf{(E) }6}$ .
The ones with lines over the shapes have at least one line of symmetry. Counting the number of shapes that have line(s) on them, we find $boxed{textbf{(D)} 6}$
Denote $a$ and $b$ as the tens and units digit of $N$ , respectively. Then $N = 10a+b$ . It follows that $10a+b=ab+a+b$ , which implies that $9a=ab$ . Since $aneq0$ , $b=9$ . So the units digit of $N$ is $boxed{(text{E}) 9}$ .
If $x$ is the number, then moving the decimal point four places to the right is the same as multiplying $x$ by $10000$ . This gives us the equation $[10000x=4cdotfrac{1}{x}]$ This is equivalent to $[x^2=frac{4}{10000}]$ Since $x$ is positive, it follows that $x=frac{2}{100}=boxed{textbf{(C)} 0.02}$ Alternatively, we could try each solution and see if it fits the problems given statements.After testing, we find that $boxed{textbf{(C)} 0.02}$ is the correct answer.
We need to find the least common multiple of the four numbers given. That is, the next time they will be together. First, find the least common multiple of $3$ and $4$ . $3 times 4 = 12$ .Find the least common multiple of $12$ and $6$ .Since $12$ is a multiple of $6$ , the least common multiple is $12$ .Lastly, the least common multiple of $12$ and $7$ is $12 times 7 = boxed{textbf{(B) }84}$ .
The first two equations in the problem are $xy=24$ and $xz=48$ . Since $xyz ne 0$ , we have $frac{xy}{xz}=frac{24}{48} implies 2y=z$ . We can substitute $z = 2y$ into the third equation $yz = 72$ to obtain $2y^2=72 implies y=6$ and $2y=z=12$ . We replace $y$ into the first equation to obtain $x=4$ .Since we know every variable's value, we can substitute them in to find $x+y+z = 4+6+12 = boxed{textbf{(D) }22}$ .These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives $(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 times 12)^2 implies xyz = 288$ . We divide $xyz = 288$ by each of the given equations, which yields $x = 4$ , $y = 6$ , and $z = 12$ . The desired sum is $4+6+12 = boxed{22}$ , so the answer is $boxed{textbf{(D)}}$ .
We can partition the $n^text{th}$ ring into $4$ rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares.There are $2(2n+1)+2(2n-1)=4n+2+4n-2=8n$ unit squares in the $n^text{th}$ ring.Thus, the $100^text{th}$ ring has $8 times 100 = boxed{textbf{(C)} 800}$ unit squares.We can make the $n^text{th}$ ring by removing a square of side length $2n-1$ from a square of side length $2n+1$ .This ring contains $(2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$ unit squares.Thus, the $100^text{th}$ ring has $8 times 100 = boxed{textbf{(C)} 800}$ unit squares.
Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$ , meaning it is divisible by $6$ .It also mentions that it is divisible by $7$ , so the number is definitely divisible by all the factors of $42$ .In our answer choices, the one that is not a factor of $42$ is $boxed{textbf{(D)} 28}$ .We can look for counterexamples. For example, letting $n = 13 cdot 14 cdot 15$ , we see that $n$ is not divisible by 28, so (D) is our answer.
The last four digits $text{GHIJ}$ are either $9753$ or $7531$ , and the other odd digit ( $1$ or $9$ ) must be $A$ , $B$ , or $C$ . Since $A + B + C = 9$ , that digit must be $1$ . Thus the sum of the two even digits in $text{ABC}$ is $8$ . $text{DEF}$ must be $864$ , $642$ , or $420$ , which respectively leave the pairs $2$ and $0$ , $8$ and $0$ , or $8$ and $6$ , as the two even digits in $text{ABC}$ . Only $8$ and $0$ has sum $8$ , so $text{ABC}$ is $810$ , and the required first digit is $boxed{(text{E})8}$ .
Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.Let $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.Hence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single half price ticket costs $frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0leq hleq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.The prime factorization of $4002$ is $4002=2cdot 3cdot 23cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2cdot 3cdot 29 = 174$ .This gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.In our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23cdot 34 = boxed{(text{A})782}$ dollars are raised by the full price tickets.Let the cost of the full price ticket be $x$ , let the number of full price tickets be $A$ and half price tickets be $B$ Multiplying everything by two first to make cancel out fractions.We have $2Ax+Bx=4002$
And we have $A+B=140implies B=140-A$

Plugging in, we get $implies 2Ax+(140-A)(x)=4002$

Simplifying, we get $Ax+140x=4002$

Factoring out the $x$ , we get $x(A+140)=4002implies x=frac{4002}{A+140}$

Obviously, we see that the fraction has to simplify to an integer.

Hence, $A+140$ has to be a factor of 4002.

By inspection, we see that the prime factorization of $4002=2times3times23times29$

We see that $A=34$ through inspection. We also find that $x=23$

Hence, the price of full tickets out of $2001$ is $23times34=boxed{782}implies boxed{A}$ .
Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$ , giving an area of $600$ . $[asy] draw((0,0)--(5,0),linewidth(2)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5)); draw((2.5,5)--(2.5,0),dashed);[/asy]$ If we look at it the other way, we see the distance between the stripes is the height and the base is $50$ . $[asy] draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed);[/asy]$ The area is still the same, so the distance between the stripes is $600/50 = boxed{textbf{(C)} 12}$ .Alternatively, we could use similar triangles--the $30-40-50$ triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the $x-y-15$ triangle, where we are trying to find $y$ (the shortest distance between the two stripes). Therefore, $y$ would have to be $boxed{textbf{(C)} 12}$ .The area of the parallelogram in the first picture is base * height which is 15 * 40 = 600. The goal of the problem is to find the minimum distance between the stripes (it's not really said in the problem but that's the goal). If you look at the parallelogram like in diagram two, the base would be 50. Since it's still the same parallelogram, and the shortest distance between two parallel lines is the line that is perpendicular to both lines, then the height would be 50*h = 600, so h = 12.It's easy to see why the minimum height would be at its perpendicular line. If you try and move the point away from the perpendicular line, you would see that the only thing changing is the distance between the original point and the new point, which would increase distance, not minimize it. The point at the perpendicular line is the only place where you would have zero distance between it and the original point (because it is the original point), and simple observation or Pythagorean theorem would tell you that that is the min distance.
Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, 5. So $dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which implies that $m=10$ . Hence, the sum of the three numbers is $3(10) = boxed{(text{D})30}$ .
The blue lines will be joined together to form a single blue line on the surface of the cone, hence $boxed{10}$ will be the $boxed{text{slant height}}$ of the cone.The red line will form the circumference of the base. We can compute its length and use it to determine the radius.The length of the red line is $dfrac{252}{360}cdot 2pi cdot 10 = 14pi$ . This is the circumference of a circle with radius $boxed{7}$ .Therefore the correct answer is $boxed{text{C}}$ .
Consider any single tile: $[asy] unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); [/asy]$ If the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons. Hence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.overline{5}%$ , and the closest integer to this value is $boxed{(mathrm{D})56}$ .
Let's use stars and bars. Let the donuts be represented by $O$ s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us $4$ in all. The four donuts we want can be represented as $OOOO$ . Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, $O|OO|O$ represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in $binom{6}{2}=15$ ways. Our answer is hence $boxed{textbf{(D)} 15}$ . Notice that this can be generalized to get the balls and urn (stars and bars) identity.Simple casework works here as well: Set up the following ratios: $[4:0:0]$ $[3:1:0]$ $[2:2:0]$ $[2:1:1]$ In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get $frac{4!}{3!2!}$ for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is $3*3+6 = boxed{15}$
$2000 - 2x = xsqrt2$ $2000 = x(2 + sqrt2)$ $x = frac {2000}{2 + sqrt2} =x = frac {2000(2 - sqrt2)}{(2 + sqrt2)(2 - sqrt2)}= frac {2000(2 - sqrt2)}{2} = 1000(2 - sqrt2)$ $xsqrt2 = 1000(2sqrt {2} - 2) = boxed{textbf{(B)} 2000(sqrt2-1)}$ .
Let the diameter of the cylinder be $2r$ . Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, $frac{12-2r}{12}=frac{2r}{10}$ which we solve to find $r=frac{30}{11}$ . Our answer is $boxed{textbf{(B)} frac{30}{11}}$ .
We know that $y+z=2v$ , so we could find one variable rather than two. $v+24+w=43+v$ $24+w=43$ $w=19$ $[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]$ $44+x=24+x+z implies z=20$ $[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]$ The sum per row is $25+21+20=66$ .Thus $66-18-25=66-43=v=23$ .Since we needed $2v$ and we know $v=23$ , $23 times 2 = boxed{textbf{(D)} 46}$ .
$v+24+w=43+v$

$24+w=43$

$w=19$

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]$

$44+x=24+x+z implies z=20$

$[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]$

The magic sum is determined by the bottom row. $25+20+21=66$ .

Solving for $y$ :

$y=66-19-21=66-40=26$ .

To find our answer, we need to find $y+z$ . $y+z=20+26 = boxed{textbf{(D)} 46}$ .

A nice thing to know is that any $3$ numbers that goes through the middle forms an arithmetic sequence.

Using this, we know that $x=24+z/2$ , or $2x=24+z$ because $x$ would be the average.

We also know that because $x$ is the average the magic sum would be $3x$ , so we can also write the equation $3x-46=z$ using the bottom row.

Solving for x in this system we get $x=22$ , so now using the arithmetic sequence knowledge we find that $y=26$ and $z=20$ .

Adding these we get $boxed{textbf{(D)} 46}$ .
Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probability that the last chip drawn is white) is $boxed{(text{D}) frac {3}{5}}$ .Let's assume we don't stop picking until all of the balls are picked. To satisfy this condition, we have to arrange the letters: W, W, R, R, R such that both R's appear in the first 4. We find the number of ways to arrange the red balls in the first 4 and divide that by the total ways to choose all the balls. The probability of this occurring is $dfrac{dbinom{4}{2}}{dbinom{5}{2}} = boxed{text{(D)}dfrac35}$
If $AB=x$ and $CD=y$ , then $BC=x+y$ . By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49.$ Solving the equation, we get $4xy=49 implies xy = boxed{textbf{(B)} 12.25}$ .Simpler is just drawing the trapezoid and then using what is given to solve. Draw a line perpendicular to AD that connects the longer side to the corner of the shorter side. Name the bottom part x and top part a. By the Pythagorean theorem, it is obvious that $a^{2} + 49 = (2x+a)^{2}$ (the RHS is the fact the two sides added together equals that). Then, we get $a^2 + 49 = 4x^2 + 4ax + a^2$ , cancel out and factor and we get $49 = 4x(x+a)$ . Notice that $x(x+a)$ is what the question asks, so the answer is $boxed{frac{49}{4}}$
Out of the numbers $1$ to $12$ four are divisible by $3$ and three by $4$ , counting $12$ twice. Hence $6$ out of these $12$ numbers are multiples of $3$ or $4$ .The same is obviously true for the numbers $12k+1$ to $12k+12$ for any positive integer $k$ .Hence out of the numbers $1$ to $60=5cdot 12$ there are $5cdot 6=30$ numbers that are divisible by $3$ or $4$ . Out of these $30$ , the numbers $15$ , $20$ , $30$ , $40$ , $45$ and $60$ are divisible by $5$ . Therefore in the set ${1,dots,60}$ there are precisely $30-6=24$ numbers that satisfy all criteria from the problem statement.Again, the same is obviously true for the set ${60k+1,dots,60k+60}$ for any positive integer $k$ .We have $1980/60 = 33$ , hence there are $24cdot 33 = 792$ good numbers among the numbers $1$ to $1980$ . At this point we already know that the only answer that is still possible is $boxed{text{(B)}}$ , as we only have $20$ numbers left.By examining the remaining $20$ by hand we can easily find out that exactly $9$ of them match all the criteria, giving us $792+9=boxed{801}$ good numbers. This is correct.We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $leftlfloor frac{2001}{3} rightrfloor$ , $leftlfloor frac{2001}{4} rightrfloor$ , and $leftlfloor frac{2001}{12} rightrfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ , $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.The second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $leftlfloor frac{2001}{3*5} rightrfloor$ , $leftlfloor frac{2001}{4*5} rightrfloor$ , and $leftlfloor frac{2001}{3*4*5} rightrfloor$ yields the numbers $133$ , $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $boxed{textbf{(B)} 801}$ .First find the number of such numbers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of 3.There are $frac45*2000=1600$ numbers that are not multiples of $5$ . $frac23*frac34*1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=boxed{textbf{(B)} 801}$
Take a good-sized sample of consecutive integers; for example, the first 25 positive integers. Determine that the numbers 3, 4, 6, 8, 9, 12, 16, 18, 21, and 24 exhibit the properties given in the question. 25 is a divisor of 2000, so there are $frac{10}{25}*2000=800$ numbers satisfying the given conditions between 1 and 2000. Since 2001 is a multiple of 3, add 1 to 800 to get $800+1=boxed{textbf{(B)} 801}$ .