1986 AMC8真题及答案详解

1986 AMC8真题及答案详解 Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月4日上午10:31 2024110706374851

1986 AMC 8 真题

答案详细解析请参考文末

Problem 1

In July 1861, $366$ inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

$text{(A)} frac{366}{31times 24} qquad text{(B)} frac{366times 31}{24}qquad text{(C)} frac{366times 24}{31}qquad text{(D)} frac{31times 24}{366}qquad text{(E)} 366times 31times 24$

Problem 2

Which of the following numbers has the largest reciprocal?

$text{(A)} frac{1}{3} qquad text{(B)} frac{2}{5} qquad text{(C)} 1 qquad text{(D)} 5 qquad text{(E)} 1986$

Problem 3

The smallest sum one could get by adding three different numbers from the set ${ 7,25,-1,12,-3 }$ is

$text{(A)} -3 qquad text{(B)} -1 qquad text{(C)} 3 qquad text{(D)} 5 qquad text{(E)} 21$

Problem 4

The product $(1.8)(40.3+.07)$ is closest to

$text{(A)} 7 qquad text{(B)} 42 qquad text{(C)} 74 qquad text{(D)} 84 qquad text{(E)} 737$

Problem 5

A contest began at noon one day and ended $1000$ minutes later. At what time did the contest end?

$text{(A)} text{10:00 p.m.} qquad text{(B)} text{midnight} qquad text{(C)} text{2:30 a.m.} qquad text{(D)} text{4:40 a.m.} qquad text{(E)} text{6:40 a.m.}$

Problem 6

$frac{2}{1-frac{2}{3}}=$

$text{(A)} -3 qquad text{(B)} -frac{4}{3} qquad text{(C)} frac{2}{3} qquad text{(D)} 2 qquad text{(E)} 6$

Problem 7

How many whole numbers are between $sqrt{8}$ and $sqrt{80}$ ?

$text{(A)} 5 qquad text{(B)} 6 qquad text{(C)} 7 qquad text{(D)} 8 qquad text{(E)} 9$

Problem 8

In the product shown, $text{B}$ is a digit. The value of $text{B}$ is

$[begin{array}{rr} &text{B}2 \ times &7text{B} \ hline &6396 \ end{array}]$

$text{(A)} 3 qquad text{(B)} 5 qquad text{(C)} 6 qquad text{(D)} 7 qquad text{(E)} 8$

Problem 9

Using only the paths and the directions shown, how many different routes are there from $text{M}$ to $text{N}$ ?

$[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(6,0),MidArrow); draw(6*dir(60)--5.1961524227066318805823390245176*dir(30),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--3*dir(60),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(3,0),MidArrow); label("M",6*dir(60),N); label("N",(6,0),SE); label("A",3*dir(60),NW); label("B",5.1961524227066318805823390245176*dir(30),NE); label("C",(3,0),S); label("D",(0,0),SW); [/asy]$

$text{(A)} 2 qquad text{(B)} 3 qquad text{(C)} 4 qquad text{(D)} 5 qquad text{(E)} 6$

Problem 10

A picture $3$ feet across is hung in the center of a wall that is $19$ feet wide. How many feet from the end of the wall is the nearest edge of the picture?

$text{(A)} 1frac{1}{2} qquad text{(B)} 8 qquad text{(C)} 9frac{1}{2} qquad text{(D)} 16 qquad text{(E)} 22$

Problem 11

If $text{A}*text{B}$ means $frac{text{A}+text{B}}{2}$ , then $(3*5)*8$ is

$text{(A)} 6 qquad text{(B)} 8 qquad text{(C)} 12 qquad text{(D)} 16qquad text{(E)} 30$

Problem 12

The table below displays the grade distribution of the $30$ students in a mathematics class on the last two tests. For example, exactly one student received a 'D' on Test 1 and a 'C' on Test 2 (see circled entry). What percent of the students received the same grade on both tests?

$[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label("0",(2.5,.2),N); label("0",(3.5,.2),N); label("2",(4.5,.2),N); label("1",(5.5,.2),N); label("0",(6.5,.2),N); label("0",(2.5,1.2),N); label("0",(3.5,1.2),N); label("1",(4.5,1.2),N); label("1",(5.5,1.2),N); label("1",(6.5,1.2),N); label("1",(2.5,2.2),N); label("3",(3.5,2.2),N); label("5",(4.5,2.2),N); label("2",(5.5,2.2),N); label("0",(6.5,2.2),N); label("1",(2.5,3.2),N); label("4",(3.5,3.2),N); label("3",(4.5,3.2),N); label("0",(5.5,3.2),N); label("0",(6.5,3.2),N); label("2",(2.5,4.2),N); label("2",(3.5,4.2),N); label("1",(4.5,4.2),N); label("0",(5.5,4.2),N); label("0",(6.5,4.2),N); label("F",(1.5,.2),N); label("D",(1.5,1.2),N); label("C",(1.5,2.2),N); label("B",(1.5,3.2),N); label("A",(1.5,4.2),N); label("A",(2.5,5.2),N); label("B",(3.5,5.2),N); label("C",(4.5,5.2),N); label("D",(5.5,5.2),N); label("F",(6.5,5.2),N); label("Test 1",(-.5,5.2),N); label("Test 2",(2.6,6),N); [/asy]$

$text{(A)} 12% qquad text{(B)} 25% qquad text{(C)} 33frac{1}{3}% qquad text{(D)} 40% qquad text{(E)} 50%$

Problem 13

The perimeter of the polygon shown is

$[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $text{(A)} 14 qquad text{(B)} 20 qquad text{(C)} 28 qquad text{(D)} 48$

$text{(E)} text{cannot be determined from the information given}$

Problem 14

If $200leq a leq 400$ and $600leq bleq 1200$ , then the largest value of the quotient $frac{b}{a}$ is

$text{(A)} frac{3}{2} qquad text{(B)} 3 qquad text{(C)} 6 qquad text{(D)} 300 qquad text{(E)} 600$

Problem 15

Sale prices at the Ajax Outlet Store are $50%$ below original prices. On Saturdays an additional discount of $20%$ off the sale price is given. What is the Saturday price of a coat whose original price is $ $180$ ?

$text{(A)}$ $ $54$

$text{(B)}$ $ $72$

$text{(C)}$ $ $90$

$text{(D)}$ $ $108$

$text{(D)}$ $ $110$

Problem 16

A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar indicating the number sold during the winter is covered by a smudge. If exactly $25%$ of the chain's hamburgers are sold in the fall, how many million hamburgers are sold in the winter?

$[asy] size(250); void bargraph(real X, real Y, real ymin, real ymax, real ystep, real tickwidth, string yformat, Label LX, Label LY, Label[] LLX, real[] height,pen p=nullpen) { draw((0,0)--(0,Y),EndArrow); draw((0,0)--(X,0),EndArrow); label(LX,(X,0),plain.SE,fontsize(9)); label(LY,(0,Y),plain.NW,fontsize(9)); real yscale=Y/(ymax+ystep); for(real y=ymin; yymax; y+=ystep) { draw((-tickwidth,yscale*y)--(0,yscale*y)); label(format(yformat,y),(-tickwidth,yscale*y),plain.W,fontsize(9)); } int n=LLX.length; real xscale=X/(2*n+2); for(int i=0;in;++i) { real x=xscale*(2*i+1); path P=(x,0)--(x,height[i]*yscale)--(x+xscale,height[i]*yscale)--(x+xscale,0)--cycle; fill(P,p); draw(P); label(LLX[i],(x+xscale/2),plain.S,fontsize(10)); } for(int i=0;in;++i) draw((0,height[i]*yscale)--(X,height[i]*yscale),dashed); } string yf="%#.1f"; Label[] LX={"Spring","Summer","Fall","Winter"}; for(int i=0;iLX.length;++i) LX[i]=rotate(90)*LX[i]; real[] H={4.5,5,4,4}; bargraph(60,50,1,5.1,0.5,2,yf,"season","hamburgers (millions)",LX,H,yellow); fill(ellipse((45,30),7,10),brown); [/asy]$

$text{(A)} 2.5 qquad text{(B)} 3 qquad text{(C)} 3.5 qquad text{(D)} 4 qquad text{(E)} 4.5$

Problem 17

Let $text{o}$ be an odd whole number and let $text{n}$ be any whole number. Which of the following statements about the whole number $(text{o}^2+text{no})$ is always true?

$text{(A)} text{it is always odd} qquad text{(B)} text{it is always even}$

$text{(C)} text{it is even only if n is even} qquad text{(D)} text{it is odd only if n is odd}$

$text{(E)} text{it is odd only if n is even}$

Problem 18

A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side. Fence posts are to be placed every $12$ meters along the fence including the two posts where the fence meets the rock wall. What is the fewest number of posts required to fence an area $36$ m by $60$ m?

$[asy] draw((0,0)--(16,12)); draw((5.33333,4)--(10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--cycle); label("WALL",(7,4),SE); [/asy]$

$text{(A)} 11 qquad text{(B)} 12 qquad text{(C)} 13 qquad text{(D)} 14 qquad text{(E)} 16$

Problem 19

At the beginning of a trip, the mileage odometer read $56,200$ miles. The driver filled the gas tank with $6$ gallons of gasoline. During the trip, the driver filled his tank again with $12$ gallons of gasoline when the odometer read $56,560$ . At the end of the trip, the driver filled his tank again with $20$ gallons of gasoline. The odometer read $57,060$ . To the nearest tenth, what was the car's average miles-per-gallon for the entire trip?

$text{(A)} 22.5 qquad text{(B)} 22.6 qquad text{(C)} 24.0 qquad text{(D)} 26.9 qquad text{(E)} 27.5$

Problem 20

The value of the expression $frac{(304)^5}{(29.7)(399)^4}$ is closest to

$text{(A)} .003 qquad text{(B)} .03 qquad text{(C)} .3 qquad text{(D)} 3 qquad text{(E)} 30$

Problem 21

Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?

$[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]$

$text{(A)} 2 qquad text{(B)} 3 qquad text{(C)} 4 qquad text{(D)} 5 qquad text{(E)} 6$

Problem 22

Alan, Beth, Carlos, and Diana were discussing their possible grades in mathematics class this grading period. Alan said, "If I get an A, then Beth will get an A." Beth said, "If I get an A, then Carlos will get an A." Carlos said, "If I get an A, then Diana will get an A." All of these statements were true, but only two of the students received an A. Which two received A's?

$text{(A)} text{Alan, Beth} qquad text{(B)} text{Beth, Carlos} qquad text{(C)} text{Carlos, Diana}$

$text{(D)} text{Alan, Diana} qquad text{(E)} text{Beth, Diana}$

Problem 23

The large circle has diameter $text{AC}$ . The two small circles have their centers on $text{AC}$ and just touch at $text{O}$ , the center of the large circle. If each small circle has radius $1$ , what is the value of the ratio of the area of the shaded region to the area of one of the small circles?

$[asy] pair A=(-2,0), O=origin, C=(2,0); path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), M=X..Y..Z..cycle; filldraw(M, black, black); draw(reflect(A,C)*M); draw(A--C, dashed); label("A",A,W); label("C",C,E); label("O",O,SE); dot((-1,0)); dot(O); dot((1,0)); label("$1$",(-.5,0),N); label("$1$",(1.5,0),N); [/asy]$

$text{(A)} text{between }frac{1}{2}text{ and 1} qquad text{(B)} 1 qquad text{(C)} text{between 1 and }frac{3}{2}$

$text{(D)} text{between }frac{3}{2}text{ and 2} qquad text{(E)} text{cannot be determined from the information given}$

Problem 24

The $600$ students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately

$text{(A)} frac{1}{27} qquad text{(B)} frac{1}{9} qquad text{(C)} frac{1}{8} qquad text{(D)} frac{1}{6} qquad text{(E)} frac{1}{3}$

Problem 25

Which of the following sets of whole numbers has the largest average?

$text{(A)} text{multiples of 2 between 1 and 101} qquad text{(B)} text{multiples of 3 between 1 and 101}$

$text{(C)} text{multiples of 4 between 1 and 101} qquad text{(D)} text{multiples of 5 between 1 and 101}$

$text{(E)} text{multiples of 6 between 1 and 101}$

1986 AMC8真题答案详细解析1.We can express the number of inches of rain in a ratio, $frac{366 text{ inches}}{1text{ month}}$ . We need to find to find the number of hours, but first we need to find the number of days. We know there are 31 days in July, so we get $[frac{366 text{ inches}}{1 text{ month}} times frac{1 text{ month}}{31 text{ days}}]$

We know there are 24 hours in a day, so we get $[frac{366 text{ inches}}{1 text{ month}} times frac{1 text{ month}}{31 text{ days}} times frac{1 text{ day}}{24 text{ hours}}]$

And simplifying gets us $dfrac{366 text{ inches}}{31 times 24 text{ hours}}$ , which is .

Note that in the simplifying step, we noted that the month and day units cancel, leaving us with only inches per hour.

2.For positive numbers, the larger the number, the smaller its reciprocal. Likewise, smaller numbers have larger reciprocals.

Thus, all we have to do is find the smallest number.

But which one is it? $frac{1}{3}$ ? or $frac{2}{5}$ ? We see that $frac{1}{3} = frac{5}{15}$ , and $frac{2}{5} = frac{6}{15}$ , so obviously $frac{1}{3}$ is smaller.

3.To find the smallest sum, we just have to find the smallest 3 numbers and add them together.

Obviously, the numbers are $-3, -1, 7$ , and adding them gets us $3$ .

$boxed{text{C}}$

$(1.8)(40.37)approx (1.8)(40)=72.$ Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ is the closest to $72$ , so the answer is $boxed{text{C}}$ .

5.There are $60$ minutes in an hour. So, we can easily eliminate some of the choices by noting that noon is exactly $720$ minutes away from midnight. Since $720 1000$ , we know that it cannot be A or B. Because midnight is $720$ minutes away, we know that the contest ended $1000 - 720 = 280$ minutes after midnight. The highest multiple of 60 that will fit into $280$ is $240$ , which is $4 times 60$ , and the remainder is $40$ minutes, meaning that the contest ended at $4:40 text{ a.m.}$

$4:40$ is $boxed{text{D}}$

6.Just simplify the bottom as $frac{3}{3}-frac{2}{3}=frac{1}{3}$ , getting us $frac{2}{frac{1}{3}}$ , with which we multiply top and bottom by 3, we get $frac{6}{1}$ , or $6$

$boxed{text{E}}$

7.No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.

Clearly it must be true that for any positive integers $a$ , $b$ , and $c$ with $abc$ , $[sqrt{a}sqrt{b}sqrt{c}]$

If we let $a=9$ , $b=8$ , and $c=4$ , then we get $[sqrt{9}sqrt{8}sqrt{4}]$ $[3sqrt{8}2]$

Therefore, the smallest whole number between $sqrt{8}$ and $sqrt{80}$ is $3$ .

Similarly, if we let $a=81$ , $b=80$ , and $c=64$ , we get $[sqrt{81}sqrt{80}sqrt{64}]$ $[9sqrt{80}8]$

So $8$ is the largest whole number between $sqrt{8}$ and $sqrt{80}$ .

So we know that we just have to find the number of integers from 3 to 8 inclusive. If we subtract 2 from every number in this set (which doesn't change the number of integers in the set at all), we find that now all we need to do is find the number of integers there are from 1 to 6, which is obviously 6.

$boxed{text{B}}$

8.Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.

So, $2 times text{B}$ has a units digit of $6$ , so $text{B}$ is either $3$ or $8$ . If $text{B}=3$ , then the product is $32times 73$ , which is clearly too small, so $text{B}=8$

$boxed{text{E}}$

9.There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.

Since A can only go to C or D, which each only have 1 way to get to N each, there are $1+1=2$ ways to get from A to N.

Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are $2+1+1=4$ ways to get from B to N.

M can only go to either B or A, A has 2 ways and B has 4 ways, so M has $4+2=6$ ways to get to N.

6 is $boxed{text{E}}$ .

A diagram labeled with the number of ways to get to $text{N}$ from each point might look like

10.Let's say that the distance from the picture to the wall is $x$ . Since that distance will be on both sides of the picture (it's in the exact middle), we can say that $begin{align*} x + 3 + x = 19 &Rightarrow 2x+3=19 \ &Rightarrow 2x=16 \ &Rightarrow x=8 \ end{align*}$

$boxed{text{B}}$

11.

We just plug in and evaluate: $begin{align*} (3*5)*8 &= left( frac{3+5}{2}right) *8 \ &= 4*8 \ &= frac{4+8}{2} \ &= 6 \ end{align*}$

12.We need to find the number of those who did get the same on both tests over 30 (the number of students in the class).

So, we have $[frac{2 + 4 + 5 + 1}{30}]$

Which simplifies to $[frac{12}{30} = frac{4}{10} = frac{40}{100} = 40 %]$

$boxed{text{D}}$

13.

Solution 1

For the segments parallel to the side with side length 8, let's call those two segments $a$ and $b$ , the longer segment being $b$ , the shorter one being $a$ .

For the segments parallel to the side with side length 6, let's call those two segments $c$ and $d$ , the longer segment being $d$ , the shorter one being $c$ .

So the perimeter of the polygon would be...

$8 + 6 + a + b + c + d$

Note that $a + b = 8$ , and $c + d = 6$ .

Now we plug those in: $begin{align*} 8 + 6 + a + b + c + d &= 8 + 6 + 8 + 6 \ &= 14 times 2 \ &= 28 \ end{align*}$

28 is $boxed{text{C}}$ .

Solution 2

$[asy] unitsize(12); draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((8,3)--(8,0)--(2.7,0),dashed); [/asy]$

The perimeter of the requested region is the same as the perimeter of the rectangle with the dashed portion. This makes the answer $2(6+8)=28rightarrow boxed{text{C}}$

14. $frac{b}{a}$ will be largest if $b$ is the largest it can be, and $a$ is the smallest it can be.

Since $b$ can be no larger than $1200$ , $b = 1200$ . Since $a$ can be no less than $200$ , $a = 200$ . $frac{1200}{200} = 6$

$6$ is $boxed{text{C}}$

15.First we need to do the first discount, which, at Ajax Outlet Store, would be there any day of the week. $begin{align*} 180 times 50% &= 180 times frac{1}{2} \ &= 90 \ end{align*}$

If we discount $20%$ , then $80 %$ will be left, so after the second discount, we get $begin{align*} 90 times 80% &= 90 times frac{8}{10} \ &= 9 times 8 \ &= 72 \ end{align*}$

$boxed{text{B}}$

16.What we want to find is the number of hamburgers sold in the winter. Since we don't know what it is, let's call it $x$ . From the graph, we know that in Spring, 4.5 million hamburgers were sold, in the Summer was 5 million and in the Fall was 4 million. We know that the number of hamburgers sold in Fall is exactly $frac{1}{4}$ of the total number of hamburgers sold, so we can say that...

$4 times text{Fall} = text{Spring} + text{Winter} + text{Fall} + text{Summer}$

$4 times 4 = 4.5 + 4 + x + 5$

$16 = x + 13.5$

$2.5 = x$

The answer is 2.5, or

17.

Solution 1

We can solve this problem using logic.

Let's say that $text{n}$ is odd. If $text{n}$ is odd, then obviously $text{no}$ will be odd as well, since $text{o}$ is odd, and the product of two odd numbers is odd. Since $text{o}$ is odd, $text{o}^2$ will also be odd. And adding two odd numbers makes an even number, so if $text{n}$ is odd, the entire expression is even.

Let's say that $text{n}$ is even. If $text{n}$ is even, then $text{no}$ will be even as well, because the product of an odd and an even is even. $text{o}^2$ will still be odd. That means that the entire expression will be odd, since the sum of an odd and an even is odd.

Looking at the multiple choices, we see that our second case fits choice E exactly.

$boxed{text{E}}$

Solution 2

We are given that $text{o}equiv 1pmod{2}$ , so in mod $2$ we have $[1^2+1(n) = n+1]$ which is odd only if $text{n}$ is even $rightarrow boxed{text{E}}$

18.The shortest possible rectangle that has sides 36 and 60 would be if the side opposite the wall was 60.

Each of the sides of length 36 contribute $frac{36}{12}+1=4$ fence posts and the side of length 60 contributes $frac{60}{12}+1=6$ fence posts, so there are $4+4+6=14$ fence posts.

However, the two corners where a 36 foot fence meets an 60 foot fence are counted twice, so there are actually $14-2=12$ fence posts.

$boxed{text{B}}$

19.The first six gallons are irrelevant. We start with the odometer at $56,200$ miles, and a full gas tank. The total gas consumed by the car during the trip is equal to the total gas the driver had to buy to make the tank full again, i.e., $12+20=32$ gallons. The distance covered is $57,060 - 56,200 = 860$ miles. Hence the average MPG ratio is $860 / 32 approx 26.9 rightarrow boxed{text{D}}$ .

20.

$[frac{(304)^5}{(29.7)(399)^4} approx frac{300^5}{30cdot400^4} = frac{3^5 cdot 10^{10}}{3cdot 4^4 cdot 10^9} = frac{3^4cdot 10}{4^4} = frac{810}{256}]$ Which is closest to $3rightarrowboxed{text{D}}$ .

(The original expression is approximately equal to $3.44921198$ .)

21.

Solution 1

The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.

$A$ would be the top side - OK

$B$ would be the left side - OK

$C$ would cause the figure to not be foldable at all

$D$ would be the left side - OK

$E$ would be the top side - OK

$F$ is the same case as $B$ - OK

$G$ is the same case as $C$

$H$ is the same case as $A$ - OK

In total, there are $6rightarrowboxed{text{E}}$ good possibilities.

Solution 2

Fold the four squares into the four sides of a cube. Then, there are six edges "open" (for lack of better term). For each open edge, we can add a square/side, so the answer is $6rightarrowboxed{text{E}}$ .

22.Let's say that Alan gets an A. Well, from his statement, then Beth would also get an A. But from her statement, Carlos would get an A. And from his statement, Diana would also get an A. So all 4 would get A's, but the problem said only 2 got A's.

Let's say that Beth gets an A. From her statement, we know that Carlos get an A, and from his statement we know that Diana gets an A. But that makes 3, which is not 2.

If Carlos gets an A, then Diana gets an A. That makes 2, so $boxed{text{C}}$ is the right answer. Note that although Beth said "If I get an A, then Carlos will get an A.", that does NOT mean that "If Carlos gets an A, then I will get an A."

23.

The small circle has radius $1$ , thus its area is $pi$ .

The large circle has radius $2$ , thus its area is $4pi$ .

The area of the semicircle above $AC$ is then $2pi$ .

The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $pi$ . This means that the area of the shaded part is $2pi-pi=pi$ . This is equal to the area of a small circle, hence the correct answer is $boxed{text{(B)} 1}$ .

24.

Imagine that we run the computer many times. In roughly $1/3$ of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly $1/3$ cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is $frac{1}{3}cdot frac{1}{3} = frac{1}{9}$ , or $boxed{text{B}}$ .

(The exact value is $frac{199}{599} cdot frac{198}{598}$ , which is $sim 0.11%$ less than our approximate answer.)

Solution 2

There are $3C1$ ways to choose which group the three kids are in and the chance that all three are in the same group is $frac{1}{27}$ . Hence $frac{1}{9}$ or $boxed {B}$ .

Solution 3

One of the statements, that there are $600$ students in the school is redundant. Taking that there are $3$ students and there are $3$ groups, we can easily deduce there are $81$ ways to group the $3$ students, and there are $3$ ways to group them in the same $1$ group, so we might think $frac{3}{54}=frac{1}{27}$ is the answer but as there are 3 groups we do $frac{1}{27} (3)=frac{1}{9}$ which is $boxed{text{(B)}}$ .

25.

Solution 1

From $1$ to $101$ there are $leftlfloor frac{101}{2} rightrfloor = 50$ (see floor function) multiples of $2$ , and their average is

$frac{2cdot 1+2cdot 2+2cdot 3+cdots + 2cdot 50}{50} \ \ = frac{2(1+2+3+cdots +50)}{50} \ \ = frac{2cdot frac{50cdot 51}{2}}{50} \ \ = frac{2cdot 51}{2} \ \ = 51$ Similarly, we can find that the average of the multiples of $3$ between $1$ and $101$ is $51$ , the average of the multiples of $4$ is $52$ , the average of the multiples of $5$ is $52.5$ , and the average of the multiples of $6$ is $51$ , so the one with the largest average is $boxed{text{D}}$ Solution 2

The multiples of any number in any range form an arithmetic sequence. It can be proven that the average of the numbers in an arithmetic sequence is simply the average of their highest and lowest entries, so you can just add the first term and the last term, and see which one is the largest (since the sum of two numbers is twice their average). 2+100=102, 3+99=102, 4+100=104, 5+100=105, 6+96=102. Therefore, the answer is multiples of five because it has the largest number.

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