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2009 AMC 8 真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2017年7月25日下午1:54

2009 AMC 8 真题

答案详细解析请参考文末

Problem 1

Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie 3 apples, keeping 4 apples for herself. How many apples did Bridget buy?

$textbf{(A)} 3qquadtextbf{(B)} 4qquadtextbf{(C)} 7qquadtextbf{(D)} 11qquadtextbf{(E)} 14$

Problem 2

On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?

$textbf{(A)} 7qquadtextbf{(B)} 32qquadtextbf{(C)} 35qquadtextbf{(D)} 49qquadtextbf{(E)} 112$

Problem 3

The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?

$[asy] import graph; /* this is a label */ Label f; f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0)); yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x) { return x; } draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black); label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10),black+linewidth(5)); dot((15,15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]$

$textbf{(A)} 5qquadtextbf{(B)} 5.5qquadtextbf{(C)} 6qquadtextbf{(D)} 6.5qquadtextbf{(E)} 7$

Problem 4

The five pieces shown below can be arranged to form four of the five figures shown in the choices. Which figure cannot be formed?

$[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p);[/asy]$

$[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]$

$[textbf{(A)}qquadqquadqquadtextbf{(B)}quadqquadqquadtextbf{(C)}:qquadqquadqquadtextbf{(D)}quadqquadqquadtextbf{(E)}]$

Problem 5

A sequence of numbers starts with $1$ , $2$ , and $3$ . The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$ . In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?

$textbf{(A)} 11qquadtextbf{(B)} 20qquadtextbf{(C)} 37qquadtextbf{(D)} 68qquadtextbf{(E)} 99$

Problem 6

Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?

$textbf{(A)} 40qquadtextbf{(B)} 42qquadtextbf{(C)} 44qquadtextbf{(D)} 46qquadtextbf{(E)} 48$

Problem 7

The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?

$[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW);[/asy]$

$textbf{(A)} 2qquadtextbf{(B)} 3qquadtextbf{(C)} 4.5qquadtextbf{(D)} 6qquadtextbf{(E)} 9$

Problem 8

The length of a rectangle is increased by 10% and the width is decreased by 10%. What percent of the old area is the new area?

$textbf{(A)} 90qquadtextbf{(B)} 99qquadtextbf{(C)} 100qquadtextbf{(D)} 101qquadtextbf{(E)} 110$

Problem 9

Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?

$[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]$

$textbf{(A)} 21qquadtextbf{(B)} 23qquadtextbf{(C)} 25qquadtextbf{(D)} 27qquadtextbf{(E)} 29$

Problem 10

On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

$[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$

$textbf{(A)} frac{1}{16}qquadtextbf{(B)} frac{7}{16}qquadtextbf{(C)} frac{1}2qquadtextbf{(D)} frac{9}{16}qquadtextbf{(E)} frac{49}{64}$

Problem 11

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ . Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ . How many more sixth graders than seventh graders bought a pencil?

$textbf{(A)} 1qquadtextbf{(B)} 2qquadtextbf{(C)} 3qquadtextbf{(D)} 4qquadtextbf{(E)} 5$

Problem 12

The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?

$[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]$ $[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$2$",(0,.5)); label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]$

$textbf{(A)} frac{1}{2}qquadtextbf{(B)} frac{2}{3}qquadtextbf{(C)} frac{3}{4}qquadtextbf{(D)} frac{7}{9}qquadtextbf{(E)} frac{5}{6}$

Problem 13

A three-digit integer contains one of each of the digits $1$ , $3$ , and $5$ . What is the probability that the integer is divisible by $5$ ?

$textbf{(A)} frac{1}{6}qquadtextbf{(B)} frac{1}{3}qquadtextbf{(C)} frac{1}{2}qquadtextbf{(D)} frac{2}{3}qquadtextbf{(E)} frac{5}{6}$

Problem 14

Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?

$textbf{(A)} 46qquadtextbf{(B)} 48qquadtextbf{(C)} 50qquadtextbf{(D)} 52qquadtextbf{(E)} 54$

Problem 15

A recipe that makes 5 servings of hot chocolate requires 2 squares of chocolate, $1/4$ cup sugar, 1 cup water and 4 cups milk. Jordan has 5 squares of chocolate, 2 cups of sugar, lots of water and 7 cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make?

$textbf{(A)} 5frac{1}8qquadtextbf{(B)} 6frac{1}4qquadtextbf{(C)} 7frac{1}2qquadtextbf{(D)} 8frac{3}4qquadtextbf{(E)} 9frac{7}8$

Problem 16

How many 3-digit positive integers have digits whose product equals $24$ ?

$textbf{(A)} 12qquadtextbf{(B)} 15qquadtextbf{(C)} 18qquadtextbf{(D)} 21qquadtextbf{(E)} 24$

Problem 17

The positive integers $x$ and $y$ are the two smallest positive integers for which the product of $360$ and $x$ is a square and the product of $360$ and $y$ is a cube. What is the sum of $x$ and $y$ ?

$textbf{(A)} 80qquadtextbf{(B)} 85qquadtextbf{(C)} 115qquadtextbf{(D)} 165qquadtextbf{(E)} 610$

Problem 18

The diagram represents a 7-foot-by-7-foot floor that is tiled with 1-square-foot black tiles and white tiles. Notice that the corners have white tiles. If a 15-foot-by-15-foot floor is to be tiled in the same manner, how many white tiles will be needed?

$[asy] unitsize(10); draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); draw((1,7)--(1,0)); draw((6,7)--(6,0)); draw((5,7)--(5,0)); draw((4,7)--(4,0)); draw((3,7)--(3,0)); draw((2,7)--(2,0)); draw((0,1)--(7,1)); draw((0,2)--(7,2)); draw((0,3)--(7,3)); draw((0,4)--(7,4)); draw((0,5)--(7,5)); draw((0,6)--(7,6)); fill((1,0)--(2,0)--(2,7)--(1,7)--cycle,black); fill((3,0)--(4,0)--(4,7)--(3,7)--cycle,black); fill((5,0)--(6,0)--(6,7)--(5,7)--cycle,black); fill((0,5)--(0,6)--(7,6)--(7,5)--cycle,black); fill((0,3)--(0,4)--(7,4)--(7,3)--cycle,black); fill((0,1)--(0,2)--(7,2)--(7,1)--cycle,black);[/asy]$

$textbf{(A)} 49qquadtextbf{(B)} 57qquadtextbf{(C)} 64qquadtextbf{(D)} 96qquadtextbf{(E)} 126$

Problem 19

Two angles of an isosceles triangle measure $70^circ$ and $x^circ$ . What is the sum of the three possible values of $x$ ?

$textbf{(A)} 95qquadtextbf{(B)} 125qquadtextbf{(C)} 140qquadtextbf{(D)} 165qquadtextbf{(E)} 180$

Problem 20

How many non-congruent triangles have vertices at three of the eight points in the array shown below?

$[asy] dot((0,0)); dot((0,.5)); dot((.5,0)); dot((.5,.5)); dot((1,0)); dot((1,.5)); dot((1.5,0)); dot((1.5,.5));[/asy]$

$textbf{(A)} 5qquadtextbf{(B)} 6qquadtextbf{(C)} 7qquadtextbf{(D)} 8qquadtextbf{(E)} 9$

Problem 21

Andy and Bethany have a rectangular array of numbers with $40$ rows and $75$ columns. Andy adds the numbers in each row. The average of his $40$ sums is $A$ . Bethany adds the numbers in each column. The average of her $75$ sums is $B$ . What is the value of $frac{A}{B}$ ?

$textbf{(A)} frac{64}{225}qquadtextbf{(B)} frac{8}{15}qquadtextbf{(C)} 1qquadtextbf{(D)} frac{15}{8}qquadtextbf{(E)} frac{225}{64}$

Problem 22

How many whole numbers between 1 and 1000 do not contain the digit 1?

$textbf{(A)} 512qquadtextbf{(B)} 648qquadtextbf{(C)} 720qquadtextbf{(D)} 728qquadtextbf{(E)} 800$

Problem 23

On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

$textbf{(A)} 26qquadtextbf{(B)} 28qquadtextbf{(C)} 30qquadtextbf{(D)} 32qquadtextbf{(E)} 34$

Problem 24

The letters $A$ , $B$ , $C$ and $D$ represent digits. If $begin{tabular}{ccc}&A&B +&C&A hline &D&Aend{tabular}$ and $begin{tabular}{ccc}&A&B -&C&A hline &&Aend{tabular}$ , what digit does $D$ represent?

$textbf{(A)} 5qquadtextbf{(B)} 6qquadtextbf{(C)} 7qquadtextbf{(D)} 8qquadtextbf{(E)} 9$

Problem 25

A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $1/2$ foot from the top face. The second cut is $1/3$ foot below the first cut, and the third cut is $1/17$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?

$[asy] size(120); import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label("A", (1,0,3/4), W); label("B", (1,0,1/3), W); label("C", (1,0,1/6-d/4), W); label("D", (1,0,d/2), W); label("1/2", (1,1,3/4), E); label("1/3", (1,1,1/3), E); label("1/17", (0,1,1/6-d/4), E);[/asy]$ $[asy] size(120); import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label("D", (1/2, 1, 0), SE); label("A", (1+1/2, 1, 0), SE); label("B", (2+1/2, 1, 0), SE); label("C", (3+1/2, 1, 0), SE);[/asy]$

$textbf{(A)}:6qquadtextbf{(B)}:7qquadtextbf{(C)}:frac{419}{51}qquadtextbf{(D)}:frac{158}{17}qquadtextbf{(E)}:11$

Let us work backwards. We know that Cassie had 4 apples for herself at the end, and we know she gave away 3 apples before. Therefore, she had 7 apples before giving half of her original amount of apples to someone else. Since half of the amount of original apples is equal to seven, then the original amount of apples Bridget had is $7cdot 2$ , giving us the answer $boxed{textbf{(E)} 14}$ .
This means the ratio is 4:7. If the ratio now is 28:x, then that means $x= boxed{textbf{(D) }49}$ .
Suzanna's speed is $frac{1}{5}$ . This means she runs $frac{1}{5} cdot 30 = boxed{ textbf{(C) }6 }$
The answer is $boxed{textbf{(B)}}$ because the longest piece can't fit in the figure.
List them out, adding the three previous numbers to get the next number, $[1,2,3,6,11,20,37,boxed{textbf{(D)} 68}]$
Each of the four hoses hose fills $24,000/4 = 6,000$ gallons of water. At the rate it goes at it will take $6,000/2.5 = 2400$ minutes, or $boxed{textbf{(A)} 40}$ hours.
The area of a triangle is $frac12 bh$ . If we let $CD$ be the base of the triangle, then the height is $AB$ , and the area is $frac12 cdot 3 cdot 3 = boxed{textbf{(C)} 4.5}$ .
In a rectangle with dimensions $10 times 10$ , the new rectangle would have dimensions $11 times 9$ . The ratio of the new area to the old area is $99/100 = boxed{textbf{(B)} 99}$ .
Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon $3+8-2(1)=9$ sides are on the outside of the final polygon. In the other shapes $4+5+6+7-4(2) = 14$ sides are on the outside. The resulting polygon has $9+14 = boxed{textbf{(B)} 23}$ sides.
There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $frac{36}{64} = boxed{textbf{(D)} frac{9}{16}}$ .
Because the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$ . They can be factored into $11cdot13$ and $3cdot5cdot13$ . The common factor cannot be $1$ or there would have to be more than $30$ sixth graders, so the pencil costs $13$ cents. The difference in costs that the sixth and seventh graders paid is $195-143=52$ cents, which is equal to $52/13 = boxed{textbf{(D)} 4}$ sixth graders.
The possible sums are $[begin{tabular}{c|ccc} & 1 & 3 & 5 hline 2 & 3 & 5 & 7 4 & 5 & 7 & 9 6 & 7 & 9 & 11 end{tabular}]$ Only $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $boxed{textbf{(D)} frac79}$ .
The three digit numbers are $135,153,351,315,513,531$ . The numbers that end in $5$ are divisible are $5$ , and the probability of choosing those numbers is $boxed{textbf{(B)} frac13}$ .
The way to Temple took $frac{50}{60}=frac56$ hours, and the way back took $frac{50}{40}=frac54$ for a total of $frac56 + frac54 = frac{25}{12}$ hours. The trip is $50cdot2=100$ miles. The average speed is $frac{100}{25/12} = boxed{textbf{(B)} 48}$ miles per hour.
Asumming excesses of the other ingredients, the chocolate can make $frac52 cdot 5=12.5$ servings, the sugar can make $frac{2}{1/4} cdot 5 = 40$ servings, the water can make unlimited servings, and the milk can make $frac74 cdot 5 = 8.75$ servings. Limited by the amount of milk, Jordan can make at most $boxed{textbf{(D)} 8 frac34}$ servings.
With the digits listed from least to greatest, the $3$ -digit integers are $138,146,226,234$ . $226$ can be arranged in $frac{3!}{2!} = 3$ ways, and the other three can be arranged in $3!=6$ ways. There are $3+6(3) = boxed{textbf{(D)} 21}$ $3$ -digit positive integers.
The prime factorization of $360=2^3*3^2*5$ . If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is the minimum possible value of x. Similarly, y can be found by making all the exponents divisible by 3, so the minimum possible value of $y$ is $3*5^2=75$ . Thus, our answer is $x+y=10+75=boxed{textbf{(B)} 85}$ .
In a $1$ -foot-by- $1$ -foot floor, there is $1$ white tile. In a $3$ -by- $3$ , there are $4$ . Continuing on, you can deduce the $n^{th}$ positive odd integer floor has $n^2$ white tiles. $15$ is the $8^{th}$ odd integer, so there are $boxed{textbf{(C)} 64}$ white tiles.
There are 3 cases: where $x^circ$ is a base angle with the $70^circ$ as the other angle, where $x^circ$ is a base angle with $70^circ$ as the vertex angle, and where $x^circ$ is the vertex angle with $70^circ$ as a base angle.Case 1: $x^circ$ is a base angle with the $70^circ$ as the other angle: Here, $x=70$ , since base angles are congruent.Case 2: $x^circ$ is a base angle with $70^circ$ as the vertex angle: Here, the 2 base angles are both $x^circ$ , so we can use the equation $2x+70=180$ , which simplifies to $x=55$ .Case 3: $x^circ$ is the vertex angle with $70^circ$ as a base angle: Here, both base angles are $70^circ$ , since base angles are congruent. Thus, we can use the equation $x+140=180$ , which simplifies to $x=40$ .Adding up all the cases, we get $70+55+40=165$ , so the answer is $boxed{textbf{(D)} 165}$ .
Assume the base of the triangle is on the bottom four points, because a congruent triangle can be made by reflecting the base on the top four points. For a triangle with a base of length $1$ , there are $3$ triangles. For a triangle with a base of length $2$ , there are $3$ triangles. For length $3$ , there are $2$ . In total, the number of non-congruent triangles is $3+3+2=boxed{textbf{(D)} 8}$ .
Note that $40A=75B=text{sum of the numbers in the array}$ . This means that $frac{A}{B}=boxed{text{(D) } frac{15}{8}}$ using basic algebraic manipulation.
Note that this is the same as finding how many numbers with up to three digits do not contain 1.Since there are 10 total possible digits, and only one of them is not allowed (1), each digit has its choice of 9 digits, for a total of $9*9*9=729$ such numbers. However, we over counted by one; 0 is not between 1 and 1000, so there are $boxed{textbf{(D)} 728}$ numbers.
If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans. $begin{align*} x^2+(x+2)^2 &= 394 x^2+x^2+4x+4 &= 394 2x^2 + 4x - 390 &= 0 x^2 + 2x - 195 &= 0 (x+15)(x-13) &=0 end{align*}$ Because $x=-15,13$ , there are $13$ girls, $15$ boys, and $13+15=boxed{textbf{(B)} 28}$ students.
Because $B+A=A$ , $B$ must be $0$ . Next, because $B-A=Aimplies0-A=A,$ we get $A=5$ Now we can rewrite this as $begin{tabular}{ccc}&5&0 +&C&5 hline &D&5end{tabular}$ . Therefore, $D=5+C$ Finally, $A-1-C=0implies{A=C+1}implies{C=4}$ , So we have $D=5+Cimplies{D=5+4}=boxed{textbf{(E)} 9 }$ .
The tops of $A$ , $B$ , $C$ , and $D$ in the figure formed has sum $1+1+1+1 = 4$ as do the bottoms. Thus, the total so far is $8$ . Now, one of the sides has area one, since it combines all of the heights of $A$ , $B$ , $C$ , and $D$ , which is $1$ . The other side also satisfies this. Thus the total area now is $10$ . From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like $A$ , with surface area half. From the back it is the same thing. Thus, the total is $10+frac{1}{2}+frac{1}{2}= 11$ , or $boxed{textbf{(E)}:11 }$ .