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2970951d406c3e91dbf40ac93bad62e6 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_5 | Hammie is in $6^\text{th}$ grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?
$\textbf{(A)}\ \text{median, by 60} \qquad \textbf{(B)}\ \text{median, by 20} \qquad \textbf{(C)}\ \text{average, by 5}\qquad \textbf{(D)}\ \text{average, by 15} \qquad \textbf{(E)}\ \text{average, by 20}$ | Listing the elements from least to greatest, we have $(5, 5, 6, 8, 106)$ , we see that the median weight is 6 pounds.
The average weight of the five kids is $\frac{5+5+6+8+106}{5} = \frac{130}{5} = 26$
Hence, \[26-6=\boxed{20}.\] | E | 20 |
c78d8106011138b0100f9f9a1293bd76 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_6 | The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$ . What is the missing number in the top row?
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); [/asy]
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | Let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("$x$",(1,1)); label("$y$",(2,3)); [/asy]
From the diagram, $600 = 30x$ , making $x = 20$
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("20",(1,1)); label("$y$",(2,3)); [/asy]
It follows that $20 = 5y$ , so $y = \boxed{4}$ | C | 4 |
c78d8106011138b0100f9f9a1293bd76 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_6 | The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, $30 = 6\times5$ . What is the missing number in the top row?
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); [/asy]
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | Another way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. Again, let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.
[asy] unitsize(0.8cm); draw((-1,0)--(1,0)--(1,-2)--(-1,-2)--cycle); draw((-2,0)--(0,0)--(0,2)--(-2,2)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((-3,2)--(-1,2)--(-1,4)--(-3,4)--cycle); draw((-1,2)--(1,2)--(1,4)--(-1,4)--cycle); draw((1,2)--(1,4)--(3,4)--(3,2)--cycle); label("600",(0,-1)); label("30",(-1,1)); label("6",(-2,3)); label("5",(0,3)); label("$x$",(1,1)); label("$y$",(2,3)); [/asy]
We can write some equations:
$600 = 30x\\ 30 = 6\cdot 5\\ x = 5y$
Now we can substitute into the first equation using the two others:
$600 = (6\cdot5)(5y)\\ 600= 6\cdot5\cdot5\cdot y\\ 600=6\cdot25\cdot y\\ 600 = 150y\\ \boxed{4} = y$ | C | 4 |
0c28121cf9adf6b50cf209747ebc2c34 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_7 | Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$ | Clearly, for every $5$ seconds, $3$ cars pass. It's more convenient to have everything in seconds: $2$ minutes and $45$ seconds $=2\cdot60 + 45 = 165$ seconds. We then set up a ratio: \[\frac{3}{5}=\frac{x}{165}\] \[3(165)=5x\] \[x=3(33)=99\approx\boxed{100}.\] | C | 100 |
0c28121cf9adf6b50cf209747ebc2c34 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_7 | Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$ | $2$ minutes and $45$ seconds is equal to $120+45=165\text{ seconds}$
Since $6$ cars pass at around $10$ seconds, there are about $\left \lfloor{\dfrac{165}{10}}\right \rfloor =16$ groups of $6$ cars. There are about $16\cdot6=96\text{ cars}$ , so the closest answer choice is $\boxed{100}$ | C | 100 |
333f6568c77b3eb090f5d04390b77e0f | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_9 | The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer (1,000 meters)?
$\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}$ | This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that $2^{10}=1024$
However, because the first term is $2^0=1$ and not $2^1=2$ , the solution to the problem is $10-0+1=\boxed{11}$ | C | 11 |
c2ce56e25aa82757ee6d584adc1ba0dd | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_10 | What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$ | To find either the LCM or the GCF of two numbers, always prime factorize first.
The prime factorization of $180 = 3^2 \times 5 \times 2^2$
The prime factorization of $594 = 3^3 \times 11 \times 2$
Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$ ). Multiply all of these to get 5940.
For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. $3^2 \times 2$ = 18.
Thus the answer = $\frac{5940}{18}$ $\boxed{330}$ | C | 330 |
c2ce56e25aa82757ee6d584adc1ba0dd | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_10 | What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$ | We start off with a similar approach as the original solution. From the prime factorizations, the GCF is $18$
It is a well known fact that $\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|$ . So we have, $18\times \operatorname{lcm} (180,594)=594\times 180$
Dividing by $18$ yields $\operatorname{lcm} (180,594)=594\times 10=5940$
Therefore, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{330}$ | C | 330 |
c2ce56e25aa82757ee6d584adc1ba0dd | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_10 | What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$ | From Solution 1,
the prime factorization of $180 = 2^2 \cdot 3^2 \cdot 5$ .
The prime factorization of $594 = 2 \cdot 3^3 \cdot 11$ .
Hence, $\operatorname{lcm} (180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11$ , and $\operatorname{gcf} (180,594) = 2 \cdot 3^2$ .
Therefore, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330 \Longrightarrow \boxed{330}$ | C | 330 |
243b55e69c8d1bf51a4f2a09e1cace45 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_11 | Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | We use that fact that $d=rt$ . Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.
On Monday, he was at a rate of $5 \text{ m.p.h}$ . So, $5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}$
For Wednesday, he walked at a rate of $3 \text{ m.p.h}$ . Therefore, $3x = 2 \text{ miles}\implies x = \frac{2}{3} \text { hours}$
On Friday, he walked at a rate of $4 \text{ m.p.h}$ . So, $4x = 2 \text{ miles}\implies x=\frac{2}{4}=\frac{1}{2} \text {hours}$
Adding up the hours yields $\frac{2}{5} \text { hours}$ $\frac{2}{3} \text { hours}$ $\frac{1}{2} \text { hours}$ $\frac{47}{30} \text { hours}$
We now find the amount of time Grandfather would have taken if he walked at $4 \text{ m.p.h}$ per day. Set up the equation, $4x = 2 \text{ miles} \times 3 \text{ days}\implies x = \frac{3}{2} \text { hours}$
To find the amount of time saved, subtract the two amounts: $\frac{47}{30} \text { hours}$ $\frac{3}{2} \text { hours}$ $\frac{1}{15} \text { hours}$ . To convert this to minutes, we multiply by $60$
Thus, the solution to this problem is $\dfrac{1}{15}\times 60=\boxed{4}$ | D | 4 |
2d18c5e8e2b2cb8a3076da1b50461b74 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_12 | At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?
$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 40 \qquad \textbf{(E)}\ 45$ | First, find the amount of money one will pay for three sandals without the discount. We have $\textdollar 50\times 3 \text{ sandals} = \textdollar 150$
Then, find the amount of money using the discount: $50 + 0.6 \times 50 + \frac{1}{2} \times 50 = \textdollar 105$
Finding the percentage yields $\frac{105}{150} = 70 \%$
To find the percent saved, we have $100 \% -70 \%= \boxed{30}$ | B | 30 |
e9966e15d15e837cc871e921fb7a5319 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_14 | Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$ | The favorable responses are either they both show a green bean or they both show a red bean. The probability that both show a green bean is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$ . The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$ . Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{38}$ | C | 38 |
e9966e15d15e837cc871e921fb7a5319 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_14 | Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$ | We can list out all the combinations and we get this: $GG, GY, GR_1, GR_2, RG, RY, RR_1, RR_2$ . There is a total of 8 combinations and 3 that are the same. Hence, we yield the answer $\boxed{38}$ | C | 38 |
5d77f13b2b5103309b57cf5acd9c9aac | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_16 | A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$ | We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:
$5:3 = 5(8):3(8) = 40:24$
$8:5 = 8(5):5(5) = 40:25$
Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$ . Since the ratio is in lowest terms, the smallest number of students participating in the project is $40+25+24 = \boxed{89}$ | E | 89 |
5d77f13b2b5103309b57cf5acd9c9aac | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_16 | A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$ | The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are $40*\frac{3}{5}=24$ 6th graders and $40*\frac{5}{8}=25$ 7th graders. The numbers of students is $40+24+25=\boxed{89}$ | E | 89 |
e1c4105b55936ff2e8b7e26d5a3f5ab0 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$ | The arithmetic mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$ . Therefore the numbers are $333$ $334$ $335$ $336$ $337$ $338$ , so the answer is $\boxed{338}$ | B | 338 |
e1c4105b55936ff2e8b7e26d5a3f5ab0 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$ | Let the $4^{\text{th}}$ number be $x$ . Then our desired number is $x+2$
Our integers are $x-3,x-2,x-1,x,x+1,x+2$ , so we have that $6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{338}$ | B | 338 |
e1c4105b55936ff2e8b7e26d5a3f5ab0 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$ | Let the first term be $x$ . Our integers are $x,x+1,x+2,x+3,x+4,x+5$ . We have, $6x+15=2013\implies x=333\implies x+5=\boxed{338}$ | B | 338 |
e1c4105b55936ff2e8b7e26d5a3f5ab0 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$ | Since there are $6$ numbers, we divide $2013$ by $6$ to find the mean of the numbers. $\frac{2013}{6} = 335 \frac{1}{2}$ .
Then, $335 \frac{1}{2} + \frac{1}{2} = 336$ (the fourth number). Fifth: $337$ ; Sixth: $\boxed{338}$ | null | 338 |
e1c4105b55936ff2e8b7e26d5a3f5ab0 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_17 | The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$ | Let the $6th$ number be $x$ . Then our list is: $x-6+x-5+x-4+x-3+x-x-1=2013$ . Simplifying this gets you $6x-21=2013\implies 6x=2034$ , which means that $x = \boxed{338}$ | B | 338 |
830165c4d781fc913031665762f3630a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_18 | Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label("10",(A--B),SE); label("12",(C--B),SW); label("5",(F--B),W);[/asy]
$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$ | There are $10 \cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11 + 9 + 11 = 40$ cubes. Hence, the answer is $120 + 4 \cdot 40 = \boxed{280}$ | B | 280 |
830165c4d781fc913031665762f3630a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_18 | Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label("10",(A--B),SE); label("12",(C--B),SW); label("5",(F--B),W);[/asy]
$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$ | We can just calculate the volume of the prism that was cut out of the original $12\times 10\times 5$ box. Each interior side of the fort will be $2$ feet shorter than each side of the outside. Since the floor is $1$ foot, the height will be $4$ feet. So the volume of the interior box is $10\times 8\times 4=320\text{ ft}^3$
The volume of the original box is $12\times 10\times 5=600\text{ ft}^3$ . Therefore, the number of blocks contained in the fort is $600-320=\boxed{280}$ | B | 280 |
5d6d92c0a1eb0074cc8690cd77d3e739 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_1 | Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighbourhood picnic?
$\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf{(E)}\hspace{.05in}9$ | Since Rachelle uses $3$ pounds of meat to make $8$ hamburgers, she uses $\frac{3}{8}$ pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or $\frac{3}{8} \cdot 24 = \boxed{9}$ | E | 9 |
311067f9ebcb05380dbbd3e0579c50a1 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_2 | In the country of East Westmore, statisticians estimate there is a baby born every $8$ hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?
$\textbf{(A)}\hspace{.05in}600\qquad\textbf{(B)}\hspace{.05in}700\qquad\textbf{(C)}\hspace{.05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000$ | There are $24\text{ hours}\div8\text{ hours} = 3$ births and one death everyday in East Westmore. Therefore, the population increases by $3$ $1$ $2$ people everyday. Thus, there are $2 \times 365 = 730$ people added to the population every year. Rounding, we find the answer is $\boxed{700}$ | B | 700 |
d1fdfbca4170bee661e145fdb491b6fb | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_5 | In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label("$X$",(3.4,1.5)); label("6",(7.6,1.5)); label("1",(7.6,3.5)); label("1",(8.4,4.6)); label("2",(9.4,4.6)); label("2",(10.4,6)); label("3",(8.4,7.4)); label("1",(7.5,7.8)); label("2",(6,8.5)); label("1",(4.7,7.8)); label("1",(4.3,7.5)); label("1",(3.5,6.5)); label("4",(1.8,6.5)); label("1",(-0.5,5.5)); label("2",(0.8,4.5)); label("2",(1.5,3.8)); label("2",(2.8,2.6));[/asy]
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$ | 2012amc85.png
$1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\ 5 + X = 10\\ X = 5$
Thus, the answer is $\boxed{5}$ | E | 5 |
d1fdfbca4170bee661e145fdb491b6fb | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_5 | In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label("$X$",(3.4,1.5)); label("6",(7.6,1.5)); label("1",(7.6,3.5)); label("1",(8.4,4.6)); label("2",(9.4,4.6)); label("2",(10.4,6)); label("3",(8.4,7.4)); label("1",(7.5,7.8)); label("2",(6,8.5)); label("1",(4.7,7.8)); label("1",(4.3,7.5)); label("1",(3.5,6.5)); label("4",(1.8,6.5)); label("1",(-0.5,5.5)); label("2",(0.8,4.5)); label("2",(1.5,3.8)); label("2",(2.8,2.6));[/asy]
$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$ | Note that we only need to consider the value below the marked red line, so we have the equation: \[X + 2 = 6 + 1\] \[X = 5\]
Hence, the answer is $\boxed{5}$ | E | 5 |
4199157ded7b381e90118019b3d5c069 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_6 | A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures $8$ inches high and $10$ inches wide. What is the area of the border, in square inches?
$\textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88$ | In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is $8 \times 10 = 80$ square inches. The height of the whole frame (including the photograph) would be $8+2+2 = 12$ , and the width of the whole frame, $10+2+2 = 14$ . Therefore, the area of the whole figure would be $12 \times 14 = 168$ square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be $168-80 = \boxed{88}$ | E | 88 |
c6f37829b48d77339cf57f3e375e50f3 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_7 | Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
$\textbf{(A)}\hspace{.05in}90\qquad\textbf{(B)}\hspace{.05in}92\qquad\textbf{(C)}\hspace{.05in}95\qquad\textbf{(D)}\hspace{.05in}96\qquad\textbf{(E)}\hspace{.05in}97$ | Isabella wants an average grade of $95$ on her 4 tests; this also means that she wants the sum of her test scores to be at least $95 \times 4 = 380$ (if she goes over this number, she'll be over her goal!). She's already taken two tests, which sum to $97+91 = 188$ , which means she needs $192$ more points to achieve her desired average. In order to minimize the score on the third test, we assume that Isabella will receive all $100$ points on the fourth test. Therefore, the lowest Isabella could have scored on the third test would be $192-100 = \boxed{92}$ | B | 92 |
c6f37829b48d77339cf57f3e375e50f3 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_7 | Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?
$\textbf{(A)}\hspace{.05in}90\qquad\textbf{(B)}\hspace{.05in}92\qquad\textbf{(C)}\hspace{.05in}95\qquad\textbf{(D)}\hspace{.05in}96\qquad\textbf{(E)}\hspace{.05in}97$ | Isabella needs to lose a max of $20$ . When she got $97$ and $91$ , the max goes down to $20 - 3 - 9 = 8$ . If we get $100$ on her last test, then she will get $\boxed{92}$ on the third test. | B | 92 |
b81007ad053def7172abcecdcb564526 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_8 | A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$ | Let the original price of an item be $x$
First, everything is half-off, so the price is now $\frac{x}{2} = 0.5x$
Next, the extra coupon applies 20% off on the sale price , so the price after this discount will be $100\% - 20\% = 80\%$ of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)
$80\% \cdot 0.5 x = \frac{4}{5} \cdot 0.5 x = 0.4x$
The price of the item after all discounts have been applied is $0.4x = 40\% \cdot x$ . However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract $40\% x$ from $100\% x$ (the original price of the item), to find the answer, $\boxed{60}$ | D | 60 |
b81007ad053def7172abcecdcb564526 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_8 | A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$ | Since the problem implies that the percentage off the original price will be the same for every item in the store, fakesolving is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a total of $\frac{10-4}{10} = \frac{6}{10} = \frac{60}{100} = \boxed{60} \%$ | D | 60 |
b81007ad053def7172abcecdcb564526 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_8 | A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$ | The price is $50\%$ off, which means the price would be $100\% - 50\% = 50\%$ of what it was before. Then, you get $20\%$ off. Sidenote: A common mistake is finding $20\%$ of the price, but forgetting to subtract it. $20\%$ off the price is $100\% - 20\% = 80\%.$ Now, we can multiply the two discounts as decimals. $0.5 \times 0.8 = 0.4 = 40\%.$ Our final answer is $100\% - 40\% = \boxed{60}$ | D | 60 |
4693961df8f8e7298d7ff82c5388dbcf | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_9 | The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$ | Let the number of two-legged birds be $x$ and the number of four-legged mammals be $y$ . We can now use systems of equations to solve this problem.
Write two equations:
$2x + 4y = 522$
$x + y = 200$
Now multiply the latter equation by $2$
$2x + 4y = 522$
$2x + 2y = 400$
By subtracting the second equation from the first equation, we find that $2y = 122 \implies y = 61$ . Since there were $200$ heads, meaning that there were $200$ animals, there were $200 - 61 = \boxed{139}$ two-legged birds. | C | 139 |
4693961df8f8e7298d7ff82c5388dbcf | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_9 | The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$ | Let's add 2 legs for each bird and assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be $4*200=800$ . Actually, there were only $522$ legs. The difference between these two numbers exactly gives us the number of all the legs we added for all birds: $800 - 522 = 278$ . Because each bird was added by 2 legs, so the total number of birds would be $278/2 = \boxed{139}$ two-legged birds. ---LarryFlora | C | 139 |
90dba02ecdb2aa188c943e616dce9da0 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_10 | How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$ | For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$ , since all of the valid 4-digit number will always be greater than $1000$ . The best way to solve this problem is by using casework.
There can be only two leading digits, namely $1$ or $2$
When the leading digit is $1$ , you can make $\frac{3!}{2!1!} \implies 3$ such numbers.
When the leading digit is $2$ , you can make $3! \implies 6$ such numbers.
Summing the amounts of numbers, we find that there are $\boxed{9}$ such numbers. | D | 9 |
90dba02ecdb2aa188c943e616dce9da0 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_10 | How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$ | Notice that the first digit cannot be $0$ , as the number is greater than $1000$ . Therefore, there are three digits that can be in the thousands.
The rest three digits of the number have no restrictions, and therefore there are $3! \implies 6$ for each leading digit.
Since the two $2$ 's are indistinguishable, there are $\frac {3\cdot6}{2}$ such numbers $\implies \boxed{9}$ | D | 9 |
904d4f9b7ca768cc62644aefebcea21d | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_11 | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$ | We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$ , because of the above statement. Now we need to test the remaining answer choices.
Case 1: $x = 6$
Mode: $6$
Median: $6$
Mean: $\frac{37}{7}$
Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated.
Case 2: $x = 11$
Mode: $6$
Median: $6$
Mean: $6$
We are done with this problem, because we have found when $x = 11$ , the condition is satisfied. Therefore, the answer is $\boxed{11}$ | D | 11 |
904d4f9b7ca768cc62644aefebcea21d | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_11 | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$ | Notice that the mean of this set of numbers, in terms of $x$ , is:
$\frac{3+4+5+6+6+7+x}{7} = \frac{31+x}{7}$
Because we know that the mode must be $6$ (it can't be any of the numbers already listed, as shown above, and no matter what $x$ is, either $6$ or a new number, it will not affect $6$ being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and $6$ equal:
\[\dfrac{31+x}{7}=6\] \[31+x=42\] \[x=\boxed{11}\] | D | 11 |
904d4f9b7ca768cc62644aefebcea21d | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_11 | The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$ | We know the unique mode must be $6$ , so the mean must be the same number $6$ . Let's imagine a scale. $6$ exactly stands the mid-point of the scale. Numbers of $3,4,5$ represent the left side "weights" of the scale. Numbers of $6,7, x$ represent the right side "weights" of the scale. On the left side, the difference of the three "weights" between $6$ are $-3, -2, -1$ , respectively. It gives us the total difference is $-6$ . In order to allow the scale to keep balance, on the right side, the total difference must be $+6$ . Because we have already known the difference of the right side "weights" between $6$ is $0+1=1$ , partially, so the difference between $6$ and unknown $x$ must be $+6-1=+5$ . It exactly gives us the answer: $x=6+5= \boxed{11}$ . ---LarryFlora | D | 11 |
6b1c7d226d6e32eb22916acb6d818b3f | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_13 | Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$ . Sharona bought some of the same pencils and paid $\textdollar 1.87$ . How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6$ | We assume that the price of the pencils remains constant. Convert $\textdollar 1.43$ and $\textdollar 1.87$ to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of $143$ and $187$ , which is $11$ . Therefore, Jamar bought $\frac{143}{11} \implies 13$ pencils and Sharona bought $\frac{187}{11} \implies 17$ pencils. Thus, Sharona bought $17-13 = \boxed{4}$ more pencils than Jamar. | C | 4 |
6b1c7d226d6e32eb22916acb6d818b3f | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_13 | Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$ . Sharona bought some of the same pencils and paid $\textdollar 1.87$ . How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6$ | We find the difference between $1.43$ and $1.87$ is $1.87-1.43 = 0.44$ , which is the extra cost of Sharona's pencils than Jamar's pencils. Because the difference between the amounts of the pencils they bought must be divided evenly by $0.44$ , looking into the answers, $2$ or $4$ is possibly correct. It gives us the price of each pencil should be $0.44/2=0.22$ or $0.44/4=0.11$ , respectively. Then we find only $0.11$ can be divided evenly by $1.43$ and $1.87$ . So the answer is $\boxed{4}$ | C | 4 |
6b1c7d226d6e32eb22916acb6d818b3f | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_13 | Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\textdollar 1.43$ . Sharona bought some of the same pencils and paid $\textdollar 1.87$ . How many more pencils did Sharona buy than Jamar?
$\textbf{(A)}\hspace{.05in}2\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}4\qquad\textbf{(D)}\hspace{.05in}5\qquad\textbf{(E)}\hspace{.05in}6$ | You can quickly tell, if you have memorized your times tables, that $187$ and $143$ are both multiples of $11$ . Additionally, the difference is 44 cents. That means that there are $\boxed{4}$ more pencils purchased by Sharona because $11*4 = 44$ | C | 4 |
d2156e43045e30a5e15469d3206f0a4b | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_14 | In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10$ | This problem is very similar to a handshake problem. We use the formula $\frac{n(n-1)}{2}$ to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.
So we have the equation $\frac{n(n-1)}{2} = 21$ . Solving, we find that the number of teams in the BIG N conference is $\boxed{7}$ | B | 7 |
d2156e43045e30a5e15469d3206f0a4b | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_14 | In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10$ | (If someone understands what I'm trying to do here and can explain it better, please edit it)We know that every team has to play a game with every other team, so we just need to find out how many consecutive numbers, $1$ to $x$ , can fit into 21. We know that $6+5+4+3+2+1=21$ , and since this doesn't count to $7th$ team that shook hands with the other $6$ , we know that there are $\boxed{7}$ teams in the BIG N conference. | B | 7 |
6b35272ddb34481e91612079fa6304a6 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_16 | Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403$ | In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: $76531$ and $87431$ . To determine the answer we will have to use estimation and the first two digits of the numbers.
For $76531$ the number that would maximize the sum would start with $98$ . The first two digits of $76531$ (when rounded) are $77$ . Adding $98$ and $77$ , we find that the first three digits of the sum of the two numbers would be $175$
For $87431$ the number that would maximize the sum would start with $96$ . The first two digits of $87431$ (when rounded) are $87$ . Adding $96$ and $87$ , we find that the first three digits of the sum of the two numbers would be $183$
From the estimations, we can say that the answer to this problem is $\boxed{87431}$ | C | 87431 |
6b35272ddb34481e91612079fa6304a6 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_16 | Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403$ | In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of this are $97531$ and $86420$ . The digits can be interchangeable between numbers because we only care about the actual digits.
The first digit must be either $9$ or $8$ . This immediately knocks out $\textbf{(A)}\ 76531$
The second digit must be either $7$ or $6$ . This doesn't cancel any choices.
The third digit must be either $5$ or $4$ . This knocks out $\textbf{(B)}\ 86724$ and $\textbf{(D)}\ 96240$
The fourth digit must be $3$ or $2$ . This cancels out $\textbf{(E)}\ 97403$
This leaves us with $\boxed{87431}$ | C | 87431 |
6b35272ddb34481e91612079fa6304a6 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_16 | Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403$ | If we use intuition, we know that the digits will be IN ORDER, to maximze the number. This eliminates $\textbf{(B)}$ $\textbf{(D)}$ ,and $\textbf{(E)}$ . Additionally, the 2 two numbers must have 9 and 8 for the first digit to maximize the sum, eliminating $\textbf{(A)}$ . This leaves $\boxed{87431}$ .
-written by litttle_master purely, not copied from anywhere | C | 87431 |
73945c82bf744c036446e98e96903933 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_17 | A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
$\text{(A)}\hspace{.05in}3\qquad\text{(B)}\hspace{.05in}4\qquad\text{(C)}\hspace{.05in}5\qquad\text{(D)}\hspace{.05in}6\qquad\text{(E)}\hspace{.05in}7$ | The first answer choice ${\textbf{(A)}\ 3}$ , can be eliminated since there must be $10$ squares with integer side lengths. We then test the next smallest sidelength which is $4$ . The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$ , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is $\boxed{4}$ | B | 4 |
161e30a9cc3ab2e932a43ae9319b5d2e | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_18 | What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
$\textbf{(A)}\hspace{.05in}3127\qquad\textbf{(B)}\hspace{.05in}3133\qquad\textbf{(C)}\hspace{.05in}3137\qquad\textbf{(D)}\hspace{.05in}3139\qquad\textbf{(E)}\hspace{.05in}3149$ | The problem states that the answer cannot be a perfect square or have prime factors less than $50$ . Therefore, the answer will be the product of at least two different primes greater than $50$ . The two smallest primes greater than $50$ are $53$ and $59$ . Multiplying these two primes, we obtain the number $3127$ , which is also the smallest number on the list of answer choices.
So we are done, and the answer is $\boxed{3127}$ | A | 3127 |
a51568069b105b6527c173091b34a261 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | $6$ are blue and green - $b+g=6$
$8$ are red and blue - $r+b=8$
$4$ are red and green- $r+g=4$
We can do trial and error. Let's make blue $5$ . That makes green $1$ and red $3$ because $6-5=1$ and $8-5=3$ . To check this, let's plug $1$ and $3$ into $r+g=4$ , which works. Now count the number of marbles - $5+3+1=9$ . So the answer is $\boxed{9}.$ | C | 9 |
a51568069b105b6527c173091b34a261 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | We already knew the facts: $6$ are blue and green, meaning $b+g=6$ $8$ are red and blue, meaning $r+b=8$ $4$ are red and green, meaning $r+g=4$ . Then we need to add these three equations: $b+g+r+b+r+g=2(r+g+b)=6+8+4=18$ . It gives us all of the marbles are $r+g+b = 18/2 = 9$ . So the answer is $\boxed{9}$ .
~LarryFlora | C | 9 |
a51568069b105b6527c173091b34a261 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | We may draw three Venn diagrams to represent these three cases, respectively.
Screen Shot 2021-08-29 at 9.14.51 AM.png
Let the amount of all the marbles be $x$ , meaning $R+G+B = x$
The Venn diagrams give us the equation: $x = (x-6)+(x-8)+(x-4)$ .
So $x = 3x-18$ $x = 18/2 =9$ .
Thus, the answer is $\boxed{9}$ .
~LarryFlora | C | 9 |
a51568069b105b6527c173091b34a261 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | Since we know all but $8$ marbles in the jar are green, the jar must have at least $9$ marbles. Then we can just start from $C$ and keep going. If there are $9$ marbles total, there are $3$ red marbles $(9-6)$ $1$ green marble $(9-8)$ , and $5$ blue marbles $(9-4)$ . Since we assumed there were $9$ marbles and $3+1+5=9$ , the answer is $\boxed{9}$ | C | 9 |
a51568069b105b6527c173091b34a261 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | Let $x$ be the number of total marbles. There are $x – 6$ red marbles, $x – 8$ green marbles, and $x – 4$ blue marbles.
We can create an equation: $(x – 6)+(x – 8)+(x – 4)=x$ Solving, we get $x=9$ , which means the total number of marbles is $\boxed{9}$ .
-J.L.L
(Feel free to edit) | C | 9 |
a51568069b105b6527c173091b34a261 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_19 | In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$ | Let $x$ be the number of total marbles, $r$ be the number of red marbles, $g$ be the number of green marbles, and $b$ be the number of blue marbles. Then we have $x - r = 6$ $x - g = 8$ $x - b = 4$ , and $r + g + b = x$ . Adding the first three equations together, we get $3x - r - g - b = 18$ or $3x - (r + g + b) = 18$ . Substituting in the fourth equation, we have $3x - x = 18$ $\implies$ $\boxed{9}$ | C | 9 |
a8cab20a7338028ed9a6d1bca92f50f8 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_21 | Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet? $\textbf{(A)}\hspace{.05in}5\sqrt2\qquad\textbf{(B)}\hspace{.05in}10\qquad\textbf{(C)}\hspace{.05in}10\sqrt2\qquad\textbf{(D)}\hspace{.05in}50\qquad\textbf{(E)}\hspace{.05in}50\sqrt2$ | If Marla evenly distributes her $300$ square feet of paint between the 6 faces, each face will get $300\div6 = 50$ square feet of paint. The surface area of one of the faces of the cube is $10^2 = 100$ square feet. Therefore, there will be $100-50 = \boxed{50}$ square feet of white on each side. | D | 50 |
4cfb1d5cbf1040d9c3177e27aea14dbb | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_23 | An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$ | Let the perimeter of the equilateral triangle be $3s$ . The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$ , since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is $1$ . The area of the hexagon is then $1 \times 6 = \boxed{6}$ | C | 6 |
4cfb1d5cbf1040d9c3177e27aea14dbb | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_23 | An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$ | Let the side length of the equilateral triangle be $s$ and the side length of the hexagon be $y$ . Since the perimeters are equal, we must have $3s=6y$ which reduces to $s=2y$ . Substitute this value in to the area of an equilateral triangle to yield $\dfrac{(2y)^2\sqrt{3}}{4}=\dfrac{4y^2\sqrt{3}}{4}$
Setting this equal to $4$ gives us $\dfrac{4y^2\sqrt{3}}{4}=4\implies 4y^2\sqrt{3}=16\implies y^2\sqrt{3}=4$
Substitute $y^2\sqrt{3}$ into the area of a regular hexagon to yield $\dfrac{3(4)}{2}=6$
Therefore, our answer is $\boxed{6}$ | C | 6 |
4cfb1d5cbf1040d9c3177e27aea14dbb | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_23 | An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$ | Let the side length of the triangle be $s$ and the side length of the hexagon be $t$ . As explained in Solution 1, $s=2t$ , or $t=\frac{s}{2}$ . The area of the triangle is $\frac{s^2\sqrt3}{4}=4$ and the area of the hexagon is $\frac{t^2\sqrt3}{4} \cdot 6=\frac{3t^2\sqrt3}{2}$ . Substituting $\frac{s}{2}$ in for $t$ , we get \[\frac{\frac{3s^2\sqrt3}{4}}{2}=\frac{3s^2\sqrt3}{8}.\] $\frac{s^2\sqrt3}{4}=4 \implies \frac{s^2\sqrt3}{8}=2 \implies \frac{3s^2\sqrt3}{8}=\boxed{6}$ | C | 6 |
232276e084901140b445eb5df357bf82 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_24 | A circle of radius $2$ is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?
[asy] size(0,50); draw((-1,1)..(-2,2)..(-3,1)..(-2,0)..cycle); dot((-1,1)); dot((-2,2)); dot((-3,1)); dot((-2,0)); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy]
$\textbf{(A)}\hspace{.05in}\frac{4-\pi}{\pi}\qquad\textbf{(B)}\hspace{.05in}\frac{1}\pi\qquad\textbf{(C)}\hspace{.05in}\frac{\sqrt2}{\pi}\qquad\textbf{(D)}\hspace{.05in}\frac{\pi-1}{\pi}\qquad\textbf{(E)}\hspace{.05in}\frac{3}\pi$ | [asy] dot((0,0),red); dot((0,2),red); dot((2,0),red); dot((2,2),red); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,red); size(0,50); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy]
Draw a square around the star figure. The side length of this square is $4$ , because the side length is the diameter of the circle. The square forms $4$ -quarter circles around the star figure. This is the equivalent of one large circle with radius $2$ , meaning that the total area of the quarter circles is $4\pi$ . The area of the square is $16$ . Thus, the area of the star figure is $16 - 4\pi$ . The area of the circle is $4\pi$ . Taking the ratio of the two areas, we find the answer is $\boxed{4}$ | A | 4 |
2a7e8708ed4a45a5c53ea289159e908b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_1 | Margie bought $3$ apples at a cost of $50$ cents per apple. She paid with a 5-dollar bill. How much change did Margie receive?
$\textbf{(A)}\ \textdollar 1.50 \qquad \textbf{(B)}\ \textdollar 2.00 \qquad \textbf{(C)}\ \textdollar 2.50 \qquad \textbf{(D)}\ \textdollar 3.00 \qquad \textbf{(E)}\ \textdollar 3.50$ | $50$ cents is equivalent to $\textdollar 0.50.$ Then the three apples cost $3 \times \textdollar 0.50 = \textdollar 1.50.$ The change Margie receives is $\textdollar 5.00 - \textdollar 1.50 = \boxed{3.50}$ | E | 3.50 |
5687e6d1deed26c82fbc4866e85bad5b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_2 | Karl's rectangular vegetable garden is $20$ feet by $45$ feet, and Makenna's is $25$ feet by $40$ feet. Whose garden is larger in area?
$\textbf{(A)}\ \text{Karl's garden is larger by 100 square feet.}$
$\textbf{(B)}\ \text{Karl's garden is larger by 25 square feet.}$
$\textbf{(C)}\ \text{The gardens are the same size.}$
$\textbf{(D)}\ \text{Makenna's garden is larger by 25 square feet.}$
$\textbf{(E)}\ \text{Makenna's garden is larger by 100 square feet.}$ | The area of a rectangle is given by the formula length times width. Karl's garden is $20 \times 45 = 900$ square feet and Makenna's garden is $25 \times 40 = 1000$ square feet. Since $1000 > 900,$ Makenna's garden is larger by $1000-900=100$ square feet. $\Rightarrow \boxed{100}$ | E | 100 |
e69a7d3fbd32bdd67a5dd7cb0b6f88b3 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_6 | In a town of $351$ adults, every adult owns a car, motorcycle, or both. If $331$ adults own cars and $45$ adults own motorcycles, how many of the car owners do not own a motorcycle?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 306 \qquad \textbf{(E)}\ 351$ | By PIE , the number of adults who own both cars and motorcycles is $331+45-351=25.$ Out of the $331$ car owners, $25$ of them own motorcycles and $331-25=\boxed{306}$ of them don't. | D | 306 |
e69a7d3fbd32bdd67a5dd7cb0b6f88b3 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_6 | In a town of $351$ adults, every adult owns a car, motorcycle, or both. If $331$ adults own cars and $45$ adults own motorcycles, how many of the car owners do not own a motorcycle?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 306 \qquad \textbf{(E)}\ 351$ | There are $351$ total adults, and $45$ own a motorcycle. The number of adults that don't own a motorcycle is $351 - 45 = 306$ . Since everyone owns a car or motorcycle and one who doesn't own a motorcycle owns a car, the answer is $\boxed{306}$ | D | 306 |
e69a7d3fbd32bdd67a5dd7cb0b6f88b3 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_6 | In a town of $351$ adults, every adult owns a car, motorcycle, or both. If $331$ adults own cars and $45$ adults own motorcycles, how many of the car owners do not own a motorcycle?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 306 \qquad \textbf{(E)}\ 351$ | Clearly, we can eliminate answer choice $E$ , for at least one adult owns motorcycles. It is also fairly obvious that the answer must be in the 300 range, giving us $\boxed{306}$ | D | 306 |
43cf73749c5588419c66678a4963937a | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_7 | Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded . What percent of the total area is partially bolded? [asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] $\textbf{(A) }12\frac{1}{2} \qquad\textbf{(B) }20 \qquad\textbf{(C) }25 \qquad\textbf{(D) }33\frac{1}{3} \qquad\textbf{(E) }37\frac{1}{2}$ | Assume that the area of each square is $1$ . Then, the area of the bolded region in the top left square is $\dfrac{1}{4}$ . The area of the top right bolded region is $\dfrac{1}{8}$ . The area of the bottom left bolded region is $\dfrac{3}{8}$ . And the area of the bottom right bolded region is $\dfrac{1}{4}$ . Add the four fractions: $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{1}{4} = 1$ . The four squares together have an area of $4$ , so the percentage bolded is $\dfrac{1}{4} \cdot 100 = \boxed{25}$ | C | 25 |
709dc22c33080a7a2e641c528a9b038e | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_8 | Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
$\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$ | By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\boxed{5}$ | B | 5 |
709dc22c33080a7a2e641c528a9b038e | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_8 | Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
$\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$ | The sum of an even number added to an odd number is always odd. The smallest possible sum is $3$ , and the largest possible sum is $11$ . The odd numbers in between can be achieved by replacing chips with ${\displaystyle \pm }2$ within the same bag. Therefore, we can conclude that there are $(11-3)/2(*)+1=\boxed{5}$ possible sums. | B | 5 |
d9847efaec5a1e9d9ae9d3ff6cba47a0 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_9 | Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?
[asy] import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; draw((0,0)--(0,8)); draw((0,0)--(8,0)); 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)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8)); draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label("$1$",(0.95,-0.24),SE*lsf); label("$2$",(1.92,-0.26),SE*lsf); label("$3$",(2.92,-0.31),SE*lsf); label("$4$",(3.93,-0.26),SE*lsf); label("$5$",(4.92,-0.27),SE*lsf); label("$6$",(5.95,-0.29),SE*lsf); label("$7$",(6.94,-0.27),SE*lsf); label("$5$",(-0.49,1.22),SE*lsf); label("$10$",(-0.59,2.23),SE*lsf); label("$15$",(-0.61,3.22),SE*lsf); label("$20$",(-0.61,4.23),SE*lsf); label("$25$",(-0.59,5.22),SE*lsf); label("$30$",(-0.59,6.2),SE*lsf); label("$35$",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6)); draw((4,4)--(7,7),linewidth(1.6)); label("HOURS",(3.41,-0.85),SE*lsf); label("M",(-1.39,5.32),SE*lsf); label("I",(-1.34,4.93),SE*lsf); label("L",(-1.36,4.51),SE*lsf); label("E",(-1.37,4.11),SE*lsf); label("S",(-1.39,3.7),SE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}2\qquad\textbf{(B)}2.5\qquad\textbf{(C)}4\qquad\textbf{(D)}4.5\qquad\textbf{(E)}5$ | We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is $\boxed{5}$ | E | 5 |
cf6642baf3bca2d06469c8b0305a60d6 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_10 | The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$ | Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation:
\begin{align*} 10 &= 2.4 + 0.2 \times \frac{x-0.5}{0.1} + 2\\ 5.6 &= 2(x-0.5)\\ 2.8 &= x-0.5\\ x &= \boxed{3.3} | C | 3.3 |
cf6642baf3bca2d06469c8b0305a60d6 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_10 | The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$ | \begin{array}{|c|c|c|}
Miles & Money & Remark\\
\hline
0 & 8 & \text{2 for tip} \\
\hline
0.5 & 5.6 & \text{-2.4 for first 0.5 mi} \\
\hline
3.3 & 0 & \text{-5.6 for 2.8 mi}
\end{array}
Miles travelled = $\boxed{3.3}$ | C | 3.3 |
b77122d66422e1435be10bb4a9cd445b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_11 | The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?
[asy] size(300); real i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label("$0$",origin,W); label("$20$",(0,20),W); label("$40$",(0,40),W); label("$60$",(0,60),W); label("$80$",(0,80),W); label("$100$",(0,100),W); label("$120$",(0,120),W); path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--(70,90)--(70,0)--cycle,TuesL=(70,0)--(70,80)--(80,80)--(80,0)--cycle,WedD=(100,0)--(100,100)--(110,100)--(110,0)--cycle,WedL=(110,0)--(110,120)--(120,120)--(120,0)--cycle,ThurD=(140,0)--(140,80)--(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle; fill(MonD,grey); fill(MonL,lightgrey); fill(TuesD,grey); fill(TuesL,lightgrey); fill(WedD,grey); fill(WedL,lightgrey); fill(ThurD,grey); fill(ThurL,lightgrey); fill(FriD,grey); fill(FriL,lightgrey); draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL); label("M",(30,-5),S); label("Tu",(70,-5),S); label("W",(110,-5),S); label("Th",(150,-5),S); label("F",(190,-5),S); label("M",(-25,85),W); label("I",(-27,75),W); label("N",(-25,65),W); label("U",(-25,55),W); label("T",(-25,45),W); label("E",(-25,35),W); label("S",(-26,25),W);[/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | Average the differences between each day. We get $10, -10,\text{ } 20,\text{ } 30,-20$ . We find the average of this list to get $\boxed{6}$ | A | 6 |
b77122d66422e1435be10bb4a9cd445b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_11 | The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?
[asy] size(300); real i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label("$0$",origin,W); label("$20$",(0,20),W); label("$40$",(0,40),W); label("$60$",(0,60),W); label("$80$",(0,80),W); label("$100$",(0,100),W); label("$120$",(0,120),W); path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--(70,90)--(70,0)--cycle,TuesL=(70,0)--(70,80)--(80,80)--(80,0)--cycle,WedD=(100,0)--(100,100)--(110,100)--(110,0)--cycle,WedL=(110,0)--(110,120)--(120,120)--(120,0)--cycle,ThurD=(140,0)--(140,80)--(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle; fill(MonD,grey); fill(MonL,lightgrey); fill(TuesD,grey); fill(TuesL,lightgrey); fill(WedD,grey); fill(WedL,lightgrey); fill(ThurD,grey); fill(ThurL,lightgrey); fill(FriD,grey); fill(FriL,lightgrey); draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^^ThurD^^ThurL^^FriD^^FriL); label("M",(30,-5),S); label("Tu",(70,-5),S); label("W",(110,-5),S); label("Th",(150,-5),S); label("F",(190,-5),S); label("M",(-25,85),W); label("I",(-27,75),W); label("N",(-25,65),W); label("U",(-25,55),W); label("T",(-25,45),W); label("E",(-25,35),W); label("S",(-26,25),W);[/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | This solution may take longer to do than the first solution. In total, Asha studied for 400 minutes a week (80 minutes per day) and Sasha studied for 430 minutes a week (86 minutes per day). 86 - 80 = 6. Therefore, the answer is $\boxed{6}$ | A | 6 |
a85279659ae9466c58afe1b773cc9c73 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_12 | Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?
$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$ | If we designate a person to be on a certain side, then all placements of the other people can be considered unique. WLOG, assign Angie to be on the side. There are then $3!=6$ total seating arrangements. If Carlos is across from Angie, there are only $2!=2$ ways to fill the remaining two seats. Then the probability Angie and Carlos are seated opposite each other is $\frac26=\boxed{13}$ | B | 13 |
a85279659ae9466c58afe1b773cc9c73 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_12 | Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?
$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$ | If we seat Angie first, there would be only one out of three ways Carlos can sit across from Angie. So the final answer is $\boxed{13}$ | B | 13 |
495a4af421111de74d8f677b30308977 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_13 | Two congruent squares, $ABCD$ and $PQRS$ , have side length $15$ . They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What percent of the area of rectangle $AQRD$ is shaded?
[asy] filldraw((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label("D",(0,0),S); label("R",(25,0),S); label("Q",(25,15),N); label("A",(0,15),N); filldraw((10,0)--(15,0)--(15,15)--(10,15)--cycle,mediumgrey,black); label("S",(10,0),S); label("C",(15,0),S); label("B",(15,15),N); label("P",(10,15),N);[/asy]
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25$ | The overlap length is $2(15)-25=5$ , so the shaded area is $5 \cdot 15 =75$ . The area of the whole shape is $25 \cdot 15 = 375$ . The fraction $\dfrac{75}{375}$ reduces to $\dfrac{1}{5}$ or 20%. Therefore, the answer is $\boxed{20}$ | C | 20 |
72795227200120a57748765bcd9d790b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_17 | Let $w$ $x$ $y$ , and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$ , then what does $2w + 3x + 5y + 7z$ equal?
$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$ | The prime factorization of $588$ is $2^2\cdot3\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$
\[2w+3x+5y+7z=4+3+0+14=\boxed{21}\] | A | 21 |
e76d455a95be02426d92aeb23ac5854b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_19 | How many rectangles are in this figure?
[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$
$3+5+3=\boxed{11}$ | D | 11 |
4c18dde81f3c229dca49f4e9795e36f0 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_20 | Quadrilateral $ABCD$ is a trapezoid, $AD = 15$ $AB = 50$ $BC = 20$ , and the altitude is $12$ . What is the area of the trapezoid?
[asy] pair A,B,C,D; A=(3,20); B=(35,20); C=(47,0); D=(0,0); draw(A--B--C--D--cycle); dot((0,0)); dot((3,20)); dot((35,20)); dot((47,0)); label("A",A,N); label("B",B,N); label("C",C,S); label("D",D,S); draw((19,20)--(19,0)); dot((19,20)); dot((19,0)); draw((19,3)--(22,3)--(22,0)); label("12",(21,10),E); label("50",(19,22),N); label("15",(1,10),W); label("20",(41,12),E);[/asy]
$\textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800$ | [asy] unitsize(1.5mm); defaultpen(linewidth(.9pt)+fontsize(10pt)); dotfactor=3; pair A,B,C,D,X,Y; A=(9,12); B=(59,12); C=(75,0); D=(0,0); X=(9,0); Y=(59,0); draw(A--B--C--D--cycle); draw(A--X); draw(B--Y); pair[] ps={A,B,C,D,X,Y}; dot(ps); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$X$",X,SE); label("$Y$",Y,S); label("$a$",D--X,S); label("$b$",Y--C,S); label("$15$",D--A,NW); label("$50$",B--A,N); label("$20$",B--C,NE); label("$12$",X--A,E); label("$12$",Y--B,W); [/asy]
If you draw altitudes from $A$ and $B$ to $CD,$ the trapezoid will be divided into two right triangles and a rectangle. You can find the values of $a$ and $b$ with the Pythagorean theorem
\[a=\sqrt{15^2-12^2}=\sqrt{81}=9\]
\[b=\sqrt{20^2-12^2}=\sqrt{256}=16\]
$ABYX$ is a rectangle so $XY=AB=50.$
\[CD=a+XY+b=9+50+16=75\]
The area of the trapezoid is
\[12\cdot \frac{(50+75)}{2} = 6(125) = \boxed{750}\] | D | 750 |
77b98baa408e3746fbfccd42b1b560ac | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_21 | Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$ , and $49$ . Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$ | If at least half the guesses are too low, then Norb's age must be greater than $36.$
If two of the guesses are off by one, then his age is in between two guesses whose difference is $2.$ It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48.$
Lastly, Norb's age is a prime number so the answer must be $\boxed{37}$ | C | 37 |
77b98baa408e3746fbfccd42b1b560ac | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_21 | Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$ , and $49$ . Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?
$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$ | Since two guesses are off by one, we know that both $x+1$ and $x-1$ are in the list where $x$ is the age of Norb. Now, we know that $x+1$ and $x-1$ are $28$ and $30$ $30$ and $32$ $36$ and $38$ and $47$ and $49$ . From these values, we know that $x$ must be $29$ $31$ , and $37$ . Since half of the guesses are too low, $24, 28, 30, 32,$ and $36$ are all too low so we can eliminate all numbers in our list lesser than or equal to $36$ . Therefore, our list has only $37$ left so the answer is $\boxed{37}$ | C | 37 |
920ba6d2072f83664c844629b8e524d3 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_22 | What is the tens digit of $7^{2011}$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$ | Since we want the tens digit, we can find the last two digits of $7^{2011}$ . We can do this by using modular arithmetic. \[7^1\equiv 07 \pmod{100}.\] \[7^2\equiv 49 \pmod{100}.\] \[7^3\equiv 43 \pmod{100}.\] \[7^4\equiv 01 \pmod{100}.\] We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$ . Using this, we can say: \[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\] From the above, we can conclude that the last two digits of $7^{2011}$ are 43. Since they have asked us to find the tens digit, our answer is $\boxed{4}$ | D | 4 |
920ba6d2072f83664c844629b8e524d3 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_22 | What is the tens digit of $7^{2011}$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$ | We can use patterns to figure out the answer.
7 to the power of 2 is 49. So, the tens digit is 4.
7 to the power of 3 is 343. So, the tens digit is 4.
7 to the power of 4 is 2401.So, the tens digit is 0.
7 to the power of 5 is 16807. So, the tens digit is 0.
By now, we can notice the pattern. The tens digit for 7 to the power of 2 is 4, then 4, then 0, then 0. It keeps on going, 2 fours, and then 2 zeros in 4 numbers. If we round up for 2011/4, we get 503. 503 * 4 is 2012. So 7 to the 2012 power has a tens digit of 0, since 2012 is a mutiple of 4, and 7 to the power of 4 has a tens digit of 0. We have to subtract a power from 7 to the 2012 power, so the tens digit goes back from 0 to 4 because if we subtract a power from 7 to the power of 4, we have 7 to the power of 3, which has a tens digit of 4. Hence the answer is $\boxed{4}$ | D | 4 |
63ee083be38943469dffd250a21884e4 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_23 | How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$ | We can separate this into two cases. If an integer is a multiple of $5,$ the last digit must be either $0$ or $5.$
Case 1: The last digit is $5.$ The leading digit can be $1,2,3,$ or $4.$ Because the second digit can be $0$ but not the leading digit, there are also $4$ choices. The third digit cannot be the leading digit or the second digit, so there are $3$ choices. The number of integers is this case is $4\cdot4\cdot3\cdot1=48.$
Case 2: The last digit is $0.$ Because $5$ is the largest digit, one of the remaining three digits must be $5.$ There are $3$ ways to choose which digit should be $5.$ The remaining digits can be $1,2,3,$ or $4,$ but since they have to be different there are $4\cdot3$ ways to choose. The number of integers in this case is $1\cdot3\cdot4\cdot3=36.$
Therefore, the answer is $48+36=\boxed{84}$ | D | 84 |
63ee083be38943469dffd250a21884e4 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_23 | How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$ | We make four cases based off where the multiple of $5$ digit ( $0$ or $5$ ) is. The number has to end with either $5$ or $0$ since it's a multiple of $5$ . In all but the last case, the $5$ and $0$ are used at the end and in another spot which separates the cases.
Case 1: The first digit can't be $0$ , so it must be $5$ . There are $4\cdot3$ to choose the middle two digits. After that, the last digit has to be $0$ , so there are a total of $1\cdot4\cdot3\cdot1=12$ numbers.
Case 2: The second digit can be $0$ or $5$ , leaving $2$ choices. The first and third numbers can be chosen in $4\cdot3$ ways, like last time. The last digit has to be $0$ or $5$ , but not the one we already used. There are a total of $4\cdot2\cdot3\cdot1=24$ numbers.
Case 3: There are the same choices, but the digits $0$ and $5$ are at the last and second-to-last spots. So there are $4\cdot3\cdot2\cdot1=24$ numbers again.
Case 4: There are $4\cdot3\cdot2$ ways to choose the first three numbers. There has to be a $5$ in the number because the largest digit is $5$ . Coincidentally, there are $4\cdot3\cdot2\cdot1$ numbers again.
There are a total of $12 + 24 \cdot 3=\boxed{84}$ numbers. | D | 84 |
b236974ec921383775db7e63efbfbdd5 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_24 | In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | For the sum of two numbers to be odd, one must be odd and the other must be even, because all odd numbers are of the form $2n+1$ where n is an integer, and all even numbers are of the form $2m$ where m is an integer. \[2n + 1 + 2m = 2m + 2n + 1 = 2(m+n) + 1\] and $m+n$ is an integer because $m$ and $n$ are both integers.
The only even prime number is $2,$ so our only combination could be $2$ and $9999.$ However, $9999$ is clearly divisible by $3$ ,
so the number of ways $10001$ can be written as the sum of two primes is $\boxed{0}$ | A | 0 |
b236974ec921383775db7e63efbfbdd5 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_24 | In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | First, we noticed that 10001 is equal to 5000+5001, if you subtract n to 5000 and add n to 5001, you always get an even number, even number is never a prime number except 2. We also see that whenever an addend is an odd number, the other addend will be even, so having an odd number as an addend is not possible, other than 9999 and 2, because 2 is a prime. We try 2 and 9999 but we can see 9999 is divisible by 3 and 9 clearly. So the answer is $\boxed{0}$ | A | 0 |
b236974ec921383775db7e63efbfbdd5 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_24 | In how many ways can $10001$ be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | It is helpful to know and understand the Goldbach Conjecture - that every even number can be written as the sum of $2$ primes - and also, that the ${\textbf{odd numbers}\ }$ that are the sum of two primes are exactly two more than a prime. This is because to make the sum of two numbers odd, you must have one even and one odd. There is only one even prime, which is two, so the sum will be of the form $2+p$ . Hence, the odd numbers that are the sum of two primes are exactly $2$ more than a prime. Relating to the problem, $10001$ is not $2$ more than a prime, because $10001-2=9999$ and we can easily see that $9999$ is divisible by $3$ . Therefore, $10001$ cannot be written as the sum of two primes, and the answer is $\boxed{0}$ ~mk | A | 0 |
c07e0f361226ee2ecaf0cba4194e73f0 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_25 | A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$ | The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: $2 \cdot 2 \cdot \frac{1}{2}=2.$
The circle's shaded area is the area of the smaller square subtracted from the area of the circle: $\pi - 2.$
If you draw the diagonals of the smaller square, you will see that the larger square is split $4$ congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: $2.$
Approximating $\pi$ to $3.14,$ the ratio of the circle's shaded area to the area between the two squares is about
\[\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{12}\] | A | 12 |
c07e0f361226ee2ecaf0cba4194e73f0 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_25 | A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$ | For the ratio of the circle's shaded area to the area between the squares to be $1,$ they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately $\boxed{12}$ | A | 12 |
c07e0f361226ee2ecaf0cba4194e73f0 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_8_Problems/Problem_25 | A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$ | Set the side length of the bigger square to be $8$ .
Then the area of the big square is $8^2 =64$ and
the area of the small square $(4\sqrt{2})^2 = 32$ .
The difference is $32$ .
The area of the circle is $4^2$ times $\pi$ which is $16 \pi$ or about $48$ .
Knowing the area of the small square is $32$ $48-32$ is $16$ .
The area of the big square is $64$ .
So $32/64$ is $1/2$ , or $\boxed{12}$ | A | 12 |
f1d824db974d1c816854b468737ced3e | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_1 | At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School are taking the contest?
$\textbf{(A)}\ 26 \qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$ | Given that these are the only math teachers at Euclid Middle School and we are told how many from each class are taking the AMC 8, we simply add the three numbers to find the total. $11+8+9=\boxed{28}$ | C | 28 |
9aecc5b4ddf581b36f328110f54eb013 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_3 | The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?
[asy] import graph; size(16.38cm); real lsf=2; pathpen=linewidth(0.7); pointpen=black; pen fp = fontsize(10); pointfontpen=fp; real xmin=-1.33,xmax=11.05,ymin=-9.01,ymax=-0.44; pen ycycyc=rgb(0.55,0.55,0.55); pair A=(1,-6), B=(1,-2), D=(1,-5.8), E=(1,-5.6), F=(1,-5.4), G=(1,-5.2), H=(1,-5), J=(1,-4.8), K=(1,-4.6), L=(1,-4.4), M=(1,-4.2), N=(1,-4), P=(1,-3.8), Q=(1,-3.6), R=(1,-3.4), S=(1,-3.2), T=(1,-3), U=(1,-2.8), V=(1,-2.6), W=(1,-2.4), Z=(1,-2.2), E_1=(1.4,-2.6), F_1=(1.8,-2.6), O_1=(14,-6), P_1=(14,-5), Q_1=(14,-4), R_1=(14,-3), S_1=(14,-2), C_1=(1.4,-6), D_1=(1.8,-6), G_1=(2.4,-6), H_1=(2.8,-6), I_1=(3.4,-6), J_1=(3.8,-6), K_1=(4.4,-6), L_1=(4.8,-6), M_1=(5.4,-6), N_1=(5.8,-6), T_1=(6.4,-6), U_1=(6.8,-6), V_1=(7.4,-6), W_1=(7.8,-6), Z_1=(8.4,-6), A_2=(8.8,-6), B_2=(9.4,-6), C_2=(9.8,-6), D_2=(10.4,-6), E_2=(10.8,-6), L_2=(2.4,-3.2), M_2=(2.8,-3.2), N_2=(3.4,-4), O_2=(3.8,-4), P_2=(4.4,-3.6), Q_2=(4.8,-3.6), R_2=(5.4,-3.6), S_2=(5.8,-3.6), T_2=(6.4,-3.4), U_2=(6.8,-3.4), V_2=(7.4,-3.8), W_2=(7.8,-3.8), Z_2=(8.4,-2.8), A_3=(8.8,-2.8), B_3=(9.4,-3.2), C_3=(9.8,-3.2), D_3=(10.4,-3.8), E_3=(10.8,-3.8); filldraw(C_1--E_1--F_1--D_1--cycle,ycycyc); filldraw(G_1--L_2--M_2--H_1--cycle,ycycyc); filldraw(I_1--N_2--O_2--J_1--cycle,ycycyc); filldraw(K_1--P_2--Q_2--L_1--cycle,ycycyc); filldraw(M_1--R_2--S_2--N_1--cycle,ycycyc); filldraw(T_1--T_2--U_2--U_1--cycle,ycycyc); filldraw(V_1--V_2--W_2--W_1--cycle,ycycyc); filldraw(Z_1--Z_2--A_3--A_2--cycle,ycycyc); filldraw(B_2--B_3--C_3--C_2--cycle,ycycyc); filldraw(D_2--D_3--E_3--E_2--cycle,ycycyc); D(B--A,linewidth(0.4)); D(H--(8,-5),linewidth(0.4)); D(N--(8,-4),linewidth(0.4)); D(T--(8,-3),linewidth(0.4)); D(B--(8,-2),linewidth(0.4)); D(B--S_1); D(T--R_1); D(N--Q_1); D(H--P_1); D(A--O_1); D(C_1--E_1); D(E_1--F_1); D(F_1--D_1); D(D_1--C_1); D(G_1--L_2); D(L_2--M_2); D(M_2--H_1); D(H_1--G_1); D(I_1--N_2); D(N_2--O_2); D(O_2--J_1); D(J_1--I_1); D(K_1--P_2); D(P_2--Q_2); D(Q_2--L_1); D(L_1--K_1); D(M_1--R_2); D(R_2--S_2); D(S_2--N_1); D(N_1--M_1); D(T_1--T_2); D(T_2--U_2); D(U_2--U_1); D(U_1--T_1); D(V_1--V_2); D(V_2--W_2); D(W_2--W_1); D(W_1--V_1); D(Z_1--Z_2); D(Z_2--A_3); D(A_3--A_2); D(A_2--Z_1); D(B_2--B_3); D(B_3--C_3); D(C_3--C_2); D(C_2--B_2); D(D_2--D_3); D(D_3--E_3); D(E_3--E_2); D(E_2--D_2); label("0",(0.88,-5.91),SE*lsf,fp); label(" 5",(0.3,-4.84),SE*lsf,fp); label(" 10",(0.2,-3.84),SE*lsf,fp); label(" 15",(0.2,-2.85),SE*lsf,fp); label(" 20",(0.2,-1.85),SE*lsf,fp); label("$\mathrm{Price}$",(0.16,-3.45),SE*lsf,fp); label("$1$",(1.54,-5.97),SE*lsf,fp); label("$2$",(2.53,-5.95),SE*lsf,fp); label("$3$",(3.53,-5.94),SE*lsf,fp); label("$4$",(4.55,-5.94),SE*lsf,fp); label("$5$",(5.49,-5.95),SE*lsf,fp); label("$6$",(6.53,-5.95),SE*lsf,fp); label("$7$",(7.55,-5.95),SE*lsf,fp); label("$8$",(8.52,-5.95),SE*lsf,fp); label("$9$",(9.57,-5.97),SE*lsf,fp); label("$10$",(10.56,-5.94),SE*lsf,fp); label("Month",(7.14,-6.43),SE*lsf,fp); D(A,linewidth(1pt)); D(B,linewidth(1pt)); D(D,linewidth(1pt)); D(E,linewidth(1pt)); D(F,linewidth(1pt)); D(G,linewidth(1pt)); D(H,linewidth(1pt)); D(J,linewidth(1pt)); D(K,linewidth(1pt)); D(L,linewidth(1pt)); D(M,linewidth(1pt)); D(N,linewidth(1pt)); D(P,linewidth(1pt)); D(Q,linewidth(1pt)); D(R,linewidth(1pt)); D(S,linewidth(1pt)); D(T,linewidth(1pt)); D(U,linewidth(1pt)); D(V,linewidth(1pt)); D(W,linewidth(1pt)); D(Z,linewidth(1pt)); D(E_1,linewidth(1pt)); D(F_1,linewidth(1pt)); D(O_1,linewidth(1pt)); D(P_1,linewidth(1pt)); D(Q_1,linewidth(1pt)); D(R_1,linewidth(1pt)); D(S_1,linewidth(1pt)); D(C_1,linewidth(1pt)); D(D_1,linewidth(1pt)); D(G_1,linewidth(1pt)); D(H_1,linewidth(1pt)); D(I_1,linewidth(1pt)); D(J_1,linewidth(1pt)); D(K_1,linewidth(1pt)); D(L_1,linewidth(1pt)); D(M_1,linewidth(1pt)); D(N_1,linewidth(1pt)); D(T_1,linewidth(1pt)); D(U_1,linewidth(1pt)); D(V_1,linewidth(1pt)); D(W_1,linewidth(1pt)); D(Z_1,linewidth(1pt)); D(A_2,linewidth(1pt)); D(B_2,linewidth(1pt)); D(C_2,linewidth(1pt)); D(D_2,linewidth(1pt)); D(E_2,linewidth(1pt)); D(L_2,linewidth(1pt)); D(M_2,linewidth(1pt)); D(N_2,linewidth(1pt)); D(O_2,linewidth(1pt)); D(P_2,linewidth(1pt)); D(Q_2,linewidth(1pt)); D(R_2,linewidth(1pt)); D(S_2,linewidth(1pt)); D(T_2,linewidth(1pt)); D(U_2,linewidth(1pt)); D(V_2,linewidth(1pt)); D(W_2,linewidth(1pt)); D(Z_2,linewidth(1pt)); D(A_3,linewidth(1pt)); D(B_3,linewidth(1pt)); D(C_3,linewidth(1pt)); D(D_3,linewidth(1pt)); D(E_3,linewidth(1pt)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 62 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 89 \qquad \textbf{(E)}\ 100$ | The highest price was in Month 1, which was $17. The lowest price was in Month 3, which was $10. 17 is $\frac{17}{10}\cdot100=170\%$ of 10, and is $170-100=70\%$ more than 10.
Therefore, the answer is $\boxed{70}$ | C | 70 |
6c7c3f5178897fb6544a4ed0c21d4b76 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_4 | What is the sum of the mean, median, and mode of the numbers $2,3,0,3,1,4,0,3$
$\textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9$ | Putting the numbers in numerical order we get the list $0,0,1,2,3,3,3,4.$ The mode is $3.$ The median is $\frac{2+3}{2}=2.5.$ The average is $\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.$ The sum of all three is $3+2.5+2=\boxed{7.5}$ | C | 7.5 |
eab82210c19c1cbf121c0699123c0438 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_5 | Alice needs to replace a light bulb located $10$ centimeters below the ceiling in her kitchen. The ceiling is $2.4$ meters above the floor. Alice is $1.5$ meters tall and can reach $46$ centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?
$\textbf{(A)}\ 32 \qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 40$ | Convert everything to the same unit. Since the answer is in centimeters, change meters to centimeters by moving the decimal place two places to the right.
The ceiling is $240$ centimeters above the floor. The combined height of Alice and the light bulb when she reaches for it is $10+150+46=206$ centimeters. That means the stool's height needs to be $240-206=\boxed{34}$ | B | 34 |
bc7681fc96cfcfb1c37ada197169a8f8 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_7 | Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$ | You need $2$ dimes, $1$ nickel, and $4$ pennies for the first $25$ cents. From $26$ cents to $50$ cents, you only need to add $1$ quarter. From $51$ cents to $75$ cents, you also only need to add $1$ quarter. The same for $76$ cents to $99$ cents. Notice that instead of $100$ , it is $99$ . We are left with $3$ quarters, $1$ nickel, $2$ dimes, and $4$ pennies. Thus, the correct answer is $3+2+1+4=\boxed{10}$ | B | 10 |
584e1dd975757e63e1726d97cc073340 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_8 | As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$ | Because they are both moving in the same direction, Emily is riding relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is not moving at all [on his skateboard] and Emily is riding at $4$ mph. It takes her
\[\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}\]
to ride the $1/2$ mile to reach him, and then the same amount of time to be $1/2$ mile ahead of him. This totals to
\[2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{15}\] | D | 15 |
358e651673e49dc794457d8f98bd022b | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_9 | Ryan got $80\%$ of the problems correct on a $25$ -problem test, $90\%$ on a $40$ -problem test, and $70\%$ on a $10$ -problem test. What percent of all the problems did Ryan answer correctly?
$\textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86$ | Ryan answered $(0.8)(25)=20$ problems correct on the first test, $(0.9)(40)=36$ on the second, and $(0.7)(10)=7$ on the third. This amounts to a total of $20+36+7=63$ problems correct. The total number of problems is $25+40+10=75.$ Therefore, the percentage is $\dfrac{63}{75} = 84\% \rightarrow \boxed{84}$ | D | 84 |
ec1bd04eda2f3ecfc45d94adee4ba7bb | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_10 | Six pepperoni circles will exactly fit across the diameter of a $12$ -inch pizza when placed. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
$\textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78$ | The pepperoni circles' diameter is $2$ , since $\dfrac{12}{6} = 2$ . From that we see that the area of the $24$ circles of pepperoni is $\left ( \frac{2}{2} \right )^2 (24\pi) = 24\pi$ . The large pizza's area is $6^2\pi$ . Therefore, the ratio is $\frac{24\pi}{36\pi} = \boxed{23}$ | B | 23 |
ef8ba675f2586cac7629415709ba3b97 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_11 | The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$ | Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$ . Since the ratio of the smaller tree to the larger tree is $\frac{3}{4}$ , we have $\frac{h-16}{h}=\frac{3}{4}$ . Solving for $h$ gives us $h=64 \Rightarrow \boxed{64}$ | B | 64 |