AMC8 2014

We have $H=8-7=1$ and $T=8-2+5=11$. Clearly $1-11=-10$, so our answer is $\boxed{\textbf{(A)}-10}$.

The fewest amount of coins that can be used is $2$ (a quarter and a dime). The greatest amount is $7$, if he only uses nickels. Therefore we have $7-2=\boxed{\textbf{(E)}~5}$.

Isabella read $3\cdot 36+3\cdot 44$ pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as $3\cdot (36+44)=3\cdot 80$, which gives that she read $240$ pages. On her last day, she read $10$ more pages for a total of $240+10=\boxed{\textbf{(B)}~250}$ pages.

Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$. The other prime number is $85-2=83$, and the product of these two numbers is $83\cdot2=\boxed{\textbf{(E)}~166}$.

Margie can afford $20/4=5$ gallons of gas. She can go $32\cdot5=\boxed{\textbf{(C)}~160}$ miles on this amount of gas.

The sum of the areas is equal to $2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36$. This is equal to $2(1+4+9+16+25+36)$, which is equal to $2\cdot91$. This is equal to our final answer of $\boxed{\textbf{(D)}~182}$.

Let $g$ being the number of girls in the class. The number of boys in the class is equal to $g-4$. Since the total number of students is equal to $28$, we get $g+g-4=28$. Solving this equation, we get $g=16$. There are $16-4=12$ boys in our class, and our answer is $16:12=\boxed{\textbf{(B)}~4:3}$.

Since all the eleven members paid the same amount, that means that the total must be divisible by $11$. We can do some trial-and-error to get $A=3$, so our answer is $\boxed{\textbf{(D)}~3}$

Using angle chasing is a good way to solve this problem. $BD = DC$, so $\angle DBC = \angle DCB = 70$, because it is an isosceles triangle. Then $\angle CDB = 180-(70+70) = 40$. Since $\angle ADB$ and $\angle BDC$ are supplementary, $\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}$.

The seventh AMC 8 would have been given in $1991$. If Samantha was 12 then, that means she was born 12 years ago, so she was born in $1991-12=1979$. Our answer is $\boxed{(\text{A})1979}$ corrections made by DrDominic

We can apply complementary counting and count the paths that DO go through the blocked intersection, which is $\dbinom{2}{1}\dbinom{3}{1}=6$. There are a total of $\dbinom{5}{2}=10$ paths, so there are $10-6=4$ paths possible. $\boxed{(\text{A})4}$ is the correct answer.

For the first celebrity, you have a 1/3 chance of picking the photo. Given you get the picture right, you now have a 1/2 chance of picking the next photo. If you get both of them right, you are guarenteed to get the third right. Thus the probability is 1/2*1/3 which is 1/6.

Since $n^2+m^2$ is even, either both $n^2$ and $m^2$ are even, or they are both odd. Therefore, $n$ and $m$ are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result, $n+m$ must be even. The answer, then, is $n^2+m^2$ $\boxed{(\text{D})}$ is odd.

The area of $\bigtriangleup CDE$ is $\frac{DC\cdot CE}{2}$. The area of $ABCD$ is $AB\cdot AD=5\cdot 6=30$, which also must be equal to the area of $\bigtriangleup CDE$, which, since $DC=5$, must in turn equal $\frac{5\cdot CE}{2}$. Through transitivity, then, $\frac{5\cdot CE}{2}=30$, and $CE=12$. Then, using the Pythagorean Theorem, you should be able to figure out that $\bigtriangleup CDE$ is a $5-12-13$ triangle, so $DE=\boxed{13}$ , or $\boxed{(B)}$.

The measure of an inscribed angle is half the measure of its corresponding central angle. Since each unit arc is $\frac{1}{12}$ of the circle's circumference, each unit central angle measures $\frac{360}{12}^{\circ}=30^{\circ}$. From this, $\angle EOG = 60^{\circ}$, so $x = 30^{\circ}$. Also, $\angle AOI = 120^{\circ}$, so $y = 60^{\circ}$. The number of degrees in the sum of both angles is $30 + 60 = \boxed{(C)\ 90}.$

Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference. Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the conference. Therefore, the total number of games is $56 + 32 = \boxed{\text{(B) }88}$.

Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6 \text{ mph}}$, so $\boxed{\text{(B) }6}$ is the answer.

We'll just start by breaking cases down. The probability of A occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of B occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of C occurring is $\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$, because we need to choose 2 of the 4 slots to be girls. For D, there are two possible cases, 3 girls and 1 boy or 3 boys and 1 girl. The probability of the first case is $\dbinom{4}{1}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{4}$ because we need to choose 1 of the 4 slots to be a boy. However, the second case has the same probability because we are choosing 1 of the 4 children to be a girl, so the total probability is $\frac{1}{4} \cdot 2 = \frac{1}{2}$. So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D) 3 are of one gender and 1 is of the other gender}}.$

For the least possible surface area that is white, we should have 1 cube in the center, and the other 5 with only 1 face exposed. This gives 5 square inches of white, surface area. Since the cube has a surface area of 54 square inches, our answer is $\boxed{\textbf{(A) }\frac{5}{54}}$.

The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle. The area of the rectangle is $3\cdot5 =15$. The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$. Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$. $\pi$ is approximately $\dfrac{22}{7},$ and substituting that in will give $15-11=\boxed{\text{(B) }4.0}$

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$. To be a multiple of $3$, $A + B$ has to be either $2$ or $5$ or $8$... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$. We then add two of the selected values, $5$ to $15$, to get $20$. We then see that C = $1, 4$ or $7, 10$... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$, to get $23$, which shows us that C = $1$ or $4$ or $7$... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1, 4,$ and $7$. However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$, but there is a $1$, so $\boxed{\textbf{(A) }1}$ is our answer.

We can think of the number as $10a+b$, where a is the tens digit and b is the unit digit. Since the number is equal to the product of the digits ($ab$) plus the sum of the digits ($a+b$), we can say that $10a+b=ab+a+b$. We can simplify this to $10a=ab+a$, which factors to $(10)a=(b+1)a$. Dividing by $a$, we have that $b+1=10$. Therefore, the units digit, $b$, is $\boxed{\textbf{(E) }9}$

The maximum amount of days any given month can have is $31$, and the smallest two-digit primes are $11, 13,$ and $17$. There are a few different sums that can be deduced from the following numbers, which are $24, 30,$ and $28$, all of which represent the three days. Therefore, since Bethany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to $24$. Similarly, Caitlin says that the other two people's uniform numbers are later, so the sum must add up to $30$. This leaves $28$ as today's date. From this, Caitlin was referring to the uniform wearers $13$ and $17$, telling us that her number is $11$, giving our solution as $\boxed{(A) 11}$.

In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are $100$ people, the median will be the average of the $50\text{th}$ and $51\text{st}$ largest amount of cans per person. To minimize the first $49$, they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people. Taking $\frac{203}{51}$ gives us $3$ and a remainder of $50$. Seeing this, the largest number of cans the $50$th person could have is $3$, which leaves $4$ to the rest of the people. The average of $3$ and $4$ is $3.5$. Thus our answer is $\boxed{\text{(C) }3.5}$.

There are two possible interpretations of the problem: that the road as a whole is $40$ feet wide, or that each lane is $40$ feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be $20$ feet wide, so Robert must be riding his bike in semicircles with radius $20$ feet and diameter $40$ feet. Since the road is $5280$ feet long, over the whole mile, Robert rides $\frac{5280}{40} =132$ semicircles in total. Were the semicircles full circles, their circumference would be $2\pi\cdot 20=40\pi$ feet; as it is, the circumference of each is half that, or $20\pi$ feet. Therefore, over the stretch of highway, Robert rides a total of $132\cdot 20\pi =2640\pi$ feet, which is equivalent to $\frac{\pi}{2}$ miles. Robert rides at 5 miles per hour, so divide the $\frac{\pi}{2}$ miles by $5$ mph (because $t = \frac{d}{r}$ and time = distance/rate) to arrive at $\boxed{\textbf{(B) }\frac{\pi}{10}}$ hours. Edit: If you are confused about the lanes, watch the video below :)