AMC8 2012

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 $\frac{3}{8} \cdot 24 = \boxed{\textbf{(E)}\ 9}$.

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 $\boxed{\textbf{(B)}\ 700}$.

The problem wants us to find the time of sunset and gives us the length of daylight and time of sunrise. So all we have to do is add the length of daylight to the time of sunrise to obtain the answer. Convert 10 hours and 24 minutes into $10:24$ in order to add easier. Adding, we find that the time of sunset is $6:57 am + 10:24 \implies 17:21 \implies \boxed{\textbf{(B)}\ 5:21 pm}$.

Peter ate $1 + \frac{1}{2} = \frac{3}{2}$ slices. The pizza has $12$ slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peter ate $\dfrac{\frac{3}{2}\text{ slices}}{12\text{ slices}} = \boxed{\textbf{(C)}\ \frac{1}{8}}$ of the pizza.

Error creating thumbnail: Unable to save thumbnail to destination $1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\ 5 + X = 10\\ X = 5$ Thus, the answer is $\boxed{\textbf{(E)}\ 5}$.

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 $168-80 = \boxed{\textbf{(E)}\ 88}$.

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 $192-100 = \boxed{\textbf{(B)}\ 92}$.

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, $\boxed{\textbf{(D)}\ 60}$.

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 $200 - 61 = \boxed{\textbf{(C)}\ 139}$ two-legged birds.

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 $\boxed{\textbf{(D)}\ 9}$ such numbers.

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 $\boxed{{\textbf{(D)}\ 11}}$.

The problem wants us to find the units digit of $13^{2012}$, therefore, we can eliminate the tens digit of $13$, because the tens digit will not affect the final result. So our new expression is $3^{2012}$. Now we need to look for a pattern in the units digit. $3^1 \implies 3$ $3^2 \implies 9$ $3^3 \implies 7$ $3^4 \implies 1$ $3^5 \implies 3$ We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we can subtract 1 from 2012 and divide by 4. The remainder is the power of three that we are looking for, plus one. $2011$ divided by $4$ leaves a remainder of $3$, so the answer is the units digit of $3^{3+1}$, or $3^4$. Thus, we find that the units digit of $13^{2012}$ is $\boxed{{\textbf{(A)}\ 1}}$.

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 $17-13 = \boxed{\textbf{(C)}\ 4}$ more pencils than Jamar.

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 $\boxed{\textbf{(B)}\ 7}$.

To find the answer to this problem, we need to find the least common multiple of $3$, $4$, $5$, $6$ and add $2$ to the result. To calculate the least common multiple, we need to find the prime factorization for each of them. $3 = 3^{1}$, $4 = 2^{2}$, $5 = 5^{1}$, and $6 = 2^{1}*{3^{1}}$. So the least common multiple of the four numbers is $2^{2}*{3^{1}*{5^{1}}} = 60$, and by adding $2$, we find that that such number is $62$. Now we need to find the only given range that contains $62$. The only such range is answer $\textbf{(D)}$, and so our final answer is $\boxed{\textbf{(D)}\ 61\text{ and }65}$.

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 $\boxed{\textbf{(C)}\ 87431}$. p.s. USE INTUITION, see answer choices before solving any question -litttle_master

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 side length, 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 $\boxed{\textbf{(B)}\ 4}$.

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 $\boxed{\textbf{(A)}\ 3127}$.

$6$ 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 $\boxed{\textbf{(C)}\ 9}.$

The value of $\frac{7}{21}$ is $\frac{1}{3}$. Now we give all the fractions a common denominator. $\frac{5}{19} \implies \frac{345}{1311}$ $\frac{1}{3} \implies \frac{437}{1311}$ $\frac{9}{23} \implies \frac{513}{1311}$ Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.

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 $100-50 = \boxed{\textbf{(D)}\ 50}$ square feet of white on each side.

Let the values of the missing integers be $x, y, z$. We will find the bound of the possible medians. The smallest possible median will happen when we order the set as $\{x, y, z, 2, 3, 4, 6, 9, 14\}$. The median is $3$. The largest possible median will happen when we order the set as $\{2, 3, 4, 6, 9, 14, x, y, z\}$. The median is $9$. Therefore, the median must be between $3$ and $9$ inclusive, yielding $\boxed{\textbf{(D)}\ 7}$ possible medians.

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 side length of the large one. Thus, the area of one of the small equilateral triangles is $1$. The area of the hexagon is then $1 \times 6 = \boxed{\textbf{(C)}\ 6}$.

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 $\boxed{\textbf{(A)}\ \frac{4-\pi}{\pi}}$.

The total area of the four congruent triangles formed by the squares is $5-4 = 1$. Therefore, the area of one of these triangles is $\frac{1}{4}$. The height of one of these triangles is $a$ and the base is $b$. Using the formula for area of the triangle, we have $\frac{ab}{2} = \frac{1}{4}$. Multiply by $2$ on both sides to find that the value of $ab$ is $\boxed{\textbf{(C)}\ \frac{1}2}$.