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https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_18

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$ $\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$ [asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]

The value of $X$ is simply the average of the average values of both diagonals that contain $X$ . This is $\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{31}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20

Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy? $\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ socks. We have $x+y+z=12$ $x+3y+4z=24$ , and $x,y,z \ge 1$ Now, we subtract to find $2y+3z=12$ , and $y,z \ge 1$ . It follows that $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$ . Since sum of 2 multiples of 3 = multiple of 3, so we must have $2y=6$ Therefore, $y=3$ , and it follows that $z=2$ . Now, $x=12-y-z=12-3-2=\boxed{7}$ , as desired.

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20

Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy? $\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Since the total cost of the socks was $$24$ and Ralph bought $12$ pairs, the average cost of each pair of socks is $\frac{$24}{12} = $2$ There are two ways to make packages of socks that average to $$2$ . You can have: $\bullet$ Two $$1$ pairs and one $$4$ pair (package adds up to $$6$ $\bullet$ One $$1$ pair and one $$3$ pair (package adds up to $$4$ Now, we need to solve \[6a+4b=24,\] where $a$ is the number of $$6$ packages and $b$ is the number of $$4$ packages. We see our only solution (that has at least one of each pair of sock) is $a=2, b=3$ , which yields the answer of $2\times2+3\times1 = \boxed{7}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20

Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy? $\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are $12-3=9$ pairs of socks left. Also, the sum of the three pairs of socks is $1+3+4=8$ . This means that there are $24-8=16$ dollars left. If there are only $1$ dollar socks left, then we would have $9\cdot1=9$ dollars wasted, which leaves $7$ more dollars. If we replace one pair with a $3$ dollar pair, then we would waste an additional $2$ dollars. If we replace one pair with a $4$ dollar pair, then we would waste an additional $3$ dollars. The only way $7$ can be represented as a sum of $2$ s and $3$ s is $2+2+3$ . If we change $3$ pairs, we would have $6$ pairs left. Adding the one pair from previously, we have $\boxed{7}$ pairs.

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_20

Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy? $\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Let the amount of $1$ dollar socks be $a$ $3$ dollar socks be $b$ , and $4$ dollar socks be $c$ . We then know that $a+b+c=12$ and $a+3b+4c=24$ . We can make $a+b+c=12$ into $a=12-b-c$ and then plug that into the other equation, producing $12-b-c+3b+4c=24$ which simplifies to $2b+3c=12$ . It's not hard to see $b=3$ and $c=2$ . Now that we know $b$ and $c$ , we know that $a=7$ , meaning the number of $1$ dollar socks Ralph bought is $\boxed{7}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_21

In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$ [asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy] $\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$

Clearly, since $\overline{FE}$ is a side of a square with area $32$ $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$ . Now, since $\overline{FE} = \overline{BC}$ , we have $\overline{BC} = 4 \sqrt{2}$ Now, $\overline{JB}$ is a side of a square with area $18$ , so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$ . Since $\Delta JBK$ is equilateral, $\overline{BK} = 3 \sqrt{2}$ Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}$ , so $\Delta KBC$ is a right triangle with legs $3 \sqrt{2}$ and $4 \sqrt{2}$ . Now, its area is $\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{12}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_21

In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$ [asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy] $\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$

Since $\overline{FE} = \sqrt{32}$ , and $\overline{FE} = \overline{BC}$ $\overline{BC} = 4\sqrt{2}$ . Meanwhile, $\overline{JB} = 3\sqrt{2}$ , and since $\triangle JBK$ is equilateral, $\overline{BK} = 3\sqrt{2}$ . If $ABCDEF$ is equiangular, $\angle ABC = \frac{180 \cdot (n-2)}{n} = 120^{\circ}$ , where $n$ is the number of sides of the shape. Adding all the angles around $B$ gives $270^{\circ}$ , so $\angle KBC = 360 - 90 = 270^{\circ}$ . Because $\triangle KBC$ is right, the area of $\triangle KBC = \frac{4\sqrt{2} \cdot 3\sqrt{2}} {2} = 12$ . Therefore, the answer is $\boxed{12}$ . ~strongstephen

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_22

On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group? $\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$

The text suggests that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\cdot3\cdot5=\boxed{60}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_22

On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group? $\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$

Since we know the number must be a multiple of $15$ , we can eliminate $A$ . We also know that after $12$ days, the students can't find any more arrangements, meaning the number has $12$ factors. Now, we just list the factors of every number, starting with $30$ \[30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6\] \[60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\cdot10\] $60$ has $12$ factors, so the answer is $\boxed{60}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24

A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division? $\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

On one team they play $3N$ games in their division and $4M$ games in the other. This gives $3N+4M=76$ Since $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$ Next, $M=6$ does not work because $52$ is not divisible by $3$ We try $M=7$ does work by giving $N=16$ $~M=7$ and thus $3\times 16=\boxed{48}$ games in their division.

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24

A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division? $\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

$76=3N+4M > 10M$ , giving $M \le 7$ . Since $M>4$ , we have $M=5,6,7$ . Since $4M$ is $1$ $\pmod{3}$ , we must have $M$ equal to $1$ $\pmod{3}$ , so $M=7$ This gives $3N=48$ , as desired. The answer is $\boxed{48}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_24

A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division? $\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

Notice that each team plays $N$ games against each of the three other teams in its division. So that's $3N$ Since each team plays $M$ games against each of the four other teams in the other division, that's $4M$ So $3N+4M=76$ , with $M>4, N>2M$ Let's start by solving this Diophantine equation. In other words, $N=\frac{76-4M}{3}$ So $76-4M\equiv0 \pmod{3}$ (remember: $M$ must be divisible by 3 for $N$ to be an integer!). Therefore, after reducing $76$ to $1$ and $-4M$ to $2M$ (we are doing things in $\pmod{3}$ ), we find that $M\equiv1 \pmod{3}$ Since $M>4$ , so the minimum possible value of $M$ is $7$ . However, remember that $N>2M$ ! To find the greatest possible value of M, we assume that $N=2M$ and that is the upper limit of $M$ (excluding that value because $N>2M$ ). Plugging $N=2M$ in, $10M=76$ . So $M<7.6$ . Since you can't have $7.6$ games, we know that we can only check $M=7$ since we know that since $M>4, M<7.6, M\equiv1 \pmod{3}$ . After checking $M=7$ , we find that it works. So $M=7, N=16$ . So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals $3n = 48$ . Select $\boxed{48}$

https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_25

One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space? [asy] size(75); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy] $\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17$

Proceed as in solution 1 to get $x^2 - 5x + 5 = 0$ Notice that the side length of the square is $\sqrt{x^2+(5-x)^2}$ by the Pythagorean Theorem. Thus, the area of the square is \[\left(\sqrt{x^2+(5-x)^2}\right)^2\] \[= x^2 + x^2 - 10x + 25\] \[= (2x^2 - 10x + 10) + 15\] \[= 2(x^2 - 5x + 5) + 15\] \[= 2(0) + 15\] \[= \boxed{15}\]

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_1

Harry and Terry are each told to calculate $8-(2+5)$ . Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$ . If Harry's answer is $H$ and Terry's answer is $T$ , what is $H-T$ $\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$

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

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_2

Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

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 $7-2=\boxed{5}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_3

Isabella had a week to read a book for a school assignment. She read an average of $36$ pages per day for the first three days and an average of $44$ pages per day for the next three days. She then finished the book by reading $10$ pages on the last day. How many pages were in the book? $\textbf{(A) }240\qquad\textbf{(B) }250\qquad\textbf{(C) }260\qquad\textbf{(D) }270\qquad \textbf{(E) }280$

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 $240+10=\boxed{250}$ pages.

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_4

The sum of two prime numbers is $85$ . What is the product of these two prime numbers? $\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$

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 $83\cdot2=\boxed{166}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_5

Margie's car can go $32$ miles on a gallon of gas, and gas currently costs $ $4$ per gallon. How many miles can Margie drive on $\textdollar 20$ worth of gas? $\textbf{(A) }64\qquad\textbf{(B) }128\qquad\textbf{(C) }160\qquad\textbf{(D) }320\qquad \textbf{(E) }640$

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

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_6

Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$ , and $36$ . What is the sum of the areas of the six rectangles? $\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202$

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 $\boxed{182}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_8

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number? $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$

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 $\boxed{3}$ ~SparklyFlowers

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_10

The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born? $\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$

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 $\boxed{1979}$ corrections made by DrDominic

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$ [asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy] $\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

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 $DE=\boxed{13}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$ [asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy] $\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

The area of the rectangle is $5\times6=30.$ Since the parallel line pairs are identical, $DC=5$ . Let $CE$ be $x$ $\dfrac{5x}{2}=30$ is the area of the right triangle. Solving for $x$ , we get $x=12.$ According to the Pythagorean Theorem, we have a $5-12-13$ triangle. So, the hypotenuse $DE$ has to be $\boxed{13}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_14

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$ [asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy] $\textbf{(A) }12\qquad\textbf{(B) }13\qquad\textbf{(C) }14\qquad\textbf{(D) }15\qquad\textbf{(E) }16$

This problem can be solved with the Pythagorean Theorem ( $a^2 + b^2 = c^2$ ). We know $AB = DC$ , so $DC = 5$ $CE$ is twice the length of $AD$ , so $CE = 12$ $5^2 + 12^2 = c^2$ $5^2 = 25$ $12^2 = 144$ $25 + 144 = 169$ $169$ has a square root of $13$ , so the hypotenuse or $DE$ is $13$ . The answer is $\boxed{13}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_16

The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams? $\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$

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 $56+32 = \boxed{88}$ is our answer.

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_17

George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today? $\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$

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 $\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6}$ is the answer.

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_17

George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today? $\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$

Using the harmonic mean formula, and making the speed he needs to take to get to school for the last half $x$ , we can make the expression: $\frac{2 * 2x}{2 + x}$ Which simplifies to: $\frac{4x}{2 + x}$ Then, because we know that since he goes at $3$ mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to our expression and turn it into something familiar: $\frac{4x}{2 + x} = 3$ Solving that equation gives us: $\boxed{6}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_21

The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$ $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$

The number $\mod{3}$ is congruent to sum of a number's digits $\mod{3}$ is congruent to the number $\mod{3}$ $74A52B1 \pmod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \pmod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$ , so $A+B\equiv 2 \pmod{3}$ . As we know, $326AB4C\equiv 0 \pmod{3}$ , so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$ , and therefore $A+B+C\equiv 0 \pmod{3}$ . We can substitute 2 for $A+B$ , so $2+C\equiv 0 \pmod{3}$ , and therefore $C\equiv 1\pmod{3}$ . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\boxed{1}$

https://artofproblemsolving.com/wiki/index.php/2014_AMC_8_Problems/Problem_21

The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$ $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$

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 $\boxed{1}$ is our answer.

https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_1

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the greatest number of additional cars she must buy in order to be able to arrange all her cars this way? $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 533$

6 x 4 = 24, which is 1 more than 23. So, the answer is $\boxed{1}$

https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_2

A sign at the fish market says, "50 $\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars? $\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is $\boxed{12}$ dollars.

https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_2

A sign at the fish market says, "50 $\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars? $\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Suppose a full pound at normal price costs $x$ dollars. Then, with the 50% off deal, the full pound would cost $x/2$ dollars. A half pound of this would cost $x/4$ dollars (including the 50% off). 50% off half a pound is 3, so we form the equation $x/4=3\Rightarrow x=\boxed{12}$

https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_3

What is the value of $4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)$ $\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000$

We group the addends inside the parentheses two at a time: \begin{align*} -1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\ &= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\ &= 500. \end{align*} Then the desired answer is $4 \times 500 = \boxed{2000}$

https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_4

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill? $\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$

Let $m$ be the total bill the 8 friends split up; each person was supposed to pay $$\frac{m}{8}$ . Since Judi forgot her money, we have to add $$\frac{m}{8}+2.50$ to get the amount of money each of the $7$ friends have to pay. After multipling by 7, we have $7(\frac{m}{8}+2.50)=\frac{7m}{8} + 17.5$ . This is equal to $m$ because both sides represent the total bill. Solving for $m$ \[\frac{7m}{8} + 17.5=m\] \[\frac{m}{8}=17.5\] \[m=\boxed{140}.\]

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, \[26-6=\boxed{20}.\]

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 $y = \boxed{4}$

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: $600 = (6\cdot5)(5y)\\ 600= 6\cdot5\cdot5\cdot y\\ 600=6\cdot25\cdot y\\ 600 = 150y\\ \boxed{4} = y$

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: \[\frac{3}{5}=\frac{x}{165}\] \[3(165)=5x\] \[x=3(33)=99\approx\boxed{100}.\]

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 $\boxed{100}$

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 $10-0+1=\boxed{11}$

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 = $\frac{5940}{18}$ $\boxed{330}$

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, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{330}$

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, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330 \Longrightarrow \boxed{330}$

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 $\dfrac{1}{15}\times 60=\boxed{4}$

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 $100 \% -70 \%= \boxed{30}$

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 $\frac{1}{4}+\frac{1}{8}=\boxed{38}$

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 $\boxed{38}$

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 $40+25+24 = \boxed{89}$

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 $40+24+25=\boxed{89}$

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 $\boxed{338}$

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 $6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{338}$

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, $6x+15=2013\implies x=333\implies x+5=\boxed{338}$

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: $\boxed{338}$

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 $x = \boxed{338}$

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 $120 + 4 \cdot 40 = \boxed{280}$

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 $600-320=\boxed{280}$

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 $\frac{3}{8} \cdot 24 = \boxed{9}$

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 $\boxed{700}$

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 $\boxed{5}$

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 $\boxed{5}$

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 $168-80 = \boxed{88}$

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 $192-100 = \boxed{92}$

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.

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, $\boxed{60}$

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 $\frac{10-4}{10} = \frac{6}{10} = \frac{60}{100} = \boxed{60} \%$

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 $100\% - 40\% = \boxed{60}$

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.

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

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.

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 $\implies \boxed{9}$

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 $\boxed{11}$

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: \[\dfrac{31+x}{7}=6\] \[31+x=42\] \[x=\boxed{11}\]

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

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.

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 $\boxed{4}$

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 $\boxed{7}$

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.

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 $\boxed{87431}$

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 $\boxed{87431}$

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 $\boxed{4}$

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 $\boxed{3127}$

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 $\boxed{9}.$

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 $\boxed{9}$

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 $3x - x = 18$ $\implies$ $\boxed{9}$

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.

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 $1 \times 6 = \boxed{6}$

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 $\boxed{6}$

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 \[\frac{\frac{3s^2\sqrt3}{4}}{2}=\frac{3s^2\sqrt3}{8}.\] $\frac{s^2\sqrt3}{4}=4 \implies \frac{s^2\sqrt3}{8}=2 \implies \frac{3s^2\sqrt3}{8}=\boxed{6}$

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 $\boxed{4}$

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. $\Rightarrow \boxed{100}$

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.

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 $\boxed{306}$

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 $\boxed{306}$

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 $\dfrac{1}{4} \cdot 100 = \boxed{25}$

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 $\boxed{5}$

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.

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 $\boxed{5}$

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 $\boxed{6}$

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 $\boxed{6}$

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 $\frac26=\boxed{13}$