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ef8ba675f2586cac7629415709ba3b97 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_11 | The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$ | To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers are consecutive, if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get $h=64 \Rightarrow \boxed{64}$ | B | 64 |
dd83a36aaf2e88f7ee9ac8969310ed49 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_12 | Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$ | Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or $\dfrac{1}{4}$ of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is $\boxed{100}$ | D | 100 |
dd83a36aaf2e88f7ee9ac8969310ed49 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_12 | Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$ | We could also set up a proportion. Since we know there are 400 red balls, we let the amount of red balls removed be $x$ , so $\frac{400-x}{500-x}=\frac{3}{4}$ . Cross-multiplying gives us $1600-4x=1500-3x \implies x=100$ , so our answer is $\boxed{100}$ | D | 100 |
dd83a36aaf2e88f7ee9ac8969310ed49 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_12 | Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$ | Before any balls are taken out of the bag, there are 400 red balls. Let the number of red balls we take out of the bag be $x$ . Since we want the number of red ball to be 75% of the remaining balls, we can set up the equation: $0.75(500–x)=400–x$ .
In words, 75% of the total remaining balls are equal to the remaining red balls.
Solving we get $x=100$ , so our answer is $\boxed{100}$ .
~J.L.L | D | 100 |
2069d49366b1b5b5e883a6f31192aa29 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_13 | The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Let $n$ $n+1$ , and $n+2$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $n + (n+1) + (n+2) = 3n+3$ . Using the fact that the length of the smallest side is $30\%$ of the perimeter, it follows that:
$n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9$ . The longest side is then $n+2 = 11$ . Thus, answer choice $\boxed{11}$ is correct. | E | 11 |
2069d49366b1b5b5e883a6f31192aa29 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_13 | The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Since the length of the shortest side is a whole number and is equal to $\frac{3}{10}$ of the perimeter, it follows that the perimeter must be a multiple of $10$ . Adding the two previous integers to each answer choice, we see that $11+10+9=30$ . Thus, answer choice $\boxed{11}$ is correct. | E | 11 |
1bd81768b6a04a331761e8db3c006f56 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_14 | What is the sum of the prime factors of $2010$
$\textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210$ | First, we must find the prime factorization of $2010$ $2010=2\cdot 3 \cdot 5 \cdot 67$ . We add the factors up to get $\boxed{77}$ | C | 77 |
0f174d87426f7c8f09f78f6a47b78fa8 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_15 | A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
$\textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$ | We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then $15\%$ of the jar's gumdrops are brown. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{42}$ | C | 42 |
da09e5a581bd1a4cbe8ed077f0c5669d | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_18 | A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$ . And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?
[asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi$ | We can set a proportion:
\[\dfrac{AD}{AB}=\dfrac{3}{2}\]
We substitute $AB$ with 30 and solve for $AD$
\[\dfrac{AD}{30}=\dfrac{3}{2}\]
\[AD=45\]
We calculate the combined area of semicircle by putting together semicircle $AB$ and $CD$ to get a circle with radius $15$ . Thus, the area is $225\pi$ . The area of the rectangle is $30\cdot 45=1350$ . We calculate the ratio:
\[\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{6}\] | C | 6 |
339fd06b73c9ff733aadf69424662313 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_19 | The two circles pictured have the same center $C$ . Chord $\overline{AD}$ is tangent to the inner circle at $B$ $AC$ is $10$ , and chord $\overline{AD}$ has length $16$ . What is the area between the two circles?
[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]
$\textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi$ | Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$ . Thus $AB=BD=8$ . From the Pythagorean Theorem, $CB=6$ . Thus the area between the two circles is $100\pi - 36\pi=64\pi$ $\boxed{64}$ | C | 64 |
a29e1f114cc96aea84a25f6fa85bf78d | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_20 | In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | Let $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of (4,5), but we do lcm of 4,5 = 20. Since we are trying to find the minimum $x$ , we must use the smallest possible value for the number of people in the room. Similarly, we can assume that there are no people present who are wearing neither of the two items since this would unnecessarily increase the number of people in the room. Thus, we can say that there are $20$ people in the room, all of which are wearing at least a hat or a glove.
It follows that there are $\frac{2}{5}\cdot 20 = 8$ people wearing gloves and $\frac{3}{4}\cdot 20 = 15$ people wearing hats. Then by applying the Principle of Inclusion Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $8+15-x = 23-x$ . Since we know that this equals $20$ , it follows that $23-x = 20$ , which implies that $x=3$ . Thus, $\boxed{3}$ is the correct answer. | A | 3 |
c83417f8cad31f46ba4914676eb57249 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_21 | Hui is an avid reader. She bought a copy of the best seller Math is Beautiful . On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360$ | Let $x$ be the number of pages in the book. After the first day, Hui had $\frac{4x}{5}-12$ pages left to read. After the second, she had $\left(\frac{3}{4}\right)\left(\frac{4x}{5}-12\right)-15 = \frac{3x}{5}-24$ left. After the third, she had $\left(\frac{2}{3}\right)\left(\frac{3x}{5}-24\right)-18=\frac{2x}{5}-34$ left. This is equivalent to $62.$
\begin{align*} \frac{2x}{5}-34&=62\\ 2x - 170 &= 310\\ 2x &= 480\\ x &= \boxed{240} | C | 240 |
c83417f8cad31f46ba4914676eb57249 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_21 | Hui is an avid reader. She bought a copy of the best seller Math is Beautiful . On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360$ | If Hui has $62$ pages left to read at the end of Day 3, then on Day 3, before she read those extra 18 pages, she had $80$ pages left to read. $80$ pages is $\frac{2}{3}$ of the pages she had left at the end of Day 2. So at the end of Day 2, she had $80 \cdot \frac{3}{2} = 120$ pages left to read.
Similarly, we can work backwards on Day 2 to find that at the end of Day 1 Hui had $180$ pages left to read. Again working backwards on Day 1 we find that Hui originally had $\boxed{240}$ pages to read. | C | 240 |
532daeff417a9da4c9f95a569845f046 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_22 | The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$ | Let the hundreds, tens, and units digits of the original three-digit number be $a$ $b$ , and $c$ , respectively. We are given that $a=c+2$ . The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$ . The hundreds, tens, and units digits of the reversed three-digit number are $c$ $b$ , and $a$ , respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$ . Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$ . Thus, the units digit in the final result is $\boxed{8}$ | E | 8 |
532daeff417a9da4c9f95a569845f046 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_22 | The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$ | The result must hold for any three-digit number with hundreds digit being $2$ more than the units digit. $301$ is such a number. Evaluating, we get $301-103=198$ . Thus, the units digit in the final result is $\boxed{8}$ | E | 8 |
532daeff417a9da4c9f95a569845f046 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_22 | The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$ | Set the units digit of the original number as $x$ . Thus, its hundreds digit is $x+2$ . After the digits are reversed, the hundreds digit of the original number will be the units digit of the new number. Since $x-(x+2) = -2$ , we can do regrouping and "borrow" $1$ from the tens digit and bring it to the units digit as a $10$ . Therefore, the units digit will end up as $-2 + 10 = \boxed{8}$ | E | 8 |
df037de97863e546885f43ef348f2a44 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_25 | Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$ | A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$ . There are $2$ ways to climb two stairs: $1$ $1$ or $2$ . For 3 stairs, there are $4$ ways:
( $1$ $1$ $1$ )
( $1$ $2$ )
( $2$ $1$ )
( $3$
For four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways.
Thus, there are $\boxed{24}$ ways to get to step $6.$ | E | 24 |
df037de97863e546885f43ef348f2a44 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_25 | Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$ | Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are $2^5=32$ subsets (hit steps) of the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of $4$ $5$ , or $6$ . The eight possible ways for this is ( $4$ $1$ $1$ ), ( $4$ $2$ ), ( $1$ $4$ $1$ ), ( $1$ $1$ $4$ ), ( $2$ $4$ ), ( $1$ $5$ ), ( $5$ $1$ ), and ( $6$
Altogether this makes for $32-8= \boxed{24}$ valid ways for Jo to get to step 6. | E | 24 |
df037de97863e546885f43ef348f2a44 | https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_25 | Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$ | We can set up a recursion to solve this problem. Suppose $f(n)$ represents the number of valid ways to get to the $n$ th step. $f(0)=1$ because there is 1 way for Jo to get to the " $0$ th" step (i.e. the ground). There is $1$ way to get to the first step (a $1$ -step), so $f(1)=1$ . There are $2$ ways to get to the second step (two $1$ -steps or one $2$ -step). Thus, $f(2) = 2$ . In general, $f(n) = f(n-1) + f(n-2) + f(n-3)$ . This is because from the $n-3$ th step, Jo can take a $3$ -step to get to the $n$ th step, from the $n-2$ th step, Jo can take a $2$ -step to get to the $n$ th step, and from the $n-1$ th step, Jo can take a $1$ -step to get to the $n$ th step. We now iteratively calculate values of $f(n)$
$f(0) = 1$
$f(1) = 1$
$f(2) = 2$
$f(3) = f(2) + f(1) + f(0) = 2 + 1 + 1 = 4$
$f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
$f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
$f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = \boxed{24}$ | E | 24 |
fbeec93e6cdb064beda4dbf40a463431 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_2 | On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 112$ | This means the ratio is $4:7$ . If the ratio now is $28:x$ , then that means $x= \boxed{49}$ | D | 49 |
1a0d033165a2eac9d1136bae69fd29bf | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_3 | The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
[asy] import graph; /* this is a label */ Label f; f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0)); yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x) { return x; } draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black); label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10), black+linewidth(5)); dot((15, 15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$ | Suzanna's speed is $\frac{1}{5}$ . This means she runs $\frac{1}{5} \cdot 30 = \boxed{6}$ | C | 6 |
1a0d033165a2eac9d1136bae69fd29bf | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_3 | The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
[asy] import graph; /* this is a label */ Label f; f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0)); yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x) { return x; } draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black); label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10), black+linewidth(5)); dot((15, 15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$ | From the graph, we can see that every $5$ minutes Suzanna goes, her distance increases by $1$ . Since half an hour is $10$ minutes away, she would go $2$ more miles. $4+2$ is $6$ , so the answer is $\boxed{6}$ | C | 6 |
af333970a2858459ea7ef2913e6e4059 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_5 | A sequence of numbers starts with $1$ $2$ , and $3$ . The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$ . In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 68 \qquad \textbf{(E)}\ 99$ | List them out, adding the three previous numbers to get the next number,
\[1,2,3,6,11,20,37,\boxed{68}\] | D | 68 |
55c7d272c74789656815fe398947f891 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_6 | Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 42 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 48$ | Each of the four hoses hose fills $24,000/4 = 6,000$ gallons of water. At the rate it goes at it will take $6,000/2.5 = 2400$ minutes, or $\boxed{40}$ hours. | A | 40 |
55c7d272c74789656815fe398947f891 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_6 | Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve's pool?
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 42 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 48$ | If all four hoses fill $2.5$ gallons a minute, every minute $10$ gallons would be added. Since every hour has $60$ minutes, $600$ gallons of water would be added every hour. $24000/600=\boxed{40}$ hours. | A | 40 |
c318ba1d2e4c7920b99eebad8353b3bb | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_7 | The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$ | The area of a triangle is $\frac12 bh$ . If we let $CD$ be the base of the triangle, then the height is $AB$ , and the area is $\frac12 \cdot 3 \cdot 3 = \boxed{4.5}$ | C | 4.5 |
c318ba1d2e4c7920b99eebad8353b3bb | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_7 | The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$ | We can see that there is a big triangle encasing $ACD$ . The area of that triangle is $\frac12 \cdot 3 \cdot 6 = 9$ . We can easily see that triangle $ABC$ is $\frac12 \cdot 3 \cdot 3$ , which is $4.5$ $9-4.5$ is $4.5$ , so the answer is $\boxed{4.5}$ | C | 4.5 |
c2aeffdb227e85d68bc4ca7aee207ad9 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_8 | The length of a rectangle is increased by $10\%$ percent and the width is decreased by $10\%$ percent. What percent of the old area is the new area?
$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 99 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 101 \qquad \textbf{(E)}\ 110$ | In a rectangle with dimensions $10 \times 10$ , the new rectangle would have dimensions $11 \times 9$ . The ratio of the new area to the old area is $99/100 = \boxed{99}$ | B | 99 |
ad4f47d0b81e3f02818803f7c4fa04b2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_9 | Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]
$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 29$ | Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon $3+8-2(1)=9$ sides are on the outside of the final polygon. In the other shapes $4+5+6+7-4(2) = 14$ sides are on the outside. The resulting polygon has $9+14 = \boxed{23}$ sides. | B | 23 |
ad4f47d0b81e3f02818803f7c4fa04b2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_9 | Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]
$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 29$ | We can quickly see a pattern if we draw out the other shapes. Every shape will have two of its sides taken out except the triangle and octagon. We can then make the expression $2+2+3+4+5+7$ which is $\boxed{23}$ | B | 23 |
4a2206ca319731f043d8fab93cde14f2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_11 | The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ dollars. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ dollars. How many more sixth graders than seventh graders bought a pencil?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | Because the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$ . They can be factored into $11\cdot13$ and $3\cdot5\cdot13$ . The common factor cannot be $1$ or there would have to be more than $30$ sixth graders, so the pencil costs $13$ cents. The difference in costs that the sixth and seventh graders paid is $195-143=52$ cents, which is equal to $52/13 = \boxed{4}$ sixth graders. | D | 4 |
b000b2d39930b61432030f2eeffada0c | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_12 | The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy] [asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$2$",(0,.5)); label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{7}{9}\qquad\textbf{(E)}\ \frac{5}{6}$ | The possible sums are \[\begin{tabular}{c|ccc} & 1 & 3 & 5 \\ \hline 2 & 3 & 5 & 7 \\ 4 & 5 & 7 & 9 \\ 6 & 7 & 9 & 11 \end{tabular}\]
Only $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $\boxed{79}$ | D | 79 |
80d6ca860b3ba1827e5ccb849605e785 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_13 | A three-digit integer contains one of each of the digits $1$ $3$ , and $5$ . What is the probability that the integer is divisible by $5$
$\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3} \qquad \textbf{(E)}\ \frac{5}{6}$ | The three digit numbers are $135,153,351,315,513,531$ . The numbers that end in $5$ are divisible are $5$ , and the probability of choosing those numbers is $\boxed{13}$ | B | 13 |
80d6ca860b3ba1827e5ccb849605e785 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_13 | A three-digit integer contains one of each of the digits $1$ $3$ , and $5$ . What is the probability that the integer is divisible by $5$
$\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3} \qquad \textbf{(E)}\ \frac{5}{6}$ | The number is divisible by 5 if and only if the number ends in $5$ (also $0$ , but that case can be ignored, as none of the digits are $0$ )
If we randomly arrange the three digits, the probability of the last digit being $5$ is $\boxed{13}$ | B | 13 |
abd44e0843793f53652c5cfdd75161d9 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_14 | Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
$\textbf{(A)}\ 46 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 54$ | The way to Temple took $\frac{50}{60}=\frac56$ hours, and the way back took $\frac{50}{40}=\frac54$ for a total of $\frac56 + \frac54 = \frac{25}{12}$ hours. The trip is $50\cdot2=100$ miles. The average speed is $\frac{100}{25/12} = \boxed{48}$ miles per hour. | B | 48 |
abd44e0843793f53652c5cfdd75161d9 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_14 | Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
$\textbf{(A)}\ 46 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 54$ | We calculate the harmonic mean of Austin and Temple.
Plugging in, we have: $\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{48}$ miles per hour. | B | 48 |
abd44e0843793f53652c5cfdd75161d9 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_14 | Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
$\textbf{(A)}\ 46 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 54$ | The 50 miles part of this question is redundant. We find that the average speed is skewed towards the lower speed, since you spend more time in the lower speed. So, we take the LCM of 40 and 60, getting 120. So, we do (40 + 40 + 40 + 60 + 60)/5, calculation the ratio of the parts and taking the mean. Solving, we get 240/5 = $\boxed{48} \qquad$ -themathgood | B | 48 |
13b500a59072d95c1e9d001b1aea40df | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_16 | How many $3$ -digit positive integers have digits whose product equals $24$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24$ | With the digits listed from least to greatest, the $3$ -digit integers are $138,146,226,234$ $226$ can be arranged in $\frac{3!}{2!} = 3$ ways, and the other three can be arranged in $3!=6$ ways. There are $3+6(3) = \boxed{21}$ $3$ -digit positive integers. | D | 21 |
b5d2ca91f976a5d67bafccd86c11f3de | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_18 | The diagram represents a $7$ -foot-by- $7$ -foot floor that is tiled with $1$ -square-foot black tiles and white tiles. Notice that the corners have white tiles. If a $15$ -foot-by- $15$ -foot floor is to be tiled in the same manner, how many white tiles will be needed?
[asy]unitsize(10); draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); draw((1,7)--(1,0)); draw((6,7)--(6,0)); draw((5,7)--(5,0)); draw((4,7)--(4,0)); draw((3,7)--(3,0)); draw((2,7)--(2,0)); draw((0,1)--(7,1)); draw((0,2)--(7,2)); draw((0,3)--(7,3)); draw((0,4)--(7,4)); draw((0,5)--(7,5)); draw((0,6)--(7,6)); fill((1,0)--(2,0)--(2,7)--(1,7)--cycle,black); fill((3,0)--(4,0)--(4,7)--(3,7)--cycle,black); fill((5,0)--(6,0)--(6,7)--(5,7)--cycle,black); fill((0,5)--(0,6)--(7,6)--(7,5)--cycle,black); fill((0,3)--(0,4)--(7,4)--(7,3)--cycle,black); fill((0,1)--(0,2)--(7,2)--(7,1)--cycle,black);[/asy]
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 57 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 96 \qquad \textbf{(E)}\ 126$ | In a $1$ -foot-by- $1$ -foot floor, there is $1$ white tile. In a $3$ -by- $3$ , there are $4$ . Continuing on, you can deduce the $n^{th}$ positive odd integer floor has $n^2$ white tiles. $15$ is the $8^{th}$ odd integer, so there are $\boxed{64}$ white tiles. | C | 64 |
fcaf5432cd1b168698dc87297e4a8f6e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_19 | Two angles of an isosceles triangle measure $70^\circ$ and $x^\circ$ . What is the sum of the three possible values of $x$
$\textbf{(A)}\ 95 \qquad \textbf{(B)}\ 125 \qquad \textbf{(C)}\ 140 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 180$ | There are 3 cases: where $x^\circ$ is a base angle with the $70^\circ$ as the other angle, where $x^\circ$ is a base angle with $70^\circ$ as the vertex angle, and where $x^\circ$ is the vertex angle with $70^\circ$ as a base angle.
Case 1: $x^\circ$ is a base angle with the $70^\circ$ as the other angle:
Here, $x=70$ , since base angles are congruent.
Case 2: $x^\circ$ is a base angle with $70^\circ$ as the vertex angle:
Here, the 2 base angles are both $x^\circ$ , so we can use the equation $2x+70=180$ , which simplifies to $x=55$
Case 3: $x^\circ$ is the vertex angle with $70^\circ$ as a base angle:
Here, both base angles are $70^\circ$ , since base angles are congruent. Thus, we can use the equation $x+140=180$ , which simplifies to $x=40$
Adding up all the cases, we get $70+55+40=165$ , so the answer is $\boxed{165}$ | D | 165 |
bde71874de06e99d1d3d1396ed7ae724 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_20 | How many non-congruent triangles have vertices at three of the eight points in the array shown below?
[asy]dot((0,0)); dot((.5,.5)); dot((.5,0)); dot((.0,.5)); dot((1,0)); dot((1,.5)); dot((1.5,0)); dot((1.5,.5));[/asy]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | Assume the base of the triangle is on the bottom four points because a congruent triangle can be made by reflecting the base on the top four points. For a triangle with a base of length $1$ , there are $3$ triangles. For a triangle with a base of length $2$ , there are $3$ triangles. For length $3$ , there are $2$ . In total, the number of non-congruent triangles is $3+3+2=\boxed{8}$ | D | 8 |
b494dac183ac9d7a645c2e3f31d09d40 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_22 | How many whole numbers between 1 and 1000 do not contain the digit 1?
$\textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800$ | Note that this is the same as finding how many numbers with up to three digits do not contain 1.
Since there are 10 total possible digits, and only one of them is not allowed (1), each place value has its choice of 9 digits, for a total of $9*9*9=729$ such numbers. However, we overcounted by one; 0 is not between 1 and 1000, so there are $\boxed{728}$ numbers. | D | 728 |
6628d90c726ae341fcf05bc91def445f | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_23 | On the last day of school, Mrs. Wonderful gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$ | If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans.
\begin{align*} x^2+(x+2)^2 &= 394\\ x^2+x^2+4x+4 &= 394\\ 2x^2 + 4x &= 390\\ x^2 + 2x &= 195\\ \end{align*}
From here, we can see that $x = 13$ as $13^2 + 26 = 195$ , so there are $13$ girls, $13+2=15$ boys, and $13+15=\boxed{28}$ students. | B | 28 |
faaccf9259bff468d99045641db8964e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_24 | The letters $A$ $B$ $C$ and $D$ represent digits. If $\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$ and $\begin{tabular}{ccc}&A&B\\ -&C&A\\ \hline &&A\end{tabular}$ ,what digit does $D$ represent?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | Because $B+A=A$ $B$ must be $0$ .
Next, because $B-A=A\implies0-A=A,$ we get $A=5$ as the "0" mentioned above is actually 10 in this case.
Now we can rewrite $\begin{tabular}{ccc}&A&0\\ +&C&A\\ \hline &D&A\end{tabular}$ as $\begin{tabular}{ccc}&5&0\\ +&C&5\\ \hline &D&5\end{tabular}$ . Therefore, $D=5+C$
Finally, $A-1-C=0\implies{A=C+1}\implies{C=4}$ , So we have $D=5+C\implies{D=5+4}=\boxed{9}$ | E | 9 |
9ac6279fb8677329a07cb23c030d2f15 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_25 | A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? [asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label("A", (1,0,3/4), W); label("B", (1,0,1/3), W); label("C", (1,0,1/6-d/4), W); label("D", (1,0,d/2), W); label("1/2", (1,1,3/4), E); label("1/3", (1,1,1/3), E); label("1/17", (0,1,1/6-d/4), E);[/asy]
[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label("D", (1/2, 1, 0), SE); label("A", (1+1/2, 1, 0), SE); label("B", (2+1/2, 1, 0), SE); label("C", (3+1/2, 1, 0), SE);[/asy]
$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11$ | The areas of the tops of $A$ $B$ $C$ , and $D$ in the figure formed has sum $1+1+1+1 = 4$ as do the bottoms. Thus, the total so far is $8$ . Now, one of the sides has an area of one, since it combines all of the heights of $A$ $B$ $C$ , and $D$ , which is $1$ . The other side is also the same. Thus the total area now is $10$ . From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like a side of $A$ , with a surface area of half. From the back, it is the same thing. Thus, the total surface area is $10+\frac{1}{2}+\frac{1}{2}= 11$ , or $\boxed{11}$ | E | 11 |
9ac6279fb8677329a07cb23c030d2f15 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_25 | A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? [asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label("A", (1,0,3/4), W); label("B", (1,0,1/3), W); label("C", (1,0,1/6-d/4), W); label("D", (1,0,d/2), W); label("1/2", (1,1,3/4), E); label("1/3", (1,1,1/3), E); label("1/17", (0,1,1/6-d/4), E);[/asy]
[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label("D", (1/2, 1, 0), SE); label("A", (1+1/2, 1, 0), SE); label("B", (2+1/2, 1, 0), SE); label("C", (3+1/2, 1, 0), SE);[/asy]
$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11$ | The top parts and the bottom parts sum to 8. The sides add on another 2. Therefore, the only logical answer would be $\boxed{11}$ since it is the only answer greater than 10. | E | 11 |
15cbb7bdf9a82576474b46734fb420aa | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_1 | Susan had 50 dollars to spend at the carnival. She spent 12 dollars on food and twice as much on rides. How many dollars did she have left to spend?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 38 \qquad \textbf{(E)}\ 50$ | If Susan spent 12 dollars, then twice that much on rides, then she spent $12+12 \times 2=36$ dollars in total. We subtract $36$ from $50$ to get $\boxed{14}$ | B | 14 |
a784ecf55d1e454d1667f94ad7fc141c | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_2 | The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$ , in order. What 4-digit number is represented by the code word $\text{CLUE}$
$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$ | We can derive that $C=8$ $L=6$ $U=7$ , and $E=1$ . Therefore, the answer is $\boxed{8671}$ ~edited by Owencheng | A | 8671 |
0cefbfa35be60791e5afd27ffbfda03b | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_5 | Barney Schwinn notices that the odometer on his bicycle reads $1441$ , a palindrome, because it reads the same forward and backward. After riding $4$ more hours that day and $6$ the next, he notices that the odometer shows another palindrome, $1661$ . What was his average speed in miles per hour?
$\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 22$ | Barney travels $1661-1441=220$ miles in $4+6=10$ hours for an average of $220/10=\boxed{22}$ miles per hour. | E | 22 |
000a900dba3123d56626acc0e8eef820 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_7 | If $\frac{3}{5}=\frac{M}{45}=\frac{60}{N}$ , what is $M+N$
$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 105\qquad \textbf{(E)}\ 127$ | Separate into two equations $\frac35 = \frac{M}{45}$ and $\frac35 = \frac{60}{N}$ and solve for the unknowns. $M=27$ and $N=100$ , therefore $M+N=\boxed{127}$ | E | 127 |
e8a5bfa197fe82e79e8bc90038d3030e | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_8 | Candy sales from the Boosters Club from January through April are shown. What were the average sales per month in dollars?
[asy] draw((0,0)--(36,0)--(36,24)--(0,24)--cycle); draw((0,4)--(36,4)); draw((0,8)--(36,8)); draw((0,12)--(36,12)); draw((0,16)--(36,16)); draw((0,20)--(36,20)); fill((4,0)--(8,0)--(8,20)--(4,20)--cycle, black); fill((12,0)--(16,0)--(16,12)--(12,12)--cycle, black); fill((20,0)--(24,0)--(24,8)--(20,8)--cycle, black); fill((28,0)--(32,0)--(32,24)--(28,24)--cycle, black); label("120", (0,24), W); label("80", (0,16), W); label("40", (0,8), W); label("Jan", (6,0), S); label("Feb", (14,0), S); label("Mar", (22,0), S); label("Apr", (30,0), S); [/asy]
$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 70\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 80\qquad\textbf{(E)}\ 85$ | There are a total of $100+60+40+120=320$ dollars of sales spread through $4$ months, for an average of $320/4 = \boxed{80}$ | D | 80 |
10c0d22ff79ba6147128a9e2ad0ec0d3 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_9 | In $2005$ Tycoon Tammy invested $100$ dollars for two years. During the first year
her investment suffered a $15\%$ loss, but during the second year the remaining
investment showed a $20\%$ gain. Over the two-year period, what was the change
in Tammy's investment?
$\textbf{(A)}\ 5\%\text{ loss}\qquad \textbf{(B)}\ 2\%\text{ loss}\qquad \textbf{(C)}\ 1\%\text{ gain}\qquad \textbf{(D)}\ 2\% \text{ gain} \qquad \textbf{(E)}\ 5\%\text{ gain}$ | After the $15 \%$ loss, Tammy has $100 \cdot 0.85 = 85$ dollars. After the $20 \%$ gain, she has $85 \cdot 1.2 = 102$ dollars. This is an increase in $2$ dollars from her original $100$ dollars, a $\boxed{2}$ | D | 2 |
dcec358ceee4001c810c87b6c0ed19c3 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_10 | The average age of the $6$ people in Room A is $40$ . The average age of the $4$ people in Room B is $25$ . If the two groups are combined, what is the average age of all the people?
$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$ | The total of all their ages over the number of people is
\[\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{34}.\] | D | 34 |
cba5e4d83ebfe96d7080808a1c7817de | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_11 | Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$ | The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \boxed{7}$ | A | 7 |
cba5e4d83ebfe96d7080808a1c7817de | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_11 | Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$ | We create a diagram: [asy] draw(circle((0,0),5)); draw(circle((5,0),5)); label("$x$",(2.3,0),S); label("$20$",(7,0),S); label("$26$",(-2,0),S); [/asy]
Let $x$ be the number of students with both a dog and a cat.
Therefore, we have
\[26+20-x = 39\] \[46-x = 39\] \[x = \boxed{7}\] | A | 7 |
a0a4e523b0ba57e0314961ba3d468fc3 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_12 | A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$ | Each bounce is $2/3$ times the height of the previous bounce. The first bounce reaches $2$ meters, the second $4/3$ , the third $8/9$ , the fourth $16/27$ , and the fifth $32/81$ . Half of $81$ is $40.5$ , so the ball does not reach the required height on bounce $\boxed{5}$ | C | 5 |
777c5b5fb07eb4b9e773ea7b63e03929 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_13 | Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than $100$ pounds or more than $150$ pounds. So the boxes are weighed in pairs in every possible way. The results are $122$ $125$ and $127$ pounds. What is the combined weight in pounds of the three boxes?
$\textbf{(A)}\ 160\qquad \textbf{(B)}\ 170\qquad \textbf{(C)}\ 187\qquad \textbf{(D)}\ 195\qquad \textbf{(E)}\ 354$ | Each box is weighed twice during this, so the combined weight of the three boxes is half the weight of these separate measures:
\[\frac{122+125+127}{2} = \frac{374}{2} = \boxed{187}.\] | C | 187 |
decbe96bab84065c6c1907bcf613256d | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_14 | Three $\text{A's}$ , three $\text{B's}$ , and three $\text{C's}$ are placed in the nine spaces so that each row and column contains one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?
[asy] size((80)); draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((0,3)--(9,3)); draw((0,6)--(9,6)); label("A", (1.5,7.5)); [/asy]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | There are $2$ ways to place the remaining $\text{As}$ $2$ ways to place the remaining $\text{Bs}$ , and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{4}$ | C | 4 |
66b1a6836e23259827844f631cb5c8ff | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_15 | In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
$\textbf{(A)}\ 35\qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 56\qquad \textbf{(E)}\ 72$ | The total number of points from the first $8$ games is $7+4+3+6+8+3+1+5=37$ . We have to make this a multiple of $9$ by scoring less than $10$ points. The closest multiple of $9$ is $45$ $45-37=8$ . Now we have to add a number to get a multiple of 10. The next multiple is $50$ we added $5$ , multiplying these together you get $8\cdot5$ is $40$ . The answer is $\boxed{40}$ | B | 40 |
bbc8b2bcbd811466824eda4c94d2dc46 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_17 | Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
$\textbf{(A)}\ 76\qquad \textbf{(B)}\ 120\qquad \textbf{(C)}\ 128\qquad \textbf{(D)}\ 132\qquad \textbf{(E)}\ 136$ | A rectangle's area is maximized when its length and width are equivalent, or the two side lengths are closest together in this case with integer lengths. This occurs with the sides $12 \times 13 = 156$ . Likewise, the area is smallest when the side lengths have the greatest difference, which is $1 \times 24 = 24$ . The difference in area is $156-24=\boxed{132}$ | D | 132 |
d66f05b0a1b2641cf2e2a46bac7d1759 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_19 | Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart? [asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5)); [/asy] $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7}$ | The two points are one unit apart at $8$ places around the edge of the square. There are $8 \choose 2$ $= 28$ ways to choose two points. The probability is
\[\frac{8}{28} = \boxed{27}\] | B | 27 |
d66f05b0a1b2641cf2e2a46bac7d1759 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_19 | Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart? [asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5)); [/asy] $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7}$ | Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is $\boxed{27}$ | B | 27 |
6f1df3fa5258106cb530532e15328d1b | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_20 | The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$ | Let $b$ be the number of boys and $g$ be the number of girls.
\[\frac23 b = \frac34 g \Rightarrow b = \frac98 g\]
For $g$ and $b$ to be integers, $g$ must cancel out with the denominator, and the smallest possible value is $8$ . This yields $9$ boys. The minimum number of students is $8+9=\boxed{17}$ | B | 17 |
6f1df3fa5258106cb530532e15328d1b | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_20 | The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$ | We know that $\frac23 B = \frac34 G$ or $\frac69 B = \frac68 G$ . So, the ratio of the number of boys to girls is 9:8. The smallest total number of students is $9 + 8 = \boxed{17}$ . ~DY | B | 17 |
003826ff123b785c14464f1af00c7587 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_21 | Jerry cuts a wedge from a $6$ -cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?
[asy] defaultpen(linewidth(0.65)); real d=90-63.43494882; draw(ellipse((origin), 2, 4)); fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white); draw(ellipse((-4,0), 2, 4)); draw((0,4)--(-4,4)); draw((0,-4)--(-4,-4)); draw(shift(-2,0)*rotate(-d-5)*ellipse(origin, 1.82, 4.56), linetype("10 10")); draw((-4,4)--(-8,4), dashed); draw((-4,-4)--(-8,-4), dashed); draw((-4,4.3)--(-4,5)); draw((0,4.3)--(0,5)); draw((-7,4)--(-7,-4), Arrows(5)); draw((-4,4.7)--(0,4.7), Arrows(5)); label("$8$ cm", (-7,0), W); label("$6$ cm", (-2,4.7), N);[/asy]
$\textbf{(A)} \ 48 \qquad \textbf{(B)} \ 75 \qquad \textbf{(C)} \ 151\qquad \textbf{(D)} \ 192 \qquad \textbf{(E)} \ 603$ | The slice is cutting the cylinder into two equal wedges with equal volume. The cylinder's volume is $\pi r^2 h = \pi (4^2)(6) = 96\pi$ . The volume of the wedge is half this which is $48\pi \approx \boxed{151}$ | C | 151 |
8113a433642e1634b3d9adc0881ee51a | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_22 | For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$ | Instead of finding n, we find $x=\frac{n}{3}$ . We want $x$ and $9x$ to be three-digit whole numbers. The smallest three-digit whole number is $100$ , so that is our minimum value for $x$ , since if $x \in \mathbb{Z^+}$ , then $9x \in \mathbb{Z^+}$ . The largest three-digit whole number divisible by $9$ is $999$ , so our maximum value for $x$ is $\frac{999}{9}=111$ . There are $12$ whole numbers in the closed set $\left[100,111\right]$ , so the answer is $\boxed{12}$ | A | 12 |
8113a433642e1634b3d9adc0881ee51a | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_22 | For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$ | We can set the following inequalities up to satisfy the conditions given by the question, $100 \leq \frac{n}{3} \leq 999$ ,
and $100 \leq 3n \leq 999$ .
Once we simplify these and combine the restrictions, we get the inequality, $300 \leq n \leq 333$ .
Now we have to find all multiples of 3 in this range for $\frac{n}{3}$ to be an integer. We can compute this by setting $\frac{n} {3}=x$ , where $x \in \mathbb{Z^+}$ . Substituting $x$ for $n$ in the previous inequality, we get, $100 \leq x \leq 111$ , and there are $111-100+1$ integers in this range giving us the answer, $\boxed{12}$ | A | 12 |
8113a433642e1634b3d9adc0881ee51a | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_22 | For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$ | So we know the largest $3$ digit number is $999$ and the lowest is $100$ . This means $\dfrac{n}{3} \ge 100 \rightarrow n \ge 300$ but $3n \le 999 \rightarrow n \le 333$ . So we have the set ${300, 301, 302, \cdots, 330, 331, 332, 333}$ for $n$ . Now we have to find the multiples of $3$ suitable for $n$ , or else $\dfrac{n}{3}$ will be a decimal. Only numbers ${300, 303, \cdots, 333}$ are counted. We can divide by $3$ to make the difference $1$ again, getting ${100, 101 \cdots , 111}$ . Due to it being inclusive, we have $111-100+1 =\boxed{12}$ | A | 12 |
341d345fdda6d30ac8049a7ddb094bdf | https://artofproblemsolving.com/wiki/index.php/2008_AMC_8_Problems/Problem_25 | Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Which of the following is closest to the percent of the design that is black?
[asy] real d=320; pair O=origin; pair P=O+8*dir(d); pair A0 = origin; pair A1 = O+1*dir(d); pair A2 = O+2*dir(d); pair A3 = O+3*dir(d); pair A4 = O+4*dir(d); pair A5 = O+5*dir(d); filldraw(Circle(A0, 6), white, black); filldraw(circle(A1, 5), black, black); filldraw(circle(A2, 4), white, black); filldraw(circle(A3, 3), black, black); filldraw(circle(A4, 2), white, black); filldraw(circle(A5, 1), black, black); [/asy]
$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad$ | Let the smallest circle be 1, the second smallest circle be 2, the third smallest circle be 3, etc. \[\begin{array}{c|cc} \text{circle \#} & \text{radius} & \text{area} \\ \hline 1 & 2 & 4\pi \\ 2 & 4 & 16\pi \\ 3 & 6 & 36\pi \\ 4 & 8 & 64\pi \\ 5 & 10 & 100\pi \\ 6 & 12 & 144\pi \end{array}\]
The entire circle's area is $144\pi$ . The area of the black regions is $(100-64)\pi + (36-16)\pi + 4\pi = 60\pi$ . The percentage of the design that is black is $\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{42}$ | A | 42 |
490166888ed539ce6bcaa517b65b14c7 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_1 | Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$ $11$ $7$ $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?
$\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13$ | Let $x$ be the number of hours she must work for the final week. We are looking for the average, so \[\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\] Solving gives: \[\frac{48 + x}{6} = 10\]
\[48 + x = 60\]
\[x = 12\]
So, the answer is $\boxed{12}$ | D | 12 |
490166888ed539ce6bcaa517b65b14c7 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_1 | Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$ $11$ $7$ $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?
$\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13$ | Use average deviation:
The average is 10 hour per day. If work 8 hours then it is 2 hours short; if work 11 hours then there is 1 hour surplus, the last day need to cancel out the collective deviation from the previous 5 days.
So we got
\[-2+1-3+2+0=-2\]
The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so $\boxed{12}$ | D | 12 |
df09333223832da82acb63ed2d570283 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_4 | A haunted house has six windows. In how many ways can
Georgie the Ghost enter the house by one window and leave
by a different window?
$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36$ | Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows.
So, Georgie has a total of $6 \cdot 5$ ways he can enter the house by one window and leave
by a different window.
Therefore, we have $\boxed{30}$ ways. | D | 30 |
49e4281f994175e4845197157f1f18c8 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_5 | Chandler wants to buy a $500$ dollar mountain bike. For his birthday, his grandparents
send him $50$ dollars, his aunt sends him $35$ dollars and his cousin gives him $15$ dollars. He earns $16$ dollars per week for his paper route. He will use all of his birthday money and all
of the money he earns from his paper route. In how many weeks will he be able
to buy the mountain bike?
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 25 \qquad\mathrm{(C)}\ 26 \qquad\mathrm{(D)}\ 27 \qquad\mathrm{(E)}\ 28$ | Let $x$ be the number of weeks.
Thus, we have the equation $50 + 35 + 15 + 16x = 500$
Simplify,
$100 + 16x = 500$
$16x = 400$
$x = 25$
The answer is $\boxed{25}$ | B | 25 |
bbbe100197914c04d2f9e1fe6c9d2062 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_6 | The average cost of a long-distance call in the USA in $1985$ was $41$ cents per minute, and the average cost of a long-distance
call in the USA in $2005$ was $7$ cents per minute. Find the
approximate percent decrease in the cost per minute of a long-
distance call.
$\mathrm{(A)}\ 7 \qquad\mathrm{(B)}\ 17 \qquad\mathrm{(C)}\ 34 \qquad\mathrm{(D)}\ 41 \qquad\mathrm{(E)}\ 80$ | The percent decrease is (the amount of decrease)/(original amount)
the amount of decrease is $41 - 7 = 34$
so the percent decrease is $\frac{34}{41}$ which is about $\boxed{80}$ | E | 80 |
3434dedfdf83f4f86eba9ddb8bdecaf4 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_7 | The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$ | Let $x$ be the average of the remaining $4$ people.
The equation we get is $\frac{4x + 18}{5} = 30$
Simplify,
$4x + 18 = 150$
$4x = 132$
$x = 33$
Therefore, the answer is $\boxed{33}$ | D | 33 |
3434dedfdf83f4f86eba9ddb8bdecaf4 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_7 | The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$ | Since an $18$ year old left from a group of people averaging $30$ , The remaining people must total $30 - 18 = 12$ years older than $30$ . Therefore, the average is $\dfrac{12}{4} = 3$ years over $30$ . Giving us $\boxed{33}$ | D | 33 |
3434dedfdf83f4f86eba9ddb8bdecaf4 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_7 | The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?
$\mathrm{(A)}\ 25 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 29 \qquad\mathrm{(D)}\ 33 \qquad\mathrm{(E)}\ 36$ | The total ages would be $30*5=150$ . Then, if one $18$ year old leaves, we subtract $18$ from $150$ and get $132$ . Then, we divide $132$ by $4$ to get the new average, $\boxed{33}$ | D | 33 |
81f4d9963c4245dff18d99703aa5264d | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_8 | In trapezoid $ABCD$ $\overline{AD}$ is perpendicular to $\overline{DC}$ $AD = AB = 3$ , and $DC = 6$ . In addition, $E$ is on $\overline{DC}$ , and $\overline{BE}$ is parallel to $\overline{AD}$ . Find the area of $\triangle BEC$ [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S); [/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$ | Clearly, $ABED$ is a square with side-length $3.$ By segment subtraction, we have $EC = DC - DE = 6 - 3 = 3.$
The area of $\triangle BEC$ is \[\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{4.5}.\] ~Aplus95 (Solution) | B | 4.5 |
81f4d9963c4245dff18d99703aa5264d | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_8 | In trapezoid $ABCD$ $\overline{AD}$ is perpendicular to $\overline{DC}$ $AD = AB = 3$ , and $DC = 6$ . In addition, $E$ is on $\overline{DC}$ , and $\overline{BE}$ is parallel to $\overline{AD}$ . Find the area of $\triangle BEC$ [asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, NW); label("$3$", A--D, W); label("$3$", A--B, N); label("$6$", E, S); [/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$ | Clearly, $ABED$ is a square with side-length $3.$
Let the brackets denote areas. We apply area subtraction to find the area of $\triangle BEC:$ \begin{align*} [BEC]&=[ABCD]-[ABED] \\ &=\frac{AB+CD}{2}\cdot AD - AB^2 \\ &=\frac{3+6}{2}\cdot 3 - 3^2 \\ &=\boxed{4.5} ~MRENTHUSIASM | B | 4.5 |
23fbdf41ec52bfb0f868946fa7227583 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_9 | To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}$ | The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$ . Therefore the number in the lower right-hand square is $\boxed{2}$ | B | 2 |
23fbdf41ec52bfb0f868946fa7227583 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_9 | To complete the grid below, each of the digits 1 through 4 must occur once
in each row and once in each column. What number will occupy the lower
right-hand square?
\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}$ | Note how the first and second row already contain a $2$ . Since the third row, last column already has a $4$ , the only possible place a $2$ could be in is the bottom right square. Thus our answer is $\boxed{2}$ | B | 2 |
47c3ea3fdcf88a901d746a48f3794189 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_13 | Sets $A$ and $B$ , shown in the Venn diagram, have the same number of elements.
Their union has $2007$ elements and their intersection has $1001$ elements. Find
the number of elements in $A$
[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N); label("$1001$", (2.5, -0.5), N);[/asy]
$\mathrm{(A)}\ 503 \qquad \mathrm{(B)}\ 1006 \qquad \mathrm{(C)}\ 1504 \qquad \mathrm{(D)}\ 1507 \qquad \mathrm{(E)}\ 1510$ | Let $x$ be the number of elements in $A$ and $B$ which is equal.
Then we could form equation $2x-1001 = 2007$
$2x = 3008$
$x = 1504$
The answer is $\boxed{1504}$ | C | 1504 |
47c3ea3fdcf88a901d746a48f3794189 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_13 | Sets $A$ and $B$ , shown in the Venn diagram, have the same number of elements.
Their union has $2007$ elements and their intersection has $1001$ elements. Find
the number of elements in $A$
[asy] defaultpen(linewidth(0.7)); draw(Circle(origin, 5)); draw(Circle((5,0), 5)); label("$A$", (0,5), N); label("$B$", (5,5), N); label("$1001$", (2.5, -0.5), N);[/asy]
$\mathrm{(A)}\ 503 \qquad \mathrm{(B)}\ 1006 \qquad \mathrm{(C)}\ 1504 \qquad \mathrm{(D)}\ 1507 \qquad \mathrm{(E)}\ 1510$ | Let $x$ be the number of elements in $A$ not including the intersection. $2007-1001=1006$ total elements excluding the intersection. Since we know that $A=B$ , we can find that $x=\frac{1006}2=503$ . Now we need to add the intersection. $503+1001=\boxed{1504}$ | C | 1504 |
b8bbec1086a1f231bc14b0f354956990 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_14 | The base of isosceles $\triangle ABC$ is $24$ and its area is $60$ . What is the length of one
of the congruent sides?
$\mathrm{(A)}\ 5 \qquad \mathrm{(B)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18$ | The area of a triangle is shown by $\frac{1}{2}bh$ . We set the base equal to $24$ , and the area equal to $60$ , and we get the height, or altitude, of the triangle to be $5$ . In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem, $a^2+b^2=c^2$ , we can solve for one of the legs of the triangle (it will be the the hypotenuse, $c$ ). $a = 12$ $b = 5$ $c = 13$ .
The answer is $\boxed{13}$ | C | 13 |
8f761c084e7dc0681fc407305ee0321f | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_17 | A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow
tint and $45\%$ water. Five liters of yellow tint are added to
the original mixture. What is the percent of yellow tint
in the new mixture?
$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50$ | Since $30\%$ of the original $30$ liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are $9+5=14$ liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This gives $40\%$ of yellow tint in the new mixture, which is $\boxed{40}$ | C | 40 |
625caf0edef34f8dfbe7954922d7b628 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_18 | The product of the two $99$ -digit numbers
$303,030,303,...,030,303$ and $505,050,505,...,050,505$
has thousands digit $A$ and units digit $B$ . What is the sum of $A$ and $B$
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$ | We can first make a small example to find out $A$ and $B$ . So,
$303\times505=153015$
The ones digit plus thousands digit is $5+3=8$
Note that the ones and thousands digits are, added together, $8$ . (and so on...) So the answer is $\boxed{8}$ This is a direct multiplication way. | D | 8 |
ab836a34efd76f55bc85a0fb12249ac8 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_19 | Pick two consecutive positive integers whose sum is less than $100$ . Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$ | Let the smaller of the two numbers be $x$ . Then, the problem states that $(x+1)+x<100$ $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$ $2x+1$ is obviously odd, so only answer choices C and E need to be considered.
$2x+1=131$ contradicts the fact that $2x+1<100$ , so the answer is $\boxed{79}$ | C | 79 |
ab836a34efd76f55bc85a0fb12249ac8 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_19 | Pick two consecutive positive integers whose sum is less than $100$ . Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$ | Since for two consecutive numbers $a$ and $b$ , the difference between their squares are $a^2-b^2=(a+b)(a-b)$ , which equals to $a+b$ , because $a$ and $b$ are consecutive. And because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and that the sum of $a$ and $b$ is less than 100, you can eliminate all answers expect for $\boxed{79}$ | C | 79 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. From the information given in the problem, we can say that $\frac{x}{y}=0.45.$ Next, the Unicorns win 6 more games and lose 2 more, for a total of $6+2=8$ games played during district play. We are told that they end the season having won half of their games, or $0.5$ of their games. We can write another equation: $\frac{x+6}{y+8}=0.5.$ This gives us a system of equations: $\frac{x}{y}=0.45$ and $\frac{x+6}{y+8}=0.5.$ We first multiply both sides of the first equation by $y$ to get $x=0.45y.$ Then, we multiply both sides of the second equation by $(y+8)$ to get $x+6=0.5(y+8).$ Applying the Distributive Property gives yields $x+6=0.5y+4.$ Now we substitute $0.45y$ for $x$ to get $0.45y+6=0.5y+4.$ Solving gives us $y=40.$ Since the problem asks for the total number of games, we add on the last 8 games to get the solution $\boxed{48}$ | A | 48 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | Simplifying 45% to $\frac{9}{20}$ , we see that the numbers of games before district play are a multiple of 20. After that the Aces played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is $\boxed{48}$ , which is 20(2)+8. | null | 48 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | First we simplify $45$ % to $\frac{9}{20}$ . Ratio of won to total is $\frac{9}{20}$ , but ratio of total number won to total number played is $\frac{9x}{20x}$ for some $x$ . After they won 6 more games and lost 2 more games the number of games they won is $9x+6$ , and the total number of games is $20x+8$ . Turning it into a fraction we get $\frac{9x+6}{20x+8}=\frac{1}{2}$ , so solving for $x$ we get $x=2.$ Plugging in 2 for $x$ we get $20(2)+8=\boxed{48}$ | null | 48 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | Because 45% can be simplified to $9/20$ , and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was $8$ games, so in total there must be $20x+8$ . The only answer in this format is $\boxed{48}$ | A | 48 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | Let $n$ be the number of pre-district games. Therefore, we can write the percentage of total games won as a weighted average, namely $.45(n)+.75(8)=(n+8)(.5)$ . Solving this equation for $n$ gives $40$ , but since the problem asked for all games, the answer is $n+8=40+8=\boxed{48}$ | A | 48 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | Let the number of games they won be $x$ and the number of games they lost be $y$ . We are told that $\frac{x}{x+y}=\frac{9}{20}$ , which can be manipulated to $20x=9x+9y$ which simplifies down to $11x=9y$ . Then, after district games, we are told $\frac{x+6}{x+y+8}=\frac12$ , which can be changed into $2x+12=x+y+8$ which simplifies down to $x+4=y$ . Then we can solve for $x$ using substitution: \[11x=9x+36\] \[2x=36\] \[x=18\] Now that we know $x=18$ , we can figure out that $y=22$ $18+22=40$ . Now we need to add on the district games: $40+8=\boxed{48}$ | A | 48 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | Let $x$ be the total number of games before the district play. The Unicorns have $0.45x$ wins, therefore the rest $0.55x$ are losses. But after the district play, they won 6 and lost 2 more games. We can solve for x by forming the equation $0.45x+6=0.55x+2$ . Subtracting $2$ and $0.45x$ from both sides gives us $0.10x=4$ , and from here we multiply both sides by 10 to get $x=40$ . We are not finished yet as the problem is asking for the total games (which includes the games after the district play), so we add 8 to our value of x to get our answer which is $\boxed{48}$ | A | 48 |
e54949986cda7aa74f2fef2cf43d547a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20 | Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$ | We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played $48$ games in total. Then, after district play, they would have won $24$ games. Now, consider the situation before district play. The Unicorns would have won $18$ games out of $40$ . Converting to a percentage, $18/40 = 45$ %. Thus, the answer is $\boxed{48}$ | A | 48 |
14b769020458d77a3e20a041876c60c8 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_22 | A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$ | The shortest segments would be perpendicular to the square. The lemming went $x$ meters horizontally and $y$ meters vertically. No matter how much it went, the lemming would have been $x$ and $y$ meters from the sides and $10-x$ and $10-y$ meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: $\frac {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} =$ $\boxed{5}$ | C | 5 |
14b769020458d77a3e20a041876c60c8 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_22 | A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$ | For any point in the square, the sum of its distance to the left edge and right edge is equal to $10$ , and the sum of its distance to the up edge and down edge is also equal to $10$ . Thus, the answer is $\boxed{5}$ , and the moving progress is misguide at all. | C | 5 |
d333d1937f2658257b4297254407ebdd | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_23 | What is the area of the shaded pinwheel shown in the $5 \times 5$ grid?
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$ | The area of the square around the pinwheel is 25. The area of the pinwheel is equal to $\text{the square } - \text{ the white space.}$ Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is $25-(15+4)$ which is $\boxed{6}$ | B | 6 |
d333d1937f2658257b4297254407ebdd | https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_23 | What is the area of the shaded pinwheel shown in the $5 \times 5$ grid?
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$ | We'd like to use Pick's Theorem on one of the kites, except it doesn't immediately apply since there is a single vertex (in the middle of the diagram) of each kite that does not lie on a lattice point.
We can remedy this be pretending the figure is twice as big:
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } for(i=0; i<5; i=i+1) { draw((i+.5,0)--(i+.5,5), rgb(0.6,0.6,0.6)); draw((0,i+.5)--(5,i+.5), rgb(0.6,0.6,0.6)); } [/asy]
Now we can safely use Pick's Theorem:
\[A=\frac{b}{2}+i-1=\frac{6}{2}+4-1=6\]
However since we scaled the figure's dimensions by $2$ , we scaled its area by $4$ (since the area of similar shapes scales quadratically with the scaling factor). Therefore the area of each kite is $\frac{6}{4}$ and the area of all four kites combined is $\boxed{6}$ | B | 6 |