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601
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_17
| 1
|
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 $\boxed{40}$
| 40
|
602
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_18
| 1
|
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.
| 8
|
603
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_19
| 1
|
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 $\boxed{79}$
| 79
|
604
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_19
| 2
|
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 $\boxed{79}$
| 79
|
605
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20
| 1
|
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 $\boxed{48}$
| 48
|
606
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20
| 3
|
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 $20(2)+8=\boxed{48}$
| 48
|
607
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20
| 4
|
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 $\boxed{48}$
| 48
|
608
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20
| 5
|
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 $n+8=40+8=\boxed{48}$
| 48
|
609
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20
| 6
|
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: $40+8=\boxed{48}$
| 48
|
610
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20
| 7
|
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 $\boxed{48}$
| 48
|
611
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_20
| 8
|
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 $\boxed{48}$
| 48
|
612
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_22
| 1
|
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: $\frac {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} =$ $\boxed{5}$
| 5
|
613
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_22
| 2
|
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.
| 5
|
614
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_23
| 1
|
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 $\boxed{6}$
| 6
|
615
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_23
| 2
|
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 $\boxed{6}$
| 6
|
616
|
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_23
| 3
|
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 any kite (concave OR convex) with diagonals $p$ $q$ is $\frac{1}{2}pq$ . Let $p$ be the smaller diagonal and $q$ be the longer diagonal. Then by Pythagorean Theorem $p=\sqrt{2}$ . Similarly, $q$ is $\sqrt{2}$ less than half of the diagonal of the $5 \times 5$ grid, or $q=\frac{5\sqrt{2}}{2}-\sqrt{2}=\frac{3\sqrt{2}}{2}$ . Therefore the area of the four kites is just:
\[A=4\cdot\frac{1}{2}pq=4\cdot\frac{1}{2}\cdot\sqrt{2}\cdot\frac{3\sqrt{2}}{2}=\boxed{6}\]
| 6
|
617
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_1
| 1
|
Mindy made three purchases for $\textdollar 1.98$ dollars, $\textdollar 5.04$ dollars, and $\textdollar 9.89$ dollars. What was her total, to the nearest dollar?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$
|
The three prices round to $\textdollar 2$ $\textdollar 5$ , and $\textdollar 10$ , which have a sum of $\boxed{17}$
| 17
|
618
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_2
| 1
|
On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 26$
|
As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly, $\boxed{13}$
| 13
|
619
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_5
| 1
|
Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$
|
Drawing segments $AC$ and $BD$ , the number of triangles outside square $ABCD$ is the same as the number of triangles inside the square. Thus areas must be equal so the area of $ABCD$ is half the area of the larger square which is $\frac{60}{2}=\boxed{30}$
| 30
|
620
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_5
| 2
|
Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$
|
If the side length of the larger square is $x$ , the side length of the smaller square is $\frac{\sqrt{2} \cdot x}{2}$ . Therefore the area of the smaller square is $\frac{x^2}{2}$ , half of the larger square's area, $x^2$
Thus, the area of the smaller square in the picture is $\frac{60}{2}=\boxed{30}$
| 30
|
621
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_6
| 1
|
The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
[asy] size(150); draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));[/asy]
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
|
If the two rectangles were seperate, the perimeter would be $2(2(2+4)=24$ . It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \times 2 = \boxed{20}$
| 20
|
622
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_8
| 1
|
The table shows some of the results of a survey by radiostation KACL. What percentage of the males surveyed listen to the station?
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & ? & 26 & ?\\ \hline Females & 58 & ? & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
$\textbf{(A)}\ 39\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 55\qquad\textbf{(E)}\ 75$
|
Filling out the chart, it becomes
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & 78 & 26 & 104\\ \hline Females & 58 & 38 & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
Thus, the percentage of males surveyed that listen to the station is $100 \cdot \frac{78}{104} \%= \boxed{75}$
| 75
|
623
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_9
| 1
|
What is the product of $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1002\qquad\textbf{(C)}\ 1003\qquad\textbf{(D)}\ 2005\qquad\textbf{(E)}\ 2006$
|
The numerator in each fraction cancels out with the denominator of the next fraction. There are only two numbers that didn't cancel: $\frac{2006}{2}=\boxed{1003}$
| 3
|
624
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_11
| 1
|
How many two-digit numbers have digits whose sum is a perfect square?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$
|
There is $1$ integer whose digits sum to $1$ $10$
There are $4$ integers whose digits sum to $4$ $13, 22, 31, \text{and } 40$
There are $9$ integers whose digits sum to $9$ $18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90$
There are $3$ integers whose digits sum to $16$ $79, 88, \text{and } 97$
Two digits cannot sum to $25$ or any greater square since the greatest sum of digits of a two-digit number is $9 + 9 = 18$
Thus, the answer is $1 + 4 + 9 + 3 = \boxed{17}$
| 17
|
625
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_12
| 1
|
Antonette gets $70 \%$ on a 10-problem test, $80 \%$ on a 20-problem test and $90 \%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 77\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 83\qquad\textbf{(E)}\ 87$
|
$70 \% \cdot 10=7$
$80 \% \cdot 20=16$
$90 \% \cdot 30=27$
Adding them up gets $7+16+27=50$ . The overall percentage correct would be $\frac{50}{60}=\frac{5}{6}=5 \cdot 16.\overline{6}=83.\overline{3} \approx \boxed{83}$
| 83
|
626
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_15
| 1
|
Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$
|
Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve.
Let $x$ be the number of pages that Chandra reads.
$30x = 45(760-x)$ Distribute the $45$
$30x = 45(760) - 45x$ Add $45x$ to both sides
$75x = 45(760)$ Divide both sides by $15$ to make it easier to solve
$5x = 3(760)$ Divide both sides by $5$
$x = 3(152) = \boxed{456}$
| 456
|
627
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_15
| 2
|
Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$
|
Chandra and Bob read at a rate of $30:45$ seconds per page, respectively. Simplifying that gets us Bob reads $2$ pages for every $3$ pages that Chandra reads. Therefore Chandra should read $\frac{3}{2+3}=\frac{3}{5}$ of the book. $\frac{3}{5}\cdot760$ $\boxed{456}$
| 456
|
628
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_16
| 1
|
Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?
$\textbf{(A)}\ 6400\qquad\textbf{(B)}\ 6600\qquad\textbf{(C)}\ 6800\qquad\textbf{(D)}\ 7000\qquad\textbf{(E)}\ 7200$
|
The amount of pages Bob, Chandra, and Alice will read is in the ratio 4:6:9. Therefore, Bob, Chandra, and Alice read 160, 240, and 360 pages respectively. They would also be reading for the same amount of time because the ratio of the pages read was based on the time it takes each of them to read a page. Therefore, the amount of seconds each person reads is simply $160 \cdot 45 = \boxed{7200}$
| 200
|
629
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_16
| 2
|
Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?
$\textbf{(A)}\ 6400\qquad\textbf{(B)}\ 6600\qquad\textbf{(C)}\ 6800\qquad\textbf{(D)}\ 7000\qquad\textbf{(E)}\ 7200$
|
We can solve this using equations. Let Bob read $x$ pages and Chandra read $y$ pages. Then Alice would read $760-x-y$ pages. Then we can set up the equations using the time it takes them to read a page: $45x=30y=20(760-x-y)$ . Since we only have two variables, we only need two equations. Let's use $45x=30y$ and $45x=20(760-x-y)$ . We can simplify $45x=30y$ down to $3x=2y$ and $45x=20(760-x-y)$ down to $9x=4(760-x-y)$ . We can then solve the equations accordingly:
\[9x=3040-4x-4y\] \[13x=3040-4y\]
From the equation $3x=2y$ , we know that $4y=6x$
\[13x=3040-6x\] \[19x=3040\] \[x=160\]
Now that we know $x$ , we can multiply it by $45$ (Bob's speed) to get the answer: $160\times45$ $\boxed{7200}$
| 200
|
630
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_20
| 1
|
A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
|
Since there are 6 players, a total of $\frac{6(6-1)}{2}=15$ games are played. So far, $4+3+2+2+2=13$ games finished (one person won from each game), so Monica needs to win $15-13 = \boxed{2}$
| 2
|
631
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_22
| 1
|
Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?
[asy] path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle); path sw=((0,0)--(1,sqrt(3))); path se=((5,0)--(4,sqrt(3))); draw(cell, linewidth(1)); draw(shift(2,0)*cell, linewidth(1)); draw(shift(4,0)*cell, linewidth(1)); draw(shift(1,3)*cell, linewidth(1)); draw(shift(3,3)*cell, linewidth(1)); draw(shift(2,6)*cell, linewidth(1)); draw(shift(0.45,1.125)*sw, EndArrow); draw(shift(2.45,1.125)*sw, EndArrow); draw(shift(1.45,4.125)*sw, EndArrow); draw(shift(-0.45,1.125)*se, EndArrow); draw(shift(-2.45,1.125)*se, EndArrow); draw(shift(-1.45,4.125)*se, EndArrow); label("$+$", (1.5,1.5)); label("$+$", (3.5,1.5)); label("$+$", (2.5,4.5));[/asy]
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35$
|
If the lower cells contain $A, B$ and $C$ , then the second row will contain $A + B$ and $B + C$ , and the top cell will contain $A + 2B + C$ . To obtain the
smallest sum, place $1$ in the center cell and $2$ and $3$ in the outer ones. The top
number will be $7$ . For the largest sum, place $9$ in the center cell and $7$ and $8$ in
the outer ones. This top number will be $33$ . The difference is $33 - 7 = \boxed{26}$
| 26
|
632
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_23
| 2
|
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$
|
If there were two more coins in the box, the number of coins would be divisible
by both $6$ and $5$ . The smallest number that is divisible by $6$ and $5$ is $30$ , so the
smallest possible number of coins in the box is $28$ and the remainder when divided by $7$ is $\boxed{0}$
| 0
|
633
|
https://artofproblemsolving.com/wiki/index.php/2006_AMC_8_Problems/Problem_25
| 1
|
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
[asy] path card=((0,0)--(0,3)--(2,3)--(2,0)--cycle); draw(card, linewidth(1)); draw(shift(2.5,0)*card, linewidth(1)); draw(shift(5,0)*card, linewidth(1)); label("$44$", (1,1.5)); label("$59$", shift(2.5,0)*(1,1.5)); label("$38$", shift(5,0)*(1,1.5));[/asy]
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 17$
|
Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number(common sum) would be to add another even number to 44, and a different one to 38. Since there is only one even prime (2), the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is $59+2=61$ . Thus, the first card's hidden number is $61-44=17$ , and the last card's hidden number is $61-38=23$
Since the sum of the hidden primes is $2+17+23=42$ , the average of the primes is $\dfrac{42}{3}=\boxed{14}$
| 14
|
634
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_1
| 1
|
Connie multiplies a number by $2$ and gets $60$ as her answer. However, she should have divided the number by $2$ to get the correct answer. What is the correct answer?
$\textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240$
|
If $x$ is the number, then $2x=60$ and $x=30$ . Dividing the number by $2$ yields $\dfrac{30}{2} = \boxed{15}$
| 15
|
635
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_3
| 1
|
What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$
[asy] defaultpen(linewidth(1)); for ( int x = 0; x < 5; ++x ) { draw((0,x)--(4,x)); draw((x,0)--(x,4)); } fill((1,0)--(2,0)--(2,1)--(1,1)--cycle); fill((0,3)--(1,3)--(1,4)--(0,4)--cycle); fill((2,3)--(4,3)--(4,4)--(2,4)--cycle); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle); label("$A$", (0, 4), NW); label("$B$", (4, 4), NE); label("$C$", (4, 0), SE); label("$D$", (0, 0), SW); [/asy]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
|
Rotating square $ABCD$ counterclockwise $45^\circ$ so that the line of symmetry $BD$ is a vertical line makes it easier to see that $\boxed{4}$ squares need to be colored to match its corresponding square.
| 4
|
636
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_4
| 1
|
A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$
|
The perimeter of the triangle is $6.1+8.2+9.7=24$ cm. A square's perimeter is four times its sidelength, since all its sidelengths are equal. If the square's perimeter is $24$ , the sidelength is $24/4=6$ , and the area is $6^2=\boxed{36}$
| 36
|
637
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_5
| 1
|
Soda is sold in packs of 6, 12 and 24 cans. What is the minimum number of packs needed to buy exactly 90 cans of soda?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 15$
|
Start by buying the largest packs first. After three $24$ -packs, $90-3(24)=18$ cans are left. After one $12$ -pack, $18-12=6$ cans are left. Then buy one more $6$ -pack. The total number of packs is $3+1+1=\boxed{5}$
| 5
|
638
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_6
| 1
|
Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10$
|
We see that $2.0055$ works but $2.0045$ does not. The digit $d$ can be from $5$ through $9$ , which is $\boxed{5}$ values.
| 5
|
639
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_8
| 1
|
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
$\textbf{(A)}\ m+3n\qquad\textbf{(B)}\ 3m-n\qquad\textbf{(C)}\ 3m^2 + 3n^2\qquad\textbf{(D)}\ (nm + 3)^2\qquad\textbf{(E)}\ 3mn$
|
Assume WLOG that $m$ and $n$ are both $1$ . Plugging into each of the choices, we get $4, 2, 6, 16,$ and $3$ . The only odd integer is $\boxed{3}$
| 3
|
640
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_9
| 1
|
In quadrilateral $ABCD$ , sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$ . What is the length of diagonal $\overline{AC}$
[asy]draw((0,0)--(17,0)); draw(rotate(301, (17,0))*(0,0)--(17,0)); picture p; draw(p, (0,0)--(0,10)); draw(p, rotate(115, (0,10))*(0,0)--(0,10)); add(rotate(3)*p); draw((0,0)--(8.25,14.5), linetype("8 8")); label("$A$", (8.25, 14.5), N); label("$B$", (-0.25, 10), W); label("$C$", (0,0), SW); label("$D$", (17, 0), E);[/asy]
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5$
|
Because $\overline{AD} = \overline{CD}$ $\triangle ADC$ is an isosceles triangle with $\angle DAC = \angle DCA$ . Angles in a triangle add up to $180^\circ$ , and since $\angle ADC=60^\circ$ , the other two angles are also $60^\circ$ , and $\triangle ADC$ is an equilateral triangle. Therefore $\overline{AC}=\overline{DA}=\boxed{17}$
| 17
|
641
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_9
| 2
|
In quadrilateral $ABCD$ , sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$ . What is the length of diagonal $\overline{AC}$
[asy]draw((0,0)--(17,0)); draw(rotate(301, (17,0))*(0,0)--(17,0)); picture p; draw(p, (0,0)--(0,10)); draw(p, rotate(115, (0,10))*(0,0)--(0,10)); add(rotate(3)*p); draw((0,0)--(8.25,14.5), linetype("8 8")); label("$A$", (8.25, 14.5), N); label("$B$", (-0.25, 10), W); label("$C$", (0,0), SW); label("$D$", (17, 0), E);[/asy]
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5$
|
We can divide $\overline{CD}$ in half and connect this point to A, dividing $\triangle ADC$ in half. This means the base will be $\frac{17}{2}$ and the hypotenuse will be 17. By using the Pythagorean's Theorem, we see that if the base and height are shared, the hypotenuse should be the same. This tells us that the length of $\overline{AC} = \boxed{17}$
| 17
|
642
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_9
| 3
|
In quadrilateral $ABCD$ , sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$ . What is the length of diagonal $\overline{AC}$
[asy]draw((0,0)--(17,0)); draw(rotate(301, (17,0))*(0,0)--(17,0)); picture p; draw(p, (0,0)--(0,10)); draw(p, rotate(115, (0,10))*(0,0)--(0,10)); add(rotate(3)*p); draw((0,0)--(8.25,14.5), linetype("8 8")); label("$A$", (8.25, 14.5), N); label("$B$", (-0.25, 10), W); label("$C$", (0,0), SW); label("$D$", (17, 0), E);[/asy]
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5$
|
Take an equilateral triangle with side length $17$ $\triangle ADC$ is congruent to this by $SAS$ , hence it is equilateral. The answer is $\boxed{17}$
| 17
|
643
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_10
| 1
|
Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran 3 times as fast as he walked. Joe took 6 minutes to walk half way to school. How many minutes did it take Joe to get from home to school?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 7.3\qquad\textbf{(C)}\ 7.7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 8.3$
|
We can use the equation $d=rt$ where $d$ is the distance, $r$ is the rate, and $t$ is the time. The distances he ran and walked are equal, so $r_rt_r=r_wt_w$ , where $r_r$ is the rate at which he ran, $t_r$ is the time for which he ran, $r_w$ is the rate at which he walked, and $t_w$ is the time for which he walked. Because he runs three times faster than he walks, $r_r=3r_w$ . We want to find the time he ran, $t_r = \frac{r_wt_w}{t_r} = \frac{(r_w)(6)}{3r_w} = 2$ minutes. He traveled for a total of $6+2=\boxed{8}$ minutes.
| 8
|
644
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_11
| 1
|
The sales tax rate in Rubenenkoville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?
$\textbf{(A)}\ -\textdollar 1.06\qquad\textbf{(B)}\ -\textdollar 0.53 \qquad\textbf{(C)}\ \textdollar 0\qquad\textbf{(D)}\ \textdollar 0.53\qquad\textbf{(E)}\ \textdollar 1.06$
|
The price Jack rings up is $\textdollar{(90.00)(1.06)(0.80)}$ . The price Jill rings up is $\textdollar{(90.00)(0.80)(1.06)}$ . By the commutative property of multiplication, these quantities are the same, and the difference is $\boxed{0}$
| 0
|
645
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_12
| 1
|
Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$
|
There are $5$ days from May 1 to May 5. The number of bananas he eats each day is an arithmetic sequence. He eats $n$ bananas on May 5, and $n-4(6)=n-24$ bananas on May 1. The sum of this arithmetic sequence is equal to $100$
\begin{align*} \frac{n+n-24}{2} \cdot 5 &= 100\\ n-12&=20\\ n&=\boxed{32}
| 32
|
646
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_12
| 2
|
Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$
|
Simply realize that the middle term of the arithmetic sequence is the arithmetic mean of all terms, which is simply $\frac{100}{5}=20$ . This means that the number of bananas the ape ate on May 5th is just $20+6*2=32$ . Select $\boxed{32}$
| 32
|
647
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_13
| 1
|
The area of polygon $ABCDEF$ is 52 with $AB=8$ $BC=9$ and $FA=5$ . What is $DE+EF$
[asy] pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4); draw(a--b--c--d--e--f--cycle); draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a); draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b); draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c); draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d); draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f); label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, SE); label("$D$", d, SW); label("$E$", e, SW); label("$F$", f, SW); label("5", (0,6.5), W); label("8", (4,9), N); label("9", (8, 4.5), E); [/asy]
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
|
Notice that $AF + DE = BC$ , so $DE=4$ . Let $O$ be the intersection of the extensions of $AF$ and $DC$ , which makes rectangle $ABCO$ . The area of the polygon is the area of $FEDO$ subtracted from the area of $ABCO$
\[\text{Area} = 52 = 8 \cdot 9- EF \cdot 4\]
Solving for the unknown, $EF=5$ , therefore $DE+EF=4+5=\boxed{9}$
| 9
|
648
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_14
| 1
|
The Little Twelve Basketball Conference has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many conference games are scheduled?
$\textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192$
|
Within each division, there are $\binom {6}{2} = 15$ pairings, and each of these games happens twice. The same goes for the other division so that there are $4(15)=60$ games within their own divisions. The number of games between the two divisions is $(6)(6)=36$ . Together there are $60+36=\boxed{96}$ conference games.
| 96
|
649
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_14
| 2
|
The Little Twelve Basketball Conference has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many conference games are scheduled?
$\textbf{(A)}\ 80\qquad\textbf{(B)}\ 96\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 192$
|
Each team plays 10 games in its own division and 6 games against teams in the other division. So each of the 12 teams plays 16 conference games. Because each game involves two teams, there are $\frac{12\times 16}{2}=\boxed{96}$ games scheduled. ~aopsav (Credit to AoPS Alcumus)
| 96
|
650
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_15
| 1
|
How many different isosceles triangles have integer side lengths and perimeter 23?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$
|
Let $b$ be the base of the isosceles triangles, and let $a$ be the lengths of the other legs. From this, $2a+b=23$ and $b=23-2a$ . From triangle inequality, $2a>b$ , then plug in the value from the previous equation to get $2a>23-2a$ or $a>5.75$ . The maximum value of $a$ occurs when $b=1$ , in which from the first equation $a=11$ . Thus, $a$ can have integer side lengths from $6$ to $11$ , and there are $\boxed{6}$ triangles.
| 6
|
651
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_16
| 1
|
A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$
|
The Martian can pull out $12$ socks, $4$ of each color, without having $5$ of the same kind yet. However, the next one he pulls out must be the fifth of one of the colors so he must remove $\boxed{13}$ socks.
| 13
|
652
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_18
| 1
|
How many three-digit numbers are divisible by 13?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77$
|
Let $k$ be any positive integer so that $13k$ is a multiple of $13$ . For the smallest three-digit number, $13k>100$ and $k>\frac{100}{13} \approx 7.7$ . For the greatest three-digit number, $13k<999$ and $k<\frac{999}{13} \approx 76.8$ . The number $k$ can range from $8$ to $76$ so there are $\boxed{69}$ three-digit numbers.
| 69
|
653
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_19
| 1
|
What is the perimeter of trapezoid $ABCD$
[asy]size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("$A$", a, SW); label("$B$", b, N); label("$C$", c, N); label("$D$", d, SE); label("$E$", e, S);[/asy]
$\textbf{(A)}\ 180\qquad\textbf{(B)}\ 188\qquad\textbf{(C)}\ 196\qquad\textbf{(D)}\ 200\qquad\textbf{(E)}\ 204$
|
Draw altitudes from $B$ and $C$ to base $AD$ to create a rectangle and two right triangles. The side opposite $BC$ is equal to $50$ . The bases of the right triangles can be found using Pythagorean or special triangles to be $18$ and $7$ . Add it together to get $AD=18+50+7=75$ . The perimeter is $75+30+50+25=\boxed{180}$
| 180
|
654
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_20
| 1
|
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise.
In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$
|
Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$ , is a multiple of $12$ . Since this number must be a multiple of $12$ , as stated in the previous sentence, $14$ has a factor $2$ $k$ must have a factor of $6$ . The smallest number of turns that is a multiple of $6$ is $\boxed{6}$
| 6
|
655
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_21
| 1
|
How many distinct triangles can be drawn using three of the dots below as vertices?
[asy]dot(origin^^(1,0)^^(2,0)^^(0,1)^^(1,1)^^(2,1));[/asy]
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$
|
The number of ways to choose three points to make a triangle is $\binom 63 = 20$ . However, two* of these are a straight line so we subtract $2$ to get $\boxed{18}$
| 18
|
656
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_23
| 1
|
Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi$ . The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}$ . What is the area of triangle $ABC$
[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(circle(o, 2)); clip(a--b--c--cycle); draw(a--b--c--cycle); dot(o); label("$C$", c, NW); label("$A$", a, NE); label("$B$", b, SW);[/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 3\pi\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 4\pi$
|
First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be $4\cdot 4 = 16.$ Divide 16 by 2 to get the original shape and you get $\boxed{8}$
| 8
|
657
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_25
| 1
|
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]
$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$
|
Let the region within the circle and square be $a$ . In other words, it is the area inside the circle $\textbf{and}$ the square. Let $r$ be the radius. We know that the area of the circle minus $a$ is equal to the area of the square, minus $a$
We get:
$\pi r^2 -a=4-a$
$r^2=\frac{4}{\pi}$
$r=\frac{2}{\sqrt{\pi}}$
So the answer is $\boxed{2}$
| 2
|
658
|
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_25
| 2
|
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]
$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$
|
We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.
$\pi r^2=4$
$r^2=\frac{4}{\pi}$
$r=\frac{2}{\sqrt{\pi}}$
So the answer is $\boxed{2}$
| 2
|
659
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_1
| 1
|
On a map, a $12$ -centimeter length represents $72$ kilometers. How many kilometers does a $17$ -centimeter length represent?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224$
|
We set up the proportion $\frac{12 \text{cm}}{72 \text{km}}=\frac{17 \text{cm}}{x \text{km}}$ . Thus $x=102 \Rightarrow \boxed{102}$
| 102
|
660
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_2
| 1
|
How many different four-digit numbers can be formed by rearranging the four digits in $2004$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$
|
We can solve this problem easily, just by calculating how many choices there are for each of the four digits.
First off, we know there are only $2$ choices for the first digit, because $0$ isn't a valid choice, or the number would a 3-digit number, which is not what we want.
We have $3$ choices for the second digit, since we already used up one of the digits, and $2$ choices for the third, and finally just $1$ choices for the fourth and final one. $2*3*2*1$ is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get $\boxed{6}$
| 6
|
661
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_2
| 2
|
How many different four-digit numbers can be formed by rearranging the four digits in $2004$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 81$
|
Note that the four-digit number must start with either a $2$ or a $4$ . The four-digit numbers that start with $2$ are $2400, 2040$ , and $2004$ . The four-digit numbers that start with $4$ are $4200, 4020$ , and $4002$ which gives us a total of $\boxed{6}$
| 6
|
662
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_3
| 1
|
Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 18$
|
Set up the proportion $\frac{12\ \text{meals}}{18\ \text{people}}=\frac{x\ \text{meals}}{12\ \text{people}}$ . Solving for $x$ gives us $x= \boxed{8}$
| 8
|
663
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_4
| 1
|
Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament.
Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$
|
There are $\binom{4}{3}$ ways to choose three starters. Thus the answer is $\boxed{4}$
| 4
|
664
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_4
| 2
|
Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament.
Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$
|
We can choose $3$ people by eliminating one from a set of $4$ one at a time and the other three get selected. There are $4$ ways to remove a person from a group of four (without considering order), so there are $\boxed{4}$ ways to choose three people, where order doesn't matter.
| 4
|
665
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_5
| 1
|
Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$
|
The remaining team will be the only undefeated one. The other $\boxed{15}$ teams must have lost a game before getting out, thus fifteen games yielding fifteen losers.
| 15
|
666
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_5
| 2
|
Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$
|
There will be $8$ games the first round, $4$ games the second round, $2$ games the third round, and $1$ game in the final round, giving us a total of $8+4+2+1=15$ games. $\boxed{15}$
| 15
|
667
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_6
| 1
|
After Sally takes $20$ shots, she has made $55\%$ of her shots. After she takes $5$ more shots, she raises her percentage to $56\%$ . How many of the last $5$ shots did she make?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
|
Sally made $0.55*20=11$ shots originally. Letting $x$ be the number of shots she made, we have $\frac{11+x}{25}=0.56$ . Solving for $x$ gives us $x=\boxed{3}$
| 3
|
668
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_7
| 1
|
An athlete's target heart rate, in beats per minute, is $80\%$ of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete's age, in years, from $220$ . To the nearest whole number, what is the target heart rate of an athlete who is $26$ years old?
$\textbf{(A)}\ 134\qquad\textbf{(B)}\ 155\qquad\textbf{(C)}\ 176\qquad\textbf{(D)}\ 194\qquad\textbf{(E)}\ 243$
|
The maximum heart rate is $220-26=194$ beats per minute. The target heart rate is then $0.8*194 \approx \boxed{155}$ beats per minute.
| 155
|
669
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_8
| 1
|
Find the number of two-digit positive integers whose digits total $7$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$
|
The numbers are $16, 25, 34, 43, 52, 61, 70$ which gives us a total of $\boxed{7}$
| 7
|
670
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_9
| 1
|
The average of the five numbers in a list is $54$ . The average of the first two
numbers is $48$ . What is the average of the last three numbers?
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$
|
Let the $5$ numbers be $a, b, c, d$ , and $e$ . Thus $\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$ . Since $\frac{a+b}{2}=48$ $a+b=96$ . Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$ . Dividing by $3$ gives the average of $\boxed{58}$
| 58
|
671
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_10
| 1
|
Handy Aaron helped a neighbor $1 \frac14$ hours on Monday, $50$ minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid $\textdollar 3$ per hour. How much did he earn for the week?
$\textbf{(A)}\ \textdollar 8 \qquad \textbf{(B)}\ \textdollar 9 \qquad \textbf{(C)}\ \textdollar 10 \qquad \textbf{(D)}\ \textdollar 12 \qquad \textbf{(E)}\ \textdollar 15$
|
Let's convert everything to minutes and add them together. On Monday he worked for $\frac54 \cdot 60 = 75$ minutes. On Tuesday he worked $50$ minutes. On Wednesday he worked for $2$ hours $25$ minutes, or $2(60)+25=145$ minutes. On Friday he worked $\frac{60}{2}=30$ minutes. This adds up to $75+50+145+30=300$ minutes, or $300/60=5$ hours and $5\cdot 3 = \boxed{15}$
| 15
|
672
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_12
| 1
|
Niki usually leaves her cell phone on. If her cell phone is on but
she is not actually using it, the battery will last for $24$ hours. If
she is using it constantly, the battery will last for only $3$ hours.
Since the last recharge, her phone has been on $9$ hours, and during
that time she has used it for $60$ minutes. If she doesn’t use it any
more but leaves the phone on, how many more hours will the battery last?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$
|
When not being used, the cell phone uses up $\frac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\frac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ being used to talk, $8(\frac{1}{24}) + 1(\frac{1}{3}) = \frac{2}{3}$ of its battery has been used up. To drain the remaining $\frac{1}{3}$ the phone can last for $\frac{\frac{1}{3}}{\frac{1}{24}}=\boxed{8}$ more hours without being used.
| 8
|
673
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_15
| 1
|
Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?
[asy] defaultpen(linewidth(1)); real x=sqrt(3)/2; path p=rotate(30)*polygon(6); filldraw(p^^shift(0,3)*p^^shift(4x,0)*p^^shift(3x,1.5)*p^^shift(2x,3)*p^^shift(-4x,0)*p^^shift(-3x,1.5)*p^^shift(-2x,3)*p^^shift(3x,-1.5)*p^^shift(-3x,-1.5)*p^^shift(2x,-3)*p^^shift(-2x,-3)*p^^shift(0,-3)*p, black, black); draw(shift(2x,0)*p^^shift(-2x,0)*p^^shift(x,1.5)*p^^shift(-x,1.5)*p^^shift(x,-1.5)*p^^shift(-x,-1.5)*p); [/asy]
$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$
|
The first ring around the middle tile has $6$ tiles, and the second has $12$ . From this pattern, the third ring has $18$ tiles. Of these, $6+18=24$ are white and $1+12=13$ are black, with a difference of $24-13 = \boxed{11}$
| 11
|
674
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_17
| 1
|
Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
|
For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use Ball-and-urn to find the number of possibilities is $\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{10}$
| 10
|
675
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_17
| 2
|
Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
|
Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use number of non-negative integral solutions.
Let the three friends be $a, b, c$ respectively.
$a + b + c = 3$ The total being 3 and 2 plus signs, which implies $\binom{3+2}{2} = \binom{5}{2} = \boxed{10}$
| 10
|
676
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_17
| 3
|
Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
|
For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below:
for n = number of items, and s = slots:
Now we can plug in our values,
number of items = 6, and slots = 3:
$\binom{6-1}{3-1} = \binom{5}{2} = \boxed{10}$
| 10
|
677
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_17
| 4
|
Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
|
Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups use casework.
Let the three friends be $a$ $b$ $c$ repectively.
$a + b + c = 3$
Case $1:a=0$
$b + c = 3$
$b = 0,1,2,3$
$c = 3,2,1,0$
$\boxed{4}$ solutions.
| 4
|
678
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_20
| 1
|
Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$
|
Working backwards, if $3/4$ of the chairs are taken and $6$ are empty, then there are three times as many taken chairs as empty chairs, or $3 \cdot 6 = 18$ . If $x$ is the number of people in the room and $2/3$ are seated, then $\frac23 x = 18$ and $x = \boxed{27}$
| 27
|
679
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_21
| 1
|
Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label("$A$",(-1,1)); label("$B$",(2,1)); label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5)); [/asy]
$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$
|
An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from $1$ . Multiply the independent probabilities of each spinner getting an odd number together and subtract it from $1$
\[1-\frac24 \cdot \frac23 = 1- \frac13 = \boxed{23}\]
| 23
|
680
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_21
| 2
|
Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label("$A$",(-1,1)); label("$B$",(2,1)); label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5)); [/asy]
$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$
|
We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is \[\boxed{23}\]
| 23
|
681
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_21
| 3
|
Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label("$A$",(-1,1)); label("$B$",(2,1)); label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5)); [/asy]
$\textbf{(A)}\ \frac14\qquad \textbf{(B)}\ \frac13\qquad \textbf{(C)}\ \frac12\qquad \textbf{(D)}\ \frac23\qquad \textbf{(E)}\ \frac34$
|
We do a little bit of casework. In order to get a product that's even, we need at least one even number. First, we consider the probability of getting an even number on the first spinner, and then multiply by 1 because the second spinner can be anything. \[\frac12 \cdot 1 = \frac12\] Next, we look at the chance that we don't get an even on the first spinner, but get an even on the second spinner (we don't do the probability of even on the second spinner multiplied by one because we would be double counting both spinners are even). \[\frac12 \cdot \frac13 = \frac16\] Add these two together to get the total probability, and we get \[\frac12 + \frac16 = \frac46 = \boxed{23}\]
| 23
|
682
|
https://artofproblemsolving.com/wiki/index.php/2004_AMC_8_Problems/Problem_22
| 1
|
At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\frac25$ . What fraction of the people in the room are married men?
$\textbf{(A)}\ \frac13\qquad \textbf{(B)}\ \frac38\qquad \textbf{(C)}\ \frac25\qquad \textbf{(D)}\ \frac{5}{12}\qquad \textbf{(E)}\ \frac35$
|
Assume arbitrarily (and WLOG) there are $5$ women in the room, of which $5 \cdot \frac25 = 2$ are single and $5-2=3$ are married. Each married woman came with her husband, so there are $3$ married men in the room as well for a total of $5+3=8$ people. The fraction of the people that are married men is $\boxed{38}$
| 38
|
683
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_1
| 1
|
Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?
$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26$
|
On a cube, there are $12$ edges, $8$ corners, and $6$ faces. Adding them up gets $12+8+6= \boxed{26}$
| 26
|
684
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_3
| 1
|
A burger at Ricky C's weighs $120$ grams, of which $30$ grams are filler.
What percent of the burger is not filler?
$\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%$
|
There are $30$ grams of filler, so there are $120-30= 90$ grams that aren't filler. $\frac{90}{120}=\frac{3}{4}=\boxed{75}$
| 75
|
685
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_4
| 1
|
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted $7$ children and $19$ wheels. How many tricycles were there?
$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$
|
If all the children were riding bicycles, there would be $2 \times 7=14$ wheels. Each tricycle adds an extra wheel and $19-14=5$ extra wheels are needed, so there are $\boxed{5}$ tricycles.
| 5
|
686
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_4
| 2
|
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted $7$ children and $19$ wheels. How many tricycles were there?
$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$
|
Setting up an equation, we have $a+b=7$ children and $3a+2b=19$ . Solving for the variables, we get, $a=\boxed{5}$ tricycles.
| 5
|
687
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_5
| 1
|
If $20\%$ of a number is $12$ , what is $30\%$ of the same number?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30$
|
$20\%$ of a number is equal to $\frac{1}{5}$ of that number. Let $n$ =the number
$\frac{1}{5}n = 12$ Multiply both sides by 5
$n = 60$
$30\%$ of $n$ is equal to $\frac{3}{10}n = \frac{3}{10}\cdot60 = 3\cdot6 = \boxed{18}$
| 18
|
688
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_6
| 1
|
Given the areas of the three squares in the figure, what is the area of the interior triangle?
[asy] draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1)); label("$25$",(14.5,1),N); label("$144$",(6,-7.5),N); label("$169$",(3.5,7),N); [/asy]
$\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800$
|
The sides of the squares are $5, 12$ and $13$ for the square with area $25, 144$ and $169$ , respectively. The legs of the interior triangle are $5$ and $12$ , so the area is $\frac{5 \times 12}{2}=\boxed{30}$
| 30
|
689
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_7
| 1
|
Blake and Jenny each took four $100$ -point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?
$\mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40$
|
Blake scored a total of $4 \times 78=312$ points. Jenny scored $10-10+20+20=40$ points higher than Blake, so her average is $\frac{312+40}{4}=88$ .
the difference is $88-78=\boxed{10}$
| 10
|
690
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_7
| 2
|
Blake and Jenny each took four $100$ -point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?
$\mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40$
|
The total point difference between Blake's and Jenny's scores is $10-10+20+20=40$ . The average of it is $\frac{40}{4}=\boxed{10}$
| 10
|
691
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_9
| 1
|
Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy]
$\circ$ Roger's cookies are rectangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy]
$\circ$ Paul's cookies are parallelograms: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy]
$\circ$ Trisha's cookies are triangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy]
Each friend uses the same amount of dough, and Art makes exactly $12$ cookies. Art's cookies sell for $60$ cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90$
|
The area of one of Art's cookies is $3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12$ . As he has $12$ cookies in a batch, the amount of dough each person used is $12 \cdot 12=144$ . Roger's cookies have an area of $\frac{144}{2 \cdot 4}=\frac{144}{8}= 18$ cookies in a batch. In total, the amount of money Art will earn is $12 \cdot 60=720$ . Thus, the amount Roger would need to charge per cookie is $\frac{720}{18}=\boxed{40}$
| 40
|
692
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_10
| 1
|
Problems 8, 9 and 10 use the data found in the accompanying paragraph and figures
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.
$\circ$ Art's cookies are trapezoids: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(5,0)--(5,3)--(2,3)--cycle); draw(rightanglemark((5,3), (5,0), origin)); label("5 in", (2.5,0), S); label("3 in", (5,1.5), E); label("3 in", (3.5,3), N);[/asy]
$\circ$ Roger's cookies are rectangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(4,0)--(4,2)--(0,2)--cycle); draw(rightanglemark((4,2), (4,0), origin)); draw(rightanglemark((0,2), origin, (4,0))); label("4 in", (2,0), S); label("2 in", (4,1), E);[/asy]
$\circ$ Paul's cookies are parallelograms: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle); draw((2.5,2)--(2.5,0), dashed); draw(rightanglemark((2.5,2),(2.5,0), origin)); label("3 in", (1.5,0), S); label("2 in", (2.5,1), W);[/asy]
$\circ$ Trisha's cookies are triangles: [asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(origin--(3,0)--(3,4)--cycle); draw(rightanglemark((3,4),(3,0), origin)); label("3 in", (1.5,0), S); label("4 in", (3,2), E);[/asy]
How many cookies will be in one batch of Trisha's cookies?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24$
|
Art's cookies have areas of $3 \cdot 3 + \frac{2 \cdot 3}{2}=9+3=12$ . There are 12 cookies in one of Art's batches so everyone used $12 \cdot 12=144 \text{ in}^2$ of dough. Trisha's cookies have an area of $\frac{3 \cdot 4}{2}=6$ so she has $\frac{144}{6}=\boxed{24}$ cookies per batch.
| 24
|
693
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_13
| 1
|
Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces
[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight); [/asy]
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$
|
This is the number cubes that are adjacent to another cube on two sides. The bottom corner cubes are connected on three sides, and the top corner cubes are connected on one. The number we are looking for is the number of middle cubes, which is $\boxed{6}$
| 6
|
694
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_14
| 1
|
In this addition problem, each letter stands for a different digit.
$\setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$
If T = 7 and the letter O represents an even number, what is the only possible value for W?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$
|
Since both T's are 7, then O has to equal 4, because 7 + 7 = 14. Then, F has to equal 1. To get R, we do 4 + 4 (since O = 4) to get R = 8. The value for W then has to be a number less than 5, otherwise it will change the value of O, and can't be a number that has already been used, like 4 or 1. The only other possibilities are 2 and 3. 2 doesn't work because it makes U = 4, which is what O already equals. So, the only possible value of W is 3 $\boxed{3}$
| 3
|
695
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_16
| 1
|
Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has $4$ seats: $1$ Driver seat, $1$ front passenger seat, and $2$ back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24$
|
There are only $2$ people who can go in the driver's seat--Bonnie and Carlo. Any of the $3$ remaining people can go in the front passenger seat. There are $2$ people who can go in the first back passenger seat, and the remaining person must go in the last seat. Thus, there are $2\cdot3\cdot2$ or $12$ ways. The answer is then $\boxed{12}$
| 12
|
696
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_16
| 2
|
Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has $4$ seats: $1$ Driver seat, $1$ front passenger seat, and $2$ back passenger seat. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24$
|
If there weren't any extra requirements, there would be 24 combinations. However, there are only 2, which is half of 4, ways to put the people. Therefore, half of 24 is $\boxed{12}$
| 12
|
697
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_18
| 1
|
Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party? [asy]/* AMC8 2003 #18 Problem */ pair a=(102,256), b=(68,131), c=(162,101), d=(134,150); pair e=(269,105), f=(359,104), g=(303,12), h=(579,211); pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501); pair m=(282,411), n=(147,451), o=(103,437), p=(31,373); pair q=(419,175), r=(462,209), s=(477,288), t=(443,358); pair oval=(282,303); draw(l--m--n--cycle); draw(p--oval); draw(o--oval); draw(b--d--oval); draw(c--d--e--oval); draw(e--f--g--h--i--j--oval); draw(k--oval); draw(q--oval); draw(s--oval); draw(r--s--t--oval); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h); dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p); dot(q); dot(r); dot(s); dot(t); filldraw(yscale(.5)*Circle((282,606),80),white,black); label(scale(0.75)*"Sarah", oval);[/asy]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
|
There are $3$ people who are friends with only each other who won't be invited, plus $1$ person who has no friends, and $2$ people who are friends of friends of friends who won’t be invited. So the answer is $\boxed{6}$
| 6
|
698
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_19
| 1
|
How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
|
Find the least common multiple of $15, 20, 25$ by turning the numbers into their prime factorization. \[15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2\] Gather all necessary multiples $3, 2^2, 5^2$ when multiplied gets $300$ . The multiples of $300 - 300, 600, 900, 1200, 1500, 1800, 2100$ . The number of multiples between 1000 and 2000 is $\boxed{3}$
| 3
|
699
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_20
| 1
|
What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$
|
Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$ , the hour hand would have moved $\frac{1}{3}$ way from 4 to 5 since $\frac{20}{60}$ is reducible to $\frac{1}{3}$ . One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is $\boxed{10}$
| 10
|
700
|
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_21
| 1
|
The area of trapezoid $ABCD$ is $164\text{ cm}^2$ . The altitude is 8 cm, $AB$ is 10 cm, and $CD$ is 17 cm. What is $BC$ , in centimeters?
[asy]/* AMC8 2003 #21 Problem */ size(4inch,2inch); draw((0,0)--(31,0)--(16,8)--(6,8)--cycle); draw((11,8)--(11,0), linetype("8 4")); draw((11,1)--(12,1)--(12,0)); label("$A$", (0,0), SW); label("$D$", (31,0), SE); label("$B$", (6,8), NW); label("$C$", (16,8), NE); label("10", (3,5), W); label("8", (11,4), E); label("17", (22.5,5), E);[/asy]
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$
|
Using the formula for the area of a trapezoid, we have $164=8(\frac{BC+AD}{2})$ . Thus $BC+AD=41$ . Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$ . From the Pythagorean Theorem, $AE=6$ and $DF=15$ thus $AD=BC+21$ . Substituting back into our original equation we have $BC+BC+21=41$ thus $BC=\boxed{10}$
| 10
|
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