id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
2,901 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | 4 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\... | Imagine Points $x$ $y$ $z$ as the "starting points" of three "blocks" of real numbers that have length $1$ . We are just trying to find the probability that those three "blocks" do not overlap. To do this we can set each unit of $1$ into $\mu$ equal little increments, and take the limit of the probability as $\mu$ appr... | 10 |
2,902 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_1 | 1 | Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$ | In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \times 4 = \boxed{64}$ | 64 |
2,903 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_2 | 1 | A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?
[asy] draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); draw(circle((10,5),5));[/asy]
$\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)... | Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is $5$ , the diameter is $2\cdot 5 = 10$ .
Since the sides of the rectangle are in a $2:1$ ratio, the longer side has length $2\cdot 10 = 20$ .
Therefore the area is $20\cdot 10 = 200$ or $\boxed{200}$ | 200 |
2,904 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_4 | 1 | When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be leftover?
$\textbf{(A)}\ 1... | In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \equiv 1 \pmod{6}$ . This means that there would be $1$ marble leftover, or $\boxed{1}$ | 1 |
2,905 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_4 | 2 | When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be leftover?
$\textbf{(A)}\ 1... | Let $r$ be the number of marbles Ringo has and let $p$ be the number of marbles Paul has. we have the following equations: \[r \equiv 4 \mod{6}\] \[p \equiv 3 \mod{6}\] Adding these equations we get: \[p + r \equiv 7 \mod{6}\] We know that $7 \equiv 1 \mod{6}$ so therefore: \[p + r \equiv 7 \equiv 1 \mod{6} \implies p ... | 1 |
2,906 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_5 | 1 | Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in ... | Let $x$ be the cost of her dinner.
$27.50=x+\frac{1}{10}*x+\frac{3}{20}*x$
$27+\frac{1}{2}=\frac{5}{4}*x$
$\frac{55}{2}=\frac{5}{4}x$
$\frac{55}{2}*\frac{4}{5}=x$
$x=22$
$\boxed{22}$ | 22 |
2,907 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_7 | 1 | For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hi... | Let $x$ be the number of acorns that both animals had.
So by the info in the problem:
$\frac{x}{3}=\left( \frac{x}{4} \right)+4$
Subtracting $\frac{x}{4}$ from both sides leaves
$\frac{x}{12}=4$
$\boxed{48}$ | 48 |
2,908 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_8 | 1 | What is the sum of all integer solutions to $1<(x-2)^2<25$
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$ | $(x-2)^2$ = perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for $x$
$(x-2)^2=4$
$x=4,0$
and
$(x-2)^2=9$
$x=5,-1$
and
$(x-2)^2=16$
$x=6,-2$
The sum of all integer solutions is
$4+5+6+0+(-1)+(-2)=\boxed{12}$ | 12 |
2,909 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_9 | 1 | Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C... | Out of the first two integers, it's possible for both to be even: for example, $10 + 16 = 26.$ But the next two integers, when added, increase the sum by $15,$ which is odd, so one of them must be odd and the other must be even: for example, $3 + 12 = 15.$ Finally, the next two integers increase the sum by $16,$ which ... | 1 |
2,910 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_10 | 1 | How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers: \begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*} The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the... | 9 |
2,911 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_10 | 2 | How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Cross-multiplying gives $MN=36.$ From the prime factorization \[36=2^2\cdot3^2,\] we conclude that $36$ has $(2+1)(2+1)=9$ positive divisors. There are $9$ values of $M,$ and each value generates $1$ ordered pair $(M,N).$ So, there are $\boxed{9}$ ordered pairs $(M,N)$ in total. | 9 |
2,912 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_11 | 1 | A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textb... | Desserts must be chosen for $7$ days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
There are $3$ choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are $3$ choices for dessert on Thursd... | 729 |
2,913 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_11 | 2 | A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textb... | There are $4 \cdot 3^6$ ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share $\frac{1}{4}$ of the total arrangements. Therefore our answer is $\frac{4\cdot3^6}{4} = 3^6 = \boxed{729}.$ | 729 |
2,914 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_13 | 1 | It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf... | Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Using $d = v t$ , the first statement can be translated to the equation $d = 60c$ . The second statement can be translated to $d = 24(c+s)$ . Since the same distance is being covered in each scenario, we can set the two equations equal and solve for $s... | 40 |
2,915 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_13 | 2 | It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf... | Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Then without loss of generality , assume that the length of the escalator be 1. Then $c=\dfrac{1}{60}$ and $c+s=\dfrac{1}{24}$ , so $s=\dfrac{1}{24}-\dfrac{1}{60}=\dfrac{1}{40}$ . Thus the time it takes for Clea to ride down the operating escalator whe... | 40 |
2,916 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_13 | 3 | It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf... | WLOG, let the length of the escalator be $120$ ft. So, Clea's rate is $120/60=2$ feet per second (fps $*$ ), and the escalator and Clea's rate is $120/24=5$ fps. So, the escalator's rate is $5-2=3$ fps. Therefore, $120/3=\boxed{40}$ | 40 |
2,917 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_15 | 1 | In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of... | The total number of games (and wins) in the tournament is $\frac{6 \times 5}{2}= 15$ . A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:
The "X's" are for when it is where a team is set again... | 5 |
2,918 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | 1 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | The last number that Bernardo says has to be between 950 and 999. Note that $1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \right... | 7 |
2,919 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | 2 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | Work backwards. The last number Bernardo produces must be in the range $[950,999]$ . That means that before this, Silvia must produce a number in the range $[475,499]$ . Before this, Bernardo must produce a number in the range $[425,449]$ . Before this, Silvia must produce a number in the range $[213,224]$ . Before thi... | 7 |
2,920 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | 3 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | If our first number is $N,$ then the sequence of numbers will be \[2N,~2N+50,~4N+100,~4N+150,~8N+300,~8N+350,~16N+700,~16N+750\] Note that we cannot go any further because doubling gives an extra $1500$ at the end, which is already greater than $1000.$ The smallest $N$ will be given if $16N+750>1000>16N+700 \implies 15... | 7 |
2,921 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_21 | 1 | Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$ $a$ $a$ $a$ $2a$ , and $b$ . What is the ratio of $b$ to $a$
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi$ | For any $4$ non-collinear points with the given requirement, notice that there must be a triangle with side lengths $a$ $a$ $2a$ , which is not possible as $a+a=2a$ . Thus at least $3$ of the $4$ points must be collinear.
If all $4$ points are collinear, then there would only be $3$ lines of length $a$ , which wouldn't... | 3 |
2,922 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | 1 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,... | If we have 1 as the first number, then the only possible list is $(1,2,3,4,5,6,7,8,9,10)$
If we have 2 as the first number, then we have 9 ways to choose where the $1$ goes, and the numbers ascend from the first number, $2$ , with the exception of the $1$ .
For example, $(2,3,1,4,5,6,7,8,9,10)$ , or $(2,3,4,1,5,6,7,8,9... | 512 |
2,923 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | 2 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,... | Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is $(1,2,3,4,5...10)$ . There are 9 arrows, so the answer is $2^{9}$ $512$ $\boxed{512}$ | 512 |
2,924 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | 3 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,... | Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, $(1, 2)$ and $(2, 1)$ . If we replace it with 3, there are four list... | 512 |
2,925 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | 4 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,... | If $a_1=10$ , the sequence must be $10, 9, 8,7,6,5,4,3,2,1$ . If $a_2=10$ , then $a_1=9$ , and the sequence is $9, 10, 8, 7, 6, 5,4,3,2,1$ . If $a_3=10$ , then the possible sequences are \[9,8,10,7,6,5,4,3,2,1 \text{ and}\] \[8,9,10,7,6,5,4,3,2,1.\] In general, for an $n$ -length sequence, if $a_i=n$ , then $a_1$ throu... | 512 |
2,926 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_24 | 1 | Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?
$\textbf{(A)}\ 108\qquad\te... | Let the ordered triple $(a,b,c)$ denote that $a$ songs are liked by Amy and Beth, $b$ songs by Beth and Jo, and $c$ songs by Jo and Amy. The only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$
To show this, observe these are all valid conditions. Second, note that none of $a,b,c$ can be bigger than 3. Suppose ... | 132 |
2,927 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_24 | 2 | Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?
$\textbf{(A)}\ 108\qquad\te... | Let $AB, BJ$ , and $AJ$ denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let $A, B, J,$ and $N$ denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least $1... | 132 |
2,928 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_24 | 3 | Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?
$\textbf{(A)}\ 108\qquad\te... | There are $\binom{4}{3}$ ways to choose the three songs that are liked by the three pairs of girls.
There are $3!$ ways to determine how the three songs are liked, or which song is liked by which pair of girls.
In total, there are $\binom{4}{3}\cdot3!$ possibilities for the first $3$ songs.
There are $3$ cases for the ... | 132 |
2,929 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | 1 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.732050807568... | [asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641... | 400 |
2,930 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | 2 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.732050807568... | [asy] size(6cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)); draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.... | 400 |
2,931 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | 3 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.732050807568... | We use casework.
The main thing to notice is that, if the bug does not go backwards, then every vertical set of arrows can be used, as the bug could travel straight downwards and then use any arrow to continue right.
Note: The motivation is quite weird so follow my numbers as they start from the left (point A) and go ... | 400 |
2,932 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_1 | 1 | A cell phone plan costs $20$ dollars each month, plus $5$ cents per text message sent, plus $10$ cents for each minute used over $30$ hours. In January Michelle sent $100$ text messages and talked for $30.5$ hours. How much did she have to pay?
$\textbf{(A)}\ 24.00 \qquad \textbf{(B)}\ 24.50 \qquad \textbf{(C)}\ 25.50 ... | The base price of Michelle's cell phone plan is $20$ dollars.
If she sent $100$ text messages and it costs $5$ cents per text, then she must have spent $500$ cents for texting, or $5$ dollars. She talked for $30.5$ hours, but $30.5-30$ will give us the amount of time exceeded past 30 hours. $30.5-30=.5$ hours $=30$ mi... | 28 |
2,933 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_2 | 1 | A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}... | To find how many small bottles we need, we can simply divide $500$ by $35$ . This simplifies to $\frac{100}{7}=14 \frac{2}{7}.$ Since the answer must be an integer greater than $14$ , we have to round up to $15$ bottles, or $\boxed{15}$ | 15 |
2,934 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | 1 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | We see that both sequences have equal numbers of terms, so reformat the sequence to look like:
\begin{align*} Y = \ &12 + 14 + \cdots + 100 + 102\\ X = 10 \ + \ &12 + 14 + \cdots + 100\\ \end{align*} From here it is obvious that $Y - X = 102 - 10 = \boxed{92}$ | 92 |
2,935 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | 2 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be $46\cdot 2=\boxed{92}$ | 92 |
2,936 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | 3 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | \begin{align*} X&=10+12+14+\cdots +100 \\ Y&=X-10+102 = X+92 \\ Y-X &= (X+92)-X \\ &= \boxed{92} $\blacksquare$ | 92 |
2,937 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_6 | 1 | Set $A$ has $20$ elements, and set $B$ has $15$ elements. What is the smallest possible number of elements in $A \cup B$
$\textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300$ | $A \cup B$ will be smallest if $B$ is completely contained in $A$ , in which case all the elements in $B$ would be counted for in $A$ . So the total would be the number of elements in $A$ , which is $\boxed{20}$ | 20 |
2,938 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_6 | 2 | Set $A$ has $20$ elements, and set $B$ has $15$ elements. What is the smallest possible number of elements in $A \cup B$
$\textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300$ | Assume WLOG that $A={1, 2, 3, 4, \cdots , 20}$ , and $B={6, 7, 8, 9, 10, \cdots , 20}$ . Then, all the integers $6$ through $20$ would be redundant in $A \cup B$ , so $A \cup B = 1, 2, 3, 4, \cdots, 20 \implies \boxed{20}$ | 20 |
2,939 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_8 | 1 | Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60$ | 75% of the total birds were not swans. Out of that 75%, there was $30\% / 75\% = \boxed{40}$ of the birds that were not swans that were geese. | 40 |
2,940 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_8 | 2 | Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60$ | WLOG, suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. $100-25 = 75$ of the birds are not swans and 30 of these are geese, so the answer is $\frac{30}{75} \times 100 = \boxed{40}$ | 40 |
2,941 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_10 | 1 | A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils... | The total cost of the pencils can be found by $(\text{students}\cdot\text{pencils purchased by each}\cdot\text{price of each pencil})$
Since $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$ . Since neither $(C)$ nor $(E)$ are ... | 11 |
2,942 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_10 | 2 | A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils... | We know the total cost of the pencils can be found by $(\text{students}\cdot\text{pencils purchased by each}\cdot\text{price of each pencil})$
Using prime factorization like in the solution above, we see that there is only one combination of three whole numbers whose product is equal to $1771$ $7, 11, 23$ (without usin... | 11 |
2,943 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_12 | 1 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $... | For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \[2 \times (3x) + 3 \times (2x) + 1 \times (3x+... | 13 |
2,944 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_12 | 2 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $... | Let $x$ be the number of free throws. Then the number of points scored by two-pointers is $2(x-1)$ and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is $x+4(x-1) = 61 \Rightarrow x=13$ , giving us $\boxed{13}$ for an answer. | 13 |
2,945 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_12 | 3 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $... | We let $a$ be the number of $2$ -point shots, $b$ be the number of $3$ -point shots, and $x$ be the number of free throws. We are looking for $x.$ We know that $2a=3b$ , and that $x=a+1$ . Also, $2a+3b+1x=61$ . We can see
\begin{align*} a&=x-1 \\ 2a &= 2x-2 \\ 3a &= 2x-2. \\ \end{align*}
Plugging this into $2a+3b+1x=61... | 13 |
2,946 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_13 | 1 | How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$ | We split up into cases of the hundreds digits being $2$ or $5$ . If the hundred digits is $2$ , then the units digits must be $8$ in order for the number to be even and then there are $4$ remaining choices ( $1,5,7,9$ ) for the tens digit, giving $1 \times 4 \times 1=4$ possibilities. Similarly, there are $1 \times 2 \... | 12 |
2,947 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_13 | 2 | How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$ | We see that the last digit of the $3$ -digit number must be even to have an even number. Therefore, the last digit must either be $2$ or $8$
Case $1$ -the last digit is $2$ . We must have the hundreds digit to be $5$ and the tens digit to be any $1$ of ${1,7,8,9}$ , thus obtaining $4$ numbers total.
Case $2$ -the last ... | 12 |
2,948 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_13 | 3 | How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$ | We see that there are $\dbinom63=20$ total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set ${1,2,5,7,8,9}.$ Obviously, some of these (such as $987,$ for example) will not work, and thus the answer will be less than $20.$ The only possible option is $\boxed{12}... | 12 |
2,949 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_15 | 1 | Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?
$\... | We know that $\frac{\text{total miles}}{\text{total gas}}=55$ . Let $x$ be the distance in miles the car traveled during the time it ran on gasoline, then the amount of gas used is $0.02x$ . The total distance traveled is $40+x$ , so we get $\frac{40+x}{0.02x}=55$ . Solving this equation, we get $x=400$ , so the total ... | 440 |
2,950 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_15 | 2 | Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?
$\... | Let $d$ be the length of the trip in miles. Roy used no gasoline for the 40 first miles, then used 0.02 gallons of gasoline per mile on the remaining $d - 40$ miles, for a total of $0.02 (d - 40)$ gallons. Hence, his average mileage was \[\frac{d}{0.02 (d - 40)} = 55.\] Multiplying both sides by $0.02 (d - 40)$ , we ge... | 440 |
2,951 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | 1 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Let $A=x$ . Then from $A+B+C=30$ , we find that $B=25-x$ . From $B+C+D=30$ , we then get that $D=x$ . Continuing this pattern, we find $E=25-x$ $F=5$ $G=x$ , and finally $H=25-x$ . So $A+H=x+25-x=25 \rightarrow \boxed{25}$ | 25 |
2,952 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | 2 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$ and $(B+C+D)+(E+F+G)=60$
It follows that $A+B+C+D+E+F+G+H=85$ because $C=5$
Subtracting, we have that $A+H=25\rightarrow \boxed{25}$ | 25 |
2,953 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | 3 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | From the given information, we can deduce the following equations:
$A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G =30$ , and $F+G+H=30$
We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.
$(A+B)-(B+D)=25-25 \implies (A-D)=0$
$(A-D)+(D+E)=0+25 \implies (A+E)=25$
$(A+E... | 25 |
2,954 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | 4 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Since all of the answer choices are constants, it shouldn't matter what we pick $A$ and $B$ to be, so let $A = 20$ and $B = 5$ . Then $D = 30 - B -C = 20$ $E = 30 - D - C = 5$ $F = 30 - D - E =5$ , and so on until we get $H = 5$ . Thus $A + H = \boxed{25}$ | 25 |
2,955 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | 5 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Assume the sequence is \[15,10,5,15,10,5,15,10\]
Thus, $15+10=\boxed{25}$ | 25 |
2,956 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_19 | 1 | In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the to... | Let the population of the town in $1991$ be $p^2$ . Let the population in $2001$ be $q^2+9$ . It follows that $p^2+150=q^2+9$ . Rearrange this equation to get $141=q^2-p^2=(q-p)(q+p)$ . Since $q$ and $p$ are both positive integers with $q>p$ $(q-p)$ and $(q+p)$ also must be, and thus, they are both factors of $141$ . W... | 62 |
2,957 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_19 | 2 | In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the to... | Since all the answer choices are around $50\%$ , we know the town's starting population must be around $600$ . We list perfect squares from $400$ to $1000$ \[441, 484, 529,576,625,676,729,784,841,900,961\] We see that $484$ and $784$ differ by $300$ , and we can confirm that $484$ is the correct starting number by noti... | 62 |
2,958 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_19 | 3 | In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the to... | Let the population of the town in 1991 be $a^2$ and the population in 2011 be $b^2$ . We know that $a^2+150+150=b^2 \implies a^2-b^2=-300 \implies b^2-a^2=300 \implies (b-a)(b+a)=300$ . Note that $b-a$ must be even. Testing, we see that $a=22$ and $b=28$ works, as $484+150-9=625=25^2$ , so $\frac{784-484}{484} \approx ... | 62 |
2,959 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 1 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | Let vertex $A$ be any vertex, then vertex $B$ be one of the diagonal vertices to $A$ $C$ be one of the diagonal vertices to $B$ , and so on. We consider cases for this problem.
In the case that $C$ has the same color as $A$ $D$ has a different color from $A$ and so $E$ has a different color from $A$ and $D$ . In this c... | 120 |
2,960 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 2 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | First, notice that there can be $3$ cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.
Case $1$ : There are $6!$ ways of assigning each vertex a different color. $6! = 72... | 120 |
2,961 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 3 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | There are $6$ ways to assign a color to $A$ . WLOG, give vertex $A$ a color; we can multiply by $6$ at the end. Since vertices $A$ and $C$ cannot have the same color, there are $5$ ways to assign colors to vertex $C$ . Using this same logic, there are $5$ ways to assign a color to vertices $E$ $B$ , and $D$ , giving a ... | 120 |
2,962 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 4 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | This problem is a direct application of the chromatic polynomial of a graph, represented by $P_k(G)$ , which returns a polynomial representing the number of ways to color a graph $G$ with $k$ colors such that no two adjacent vertices share the same color. In other words, it gives us a polynomial in terms of $k$ where w... | 120 |
2,963 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 5 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | In pentagon $ABCDE$ , fix any vertex $A$ . Now draw diagonal $AC$ . There are six choices for vertex $A$ and $5$ choices for vertex $C$
Now draw diagonal $CE$ . Since $E$ cannot be the same color as vertex $C$ , we have $5$ choices for $E$ . Again, we have five choices for vertex $D$ (draw diagonal $AD$ ).
Thus there a... | 120 |
2,964 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 6 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | WLOG, draw $ABCDE$ such that point $A$ is at the top, and write the letters in counterclockwise order. WLOG, fill in $A$ first. There are $6$ ways to do so. From here we proceed with casework on the color of $B$
Summing up the cases, $600+600+1920=3120 \Longrightarrow \boxed{3120}$ | 120 |
2,965 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 7 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | Notice that a minimum of, $3$ and a maximum of $5$ colours can be used.
This is straight forward and there are $6P5 = 720$ ways.
If $x$ and $y$ have same colours, they have to be adjacent. Let $x$ be directly right of $y$ $4$ consecutive points from $x$ clockwise have different colours. They can be coloured in $6P4=360... | 120 |
2,966 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 8 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | Name the pentagon $ADBEC$ . Then $A,B;$ $B,C;$ $C,D;$ $D,E$ and $E,A$ have to be of different colours.
$A$ can be coloured in $6$ ways, $B$ in $5$ ways (Without the one used for $A$ ), $C$ in $5$ ways (Without the one used for $B$ ), $D$ in $5$ ways (Without the one used for $C$ ).
Now we find the expected value of how... | 120 |
2,967 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | 9 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | First, we know that there are $6$ ways to assign a color to vertex $A$ . Then, there are $6$ ways to assign a color to vertex $B$ , because it doesn't depend on vertex $A$ . There are also $5$ ways to assign a color to vertex $C$ , because it can't be the same color as vertex $A$ . Notice that both vertices $E$ and $D$... | 120 |
2,968 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | 1 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each ... | First look at the numbers Alice says. $1, 3, 4, 6, 7, 9 \cdots$ skipping every number that is congruent to $2 \pmod 3$ . Thus, Barbara says those numbers EXCEPT every second - being $2 + 3^1 \equiv 5 \pmod{3^2=9}$ . So Barbara skips every number congruent to $5 \pmod 9$ . We continue and see:
Alice skips $2 \pmod 3$ , ... | 365 |
2,969 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | 2 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each ... | After Alice says all her numbers, the numbers not mentioned yet are \[\text{Alice: } 2,5,8,11,14,17,\cdots,998.\] After Barbara says all her numbers, the numbers that haven't been said yet are \[\text{Barbara: } 5,14,23,32,41,50,\cdots,995.\] After Candice, the list is \[\text{Candice: } 14,41,68,\cdots,986.\] Notice h... | 365 |
2,970 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | 4 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each ... | Similar to Solution 1 we find that the only number that George can say must leave a remainder of $2$ when divided by $3$ , and that it must also leave a remainder of $5$ when divided by $9$ . Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that... | 365 |
2,971 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | 5 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each ... | Every integer from 1 to 1000 can be written in the form $t_n+1$ in base 10, where $t_n$ is a trinary integer with no more than 7 significant digits. The important insight is that person $1\le{k}\le6$ will not say $n$ if and only if the $k$ th digit from the right of $t_n$ is 1. Therefore, the last 6 digits of $t_n$ mus... | 365 |
2,972 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_25 | 1 | Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textb... | There must be four rays emanating from $X$ that intersect the four corners of the square region. Depending on the location of $X$ , the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is $100$ -ray partitional (let this point be the bottom-left-mo... | 320 |
2,973 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_25 | 2 | Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textb... | Position the square region $R$ so that the bottom-left corner of the square is at the origin. Then define $s$ to be the sidelength of $R$ and $X$ to be the point $(rs, qs)$ , where $0<r,q<1$
There must be four rays emanating from $X$ that intersect the four corners of $R$ . The areas of the four triangles formed by the... | 320 |
2,974 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_25 | 3 | Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textb... | For the sake of simplicity, let $R$ be a $60 \times 60$ square and set the bottom-left point as the origin. Then, $R$ has vertices: \[(0,0), (60,0), (60,60), (0,60).\]
Now, let a point in the square have coordinates $(x, y).$
In order for the point to be $100-$ ray partitional, we must be able to make $100$ triangles w... | 320 |
2,975 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_2 | 1 | Josanna's test scores to date are $90, 80, 70, 60,$ and $85$ . Her goal is to raise here test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 82 \qquad\textbf{(C)}\ 85 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ ... | The average of her current scores is $77$ . To raise it $3$ points, she needs an average of $80$ , and so after her $6$ tests, a sum of $480$ . Her current sum is $385$ , so she needs a $480 - 385 = \boxed{95}$ | 95 |
2,976 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_4 | 1 | LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B$ . How many dollar... | The difference in how much LeRoy and Bernardo paid is $B-A$ . To share the costs equally, LeRoy must give Bernardo half of the difference, which is $\boxed{2}$ | 2 |
2,977 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_4 | 2 | LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid $A$ dollars and Bernardo had paid $B$ dollars, where $A < B$ . How many dollar... | Since there are no restrictions on cost paid besides $A<B$ , we can use an example where $A = 40$ and $B = 50$ . Quickly, we realize the only way they could pay the same amount of money is if they both pay 45 dollars. This means LeRoy must give Bernardo $50 - 45 = 5$ . Looking at the answer choices we see only $\frac{5... | 2 |
2,978 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_5 | 1 | In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161$ . What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad\textbf{(B)}\ 161 \qquad\textbf{(C)}\ 204 \qquad\textbf{(D)}\ 214 \qquad\textbf{(E)}\ 224$ | We have $161 = 7 \cdot 23.$ Since $a$ has two digits, the factors must be $23$ and $7,$ so $a = 32$ and $b = 7.$ Then, $ab = 7 \times 32 = \boxed{224}.$ | 224 |
2,979 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_6 | 1 | On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\te... | Let $x$ represent the amount of candies Casper had at the beginning.
\begin{align*} \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ \frac{2}{3} x - 2 &= 18\\ \frac{2}{3} x &= 20\\ x &= \boxed{30} | 30 |
2,980 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_6 | 2 | On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\te... | We work backwards. If he had 8 candies at the end, then before he gave candies to his sister he had 12 candies. This means that at the end of Halloween he had 18 candies, so before he gave candies to his brother he had 20 candies. Therefore, at the start he had $\boxed{30}$ | 30 |
2,981 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_6 | 3 | On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\te... | A solve by algebra is more secure and safe (and usually faster), but you can also test the answer choices. We are lucky and option A works $\Longrightarrow \boxed{30}$ | 30 |
2,982 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_7 | 1 | The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?
$\textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108$ | The sum of two angles in a triangle is $\frac{6}{5}$ of a right angle $\longrightarrow \frac{6}{5} \times 90 = 108$
If $x$ is the measure of the first angle, then the measure of the second angle is $x+30$ \[x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39\]
Now we know the measure of two angles are $39^{... | 72 |
2,983 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_10 | 1 | Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$ | The requested ratio is \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Using the formula for a geometric series, we have \[10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},\] which is very close to $\dfrac{10^{10}}{9},$ so the ratio is very close to $\boxed{9}.$ | 9 |
2,984 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_10 | 2 | Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$ | The problem asks for the value of \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Written in base 10, we can find the value of $10^9 + 10^8 + \ldots + 10 + 1$ to be $1111111111.$ Long division gives us the answer to be $\boxed{9}.$ | 9 |
2,985 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_10 | 3 | Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$ | Let $f(n)=\dfrac{10^n}{1+10+10^2+10^3+\cdots+10^{n-1}}$ . We are approximating $f(10)$ . Trying several small values of $n$ gives answers very close to $9$ , so our answer is $\boxed{9}\approx9.09.$ ~Technodoggo | 9 |
2,986 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_11 | 1 | There are $52$ people in a room. what is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$ | Pretend you have $52$ people you want to place in $12$ boxes, because there are $12$ months in a year. By the Pigeonhole Principle , one box must have at least $\left\lceil \frac{52}{12} \right\rceil$ people $\longrightarrow \boxed{5}$ | 5 |
2,987 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_12 | 1 | Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in me... | Let $s$ be Keiko's speed in meters per second, $a$ be the length of the straight parts of the track, $b$ be the radius of the smaller circles, and $b+6$ be the radius of the larger circles. The length of the inner edge will be $2a+2b \pi$ and the length of the outer edge will be $2a+2\pi (b+6).$ Since it takes $36$ sec... | 3 |
2,988 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_12 | 2 | Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in me... | It is basically the same as Solution 1 except you can completely disregard the straight edges of the track since it will take Keiko the same time to walk that length for both inner and outer circles. Instead, focus on the circular part. If the diameter of the smaller circle were $r,$ then the length of the smaller circ... | 3 |
2,989 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_13 | 1 | Two real numbers are selected independently at random from the interval $[-20, 10]$ . What is the probability that the product of those numbers is greater than zero?
$\textbf{(A)}\ \frac{1}{9} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{4}{9} \qquad\textbf{(D)}\ \frac{5}{9} \qquad\textbf{(E)}\ \frac{2}{3}... | We will use complementary counting. The probability that the product is negative can be found by finding the probability that one number is positive and the other number is negative. The probability of a positive number being selected is $\frac13$ , and the probability of a negative number being selected is $\frac23$ .... | 59 |
2,990 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_14 | 1 | A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$ | Let the sides of the rectangular parking lot be $a$ and $b$ . Then $a^2 + b^2 = 625$ and $ab = 168$ . Add the two equations together, then factor. \begin{align*} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align*} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \boxed{62}$ | 62 |
2,991 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_14 | 2 | A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$ | We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy $a^2+b^2=25^2$ . In other words, $(a,b,25)$ is a set of Pythagorean triples. Guessing and checking, we have $(7,24,25)$ as the triplet, as the area is $7 \cdot 24 = 168$ as requested. The... | 62 |
2,992 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_17 | 1 | In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 1... | We can let $\angle AEB$ be $4x$ and $\angle ABE$ be $5x$ because they are in the ratio $4 : 5$ . When an inscribed angle contains the diameter , the inscribed angle is a right angle . Therefore by triangle sum theorem, $4x+5x+90=180 \longrightarrow x=10$ and $\angle ABE = 50$
$\angle ABE = \angle BED$ because they are ... | 130 |
2,993 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_17 | 2 | In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 1... | Note $\angle ABE = \angle BED=50$ as before. The sum of the interior angles for quadrilateral $EBCD$ is $360$ . Denote the center of the circle as $P$ $\angle PDE = \angle PED = 50$ . Denote $\angle PDC = \angle PCD = x$ and $\angle PBC = \angle PCB = y$ . We wish to find $\angle BCD = x+y$ . Our equation is $(\angle P... | 130 |
2,994 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_17 | 3 | In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 1... | Note that $\overset{\Large\frown} {BE}$ intercepts $\angle BAE$ . Since, $\overset{\Large\frown} {BE}=180$ , thus $\angle BAE=90°$ (courtesy of the Inscribed Angles Theorem).
Since we know that $\angle BAE=90°$ , then $\angle AEB + \angle ABE = 90°$ , (courtesy of the Triangle Sum Theorem) and also $5\angle AEB = 4\ang... | 130 |
2,995 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | 1 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | It is given that $\angle AMD \sim \angle CMD$ . Since $\angle AMD$ and $\angle CDM$ are alternate interior angles and $\overline{AB} \parallel \overline{DC}$ $\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM$ . Use the Base Angle Theorem to show $\overline{DC} \cong \overline{MC}$ . We know that ... | 75 |
2,996 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | 2 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | After finding $MC = 6,$ we can continue using trigonometry as follows.
We know that $\angle{BMC} = 180-2x$ and so $\sin (180-2x) = \frac{3}{6} = \frac{1}{2}$
It is obvious that $\sin (30) = \frac{1}{2}$ and so $180-2x=30.$
Solving, we have $x = \boxed{75}$ | 75 |
2,997 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | 3 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | Let $\angle{DMC} = \angle{AMD} = \theta$ . If we let $AM = x$ , we have that $MD = \sqrt{x^2 + 9}$ , by the Pythagorean Theorem, and similarily, $MC = \sqrt{x^2 - 12x + 45}$ . Applying the law of cosine, we see that \[2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36\] and \[\tan (\th... | 75 |
2,998 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_18 | 4 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | We have $DC=CM=6$ . By the Pythagorean Theorem, $BM=\sqrt{6^2-3^2}=3\sqrt{3}$ , and thus $AM=6-3\sqrt{3}$ , We have $\tan(AMD)=\frac{6-3\sqrt{3}}{3}=2+\sqrt{3}$ , or $\angle AMD=\boxed{75}$ ~awsomek | 75 |
2,999 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_19 | 1 | What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$ | First, square both sides, and isolate the absolute value. \begin{align*} 5|x|+8&=x^2-16\\ 5|x|&=x^2-24\\ |x|&=\frac{x^2-24}{5}. \\ \end{align*} Solve for the absolute value and factor.
Case 1: $x=\frac{x^2-24}{5}$
Multiplying both sides by $5$ gives us \[5x=x^2-24.\] Rearranging and factoring, we have \begin{align*} x^... | 64 |
3,000 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_19 | 2 | What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$ | Square both sides, to get $5|x| + 8 = x^2-16$ . Rearrange to get $x^2 - 5|x| - 24 = 0$ . Seeing that $x^2 = |x|^2$ , substitute to get $|x|^2 - 5|x| - 24 = 0$ . We see that this is a quadratic in $|x|$ . Factoring, we get $(|x|-8)(|x|+3) = 0$ , so $|x| = \{8,-3\}$ . Since the radicand of the equation can't be negative,... | 64 |
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