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616ee74293cf497e3090950d3d87cb29 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$ | Let the mean be $\frac{(a)+(a+d)+(a+2d)+...+(a+(n-1)) \cdot d)}{n}$ $=\frac{n \cdot a}{n} + \frac{(1+2+3+...+(n-1)) \cdot d}{n}$ $=a + \frac{n \cdot (n-1) \cdot d}{2n}$ $=a+ \frac{(n-1) \cdot d}{2}$
Note that this is also equal to n
$a+ \frac{(n-1) \cdot d}{2}=n$ $\therefore 2a+ (n-1) \cdot d=2n$
1st term + nth term $=2n=2 \cdot 2008=4016$ Now note that, from previous solutions, the first term is 1, hence the 2008th term is $4016-1=\boxed{4015}$ | B | 4015 |
616ee74293cf497e3090950d3d87cb29 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_13 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008^{\text{th}}$ term of the sequence?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$ | From inspection, we see that the sum of the sequence is $n^2$ . We also notice that $n^2$ is the sum of the first $n$ odd integers. Because $4015$ is the only odd integer, $\boxed{4015}$ is the answer. | B | 4015 |
b8baf700d903f2299bcb133be3b5daa3 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_15 | How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$
This means that $a^2=2b+1$
We know that $a,b>0$ and that $b<100$
We also know that $a^2$ is odd and thus $a$ is odd, since the right side of the equation is odd. $2b$ is even. $2b+1$ is odd.
So $a=1,3,5,7,9,11,13$ , but if $a=1$ , then $b=0$ . Thus $a\neq1.$
$a=3,5,7,9,11,13$
The answer is $\boxed{6}$ | A | 6 |
1866dde08beb55f74bac383e0ec6ad9a | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_17 | A poll shows that $70\%$ of all voters approve of the mayor's work. On three separate occasions a pollster selects a voter at random. What is the probability that on exactly one of these three occasions the voter approves of the mayor's work?
$\mathrm{(A)}\ {{{0.063}}} \qquad \mathrm{(B)}\ {{{0.189}}} \qquad \mathrm{(C)}\ {{{0.233}}} \qquad \mathrm{(D)}\ {{{0.333}}} \qquad \mathrm{(E)}\ {{{0.441}}}$ | Letting Y stand for a voter who approved of the work, and N stand for a person who didn't approve of the work, the pollster could select responses in $3$ different ways: $\text{YNN, NYN, and NNY}$ . The probability of each of these is $(0.7)(0.3)^2=0.063$ . Thus, the answer is $3\cdot0.063=\boxed{0.189}$ | B | 0.189 |
1866dde08beb55f74bac383e0ec6ad9a | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_17 | A poll shows that $70\%$ of all voters approve of the mayor's work. On three separate occasions a pollster selects a voter at random. What is the probability that on exactly one of these three occasions the voter approves of the mayor's work?
$\mathrm{(A)}\ {{{0.063}}} \qquad \mathrm{(B)}\ {{{0.189}}} \qquad \mathrm{(C)}\ {{{0.233}}} \qquad \mathrm{(D)}\ {{{0.333}}} \qquad \mathrm{(E)}\ {{{0.441}}}$ | In more concise terms, this problem is an extension of the binomial distribution. We find the number of ways only 1 person approves of the mayor multiplied by the probability 1 person approves and 2 people disapprove: ${3\choose 1} \cdot(0.7)^1\cdot(1-0.7)^{(3-1)}=3\cdot0.7\cdot0.09=\boxed{0.189}$ | B | 0.189 |
7b2e0c18ea2399c5ceb288aeb17c4f00 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_18 | Bricklayer Brenda takes $9$ hours to build a chimney alone, and bricklayer Brandon takes $10$ hours to build it alone. When they work together, they talk a lot, and their combined output decreases by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?
$\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900$ | Let $x$ be the number of bricks in the chimney. The work done is the rate multiplied by the time.
Using $w = rt$ , we get $x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)$ . Solving for $x$ , we get $\boxed{900}$ | B | 900 |
271d3eb94e6cd17307bc080ae5d30a8f | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_21 | Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$ . Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?
$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$ | For the first man, there are $10$ possible seats. For each subsequent man, there are $4$ $3$ $2$ , or $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{480}$ | C | 480 |
271d3eb94e6cd17307bc080ae5d30a8f | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_21 | Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$ . Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?
$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$ | Label the seats ABCDEFGHIJ, where A is the top seat. The first man has $10$ possible seats. WLOG, assume he is in seat A in the diagram. Then, his wife can be in one of two seats, namely D or H. WLOG, assume she is in seat D. Now, in each structurally distinct solution we find, we know that there are $4! = 24$ ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be $10\cdot 2\cdot 24\cdot x = 480x$ possible seating arrangements, and x is a nonnegative integer. There is only one answer that is a multiple of $480$ . So, our answer is $\boxed{480}$ | C | 480 |
cb2cfcbc281bf0c5cdd6e7abd4241f40 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_22 | Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?
$\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2$ | There are $\frac{6!}{3!\cdot2!\cdot1!}=60$ total orderings.
Suppose we order the red and white beads first. If these two colors are ordered first, we must make sure that no neighboring beads are the same color, or only one pair of neighboring beads are the same color. There are five possible orderings:
$R\ W R\ W R$
$R\ R\ W R\ W$
$W R\ R\ W R$
$R\ W R\ R\ W$
$W R\ W R\ R$
For the first case, there are $6$ possible places we can put the blue bead. For the other 4 cases, there is only one place we can put the blue bead, which is between the pair of red beads. The desired probability is $\frac{6+4(1)}{60}=\frac{10}{60}=\boxed{16}$ | C | 16 |
9188c6efe2a616d448195fd74135b103 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_23 | A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers and $b > a$ . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of the whole floor. How many possibilities are there for the ordered pair $(a,b)$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$ . With this information we can make the equation:
\begin{eqnarray*} ab &=& 2\left((a-2)(b-2)\right) \\ ab &=& 2ab - 4a - 4b + 8 \\ ab - 4a - 4b + 8 &=& 0 \end{eqnarray*} Applying Simon's Favorite Factoring Trick , we get
\begin{eqnarray*}ab - 4a - 4b + 16 &=& 8 \\ (a-4)(b-4) &=& 8 \end{eqnarray*}
Since $b > a$ , then we have the possibilities $(a-4) = 1$ and $(b-4) = 8$ , or $(a-4) = 2$ and $(b-4) = 4$ . This allows for 2 possibilities: $(5,12)$ or $(6,8)$ which gives us $\boxed{2}$ | B | 2 |
d02bfa57011bc2d58da61773cb5f4d78 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24 | Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | To start off, draw a diagram like in solution two and label the points. Create lines $\overline{AC}$ and $\overline{BD}$ . We can call their intersection point $Y$ . Note that triangle $BCD$ is an isosceles triangle so angles $CDB$ and $CBD$ are each $5$ degrees. Since $AB$ equals $BC$ , angle $BAC$ equals $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle $CYB$ equals $120$ degrees.
Extend $\overline{CD}$ and $\overline{AB}$ and let their intersection be $E$ . Since angle $BEC$ plus angle $CYB$ equals $180$ degrees, quadrilateral $YCEB$ is a cyclic quadrilateral.
Next, draw a line from point $Y$ to point $E$ . Since angle $YBC$ and angle $YEC$ point to the same arc, angle $YEC$ is equal to $5$ degrees. Since $EYD$ is an isosceles triangle (based on angle properties) and $YAE$ is also an isosceles triangle, we can find that $YAD$ is also an isosceles triangle. Thus, each of the other angles is $\frac{180-120}{2}=30$ degrees. Finally, we have angle $BAD$ equals $30+55=\boxed{85}$ degrees. | null | 85 |
d02bfa57011bc2d58da61773cb5f4d78 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24 | Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | First, connect the diagonal $DB$ , then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$ . Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$ , the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$ . Because angle $ABC$ is $70^\circ$ , we get angle $ABD$ is $65^\circ$ . Next, noticing parallel lines $AB$ and $DE$ and transversal $DB$ , we see that angle $BDE$ is also $65^\circ$ , and subtracting off angle $CDB$ gives that angle $EDC$ is $60^\circ$
Now, because we drew $ED = DC$ , triangle $DEC$ is equilateral. We can also conclude that $EC=DC=CB$ meaning that triangle $ECB$ is isosceles, and angles $CBE$ and $CEB$ are equal.
Finally, we can set up our equation. Denote angle $BAD$ as $x^\circ$ . Then, because $ABED$ is a parallelogram, the angle $DEB$ is also $x^\circ$ . Then, $CEB$ is $(x-60)^\circ$ . Again because $ABED$ is a parallelogram, angle $ABE$ is $(180-x)^\circ$ . Subtracting angle $ABC$ gives that angle $CBE$ equals $(110-x)^\circ$ . Because angle $CBE$ equals angle $CEB$ , we get \[x-60=110-x\] , solving into $x=\boxed{85}$ | null | 85 |
d02bfa57011bc2d58da61773cb5f4d78 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24 | Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | [asy] unitsize(3 cm); pair A, B, C, D; A = (0,0); B = dir(85); C = B + dir(-25); D = C + dir(-35); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B))); draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + scale(0.02)*rotate(90)*(B - C))); draw(((C + D)/2 + scale(0.02)*rotate(90)*(D - C))--((C + D)/2 + scale(0.02)*rotate(90)*(C - D))); dot("$A$", A, SW); dot("$B$", B, N); dot("$C$", C, NE); dot("$D$", D, SE); label("$I$", 6/7*C); [/asy]
Let the unknown $\angle BAD$ be $x$
First, we draw diagonal $BD$ and $AC$ $I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\triangle BCD$ and $\triangle ABC$
Using this, we find: $\angle DBC = \angle CDB = 5^\circ$ and $\angle BAC = \angle BCA = 55^\circ$ . Expanding on this, we can fill in a couple more angles. $\angle ABD = 70^\circ - 5^\circ = 65^\circ$ $\angle ACD = 170^\circ - 55^\circ = 115^\circ$ $\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ$ $\angle BIC = \angle AID = 180^\circ - 60^\circ = 120^\circ$
We can rewrite $\angle CAD$ and $\angle BDA$ in terms of $x$ $\angle CAD = x - 55^\circ$ and $\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x$
Let us relabel $AB = BC = CD = a$ and $AD = b$
By Rule of Sines on $\triangle ACD$ and $\triangle ABD$ respectively, $\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}$ , and $\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}$
In a more convenient form, $\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
and $\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}$
$\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}$
Now, by identity $\sin(\theta) = \sin(180^\circ-\theta)$ $\sin(65^\circ) = \sin(115^\circ)$
Therefore, $\sin(115^\circ-x) = \sin(x-55^\circ).$ This equation is only satisfied by option $\boxed{85}$ | C | 85 |
d02bfa57011bc2d58da61773cb5f4d78 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_24 | Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | Using a protractor and rule, draw an accurate diagram ( Example Diagram ). $\angle BAD$ looks slightly less than $90$ degrees. Therefore the answer is $\boxed{85}$ as $85$ is slightly less than $90$ | C | 85 |
8a546975d60c01aa02dd71bfea2f06a1 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$ .
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$ .
We have $D(0)=200$
The truck always moves for $20$ seconds, then stands still for $30$ . During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$ , hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$ .
During the remaining $30$ seconds $D(t)$ decreases by $150$
From this observation it is obvious that after four full cycles, i.e. at $t=200$ , we will have $D(t)=0$ for the first time.
During the fifth cycle, $D(t)$ will first grow from $0$ to $100$ , then fall from $100$ to $-50$ . Hence Michael overtakes the truck while it is standing at the pail.
During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$ , then fall from $50$ to $-100$ . Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.
During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$ , then fall from $0$ to $-150$ . Hence the truck meets Michael at the moment when it arrives to the next pail.
Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5}$ meetings. | B | 5 |
8a546975d60c01aa02dd71bfea2f06a1 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | The truck takes $20$ seconds to go from one pail to the next and then stops for $30$ seconds at the new pail. Thus it sets off from a pail every 50 sec. Let $t$ denote the time elapsed and write $t=50k + \Delta$ , where $\Delta \in [0,50)$ . In this time Michael has traveled $5t = 250k+5\Delta$ feet. What about the truck? In the first $50k$ seconds the truck covers $k$ pails, i.e. $200k$ feet so it moves $200+200k$ feet from Michael's starting point. Then we have two cases:
(a) if $\Delta < 20$ , then the truck travels an additional $10\Delta$ feet. For them to intersect we must have $200+200k + 10\Delta = 250k + 5\Delta$ . Solving, we get $\Delta = 10k - 40$ . Since $\Delta$ must lie in the interval $[0,20)$ we get $k \in \{4, 5\}$
(b) if $\Delta \in [20, 50)$ , then the truck travels an additional $200$ feet. For them to intersect we must have $200+200k + 200 = 250k + 5\Delta$ . Solving, we get $\Delta = 10(8-k)$ . Since $\Delta$ must lie in the interval $[20,50)$ we get $k \in \{4, 5, 6\}$
Thus Michael intersects with the truck $5$ times, which is option $\boxed{5}$ | B | 5 |
8a546975d60c01aa02dd71bfea2f06a1 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | We make a chart by seconds in increments of ten.
$\begin{tabular}{c|c|c}Time (s) &Distance of Michael&Distance of Garbage\\ \hline 0&0&200\\ 10&50&300\\ 20&100&400\\ 30&150&400\\ 40&200&400\\ 50&250&400\\ 60&300&500\\ 70&350&600\\ 80&400&600\\ 90&450&600\\ 100&500&600\\ 110&550&700\\ 120&600&800\\ 130&650&800\\ 140&700&800\\ 150&750&800\\ 160&800&900\\ 170&850&1000\\ 180&900&1000\\ 190&950&1000\\ 200&1000&1000\\ 210&1050&1100\\ 220&1100&1200\\ 230&1150&1200\\ 240&1200&1200\\ 250&1250&1200\\ 260&1300&1300\\ 270&1350&1400\\ 280&1400&1400\\ 290&1450&1400\\ 300&1500&1400\\ 310&1550&1500\\ 320&1600&1600\\ 330&1650&1600\\ 340&1700&1600\\ 350&1750&1600\\ 360&1800&1700\\ 370&1850&1800\\ 380&1900&1800\\ 390&1950&1800\\ 400&2000&1800\\ \end{tabular}$
Notice at 200, 240, 260, 280, and 320 seconds, Michael and the garbage truck meet. It is clear that they met at these times, and will meet no more. Thus the answer is $\boxed{5}$ | B | 5 |
8a546975d60c01aa02dd71bfea2f06a1 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_10B_Problems/Problem_25 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | This solution might be time consuming, but it is pretty rigorous. Also, throughout the solution refer to the graph in solution 1 to understand this one more. Lets first start off by defining the position function for Michael. We let $M(t) = 5t$ , where $t$ is the amount of seconds passed. Now, lets define the position function of the truck as two independent functions. It is obvious, graphically, that the position function of the truck is a piecewise function alternating between linear lines and constant lines. Lets focus on the linear pieces of the truck's position function. Let the $kth$ linear part of the truck's position function be denoted as $L_k(t)$ . Then through algebra, it is found that $L_k(t) = 10t + 500 - 300k$ ${50k - 50 <= t <= 50k - 30}$ . Now, lets move on to the constant pieces, which is a lot easier in terms of algebra. Let the $ith$ constant part of the truck's position function be denoted as $C_i(t)$ . Then, again, through algebra we obtain $C_i(t) = 200 * (i + 1)$ ${50i - 30 <= t <= 50i}$ . Now, let me stress that $i$ and $k$ are disjoint, which is why I used different variable names. We are interested in where $C_i(t)$ and $L_k(t)$ intersect $M(t)$ or $5t$ $L_k(t)$ intersects $M$ at $(60k - 100, 300k - 500)$ . However, the only $k$ values that actually work are $5 <= k <= 7$ because of the domain restrictions on $L_k(t)$ . Similarly, we also see that for $C_i(t)$ it intersects at $(40i + 40, 200i + 200)$ . The only $i$ values that work is $4 <= i <= 7$ . However, some pair $(i, k)$ might yield the same exact intersection point. Checking this through simple algebra we see that $(4,5)$ and $(7,7)$ do indeed yield the same intersection point. Thus, our answer is $(7 - 5 + 1) + (7 - 4 + 1) - 2 = 7 - 2 = 5$ or $\boxed{5}$ | B | 5 |
97424196c74821d00a12073a4f7166a3 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_10 | The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20$ , the father is $48$ years old, and the average age of the mother and children is $16$ . How many children are in the family?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | Let $x$ be the number of the children and the mom. The father, who is $48$ , plus the sum of the ages of the kids and mom divided by the number of kids and mom plus $1$ (for the dad) = $20$ . This is because the average age of the entire family is $20.$ This statement, written as an equation, is: \[\frac{48+16x}{x+1}=20\] \[48+16x=20x+20\] \[4x=28\] \[x=7\]
$7$ people - $1$ mom = $6$ children.
Therefore, the answer is $\boxed{6}$ | E | 6 |
97424196c74821d00a12073a4f7166a3 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_10 | The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20$ , the father is $48$ years old, and the average age of the mother and children is $16$ . How many children are in the family?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | Let $m$ be the Mom's age.
Let the number of children be $x$ and their average age be $y$ . Their age totaled up is simply $xy$
We have the following two equations:
$\frac{m+48+xy}{2+x}=20$ , where $m+48+xy$ is the family's total age and $2+x$ is the total number of people in the family.
$\frac{m+48+xy}{2+x}=20$
$m+48+xy=40+20x$
The next equation is $\frac{m+xy}{1+x}=16$ , where $m+xy$ is the total age of the Mom and the children, and $1+x$ is the number of children along with the Mom.
$\frac{m+xy}{1+x}=16$
$m+xy=16+16x$
We know the value for $m+xy$ , so we substitute the value back in the first equation.
$m+48+xy=40+20x$
$(16+16x)+48=40+20x$
$x=6$
Earlier, we set $x$ to be the number of children. Therefore, there are $\boxed{6}$ children. | E | 6 |
5c98af8050611692e6f75fcfbb4284f1 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_13 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$ | Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r + \frac{x}{7}$ units of time. Because the two times are equal, we can create the following equation: $x-r = r + \frac{x}{7}$ . We get $x-2r=\frac{x}{7}$ , so $\frac{6}{7}x = 2r$ , and $\frac{x}{r} = \frac{7}{3}$ . This minus one is the reciprocal of what we want to find: $\frac{7}{3}-1 = \frac{4}{3}$ , so the answer is $\boxed{34}$ | B | 34 |
5c98af8050611692e6f75fcfbb4284f1 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_13 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$ | [asy] draw((0,0)--(7,0)); dot((0,0)); dot((3,0)); dot((6,0)); dot((7,0)); label("$H$",(0,0),S); label("$Y$",(3,0),S); label("$P$",(6,0),S); label("$S$",(7,0),S); [/asy] Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home.
Since he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$ , then $HS=7x$ and $HP=6x$ . Since Y is the midpoint of $\overline{HP}$ $HY=YP=3x$ . Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\frac{YH}{YS}=\frac{3x}{4x}=\boxed{34}$ | B | 34 |
0e77c0e705a98bcb37125278fe0a7194 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity . The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$ , since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ chance of being odd and a $\frac34$ chance of being even. Therefore, the probability that $ad-bc$ will be even is $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed{58}$ | E | 58 |
0e77c0e705a98bcb37125278fe0a7194 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_16 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | If we don't know our parity rules, we can check and see that $ad-bc$ is only even when $ad$ and $bc$ are of the same parity (as stated above). From here, we have two cases.
Case 1: $odd-odd$ (which must be $o \cdot o-o \cdot o$ ). The probability for this to occur is $\left(\frac 12\right)^4 = \frac 1{16}$ , because each integer has a $\frac 12$ chance of being odd.
Case 2: $even-even$ (which occurs in 4 cases: $(e \cdot e-e \cdot e$ ), ( $o \cdot e-o \cdot e$ ) (alternating of any kind), and ( $e \cdot e-o \cdot e$ ) with its reverse, ( $o \cdot e-e \cdot e$ ).
Our first subcase of case 2 has a chance of $\frac 1{16}$ (same reasoning as above).
Our second subcase of case 2 has a $\frac 14$ chance, since only the 2nd and 4th flip matter (or 1st and 3rd).
Our third subcase of case 2 has a $\frac 18$ chance, because the 1st, 2nd, and either 3rd or 4th flip matter.
Our fourth subcase of case 2 has a $\frac 18$ chance, because it's the same, just reversed.
We sum these, and get our answer of $\frac 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \boxed{58}.$ | E | 58 |
f6a1ca60bf24cf56f0253ec4c248e14e | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_17 | Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$ . What is the minimum possible value of $m + n$
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 5700$ | $3 \cdot 5^2m$ must be a perfect cube, so each power of a prime in the factorization for $3 \cdot 5^2m$ must be divisible by $3$ . Thus the minimum value of $m$ is $3^2 \cdot 5 = 45$ , which makes $n = \sqrt[3]{3^3 \cdot 5^3} = 15$ . The minimum possible value for the sum of $m$ and $n$ is $\boxed{60}.$ | D | 60 |
f6a1ca60bf24cf56f0253ec4c248e14e | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_17 | Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$ . What is the minimum possible value of $m + n$
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 5700$ | First, we need to prime factorize $75$ $75$ $5^2 \cdot 3$ . We need $75m$ to be in the form $x^3y^3$ . Therefore, the smallest $m$ is $5 \cdot 3^2$ $m$ = 45, and since $5^3 \cdot 3^3 = 15^3$ , our answer is $45 + 15$ $\boxed{60}$ | D | 60 |
6b797ae1712e1f7b6b345723db1b332a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20 | Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$ | Note that for all real numbers $k,$ we have $a^{2k} + a^{-2k} + 2 = (a^{k} + a^{-k})^2,$ from which \[a^{2k} + a^{-2k} = (a^{k} + a^{-k})^2-2.\] We apply this result twice to get the answer: \begin{align*} a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \\ &= [(a + a^{-1})^2 - 2]^2 - 2 \\ &= \boxed{194} ~Azjps (Fundamental Logic) | D | 194 |
6b797ae1712e1f7b6b345723db1b332a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20 | Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$ | Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$
Squaring both sides of $a^2+a^{-2}=14$ gives $a^4+a^{-4}+2=196,$ from which $a^4+a^{-4}=\boxed{194}.$ | D | 194 |
6b797ae1712e1f7b6b345723db1b332a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20 | Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$ | The detailed explanation of Solution 2 is as follows: \begin{alignat*}{8} a+a^{-1}&=4 \\ (a+a^{-1})^2&=4^2 \\ a^2+2aa^{-1}+a^{-2}&=16 \\ a^2+a^{-2}&=16-2&&=14 \\ (a^2+a^{-2})^2&=14^2 \\ a^4+2a^2a^{-2}+a^{-4}&=196 \\ a^4+a^{-4}&=196-2&&=\boxed{194} ~MathFun1000 (Solution) | D | 194 |
6b797ae1712e1f7b6b345723db1b332a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20 | Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$ | Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$
Applying the Binomial Theorem, we raise both sides of $a+a^{-1}=4$ to the fourth power: \begin{align*} \binom40a^4a^0+\binom41a^3a^{-1}+\binom42a^2a^{-2}+\binom43a^1a^{-3}+\binom44a^0a^{-4}&=256 \\ a^4+4a^2+6+4a^{-2}+a^{-4}&=256 \\ \left(a^4+a^{-4}\right)+4\left(a^2+a^{-2}\right)&=250 \\ \left(a^4+a^{-4}\right)+4(14)&=250 \\ a^4+a^{-4}&=\boxed{194} ~MRENTHUSIASM | D | 194 |
6b797ae1712e1f7b6b345723db1b332a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20 | Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$ | We multiply both sides of $4=a+a^{-1}$ by $a,$ then rearrange: \[a^2-4a+1=0.\]
We apply the Quadratic Formula to get $a=2\pm\sqrt3.$
By Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, both values of $a$ produce the same value of $a^4+a^{-4}:$ \begin{align*} a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \left(2+\sqrt{3}\right)^{-4} \\ &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 &&(*) \\ &=\boxed{194}$ | D | 194 |
6b797ae1712e1f7b6b345723db1b332a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20 | Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$ | From the first sentence of Solution 4, we conclude that $a$ and $a^{-1}$ are the roots of $x^2-4x+1=0.$ Let \begin{align*} P_1&=a+a^{-1}, \\ P_2&=a^2+a^{-2}, \\ P_3&=a^3+a^{-3}, \\ P_4&=a^4+a^{-4}. \end{align*} By Newton's Sums, we have \begin{alignat*}{12} &1\cdot P_1-4\cdot 1&&=0 &&\qquad\implies\qquad P_1&&=4, \\ &1\cdot P_2-4\cdot P_1+1\cdot2 &&=0 &&\qquad\implies\qquad P_2&&=14, \\ &1\cdot P_3-4\cdot P_2+1\cdot P_1&&=0 &&\qquad\implies\qquad P_3&&=52, \\ &1\cdot P_4-4\cdot P_3+1\cdot P_2&&=0 &&\qquad\implies\qquad P_4&&=\boxed{194} ~Albert1993 (Fundamental Logic) | D | 194 |
6b797ae1712e1f7b6b345723db1b332a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_20 | Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$ | Note that \[a^{4} + a^{-4} = (a^{2} + a^{-2})^{2} - 2.\] We guess that $a^{2} + a^{-2}$ is an integer, so the answer must be $2$ less than a perfect square. The only possibility is $\boxed{194}.$ | D | 194 |
d6c5203c21b68c5504b7b886d48c2712 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21 | sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$ | We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.
Let $S$ be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length $s\sqrt{3}$ where $s$ is a side of the cube, the ratio of a side of the inner square to a side of the outer square is $\frac{1}{\sqrt{3}}$ (since the side of the outer square = the diagonal of the inner square). So we have $\frac{S}{24} = \left(\frac{1}{\sqrt{3}}\right)^2$ . Thus $S = 8\Rightarrow \mathrm{\boxed{8}$ | C | 8 |
d6c5203c21b68c5504b7b886d48c2712 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21 | sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$ | Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives \[l^2 + l^2 + l^2 = 2^2 = 4.\] Hence each face has surface area \[l^2 = \frac{4}{3} \ \text{square meters}.\] So the surface area of the inscribed cube is $6\cdot \frac43 = \boxed{8}$ square meters. | null | 8 |
d6c5203c21b68c5504b7b886d48c2712 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_21 | sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$ | First of all, it is pretty easy to see the length of each edge of the bigger cube is $2$ so the radius of the sphere is $1$ . We know that when a cube is inscribed in a sphere, assuming the edge length of the square is $x$ the radius can be presented as $\frac {\sqrt3x}{2} = 1$ , we can solve that $x=\frac {2\sqrt3}{3}$ and now we only need to apply the basic formula to find the surface and we got our final answer as $\frac {2\sqrt3}{3}*\frac {2\sqrt3}{3}*6=8$ and the final answer is $\boxed{8}$ | C | 8 |
d6d3c0d5a98c7068c3a9ab4b9f0e8329 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_22 | A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$
$\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43$ | A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$ , so $S$ must be divisible by $37\ \mathrm{(D)}$ . To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{37}$ | D | 37 |
eb0e421da4d88af599164ae1dcf2581b | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_23 | How many ordered pairs $(m,n)$ of positive integers , with $m \ge n$ , have the property that their squares differ by $96$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$ | Find all of the factor pairs of $96$ $(1,96),(2,48),(3,32),(4,24),(6,16),(8,12).$ You can eliminate $(1,96)$ and ( $3,32)$ because you cannot have two numbers add to be an even number and have an odd difference at the same time without them being a decimal. You only have $4$ pairs left, so the answer is $\boxed{4}$ | B | 4 |
ac695a86271758fc49833bfda40e1217 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | It is well-known that $n \equiv S(n)\equiv S(S(n)) \pmod{9}.$ Substituting, we have that \[n+n+n \equiv 2007 \pmod{9} \implies n \equiv 0 \pmod{3}.\] Since $n \leq 2007,$ we must have that $\max S(n)=1+9+9+9=28.$ Now, we list out the possible vales for $S(n)$ in a table, noting that it is a multiple of $3$ because $n$ is a multiple of $3.$
Then, we compute the corresponding values of $S(S(n)).$
Finally, we may compute the corresponding values of $n$ using the fact that $n=2007-S(n)-S(S(n)).$
Notice how all conditions are designed to be satisfied except whether $S(n)$ is accurate with respect to $n.$ So, the only thing that remains is to check this. We may eliminate, for example, when $n=2007$ we have $S(n)=9$ while the table states that it is $0.$ Proceeding similarly, we obtain the following table.
It follows that there are $\boxed{4}$ possible values for $n.$ ~samrocksnature | D | 4 |
ac695a86271758fc49833bfda40e1217 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$
Proof of claim:
We examine the positive integers mod $9$ . Here are the cases.
Case 1. $n \equiv 1 \pmod 9$ . Now, we examine $S(n)$ modulo $9$ .
Case 1.1. The tens digit of $n$ is different from the tens digit of the largest multiple of $9$ under $n$ . (In other words, this means we will carry when adding from the perfect multiple of $9$ under $n$ .)
Observe that when we carry, i.e. Add $1$ onto $1989$ to obtain $1990$ , the units digit decreases by $9$ while the tens digit increases by $1$ . This means that the sum of the digits decreases by $8$ in total, and we have $-8 \equiv 1 \pmod 9$ , so the "mod 9" of the sum increases by $1$ . This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by $1$
Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by $1$ which means that the sum is also equivalent to $1 \pmod 9$
This means that $S(n) \equiv 1 \pmod 9$ and similarly letting the next $n=S(n)$ $S(S(n)) \equiv 1 \pmod 9$ . Summing these, we have $n+S(n)+S(S(n)) \equiv 3 \pmod 9$ . Clearly, no integers of this form will satisfy the condition because $2007$ is a perfect multiple of $9$
Case 2. $n \equiv 2 \pmod 9$
In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to $2$ mod $9$ . Then we can apply similar arguments to get $S(n) \equiv 2 \pmod 9$ and $S(S(n)) \equiv 2 \pmod 9$ , so adding gives $n+S(n)+S(S(n)) \equiv 6 \pmod 9$
It is trivial to see that for $n \equiv k \pmod 9$ , for $0 \leq k \leq 8$ , we must have $n+S(n)+S(S(n)) \equiv 3k \pmod 9$ . Only when $k=0, 3, 6$ is $3k$ a multiple of $9$ , which means that $n$ must be a multiple of $3$
Now, we find the integers. Again, consider two cases: Integers that are direct multiples of $9$ and integers that are multiples of $3$ but not $9$
Case 1. $n$ is a multiple of $9$ . An integer of the form $\overline{20ab}$ will not work since the least such integer is $2007$ which already exceeds our bounds. Thus, we need only consider the integers of the form $\overline{19ab}$ . The valid sums of the digits of $n$ are $18$ and $27$ in this case.
Case 1.1. The sum of the digits is $18$ . This means that $S(n)=18, S(S(n))=9$ , so $n=2007-18-9=1980$ . Clearly this number satisfies our constraints.
Case 1.2. The sum of the digits is $27$ . This means that $S(n)=27, S(S(n))=9$ , ,so $n=2007-27-9=1971$ . Since the sum of the digits of $1971$ is not $27$ , this does not work.
This means that there is $1$ integer in this case.
Case 2. $n$ is a multiple of $3$ , not $9$ .
.
Case 2.1. Integers of the form $\overline{20ab}$ . Then $S(n)=3$ or $S(n)=6$ ; it is trivial to see that $S(n)=6$ exceeds our bounds, so $S(n)=3$ and $n=2007-6=2001$
Case 2.2. Integers of the form $\overline{19ab}$ . Then $S(n)=12, 15, 21, 24$ and we consider each case separately.
Case 2.2.1. Integers with $S(n)=12$ . That means $n=2007-12-3=1992$ which clearly does not work.
Case 2.2.2. Integers with $S(n)=15$ . That means $n=2007-15-6=1986$ which also does not work
Case 2.2.3. Integers with $S(n)=21$ . That means $n=2007-21-3=1983$ which is valid.
Case 2.2.4. Integers with $S(n)=24$ . That means $n=2007-24-6=1977$ which is also valid.
We have considered every case, so there are $\boxed{4}$ integers that satisfy the given condition. | null | 4 |
ac695a86271758fc49833bfda40e1217 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10A_Problems/Problem_25 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Let the number of digits of $n$ be $m$ . If $m = 5$ $n$ will already be greater than $2007$ . Notice that $S(n)$ is always at most $9m$ . Then if $m = 3$ $n$ will be at most $999$ $S(n)$ will be at most $27$ , and $S(S(n))$ will be even smaller than $27$ . Clearly we cannot reach a sum of $2007$ , unless $m = 4$ (i.e. $n$ has $4$ digits).
Then, let $n$ be a four digit number in the form $1000a + 100b + 10c + d$ . Then $S(n) = a + b + c + d$
$S(S(n))$ is the sum of the digits of $a + b + c + d$ . We can represent $S(S(n))$ as the sum of the tens digit and the ones digit of $S(n)$ . The tens digit in the form of a decimal is
$\frac{a + b + c + d}{10}$
To remove the decimal portion, we can simply take the floor of the expression,
$\lfloor\frac{a + b + c + d}{10}\rfloor$
Now that we have expressed the tens digit, we can express the ones digit as $S(n) -10$ times the above expression, or
$a + b + c + d - 10\lfloor\frac{a + b + c + d}{10}\rfloor$
Adding the two expressions yields the value of $S(S(n))$
$= a + b + c + d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$
Combining this expression to the ones for $n$ and $S(n)$ yields
$1002a + 102b + 12c + 3d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$
Setting this equal to $2007$ and rearranging a bit yields
$12c + 3d = 2007 - 1002a - 102b + 9\lfloor\frac{a + b + c + d}{10}\rfloor$
$\Rightarrow$ $4c + d = 669 - 334a - 34b + 3\lfloor\frac{a + b + c + d}{10}\rfloor$
(The reason for this slightly weird arrangement will soon become evident)
Now we examine the possible values of $a$ . If $a \ge 3$ $n$ is already too large. $a$ must also be greater than $0$ , or $n$ would be a $3$ -digit number. Therefore, $a = 1 \, \text{or} \, 2$ . Now we examine by case.
If $a = 2$ , then $b$ and $c$ must both be $0$ (otherwise $n$ would already be greater than $2007$ ). Substituting these values into the equation yields
$d = 1 + 3\lfloor\frac{2 + d}{10}\rfloor$
$\Rightarrow$ $d=1$
Sure enough, $2001 + (2+1) + 3=2007$
Now we move onto the case where $a = 1$ . Then our initial equation simplifies to
$4c + d = 335 - 34b + 3\lfloor\frac{1 + b + c + d}{10}\rfloor$
Since $c$ and $d$ can each be at most $9$ , we substitute that value to find the lower bound of $b$ . Doing so yields
$34b \ge 290 + 3\lfloor\frac{19 + b}{10}\rfloor$
The floor expression is at least $3\lfloor\frac{19}{10}\rfloor=3$ , so the right-hand side is at least $293$ . Solving for $b$ , we see that $b \ge 9$ $\Rightarrow$ $b=9$ . Again, we substitute for $b$ and the equation becomes
$4c + d = 29 + 3\lfloor\frac{10 + c + d}{10}\rfloor$
$\Rightarrow$ $4c + d = 32 + 3\lfloor\frac{c + d}{10}\rfloor$
Just like we did for $b$ , we can find the lower bound of $c$ by assuming $d = 9$ and solving:
$4c + 9 \ge 29 + 3\lfloor\frac{c + 9}{10}\rfloor$
$\Rightarrow$ $4c \ge 20 + 3\lfloor\frac{c + 9}{10}\rfloor$
The right hand side is $20$ for $c=0$ and $23$ for $c \ge 1$ . Solving for c yields $c \ge 6$ . Looking back at the previous equation, the floor expression is $0$ for $c+d \le 9$ and $3$ for $c+d \ge 10$ . Thus, the right-hand side is $32$ for $c+d \le 9$ and $35$ for $c+d \ge 10$ . We can solve these two scenarios as systems of equations/inequalities:
$4c+d = 32$
$c+d \le 9$
and
$4c+d=35$
$c+d \ge 10$
Solving yields three pairs $(c, d):$ $(8, 0)$ $(8, 3)$ ; and $(7, 7)$ . Checking the numbers $1980$ $1983$ , and $1977$ ; we find that all three work. Therefore there are a total of $4$ possibilities for $n$ $\Rightarrow$ $\boxed{4}$ | D | 4 |
a0dc78413a36ff7c90ff8987b7a605e6 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_1 | Isabella's house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?
$\mathrm{(A)}\ 678 \qquad \mathrm{(B)}\ 768 \qquad \mathrm{(C)}\ 786 \qquad \mathrm{(D)}\ 867 \qquad \mathrm{(E)}\ 876$ | There are four walls in each bedroom (she can't paint floors or ceilings). Therefore, we calculate the number of square feet of walls there is in one bedroom: \[2\cdot(12\cdot8+10\cdot8)-60=2\cdot176-60=292\] We have three bedrooms, so she must paint $292\cdot3=\boxed{876}$ square feet of walls. | E | 876 |
6fe581997f9fcce6dca58809d836d548 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_2 | Define the operation $\star$ by $a \star b = (a+b)b.$ What is $(3 \star 5) - (5 \star 3)?$
$\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16$ | Substitute and simplify. \[(3+5)5 - (5+3)3 = (3+5)2 = 8\cdot2 = \boxed{16}\] | E | 16 |
6fe581997f9fcce6dca58809d836d548 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_2 | Define the operation $\star$ by $a \star b = (a+b)b.$ What is $(3 \star 5) - (5 \star 3)?$
$\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16$ | Note that $(a \star b) - (b \star a) = (a+b)b - (b+a)a= (a+b)(b-a)= b^2 - a^2$ . We can substitute $a=3$ and $b=5$ to get $5^2 - 3^2 = \boxed{16}$ | E | 16 |
c853b5378b2e6e67b840052209757637 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_3 | A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28$ | The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \boxed{24}$ | B | 24 |
c853b5378b2e6e67b840052209757637 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_3 | A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28$ | Alternatively, we can use the harmonic mean to get $\frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{1}{12}} = \boxed{24}$ | B | 24 |
61c3a0c6dd631693bd198d6c63ea5bb7 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_4 | The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC=120^\circ$ and $\angle AOB=140^\circ,$ as shown. What is the degree measure of $\angle ABC?$
$\textbf{(A) } 35 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 50 \qquad\textbf{(E) } 60$ | Because all the central angles of a circle add up to $360^\circ,$
\begin{align*} \angle BOC + \angle AOB + \angle AOC &= 360\\ 120 + 140 + \angle AOC &= 360\\ \angle AOC &= 100. \end{align*}
Therefore, the measure of $\text{arc}AC$ is also $100^\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, $\angle ABC = \boxed{50}$ | D | 50 |
990c24664c6cad8c3e213c6f7635d506 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_6 | The $2007 \text{ AMC }10$ will be scored by awarding $6$ points for each correct response, $0$ points for each incorrect response, and $1.5$ points for each problem left unanswered. After looking over the $25$ problems, Sarah has decided to attempt the first $22$ and leave only the last $3$ unanswered. How many of the first $22$ problems must she solve correctly in order to score at least $100$ points?
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Sarah is leaving $3$ questions unanswered, guaranteeing her $3 \times 1.5 = 4.5$ points. She will either get $6$ points or $0$ points for the rest of the questions. Let $x$ be the number of questions Sarah answers correctly. \begin{align*} 6x+4.5 &\ge 100\\ 6x &\ge 95.5\\ x &\ge 15.92 \end{align*} The number of questions she answers correctly has to be a whole number, so round up to get $\boxed{16}$ | D | 16 |
e629005d1169ac5c1b94bd191ba519c4 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_7 | All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A= \angle B = 90^\circ.$ What is the degree measure of $\angle E?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 150$ | $AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\triangle CED$ is an equilateral triangle, $\angle AEC = 90$ and $\angle CED = 60.$ Use angle addition: \[\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{150}\] | E | 150 |
cf969a74091874dc74c9d59d386d852c | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_8 | CHIKEN NUGGIEs | Case $1$ : The numbers are separated by $1$
We this case with $a=0, b=1,$ and $c=2$ . Following this logic, the last set we can get is $a=7, b=8,$ and $c=9$ . We have $8$ sets of numbers in this case.
Case $2$ : The numbers are separated by $2$
This case starts with $a=0, b=2,$ and $c=2$ . It ends with $a=5, b=7,$ and $c=9$ . There are $6$ sets of numbers in this case.
Case $3$ : The numbers start with $a=0, b=3,$ and $c=6$ . It ends with $a=3, b=6,$ and $c=9$ . This case has $4$ sets of numbers.
It's pretty clear that there's a pattern: $8$ sets, $6$ sets, $4$ sets. The amount of sets per case decreases by $2$ , so it's obvious Case $4$ has $2$ sets. The total amount of possible five-digit numbers is $8+6+4+2=\boxed{20}$ | D | 20 |
d5d1491f0655b8dd1a5d0c10f287861a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_12 | Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$ | Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{5} Note that actual values were not found. | D | 5 |
6c8ff7468e043e0181d3c38f3b2f5210 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_14 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12$ | If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\%$ of the group are girls, \begin{align*} \frac{0.4p-2}{p}&=\frac{3}{10}\\ 4p-20&=3p\\ p&=20\end{align*} The number of girls initially in the group is $0.4p=0.4(20)=\boxed{8}$ | C | 8 |
6c8ff7468e043e0181d3c38f3b2f5210 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_14 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12$ | Let $x$ be the number of people initially in the group and $g$ the number of girls. $\frac{2}{5}x = g$ , so $x = \frac{5}{2}g$ . Also, the problem states $\frac{3}{10}x = g-2$ . Substituting $x$ in terms of $g$ into the second equation yields that $g = \boxed{8}$ | C | 8 |
7467db28da35b9131dfd1d45210d40e1 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_15 | The angles of quadrilateral $ABCD$ satisfy $\angle A=2 \angle B=3 \angle C=4 \angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?
$\textbf{(A) } 125 \qquad\textbf{(B) } 144 \qquad\textbf{(C) } 153 \qquad\textbf{(D) } 173 \qquad\textbf{(E) } 180$ | The sum of the interior angles of any quadrilateral is $360^\circ.$ \begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*} \[\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \boxed{173}\] | D | 173 |
abaed81767d88aa471e0c98a53c70f1c | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_16 | A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$ | We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \cdot 83 = 747$ . The only score that has not been added to that is the junior's score, which is $840 - 747 = \boxed{93}$ | C | 93 |
abaed81767d88aa471e0c98a53c70f1c | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_16 | A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$ | Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\frac{j}{10} + \frac{83 \cdot 9}{10} = 84$ . Simplifying this equation yields $j = \boxed{93}$ | C | 93 |
3a8554ef6bb8c95114fae8814e3edf80 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_20 | A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
$\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$ | There are $25$ ways to choose the first square. The four remaining squares in its row and column and the square you chose exclude nine squares from being chosen next time.
There are $16$ remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.
Finally, the last square has $9$ remaining choices.
The number of ways to choose $3$ squares is $25 \cdot 16 \cdot 9,$ but the order in which you chose the squares does not matter as the blocks are indistinguishable, so we divide by $3!$
\[\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{600}\] | C | 600 |
3a8554ef6bb8c95114fae8814e3edf80 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_20 | A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
$\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$ | Once we choose our three squares, we will have occupied three separate columns $(A, B, C)$ and three separate rows. There are ${5 \choose 3} \times {5 \choose 3}$ ways to choose these rows and columns.
There are $3$ ways to assign the square in column $A$ to a row, $2$ ways to assign the square in column $B$ to one of the remaining two rows, and poor square in column $C$ doesn't get to choose. $:($
In total, we have \[{5 \choose 3} \times {5 \choose 3} \times 3!\] which totals out to $\boxed{600}$ | C | 600 |
3a8554ef6bb8c95114fae8814e3edf80 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_20 | A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
$\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$ | We know that there are $\binom{25}{3}=2300$ ways to choose three blocks. However, the restriction clearly limits the number of ways we can choose our blocks. Hence, only $\text{(A)}$ $\text{(B)}$ , or $\text{(C)}$ could be the correct answer. Clearly, there are more than $125$ ways, thus yielding $\boxed{600}$ ways. | C | 600 |
d97f2bbf66fc6165c7ab9e0cd4c9ebe8 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_24 | Let $n$ denote the smallest positive integer that is divisible by both $4$ and $9,$ and whose base- $10$ representation consists of only $4$ 's and $9$ 's, with at least one of each. What are the last four digits of $n?$
$\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qquad\textbf{(E) } 9944$ | For a number to be divisible by $4,$ the last two digits have to be divisible by $4.$ That means the last two digits of this integer must be $4.$
For a number to be divisible by $9,$ the sum of all the digits must be divisible by $9.$ The only way to make this happen is with nine $4$ 's. However, we also need one $9.$
The smallest integer that meets all these conditions is $4444444944$ . The last four digits are $\boxed{4944}$ | C | 4944 |
56e12a3b129d97ea4b4e367304c59640 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_25 | How many pairs of positive integers $(a,b)$ are there such that $a$ and $b$ have no common factors greater than $1$ and:
\[\frac{a}{b} + \frac{14b}{9a}\]
is an integer?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many}$ | Let $\frac{a}{b} = x$ and $\frac{14b}{9a} = y.$ Then for rational numbers $x,y$ we have $xy = \frac{14}{9}$ and $x+y \in Z.$ One can prove that $x$ and $y$ must both have the same denominator in order for their sum to be an integer. This denominator is $3$ meaning $b=3$ and $a$ is any factor of $14.$ This yields $\boxed{4}$ solutions.
~ab2024 | A | 4 |
2e80d0c28902e256af27aa3005603609 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_1 | Sandwiches at Joe's Fast Food cost $$3$ each and sodas cost $$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?
$\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35$ | The $5$ sandwiches cost $5\cdot 3=15$ dollars. The $8$ sodas cost $8\cdot 2=16$ dollars. In total, the purchase costs $15+16=\boxed{31}$ dollars. | A | 31 |
d8a7a6b1000f3bab88fa278198df80cd | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_3 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=\boxed{18}.$ | B | 18 |
d8a7a6b1000f3bab88fa278198df80cd | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_3 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\div5=6$ , and $6\cdot3=\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\cdot8=30+18$ . Therefore, we can see our answer is correct. | B | 18 |
0f0763a1eecf81fe1fb11ad22d6a5b82 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_4 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{23}$ | E | 23 |
0f0763a1eecf81fe1fb11ad22d6a5b82 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_4 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$ .
For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$ , and finally $14+9=\boxed{23}$ | E | 23 |
0f0763a1eecf81fe1fb11ad22d6a5b82 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_4 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$
Next, we look at the value of the minutes section, which varies from $00-59$ . Let this value be a number $ab$ . We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$
Adding these up, we get $9+5+9=\boxed{23}$ | E | 23 |
00a67d27f3f850289722347f2027bee1 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_5 | Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=\boxed{4}$ more dollars than Doug. | D | 4 |
62397876314eb6c102b100f48a943e14 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_6 | What non-zero real value for $x$ satisfies $(7x)^{14}=(14x)^7$
$\textbf{(A) } \frac17\qquad \textbf{(B) } \frac27\qquad \textbf{(C) } 1\qquad \textbf{(D) } 7\qquad \textbf{(E) } 14$ | Taking the seventh root of both sides, we get $(7x)^2=14x$
Simplifying the LHS gives $49x^2=14x$ , which then simplifies to $7x=2$
Thus, $x=\boxed{27}$ | B | 27 |
29eefa683717b1a890f8f6e5655b231a | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_7 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$A$",A,W); label("$B$",B,NE); label("$C$",(12.6,4)); label("$D$",D,SW); label("$12$",E--B,N); label("$12$",D--F,S); label("$4$",E--A,W); label("$4$",(12.4,-1.75),E); label("$8$",A--D,W); label("$8$",(12.4,4),E); label("$y$",A--H,S); label("$y$",G--C,N); [/asy]
As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{6}$ | A | 6 |
29eefa683717b1a890f8f6e5655b231a | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_7 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$y$",A--H,S); label("$y$",G--C,N); [/asy]
As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$y$",(9,-2),NW); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{6}$ | A | 6 |
29eefa683717b1a890f8f6e5655b231a | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_7 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\boxed{6}$ | A | 6 |
3a7382bab04e196dc30ac84761e2163f | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | Substitute the points $(2,3)$ and $(4,3)$ into the given equation for $(x,y)$
Then we get a system of two equations:
$3=4+2b+c$
$3=16+4b+c$
Subtracting the first equation from the second we have:
$0=12+2b$
$b=-6$
Then using $b=-6$ in the first equation:
$0=1+-12+c$
$c=\boxed{11}$ | E | 11 |
3a7382bab04e196dc30ac84761e2163f | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely $(3,2)$ . Thus, the form of the equation of the parabola is $y - 2 = (x - 3)^2$ . Expanding this out, we find that $c=\boxed{11}$ | E | 11 |
3a7382bab04e196dc30ac84761e2163f | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | The points given have the same $y$ -value, so the vertex lies on the line $x=\frac{2+4}{2}=3$
The $x$ -coordinate of the vertex is also equal to $\frac{-b}{2a}$ , so set this equal to $3$ and solve for $b$ , given that $a=1$
$x=\frac{-b}{2a}$
$3=\frac{-b}{2}$
$6=-b$
$b=-6$
Now the equation is of the form $y=x^2-6x+c$ . Now plug in the point $(2,3)$ and solve for $c$
$y=x^2-6x+c$
$3=2^2-6(2)+c$
$3=4-12+c$
$3=-8+c$
$c=\boxed{11}$ | E | 11 |
3a7382bab04e196dc30ac84761e2163f | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_8 | parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ . What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | Substituting y into the two equations, we get:
$3=x^2+bx+c$
Which can be written as:
$x^2+bx+c-3=0$
$4$ and $2$ are the solutions to the quadratic. Thus:
$c-3=4\times2$
$c-3=8$
$c=\boxed{11}$ | E | 11 |
0a1e95295793f5bd5857101f243d31b5 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_9 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:
If the number of integers in the list is even, then the average will have a $\frac{1}{2}$ . The only possibility is $\frac{15}{2}$ , from which we get:
Thus, the correct answer is $\boxed{3}.$ | C | 3 |
0a1e95295793f5bd5857101f243d31b5 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_9 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Any set will form a arithmetic progression with the first term say $a$ . Since the numbers are consecutive the common difference $d = 1$
The sum of the AP has to be 15. So,
\[S_n = \frac{n}{2} \cdot (2a + (n-1)d)\] \[S_n = \frac{n}{2} \cdot (2a + (n-1)1)\] \[15 = \frac{n}{2} \cdot (2a + n - 1)\] \[2an + n^2 - n = 30\] \[n^2 + n(2a - 1) - 30 = 0\]
Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now $a$ cannot be 15 as we need 2 terms. So a can only be less the 15.
Trying all the values of a from 1 to 14 we observe that $a = 4$ $a = 7$ and $a = 1$ provide the only real solutions to the above equation.The three possibilites of a and n are.
\[(a,n) = (4,3),(7, 2),(1, 6)\]
The above values are obtained by solving the following equations obtained by substituting the above mentioned values of a into $n^2 + n(2a - 1) - 30 = 0$
\[n^2 + 7n - 30 = 0\] \[n^2 + 13n - 30 = 0\] \[n^2 - n - 30 = 0\]
Since there are 3 possibilities the answer is $\boxed{3}.$ | C | 3 |
420116e9c489fd04bb00887a13398584 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_10 | For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?
$\textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square.
Since $\sqrt{x}$ can't be negative, $120-\sqrt{x} \leq 120$
The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$ , so there are $11$ values for $120-\sqrt{x}$
Since every value of $120-\sqrt{x}$ gives one and only one possible value for $x$ , the number of values of $x$ is $\boxed{11}$ | E | 11 |
bad1d129635ebabb0bb4c24989514397 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_12 | Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.
[asy] size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,-8)--(24,-8)); label('Dog', (24, -8), SE); MP('I', (8,-8), (0,0)); MP('8', (16,-4), W); MP('8', (16,-12),W); MP('8', (20,-8), N); label('Rope', (20,-8),S); D((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle); D((16,-24)--(24,-24)); MP("II", (8,-28), (0,0)); MP('4', (16,-22), W); MP('8', (20,-24), N); label("Dog",(24,-24),SE); label("Rope", (20,-24), S); [/asy]
Which of these arrangements give the dog the greater area to roam, and by how many square feet?
$\textbf{(A) } I,\,\textrm{ by }\,8\pi\qquad \textbf{(B) } I,\,\textrm{ by }\,6\pi\qquad \textbf{(C) } II,\,\textrm{ by }\,4\pi\qquad \textbf{(D) } II,\,\textrm{ by }\,8\pi\qquad \textbf{(E) } II,\,\textrm{ by }\,10\pi$ | [asy] size(150); pathpen = linewidth(0.7); defaultpen(linewidth(0.7)+fontsize(10)); filldraw(arc((16,-8),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-26),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-22),4,90,180)--(16,-22)--cycle, rgb(0.9,0.9,0.6)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,-8)--(24,-8)); label('Dog', (24, -8), SE); MP('I', (8,-8), (0,0)); MP('8', (16,-4), W); MP('8', (16,-12),W); MP('8', (20,-8), N); label('Rope', (20,-8),S); pair sD = (0,-2); D(shift(sD)*((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle)); D(shift(sD)*((16,-24)--(24,-24))); MP("II", (8,-28)+sD, (0,0)); MP('4', (16,-22)+sD, W); MP('8', (20,-24)+sD, N); label("Dog",(24,-24)+sD,SE); label("Rope", (20,-24)+sD, S); [/asy] Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement $I$ allows the dog $\frac12\cdot(\pi\cdot8^2) = 32\pi$ square feet of area. Arrangement $II$ allows $32\pi$ square feet plus a little more on the top part of the fence. So we already know that Arrangement $II$ allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is $\frac14\cdot(\pi\cdot4^2) = 4\pi$ . Thus the answer is $\boxed{4}$ | C | 4 |
6dd74213d344024995d895dc1e663584 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_13 | A player pays $\textdollar 5$ to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)
$\textbf{(A) } \textdollar12\qquad\textbf{(B) } \textdollar30\qquad\textbf{(C) } \textdollar50\qquad\textbf{(D) } \textdollar60\qquad\textbf{(E) } \textdollar 100\qquad$ | The probability of rolling an even number on the first turn is $\frac{1}{2}$ and the probability of rolling the same number on the next turn is $\frac{1}{6}$ . The probability of winning is $\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}$ . If the game is to be fair, the amount paid, $5$ dollars, must be $\frac{1}{12}$ the amount of the prize money, so the answer is $\boxed{60}.$ | D | 60 |
f61b38cd82dd6149e6ce4021a24a88b1 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_14 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]
$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$ | The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\frac{18 \cdot 19}{2} + 2 = \boxed{173}$ | B | 173 |
f61b38cd82dd6149e6ce4021a24a88b1 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_14 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]
$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$ | Alternatively, the sum of the consecutive integers from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \boxed{173}$ | B | 173 |
e193e6e46d2db81540cf168cc9645251 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
$\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad$ | Since $d = rt$ , we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 =\boxed{47}$ meeting points. | D | 47 |
e193e6e46d2db81540cf168cc9645251 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
$\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad$ | We first find the amount of minutes, $k$ , until Odell and Kershaw's next meeting. Let $a$ be the angle in radians between their starting point and the point where they first meet, measured counterclockwise.
Since Kershaw has traveled $300k$ meters at this point and the circumference of his track is $120\pi$ $a=\frac{300k}{120\pi}\cdot 2\pi$ . Similarly, $2\pi-a=\frac{250k}{100\pi}\cdot{2\pi}$ since Odell has traveled $250k$ meters in the opposite direction and the circumference of his track is $100\pi$
Solving for $a$ in the second equation, we get $a=2\pi-\frac{250k}{100\pi}\cdot 2\pi$ . Then, from the first equation, we have $\frac{300k}{120\pi}\cdot 2\pi=2\pi-\frac{250k}{100\pi}\cdot 2\pi$ . Solving for $k$ , we get $k=\frac{\pi}{5}$ . After $k$ minutes, they are back at the same position, except rotated, so they will meet again in $k$ minutes. So the total amount of meetings is $\lfloor\frac{30}{k}\rfloor=\lfloor\frac{150}{\pi}\rfloor=\boxed{47}$ | D | 47 |
e193e6e46d2db81540cf168cc9645251 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_15 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
$\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad$ | Since Odell's rate is $5/6$ that of Kershaw, but Kershaw's lap distance is $6/5$ that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs \[(30 \ \text{min})\left(250\frac{\text{m}}{\text{min}}\right)\left(\frac{1}{100\pi}\frac{\text{laps}}{\text{m}}\right)= \frac{75}{\pi}\,\text{laps}\approx 23.87\ \text{laps},\] as does Kershaw. Because $23.5 < 23.87 < 24$ , they pass each other $2(23.5)=\boxed{47}$ times. | null | 47 |
a565a47307faf7e8f75077ba58925d2b | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_17 | In rectangle $ADEH$ , points $B$ and $C$ trisect $\overline{AD}$ , and points $G$ and $F$ trisect $\overline{HE}$ . In addition, $AH=AC=2$ , and $AD=3$ . What is the area of quadrilateral $WXYZ$ shown in the figure?
[asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); D(A--D--E--H--cycle); [/asy]
$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{\sqrt{2}}{2}\qquad\textbf{(C) } \frac{\sqrt{3}}{2}\qquad\textbf{(D) } \sqrt{2} \qquad\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad$ | By symmetry $WXYZ$ is a square.
[asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); MP("1",(A+B)/2,2*N); MP("2",(A+H)/2,plain.W); D(B--Z); MP("1",(B+Z)/2,plain.W); MP("\frac{\sqrt{2}}{2}",(W+Z)/2,plain.SE); D(A--D--E--H--cycle); [/asy]
Draw $\overline{BZ}$ $BZ = \frac 12AH = 1$ , so $\triangle BWZ$ is a $45-45-90 \triangle$ . Hence $WZ = \frac{1}{\sqrt{2}}$ , and $[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 =\boxed{12}$ | A | 12 |
a565a47307faf7e8f75077ba58925d2b | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_17 | In rectangle $ADEH$ , points $B$ and $C$ trisect $\overline{AD}$ , and points $G$ and $F$ trisect $\overline{HE}$ . In addition, $AH=AC=2$ , and $AD=3$ . What is the area of quadrilateral $WXYZ$ shown in the figure?
[asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); D(A--D--E--H--cycle); [/asy]
$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{\sqrt{2}}{2}\qquad\textbf{(C) } \frac{\sqrt{3}}{2}\qquad\textbf{(D) } \sqrt{2} \qquad\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad$ | [asy] size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); D(B--D((A+H)/2)--G);D(C--D((E+D)/2)--F); D(A--D--E--H--cycle); [/asy]
Drawing lines as shown above and piecing together the triangles, we see that $ABCD$ is made up of $12$ squares congruent to $WXYZ$ . Hence $[WXYZ] = \frac{2\cdot 3}{12} =\boxed{12}$ | A | 12 |
dd718c56c9cf9e22a0a56384a99d3f77 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_19 | How many non- similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$ | The sum of the angles of a triangle is $180$ degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it $\frac{180}{3} = 60$ degrees. The minimum possible value for the smallest angle is $1$ and the highest possible is $59$ (since the numbers are distinct), so there are $\boxed{59}$ possibilities. | C | 59 |
dd718c56c9cf9e22a0a56384a99d3f77 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_19 | How many non- similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$ | Let the first angle be $x$ , and the common difference be $d$ . The arithmetic progression can now be expressed as $x + (x + d) + (x + 2d) = 180$ . Simplifiying, $x + d = 60$ . Now, using stars and bars, we have $\binom{61}{1} = 61$ .
However, we must subtract the two cases in which either $x$ or $d$ equal $0$ , so we have $61 - 2$ $\boxed{59}$ | C | 59 |
dd718c56c9cf9e22a0a56384a99d3f77 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_19 | How many non- similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$ | Consider that we have $(a+n)+(a+n+1)+(a+n+2)=180 \Longleftrightarrow 3a+3(n+1)=180 \Longleftrightarrow a=59-n$ , where $n \geq 0$ and $n$ is an integer. Since $a \neq 0$ $n=0,1,2,3,\cdots, 58$ which is $\boxed{59}$ solutions. | C | 59 |
a50e5a6e7885496c365d22ca13bfe575 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_20 | Six distinct positive integers are randomly chosen between $1$ and $2006$ , inclusive. What is the probability that some pair of these integers has a difference that is a multiple of $5$
$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{3}{5}\qquad\textbf{(C) } \frac{2}{3}\qquad\textbf{(D) } \frac{4}{5}\qquad\textbf{(E) } 1\qquad$ | For two numbers to have a difference that is a multiple of $5$ , the numbers must be congruent $\bmod{5}$ (their remainders after division by $5$ must be the same).
$0, 1, 2, 3, 4$ are the possible values of numbers in $\bmod{5}$ . Since there are only $5$ possible values in $\bmod{5}$ and we are picking $6$ numbers, by the Pigeonhole Principle , two of the numbers must be congruent $\bmod{5}$
Therefore the probability that some pair of the $6$ integers has a difference that is a multiple of $5$ is $\boxed{1}$ | E | 1 |
b36e53894fd42003fe068f1aec8a38d0 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_22 | problem_id
b36e53894fd42003fe068f1aec8a38d0 Two farmers agree that pigs are worth $300$ do...
b36e53894fd42003fe068f1aec8a38d0 Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD ( greatest common divisor ) of $a$ and $b$ . Therefore, the answer is $gcd(300,210)=\boxed{30}.$ | C | 30 |
b36e53894fd42003fe068f1aec8a38d0 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_22 | problem_id
b36e53894fd42003fe068f1aec8a38d0 Two farmers agree that pigs are worth $300$ do...
b36e53894fd42003fe068f1aec8a38d0 Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. Since our supposition can be achieved, the answer is $\boxed{30}$ | C | 30 |
6d0d7ad563b09add424eae19045b7a55 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_1 | What is $(-1)^{1} + (-1)^{2} + ... + (-1)^{2006}$
$\textbf{(A)} -2006\qquad \textbf{(B)} -1\qquad \textbf{(C) } 0\qquad \textbf{(D) } 1\qquad \textbf{(E) } 2006$ | Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$
$(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{0}.$ | C | 0 |
e3548147f8a38545c783b72ab0f4ee93 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_2 | For real numbers $x$ and $y$ , define $x \spadesuit y = (x+y)(x-y)$ . What is $3 \spadesuit (4 \spadesuit 5)$
$\mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72$ | Since $x \spadesuit y = (x+y)(x-y)$
$3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{72}$ | A | 72 |
db4cea56c885baaab2b20af6a4b2ccb7 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_3 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of $34$ points, and the Cougars won by a margin of $14$ points. How many points did the Panthers score?
$\textbf{(A) } 10\qquad \textbf{(B) } 14\qquad \textbf{(C) } 17\qquad \textbf{(D) } 20\qquad \textbf{(E) } 24$ | Let $x$ be the number of points scored by the Cougars, and $y$ be the number of points scored by the Panthers. The problem is asking for the value of $y$ \begin{align*} x+y &= 34 \\ x-y &= 14 \\ 2x &= 48 \\ x &= 24 \\ y &= \boxed{10} | A | 10 |
2fe56fbfd74b3691d3e5e8d2ef85d302 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_4 | Circles of diameter $1$ inch and $3$ inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
2006amc10b04.gif
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9$ | The area painted red is equal to the area of the smaller circle and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.
So we have:
\begin{align*} A_{red}&=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}\\ A_{blue}&=\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{1}{2}\right)^2=2\pi\\ \end{align*}
So the desired ratio is:
$\frac{A_{blue}}{A_{red}}=\frac{2\pi}{\frac{\pi}{4}}=2\cancel{\pi}\cdot \frac{4}{\cancel{\pi}}=\boxed{8}.$ | D | 8 |
ad805bfdc511371c79b36d82c15731be | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_5 | $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?
$\textbf{(A) } 16\qquad \textbf{(B) } 25\qquad \textbf{(C) } 36\qquad \textbf{(D) } 49\qquad \textbf{(E) } 64$ | By placing the $2 \times 3$ rectangle adjacent to the $3 \times 4$ rectangle with the 3 side of the $2 \times 3$ rectangle next to the 4 side of the $3 \times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2 = 25$
Since placing the two rectangles inside a $4 \times 4$ square must result in overlap, the smallest possible area of the square is $25$
So the answer is $\boxed{25}$ | B | 25 |
70cfc6095f6ecca2eb7679cfdde4892f | https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_6 | A region is bounded by semicircular arcs constructed on the side of a square whose sides measure $\frac{2}{\pi}$ , as shown. What is the perimeter of this region?
[asy] size(90); defaultpen(linewidth(0.7)); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle,gray(0.5)); filldraw(arc((1,0),1,180,0, CCW)--cycle,gray(0.7)); filldraw(arc((0,1),1,90,270)--cycle,gray(0.7)); filldraw(arc((1,2),1,0,180)--cycle,gray(0.7)); filldraw(arc((2,1),1,270,90, CCW)--cycle,gray(0.7)); [/asy]
$\textbf{(A) } \frac{4}{\pi}\qquad \textbf{(B) } 2\qquad \textbf{(C) } \frac{8}{\pi}\qquad \textbf{(D) } 4\qquad \textbf{(E) } \frac{16}{\pi}$ | Since the side of the square is the diameter of the semicircle, the radius of the semicircle is $\frac{1}{2}\cdot\frac{2}{\pi}=\frac{1}{\pi}$
Since the length of one of the semicircular arcs is half the circumference of the corresponding circle, the length of one arc is $\frac{1}{2}\cdot2\cdot\pi\cdot\frac{1}{\pi}=1$
Since the desired perimeter is made up of four of these arcs, the perimeter is $4\cdot1=\boxed{4}$ | D | 4 |
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