Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: 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?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$

Solution: Since the mean of the first $n$ terms is $n$ , the sum of the first $n$ terms is $n^2$ . Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$ . Hence, the $2008^{\text{th}}$ term of the sequence is $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{4015}$

Answer: B 4015
Problem: 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?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$

Solution: Let $a_1, a_2, a_3, \cdots, a_n$ be the terms of the sequence. We know $\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = n$ , so we must have $a_1 + a_2 + a_3 + \cdots + a_n = n^2$ . The sum of consecutive odd numbers down to $1$ is a perfect square, if you don't believe me, try drawing squares with the sum, so $a_1 = 1, a_2 = 3, a_3 = 5, \cdots , a_n = 2(n-1) + 1$ , so the answer is $a_{2008} = 2(2007) + 1 = \boxed{4015}$

Answer: B 4015
Problem: 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?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$

Solution: 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 $4016-1=\boxed{4015}$

Answer: B 4015
Problem: 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?
$\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$

Solution: 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.

Answer: B 4015
Problem: How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

Solution: 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 $\boxed{6}$

Answer: A 6
Problem: 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?
$\mathrm{(A)}\ {{{0.063}}} \qquad \mathrm{(B)}\ {{{0.189}}} \qquad \mathrm{(C)}\ {{{0.233}}} \qquad \mathrm{(D)}\ {{{0.333}}} \qquad \mathrm{(E)}\ {{{0.441}}}$

Solution: 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 $3\cdot0.063=\boxed{0.189}$

Answer: B 0.189
Problem: 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?
$\mathrm{(A)}\ {{{0.063}}} \qquad \mathrm{(B)}\ {{{0.189}}} \qquad \mathrm{(C)}\ {{{0.233}}} \qquad \mathrm{(D)}\ {{{0.333}}} \qquad \mathrm{(E)}\ {{{0.441}}}$

Solution: 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: ${3\choose 1} \cdot(0.7)^1\cdot(1-0.7)^{(3-1)}=3\cdot0.7\cdot0.09=\boxed{0.189}$

Answer: B 0.189
Problem: 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?
$\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900$

Solution: 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 $\boxed{900}$

Answer: B 900
Problem: 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?
$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$

Solution: 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 $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{480}$

Answer: C 480
Problem: 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?
$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$

Solution: 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 $\boxed{480}$

Answer: C 480
Problem: 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?
$\mathrm{(A)}\ 1/12\qquad\mathrm{(B)}\ 1/10\qquad\mathrm{(C)}\ 1/6\qquad\mathrm{(D)}\ 1/3\qquad\mathrm{(E)}\ 1/2$

Solution: 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 $\frac{6+4(1)}{60}=\frac{10}{60}=\boxed{16}$

Answer: C 16
Problem: 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 $(a,b)$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: 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 $\boxed{2}$

Answer: B 2
Problem: Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

Solution: 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.

Answer:  85
Problem: Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

Solution: 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 $x=\boxed{85}$

Answer:  85
Problem: Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

Solution: [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 $\boxed{85}$

Answer: C 85
Problem: Quadrilateral $ABCD$ has $AB = BC = CD$ $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$ . What is the degree measure of $\angle BAD$
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

Solution: 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 $90$

Answer: C 85
Problem: 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?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$

Solution: 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.

Answer: B 5
Problem: 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?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$

Solution: 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 $\boxed{5}$

Answer: B 5
Problem: 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?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$

Solution: 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 $\boxed{5}$

Answer: B 5
Problem: 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?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$

Solution: 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 $\boxed{5}$

Answer: B 5
Problem: 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?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: 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 $\boxed{6}$

Answer: E 6
Problem: 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?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: 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.

Answer: E 6