Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$

Solution: Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=410.$
The value $y$ has to be greater than $82$ , because otherwise she would receive the same score on her last test. Additionally, the greatest value for $y$ is $98$ (as $y=100$ would make $x$ as a decimal), so therefore, the greatest value $x$ can be is $98$ . As a result, only $4$ numbers work, $86, 90, 94$ and $98$ . Thus, the answer is $\boxed{4}$

Answer: A 4
Problem: Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?
$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$

Solution: The average point is $82$ leads us to suppose that Laila got all $82$ points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests is $81$ points, then the last tests should be $86$ points to keep the average point is fixed. The possible points are $86$ $90$ $94$ $98$ . The answer is $\boxed{4}$

Answer: A 4
Problem: Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$
$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

Solution: If we start off with the first digit, we know that it can't be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$ . The next place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18\implies \boxed{18}$

Answer: D 18
Problem: Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$
$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

Solution: 120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
$(5)(4)(3)(2)(1) = 120$ Making the greatest integer,
$(5)(4 \cdot 2)(3)\left(\frac{2}{2}\right)(1)$ $= (5)(8)(3)(1)(1) =120$
8 is the largest value and will go in the front.
We can express the number as $85311$
$8+5+3+1+1=\boxed{18}$

Answer: D 18
Problem: In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$

Solution: Let the radius of the large circle be $R$ . Then, the radius of the smaller circles are $\frac R2$ . The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is $\frac 14$ . This means the combined area of the 2 smaller circles is half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is $\boxed{1}$

Answer: D 1
Problem: In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$

Solution: Let the radius of the two smaller circles be $r$ . It follows that the area of one of the smaller circles is ${\pi}r^2$ . Thus, the area of the two inner circles combined would evaluate to $2{\pi}r^2$ which is $1$ . Since the radius of the bigger circle is two times that of the smaller circles (the diameter), the radius of the larger circle in terms of $r$ would be $2r$ . The area of the larger circle would come to $(2r)^2{\pi} = 4{\pi}r^2$
Subtracting the area of the smaller circles from that of the larger circle (since that would be the shaded region), we have \[4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.\]
Therefore, the area of the shaded region is $\boxed{1}$

Answer: D 1
Problem: Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880$

Solution: Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $5$ factorial. Now, we multiply this product by $2!$ because there are $2!$ ways to arrange the Arabic books within themselves, and $4!$ ways to arrange the Spanish books within themselves. Multiplying all these together, we have $2! \cdot 4! \cdot 5!=\boxed{5760}$

Answer: C 5760
Problem: Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?
$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$

Solution: Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet $10560\div60=176$ times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, $1760\div2.5=\boxed{704}$

Answer: A 704
Problem: Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?
$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$

Solution: We know that Bella goes 2.5 feet per step and since Ella rides 5 times faster than Bella she must go 12.5 feet on her bike for every step of Bella's. For Bella, it takes 4,224 steps, and for Ella, it takes 1/5th those steps since Ella goes 5 times faster than Bella, taking her 844.8 steps. The number of steps where they meet therefore must be less than 844.8. The only answer choice less than it is $\boxed{704}$

Answer: A 704
Problem: Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \tfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?
$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$

Solution: We can turn $2 \tfrac{1}{2}$ into an improper fraction. It will then become 5/2. Since Ella bikes 5 times faster, we multiply 5/2 by 5 to get 25/2. Then we add 5/2 to it in order to find the distance they walk and bike together in total. After adding, you should get 30/2 which is equal to 15. This means that after 15 times, they will meet. So you have to divide 10,560 by 15. The answer should be $\boxed{704}$

Answer: A 704
Problem: How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: We can first find the prime factorization of $23,232$ , which is $2^6\cdot3^1\cdot11^2$ . Now, we add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{42}$

Answer: E 42
Problem: How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: Observe that $69696$ $264^2$ , so this is $\frac{1}{3}$ of $264^2$ which is $88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3$ , which has $3 \cdot 7 \cdot 2 = 42$ factors. The answer is $\boxed{42}$

Answer: E 42
Problem: In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]
$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

Solution: You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\boxed{8}$

Answer: C 8
Problem: In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]
$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

Solution: The top box is fixed by the problem.
Choose the left 3 bottom-row boxes freely. There are $2^3=8$ ways.
Then the left 2 boxes on the row above are determined.
Then the left 1 box on the row above that is determined
Then the right 1 box on that row is determined.
Then the right 1 box on the row below is determined.
Then the right 1 box on the bottom row is determined, completing the diagram.
So the answer is $\boxed{8}$

Answer: C 8
Problem: In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]
$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

Solution: Let the plus sign represent 1 and the negative sign represent -1.
The four numbers on the bottom are $a$ $b$ $c$ , and $d$ , which are either 1 or -1.
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$a$",(0,0)); draw(shift(1,0)*box); label("$b$",(1,0)); draw(shift(2,0)*box); label("$c$",(2,0)); draw(shift(3,0)*box); label("$d$",(3,0)); draw(shift(0.5,0.4)*box); label("$ab$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$bc$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$cd$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$ab^2c$",(1,0.8)); draw(shift(2,0.8)*box); label("$bc^2d$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$ab^3c^3d$",(1.5,1.2)); [/asy]
Which means $ab^3c^3d$ = 1. Since $b$ and $c$ are either 1 or -1, $b^3 = b$ and $c^3 = c$ . This shows that $abcd$ = 1.
Therefore either $a$ $b$ $c$ , and $d$ are all positive or negative, or 2 are positive and 2 are negative.
There are 2 ways where $a$ $b$ $c$ , and $d$ are 1 (1, 1, 1, 1) and (-1, -1, -1, -1)
There are 6 ways where 2 variables are positive and 2 are negative: (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (-1, -1, 1, 1), (-1, 1, -1, 1), and (-1, -1, 1, 1).
So the answer is $\boxed{8}$

Answer: C 8
Problem: In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]
$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

Solution: The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a $+$ on the above, half of them have a $-$ above. So, For the lowest layer with $4$ blocks, there are $2^4=16$ possible combination and half of them will lead a $+$ (or $-$ ) on the top. The answer is $16/2=\boxed{8}$

Answer: C 8
Problem: How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: Looking at the values, we notice that $11-7=4$ $9-5=4$ and $6-2=4$ . This means we are looking for a value that is four less than a multiple of $11$ $9$ , and $6$ . The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$ , so the numbers that fulfill this can be written as $198k-4$ , where $k$ is a positive integer.  This value is only a three-digit integer when $k$ is $1, 2, 3, 4$ or $5$ , which gives $194, 392, 590, 788,$ and $986$ respectively. Thus, we have $5$ values, so our answer is $\boxed{5}$

Answer: E 5
Problem: How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: Let us create the equations: $6x+2 = 9y+5 = 11z+7$ , and we know $100 \leq 11z+7 <1000$ , it gives us $9 \leq z \leq 90$ , which is the range of the value of z. Because of $6x+2=11z+7$ , then $6x=11z+5=6z+5(z+1)$ , so $(z+1)$ must be a mutiple of 6. Because of $9y+5=11z+7$ , then $9y=11z+2=9z+2(z+1)$ , so $(z+1)$ must also be a mutiple of $9$ . Hence, the value of $(z+1)$ must be a common multiple of $6$ and $9$ , which means multiples of $18 (LCM \text{ of }\ 6, 9)$ . So, let's say $z+1 = 18p$ ; then, $9 \leq z = 18p-1 \leq 90$ ,  so $1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5$ . Thus, the answer is $\boxed{5}$

Answer: E 5
Problem: How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: By the Chinese Remainder Theorem , we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{5}.$

Answer: E 5
Problem: How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: We can use modular arithmetic. Set up the equations: $x \equiv 2 \mod 6,$ $x \equiv 5 \mod 9,$ and $x \equiv 7 \mod 11.$ These equations can also be written as $x+4 \equiv 0 \mod 6,$ $x+4 \equiv 0 \mod 9,$ and $x+4 \equiv 0 \mod 11.$ Since $x+4$ is congruent to numbers $6, 9,$ and $11,$ then it must also be congruent to their LCM. Thus, $x+4 \equiv 0 \mod 198,$ since 198 is the LCM of $6, 9,$ and $11.$ Since these numbers have to be three digits, they can only be $194, 392, 590, 788,$ and $986.$ This gives us the answer of $\boxed{5}.$

Answer: E 5
Problem: How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: Let $N$ be the three digit positive integer. $N = 6a + 2 = 9b + 5 = 11c + 7$ . Then, we add four to all sides and write $N + 4 = 6(a+1) = 9(b+1) = 11(c+1)$ . Now, we know that $N + 4$ is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so $N = 198k - 4$ . From this, we can figure out that $N$ can be 5 different three digit numbers -- $194, 392, 590, 788,$ and $986$ $\therefore$ , the answer is $\boxed{5}.$

Answer: E 5
Problem: Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Solution: Let the area of $\triangle CEF$ be $x$ . Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$
By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$ . Now, consider trapezoid $ABED$ . Its area is $45+4x$ , which is three-fourths the area of the square. We set up an equation in $x$
\[45+4x = \frac{3}{4}\left(90+2x\right)\] Solving, we get $x = 9$ . The area of square $ABCD$ is $90+2x = 90 + 2 \cdot 9 = \boxed{108}$

Answer: B 108
Problem: Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Solution: We can use analytic geometry for this problem.
Let us start by giving $D$ the coordinate $(0,0)$ $A$ the coordinate $(0,1)$ , and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving for their intersection gives point $F$ coordinates $\left(\frac{2}{3},\frac{1}{3}\right)$
Now, $\triangle$ $EFC$ ’s area is simply $\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}$ or $\frac{1}{12}$ . This means that pentagon $ABCEF$ ’s area is $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ of the entire square, and it follows that quadrilateral $AFED$ ’s area is $\frac{5}{12}$ of the square.
The area of the square is then $\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{108}$

Answer: B 108
Problem: Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Solution: $\triangle ABC$ has half the area of the square. $\triangle FEC$ has base equal to half the square side length, and by AA Similarity with $\triangle FBA$ , it has 1/(1+2)= 1/3 the height, so has $\dfrac1{12}$ th area of square. Thus, the area of the quadrilateral is $1-1/2-1/12=5/12$ th the area of the square. The area of the square is then $45\cdot\dfrac{12}{5}=\boxed{108}$

Answer: B 108