Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$

Solution: On any white square, we may choose to go left or right, as long as we do not cross over the border of the board. Call the moves $L$ and $R$ respectively. Every single legal path consists of $4$ $R's$ and $3$ $L's$ , so now all we have to find is the number of ways to order $4 R's$ and $3 L's$ in any way, which is ${7 \choose 3}=35$ . However, we originally promised that we will not go over the border, and now we have to subtract the paths that do go over the border. The paths that go over the border are any paths that start with RRR(1 path), RR(5 paths) and LRRR(1 path) so our final number of paths is $35-7=\boxed{28}.$ ~PEKKA

Answer: A 28
Problem: A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$

Solution: We label the rows starting from the bottom. At row 1, there is $1$ way: at P. We draw all the possible ways to get to Q. There are two ways to choose for row 2, and another two ways to choose for row 3. However, you can go to the "edge" or the farthest possible square westward of Q, so you can't multiply by 2 again. Notice how, at the first step, we figured that the answer was even, so choice D and E are eliminated, and after the second row, we realized it must be a multiple of 4, so choice B is eliminated. When we get to the fourth row, we do not multiply by 2 again, since we have limited possibilities rather than multiplying by 2 again. Choice C implies that there are two possibilities per row; however, we know that if you go to the farthest possible, you only have one possibility, so it is not $2^5 = 32$ so we know that the answer is choice $\boxed{28}$ . 
~hh99754539

Answer: A 28
Problem: When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$

Solution: We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$ ), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$ , and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$ -step process, and sum is $1+8+10+64=\boxed{83}$

Answer: E 83
Problem: When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$

Solution: As in Solution 1, we work backwards from $1$ , this time showing the possible cases in a tree diagram:
[asy] // Upper branches draw((-6, 1.5)--(-5, 1)--(-3, 1)--(-2,0)--(0, 0)); draw((-6, 0.5)--(-5, 1)); // Lower branches draw((-6, -1.5)--(-5, -1.5)--(-4, -1)--(-3, -1)--(-2, 0)); draw((-6, -0.5)--(-5, -0.5)--(-4, -1));  label("$1$", (0, 0),  UnFill(0.1mm)); label("$2$", (-1, 0), UnFill(0.1mm)); label("$4$", (-2, 0), UnFill(0.1mm));  // Upper branches label("$1$", (-3, 1), UnFill(0.1mm)); label("$2$", (-4, 1), UnFill(0.1mm)); label("$4$", (-5, 1), UnFill(0.1mm)); label("$\textbf{8}$", (-6, 1.5), UnFill(0.1mm)); label("$\textbf{1}$", (-6, 0.5), UnFill(0.1mm));  // Lower branches label("$8$", (-3, -1), UnFill(0.1mm)); label("$16$",(-4, -1), UnFill(0.1mm)); label("$5$", (-5, -0.5), UnFill(0.1mm)); label("$32$", (-5, -1.5), UnFill(0.1mm)); label("$\textbf{10}$", (-6, -0.5), UnFill(0.1mm)); label("$\textbf{64}$", (-6, -1.5), UnFill(0.1mm)); [/asy]
The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are $1$ $8$ $64$ , and $10$ . Thus the answer is $1+8+64+10=\boxed{83}$

Answer: E 83
Problem: Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Solution: Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$ $2$ , or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn.
If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case.
In the other case, two students receive $2$ awards and one student recieves $1$ award . We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to distribute. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{150}$

Answer: B 150
Problem: Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Solution: Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards.
In the first case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 3 awards. From here, there are $\binom{5}{3} = 10$ ways to choose the 3 awards from the 5 total awards. Now, one person has $2$ choices for awards and the other has $1$ choice for the award. Thus, the total number of ways to choose awards in this case is $3 \cdot 10 \cdot 2 \cdot 1 = 60$
In the other case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 1 award, and $5$ choices for his/her award. Then, one person has $\binom{4}{2} = 6$ ways to have his/her awards and the other person has $\dbinom{2}{2} = 1$ ways to have his/her awards. This gives $3 \cdot 5 \cdot 6 \cdot 1  = 90$ ways for this case.
Adding these cases together, we get $60 + 90 = 150$ ways to distribute the awards, or choice $\boxed{150}$

Answer: B 150
Problem: Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Solution: Without the restriction that each student receives at least one award, we could take each of the awards and choose one of the $3$ students to give it to. This would be $3^5$ ways to distribute the awards in total. Now we need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are $3$ choices for which student that is, so $2^5$ ways of choosing a student to receive each of the awards; in total, $3\cdot32=96$
However, if $2$ students both don't receive an award, then this case would be counted twice among the $96$ , so we need to add back in these cases. Said in other words, $2$ students not receiving an award is equivalent to $1$ student receiving $5$ awards, and there are $3$ choices for whom that student would be. To finish, the total number of ways to distribute the awards is $243 - 96+3$ , or $\boxed{150}$

Answer: B 150
Problem: Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Solution: Let the side length of each square $S_k$ be $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of $S_1$ $S_2$ , and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of $R_2$ will be $s_1-s_2$ , and lining this up with the left side of $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$ . We subtract the second equation from the first to obtain $2s_{2}=1302$ , and thus $s_{2}=\boxed{651}$

Answer: A 651
Problem: Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Solution: Assuming that the problem is well-posed, it should be true in the particular case where $S_1 \cong S_3$ and $R_1 \cong R_2$ . Let the sum of the side lengths of $S_1$ and $S_3$ be $x$ , and let the length of square $S_2$ be $y$ . We then have the system \[\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}\] which we solve to determine $y=\boxed{651}$

Answer: A 651
Problem: Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Solution: Since, for each pair of rectangles, the side lengths have a sum of $3322$ or $2020$ and a difference of $S_2$ , the answer must be $\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{651}$

Answer: A 651
Problem: Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Solution: Let the side length of $S_2$ be s, and the shorter side length of $R_1$ and $R_2$ be $r$ . We have
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); label("$r$",(5.2,5/2)); label("$r$",(3.2,1/2)); label("$s$",(3.2,3/2)); [/asy]
From this diagram, it is evident that $r+s+r=2020$ . Also, the side length of $S_1$ and $S_3$ is $r+s$ . Then, $r+s+s+r+s=3322$ . Now, we have 2 systems of equations.
\begin{align*}r+s+r &= 2020 \\ r+s+r+s+s &= 3322 \\ \end{align*}
We can see an $r+s+r$ in the 2nd equation, so substituting that in gives us $2020+2s=3322 \Rightarrow 2s= 1302 \Rightarrow s=\boxed{651}$

Answer: A 651
Problem: Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Solution: We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\boxed{9}$

Answer: D 9
Problem: Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can,
and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how
many items will they buy?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Solution: Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of \[4.50s+d=30\] In the question, it states that Ike and Mike buys as many sandwiches as possible. 
So, we drop the number of sodas for a while.
We have: \begin{align*} 4.50s&=30 \\ s&=\frac{30}{4.5} \\ s&=6R3 \end{align*} We don't want a remainder so the maximum number of sandwiches is $6$ .
The total money spent is $6\cdot 4.50=27$ .
The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items. 
Hence, the answer is $\boxed{9}$

Answer: D 9
Problem: Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is  5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Solution: We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\boxed{150}$ . ~avamarora

Answer: E 150
Problem: Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is  5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Solution: Using the diagram we find that the larger side of the small rectangle is $2$ times the length of the smaller side. Therefore, the longer side is $5 \cdot 2 = 10$ . So the area of the identical rectangles is $5 \cdot 10 = 50$ . We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is $50 \cdot 3 = \boxed{150}$ .    ~~fath2012

Answer: E 150
Problem: Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is  5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Solution: We see that if the short sides are 5, the long side has to be $5\cdot2=10$ because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle $ABCD$ )
is $10+5=15$ because long side + short side of the small rectangle is $15$ . The short side of rectangle $ABCD$ is $10$ because it is the long side of the short rectangle. Multiplying $15$ and $10$ together gets us $15\cdot10$ which is $\boxed{150}$ .
~~mathboy282

Answer: E 150
Problem: Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is  5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Solution: We see that the $2$ rectangles lying on top of each other give us the height of the rectangle. Using what we know, we can find out that the $2$ rectangles put together is a square. So, we can infer that the length of the rectangles is $10$ . Adding that to the width of the third rectangle which is $5$ , we get that the length of the rectangle is $15$ . Multiplying $10$ and $15$ gives us $15\cdot10$ which is $\boxed{150}$ .
~~awesomepony566

Answer: E 150
Problem: Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is  5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Solution: There are two rectangles lying on the side and one standing up. Given that one small side is 5, we can determine that two of the small sides make up a big side which means that the long side is equal to 10. The top side of the rectangle is made up of one small side and one long side, therefore the dimensions for the rectangle is 10x15. 10 multiplied by 15 is 150, hence the answer $\boxed{150}$ .
~elenafan

Answer: E 150