Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In rectangle $ABCD$ , we have $AB=8$ $BC=9$ $H$ is on $BC$ with $BH=6$ $E$ is on $AD$ with $DE=4$ , line $EC$ intersects line $AH$ at $G$ , and $F$ is on line $AD$ with $GF \perp AF$ . Find the length of $GF$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("$F$",F,W); label("$G$",G,W); label("$C$",C,WSW); label("$H$",H,NNE); label("$6$",(6,8),N); label("$B$",B,NE); label("$A$",A,SW); label("$E$",Ep,S); label("$4$",(2,0),S); label("$D$",D,S);[/asy]
$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$

Solution: We extend $BC$ such that it intersects $GF$ at $X$ . Since $ABCD$ is a rectangle, it follows that $CD=8$ , therefore, $XF=8$ . Let $GX=y$ . From the similarity of triangles $GCH$ and $GEA$ , we have the ratio $3:5$ (as $CH=9-6=3$ , and $EA=9-4=5$ ). $GX$ and $GF$ are the altitudes of $GCH$ and $GEA$ , respectively. Thus, $y:y+8 = 3:5$ , from which we have $y=12$ , thus $GF=y+8=12+8=\boxed{20}$

Answer: B 20
Problem: In rectangle $ABCD$ , we have $AB=8$ $BC=9$ $H$ is on $BC$ with $BH=6$ $E$ is on $AD$ with $DE=4$ , line $EC$ intersects line $AH$ at $G$ , and $F$ is on line $AD$ with $GF \perp AF$ . Find the length of $GF$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("$F$",F,W); label("$G$",G,W); label("$C$",C,WSW); label("$H$",H,NNE); label("$6$",(6,8),N); label("$B$",B,NE); label("$A$",A,SW); label("$E$",Ep,S); label("$4$",(2,0),S); label("$D$",D,S);[/asy]
$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$

Solution: Since $GF\perp AF$ and $AF\perp CD,$ we have $GF\parallel CD\parallel AB.$ Thus, $\triangle CDE\sim GFE.$ Suppose $GF=x$ and $FD=y.$ Thus, we have $\dfrac{x}{8}=\dfrac{y+4}{4}.$ Additionally, now note that $\triangle GAF\sim AHB,$ which is pretty obvious from insight, but can be proven by AA with extending $BH$ to meet $GF.$ From this new pair of similar triangles, we have $\dfrac{x}{8}=\dfrac{y+9}{6}.$ Therefore, we have by combining those two equations, \[\dfrac{y+9}{6}=\dfrac{y+4}{4}.\] Solving, we have $y=6,$ and therefore $x=\boxed{20}$

Answer: B 20
Problem: In rectangle $ABCD$ , we have $AB=8$ $BC=9$ $H$ is on $BC$ with $BH=6$ $E$ is on $AD$ with $DE=4$ , line $EC$ intersects line $AH$ at $G$ , and $F$ is on line $AD$ with $GF \perp AF$ . Find the length of $GF$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("$F$",F,W); label("$G$",G,W); label("$C$",C,WSW); label("$H$",H,NNE); label("$6$",(6,8),N); label("$B$",B,NE); label("$A$",A,SW); label("$E$",Ep,S); label("$4$",(2,0),S); label("$D$",D,S);[/asy]
$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$

Solution: Since there are only lines, you can resort to coordinate bashing. Let $FD=k$ . Three lines, line $GF$ , line $GE$ , and line $GA$ , intersect at $G$ . Our goal is to find the y-coordinate of that intersection point.
Line $GF$ is $x=0$
Line $GE$ passes through $(k+4, 0)$ and $(k, 8)$ . Therefore the slope is $-2$ and the line is $y-0=-2(x-k-4)$ which is $y=-2x+2k+8$
Line $GA$ passes through $(k+9, 0)$ and $(k+3, 8)$ . Therefore the slope is $\frac{-4}{3}$ and the line is $y-0=-\frac{4}{3}(x-k-9)$ which simplifies to $y=-\frac{4}{3}x+\frac{4}{3}k+12$
We solve the system of equations with these three lines. First we plug in $x=0$
$y=2k+8$
$y=\frac{4}{3}k+12$
Next, we solve for k. $k=6$ Therefore $y=20$ . The y-coordinate of this intersection point is indeed our answer. $\boxed{20}$ ~superagh

Answer: B 20
Problem: In rectangle $ABCD$ , we have $AB=8$ $BC=9$ $H$ is on $BC$ with $BH=6$ $E$ is on $AD$ with $DE=4$ , line $EC$ intersects line $AH$ at $G$ , and $F$ is on line $AD$ with $GF \perp AF$ . Find the length of $GF$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("$F$",F,W); label("$G$",G,W); label("$C$",C,WSW); label("$H$",H,NNE); label("$6$",(6,8),N); label("$B$",B,NE); label("$A$",A,SW); label("$E$",Ep,S); label("$4$",(2,0),S); label("$D$",D,S);[/asy]
$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$

Solution: Let $A$ be the origin of our coordinate system. Now line $GA$ has equation $-\frac{4}{3}x$ . We can use point-slope form to find the equation for line $GE$ . First, we know that its slope is $-2$ , and we know that it passes through $E=(-5,0)$ , so line $GE$ has equation $-2(x+5)$ . Solving for the intersection by letting $-\frac{4}{3}x=-2(x+5)$ , we get $x=-15$ . Plugging this into our equation for line $GA$ gives us $G=(-15,20)$ , so $GF= \boxed{20}$ ~chrisdiamond10

Answer: B 20
Problem: In rectangle $ABCD$ , we have $AB=8$ $BC=9$ $H$ is on $BC$ with $BH=6$ $E$ is on $AD$ with $DE=4$ , line $EC$ intersects line $AH$ at $G$ , and $F$ is on line $AD$ with $GF \perp AF$ . Find the length of $GF$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("$F$",F,W); label("$G$",G,W); label("$C$",C,WSW); label("$H$",H,NNE); label("$6$",(6,8),N); label("$B$",B,NE); label("$A$",A,SW); label("$E$",Ep,S); label("$4$",(2,0),S); label("$D$",D,S);[/asy]
$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$

Solution: First, using given information, we can find the values of some line segments in the figure. We find that $HA = 10$ (through Pythagorean Theorem), $CH = 3$ , and $EA = 5$ . 
Let Line $FD = x$ and let Line $FG = y$ . 
We find that $\triangle FGE \sim \triangle CDE$ through some angle chasing (they both have a right angle, and they both share angle $\angle CED$ . Using this information, we can write the equation $\frac{4}{8} = \frac{4+x}{y}$ . Through simplifying this equation, we get that $y=2x+8$ . Let point $I$ be the point on line $FG$ so that lines $CI$ and $FG$ are perpendicular, and we get that $GI = 2x$ and $FI =8$ . Doing some more angle chasing, we can find that $\triangle GIH \sim \triangle GFA$ , as they both share $\angle FGH$ and they both have a right angle.
With this information, we can write the equation $\frac{x+3}{y-8} = \frac{x+9}{y}.$ Simplifying this equation we get the equation $-8x+6y-72 = 0$ .
Plugging in $y=2x+8$ for $6y$ , we get $4x-24 = 0$ , so $x=6$ .
Lastly, to find the value of y, which is the value of Line $FG$ , our desired value, we plug in $6$ for $x$ in the equation $y=2x+8$ , we get $2(6)+8$ , which, finally, we get our $y$ value of $20$ , so therefore, our answer is $GF= \boxed{20}$

Answer: B 20
Problem: A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row.  How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?
[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("$1$",Cn[0]); label("$2$",Cn[1]); label("$3$",Cn[2]); label("$4$",Cn[3]); label("$5$",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]
$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

Solution: There are $1+3+5+...+2003=1002^{2}=1004004$ small equilateral triangles.
Each small equilateral triangle needs $3$ toothpicks to make it.
But, each toothpick that isn't one of the $1002\cdot3=3006$ toothpicks on the outside of the large equilateral triangle is a side for $2$ small equilateral triangles.
So, the number of toothpicks on the inside of the large equilateral triangle is $\frac{10040004\cdot3-3006}{2}=1504503$
Therefore the total number of toothpicks is $1504503+3006=\boxed{1,507,509}$ ~dolphin7

Answer: C 1,507,509
Problem: A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row.  How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?
[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("$1$",Cn[0]); label("$2$",Cn[1]); label("$3$",Cn[2]); label("$4$",Cn[3]); label("$5$",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]
$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

Solution: We just need to count upward facing triangles because if we exclude the downward-facing triangles, we won't be overcounting any toothpicks. The first row of triangles has $1$ upward-facing triangle, the second row has $2$ upward-facing triangles, the third row has $3$ upward-facing triangles, and so on having $n$ upward-facing triangles in the $n^\text{th}$ row. The last row with $2003$ small triangles has $1002$ upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now $\frac{1002\times1003}{2}$ , meaning that the number of toothpicks are $\frac{1002\times1003}{2}\times3$ , or $\boxed{1,507,509}$

Answer: C 1,507,509
Problem: A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row.  How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?
[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("$1$",Cn[0]); label("$2$",Cn[1]); label("$3$",Cn[2]); label("$4$",Cn[3]); label("$5$",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]
$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

Solution: You don't have to calculate the value of $\frac{1002\times1003}{2}\times3$ , and you can use units digits to find the answer easily. The units digit of $1002\times1003$ is $6$ , and has a unit digit of $3$ after being divided by $2$ . Then this is multiplied by $3$ , now the final number ending with a $9$ . This leaves only one answer choice possible, which is $\boxed{1,507,509}$

Answer: C 1,507,509
Problem: A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row.  How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?
[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("$1$",Cn[0]); label("$2$",Cn[1]); label("$3$",Cn[2]); label("$4$",Cn[3]); label("$5$",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]
$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

Solution: Test out some fewer cases first.
When there is just 1 equilateral triangle in the base, you need $3$ toothpicks.
When there are 3 equilateral triangles in the base, you need $9$ toothpicks in all.
When there are 5 equilateral triangles in the base, you need $18$ toothpicks in all.
When there are 7 equilateral triangles in the base, you need $30$ toothpicks in all.
Taking the finite differences, we get $6, 9, 12.$ It forms a linear equations. This means the original numbers $(3, 9, 18, 30)$ form a quadratic.
Let the quadratic be $y = ax^2 + bx + c$ where $y = 2* \text{equilateral triangles in base} - 1.$
Then, we have the following points: $(1, 3), (2, 9), (3, 18), (4, 30).$
We can plug these values into $y = ax^2 + bx + c$ , giving:
\[a + b + c = 3, 4a + b + c = 9, 9a + 3b + c = 18.\]
Solving gives $a = b = 1.5, c = 0.$ So, \[y = 1.5x^2 + 1.5x.\]
For our problem, we need it when there are $2003$ equilateral triangles in the base. For the quadratic, the corresponding $x$ -value would be $\frac{2003 + 1}{2} = 1002.$ . So, our answer is simply: \[1.5 * 1002^2 + 1.5*1002 = \boxed{1507509}.\]

Answer:  1507509
Problem: Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6$ . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?
$\mathrm{(A) \ } 8\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } 12$

Solution: Let $R_i$ and $B_j$ designate the red card numbered $i$ and the blue card numbered $j$ , respectively.
$B_5$ is the only blue card that $R_5$ evenly divides, so $R_5$ must be at one end of the stack and $B_5$ must be the card next to it.
$R_1$ is the only other red card that evenly divides $B_5$ , so $R_1$ must be the other card next to $B_5$
$B_4$ is the only blue card that $R_4$ evenly divides, so $R_4$ must be at one end of the stack and $B_4$ must be the card next to it.
$R_2$ is the only other red card that evenly divides $B_4$ , so $R_2$ must be the other card next to $B_4$
$R_2$ doesn't evenly divide $B_3$ , so $B_3$ must be next to $R_1$ $B_6$ must be next to $R_2$ , and $R_3$ must be in the middle.
This yields the following arrangement from top to bottom: $\{R_5,B_5,R_1,B_3,R_3,B_6,R_2,B_4,R_4\}$
Therefore, the sum of the numbers on the middle three cards is $3+3+6=\boxed{12}$

Answer: E 12
Problem: Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$

Solution: $11|(q+r)$ implies that $11|(99q+q+r)$ and therefore $11|(100q+r)$ , so $11|n$ . Then, $n$ can range from $10010$ to $99990$ for a total of $\boxed{8181}$ numbers.

Answer: B 8181
Problem: Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $$ $1$ more than a pink pill, and Al's pills cost a total of $\textdollar 546$ for the two weeks. How much does one green pill cost?
$\textbf{(A)}\ \textdollar 7 \qquad\textbf{(B) }\textdollar 14 \qquad\textbf{(C) }\textdollar 19\qquad\textbf{(D) }\textdollar 20\qquad\textbf{(E) }\textdollar 39$

Solution: Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\boxed{20}$

Answer: D 20
Problem: The sum of $5$ consecutive even integers is $4$ less than the sum of the first $8$ consecutive odd counting numbers. What is the smallest of the even integers?
$\textbf{(A) } 6 \qquad\textbf{(B) } 8 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 14$

Solution: It is a well-known fact that the sum of the first $n$ odd numbers is $n^2$ .  This makes the sum of the first $8$ odd numbers equal to $64$
Let $n$ be equal to the smallest of the $5$ even integers.  Then $(n+2)$ is the next highest, $(n+4)$ even higher, and so on.
This sets up the equation $n+(n+2)+(n+4)+(n+6)+(n+8)+4=64$
Now we solve:
\[5n+24=64\] \[5n=40\] \[n=8\]
Thus, the smallest integer is $\boxed{8}$

Answer: B 8
Problem: Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $$ $1$ each, begonias $$ $1.50$ each, cannas $$ $2$ each, dahlias $$ $2.50$ each, and Easter lilies $$ $3$ each. What is the least possible cost, in dollars, for her garden?
[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((6,0)--(0,0)--(0,1)--(6,1)); draw((0,1)--(0,6)--(4,6)--(4,1)); draw((4,6)--(11,6)--(11,3)--(4,3)); draw((11,3)--(11,0)--(6,0)--(6,3)); label("1",(0,0.5),W); label("5",(0,3.5),W); label("3",(11,1.5),E); label("3",(11,4.5),E); label("4",(2,6),N); label("7",(7.5,6),N); label("6",(3,0),S); label("5",(8.5,0),S);[/asy]
$\textbf{(A) } 108 \qquad\textbf{(B) } 115 \qquad\textbf{(C) } 132 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156$

Solution: The areas of the five regions from greatest to least are $21,20,15,6$ and $4$
If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1\cdot21+1.50\cdot20+2\cdot15+2.50\cdot6+3\cdot4$ , which simplifies to $$ $108$ .
Therefore the answer is $\boxed{108}$

Answer: A 108
Problem: Moe uses a mower to cut his rectangular $90$ -foot by $150$ -foot lawn. The swath he cuts is $28$ inches wide, but he overlaps each cut by $4$ inches to make sure that no grass is missed. He walks at the rate of $5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn?
$\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3$

Solution: Since the swath Moe actually mows is $24$ inches, or $2$ feet wide, he mows $10000$ square feet in one hour. His lawn has an area of $13500$ , so it will take Moe $1.35$ hours to finish mowing the lawn. Thus the answer is $\boxed{1.35}$

Answer: C 1.35
Problem: Moe uses a mower to cut his rectangular $90$ -foot by $150$ -foot lawn. The swath he cuts is $28$ inches wide, but he overlaps each cut by $4$ inches to make sure that no grass is missed. He walks at the rate of $5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow the lawn?
$\textbf{(A) } 0.75 \qquad\textbf{(B) } 0.8 \qquad\textbf{(C) } 1.35 \qquad\textbf{(D) } 1.5 \qquad\textbf{(E) } 3$

Solution: Let's assume that the swath moves back and forth; parallel to the $90$ feet side. Thus, the length of one strip is $90$ feet. Now we need to find out how many strips there are. In reality, the swath Moe mows is $24$ inches wide, which can be easily translated into $2$ feet. $\frac{150}{2}$ is the number of strips Moe needs to mow, which is equal to $75$ . 
Therefore, the total number of feet Moe mows is $75\times90$ . Since Moe's mowing rate is $5000$ feet per hour, $\frac{75\times90}{5000}$ is the number of hours it takes him to do his job. Using basic calculations, we compute the answer. $\boxed{1.35}$

Answer: C 1.35
Problem: Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4:3$ . The horizontal length of a " $27$ -inch" television screen is closest, in inches, to which of the following?
$\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22$

Solution: If you divide the television screen into two right triangles, the legs are in the ratio of $4 : 3$ , and we can let one leg be $4x$ and the other be $3x$ . Then we can use the Pythagorean Theorem.
\begin{align*}(4x)^2+(3x)^2&=27^2\\ 16x^2+9x^2&=729\\ 25x^2&=729\\ x^2&=\frac{729}{25}\\ x&=\frac{27}{5}\\ x&=5.4\end{align*}
The horizontal length is $5.4\times4=21.6$ , which is closest to $\boxed{21.5}$

Answer: D 21.5
Problem: Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $4:3$ . The horizontal length of a " $27$ -inch" television screen is closest, in inches, to which of the following?
$\textbf{(A) } 20 \qquad\textbf{(B) } 20.5 \qquad\textbf{(C) } 21 \qquad\textbf{(D) } 21.5 \qquad\textbf{(E) } 22$

Solution: One can realize that the diagonal, vertical, and horizontal lengths all make up a $3,4,5$ triangle. Therefore, the horizontal length, being the $4$ in the $4 : 3$ ratio, is simply $\frac{4}{5}$ times the hypotenuse. $\frac{4}{5}\cdot27=21.6 \approx \boxed{21.5}$

Answer: D 21.5
Problem: The symbolism $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . For example, $\lfloor 3 \rfloor = 3,$ and $\lfloor 9/2 \rfloor = 4$ . Compute \[\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor.\]
$\textbf{(A) } 35 \qquad\textbf{(B) } 38 \qquad\textbf{(C) } 40 \qquad\textbf{(D) } 42 \qquad\textbf{(E) } 136$

Solution: The first three values in the sum are equal to $1,$ the next five equal to $2,$ the next seven equal to $3,$ and the last one equal to $4.$ For example, since $2^2=4$ any square root of a number less than $4$ must be less than $2.$ Sum them all together to get
\[3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{38}\]

Answer: B 38