Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $ $1.02$ , with at least one coin of each type. How many dimes must you have?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution: It is clear that there should only be $2$ pennies; any more would take up too many coins, and the limit is $9$ . Now we have $ $1$ left, and $7$ coins to use. Looking at the quarters, we can make methodical guesses. If there is $1$ quarter, then we would have to make $ $0.75$ with $6$ coins. We take a few educated guesses for the nickel and dime combos, and see that $1$ quarter will not work. Trying values for $2$ quarters, we see that this will not work either. When we reach $3$ quarters, the remaining is $ $0.25$ made from $4$ coins. We try with $2$ dimes, which does not work (it only takes $3$ coins) and we try with $1$ dime. After trying $2$ pennies, $3$ quarters, $1$ dime, and $3$ nickels, it is evident that this combo works. Therefore, the answer is $\boxed{1}$

Answer: A 1
Problem: Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

Solution: Divide it into $2$ cases:
1) Keiko and Ephriam both get $0$ heads:
This means that they both roll all tails, so there is only $1$ way for this to happen.
2) Keiko and Ephriam both get $1$ head:
For Keiko, there is only $1$ way for this to happen because he is only flipping 1 penny, but for Ephriam, there are 2 ways since there are $2$ choices for when he can flip the head. So, in total there are $2 \cdot 1 = 2$ ways for this case.
Thus, in total there are $3$ ways that work. Since there are $2$ choices for each coin flip (Heads or Tails), there are $2^3 = 8$ total ways of flipping 3 coins.
Thus, since all possible coin flips of 3 coins are equally likely, the probability is $\boxed{38}$

Answer: B 38
Problem: A cube has edge length $2$ . Suppose that we glue a cube of edge length $1$ on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to
[asy] draw((0,0)--(2,0)--(3,1)--(3,3)--(2,2)--(0,2)--cycle); draw((2,0)--(2,2)); draw((0,2)--(1,3)); draw((1,7/3)--(1,10/3)--(2,10/3)--(2,7/3)--cycle); draw((2,7/3)--(5/2,17/6)--(5/2,23/6)--(3/2,23/6)--(1,10/3)); draw((2,10/3)--(5/2,23/6)); draw((3,3)--(5/2,3));[/asy]
$\text{(A)}\ 10\qquad\text{(B)}\ 15\qquad\text{(C)}\ 17\qquad\text{(D)}\ 21\qquad\text{(E)}\ 25$

Solution: The original cube has $6$ faces, each with an area of $2\cdot 2 = 4$ square units.  Thus the original figure had a total surface area of $24$ square units.
The new figure has the original surface, with $6$ new faces that each have an area of $1$ square unit, for a total surface area of of $6$ additional square units added to it.  But $1$ square unit of the top of the bigger cube, and $1$ square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area.
The total surface area is therefore $24 + 6 - 1 - 1 = 28$ square units.
The percent increase in surface area is $\frac{SA_{new} - SA_{old}}{SA_{old}}\cdot 100\% = \frac{28-24}{24}\cdot 100\% \approx 16.67\%$ , giving the closest answer as $\boxed{17}$

Answer: C 17
Problem: There is a list of seven numbers. The average of the first four numbers is $5$ , and the average of the last four numbers is $8$ . If the average of all seven numbers is $6\frac{4}{7}$ , then the number common to both sets of four numbers is
$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7}$

Solution: Remember that if a list of $n$ numbers has an average of $k$ , then the sum $S$ of all the numbers on the list is $S = nk$
So if the average of the first $4$ numbers is $5$ , then the first four numbers total $4 \cdot 5 = 20$
If the average of the last $4$ numbers is $8$ , then the last four numbers total $4 \cdot 8 = 32$
If the average of all $7$ numbers is $6\frac{4}{7}$ , then the total of all seven numbers is $7 \cdot 6\frac{4}{7} = 7\cdot 6 + 4 =  46$
If the first four numbers are $20$ , and the last four numbers are $32$ , then all "eight" numbers are $20 + 32 = 52$ .  But that's counting one number twice.  Since the sum of all seven numbers is $46$ , then the number that was counted twice is $52 - 46 = 6$ , and the answer is $\boxed{6}$

Answer: B 6
Problem: If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$ , then $\angle B+\angle D =$
[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,WSW); label("$G$",G,NW);[/asy]
$\text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ$

Solution: As a strategy, think of how $\angle B + \angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\angle B$ or $\angle D$ alone.  If you see $\triangle BFD$ , then you can see that the problem is solved quickly after determining $\angle BFD$
But start with $\triangle AGF$ , since that's where most of our information is.  Looking at $\triangle AGF$ , since $\angle F = \angle G$ , and $\angle A = 20$ , we can write:
$\angle A + \angle G + \angle F = 180$
$20 + 2\angle F = 180$
$\angle AFG = 80$
By noting that $\angle AFG$ and $\angle GFD$ make a straight line, we know
$\angle AFG + \angle GFD = 180$
$80 + \angle GFD = 180$
$\angle GFD = 100$
Ignoring all other parts of the figure and looking only at $\triangle BFD$ , you see that $\angle B + \angle D + \angle F = 180$ .  But $\angle F$ is the same as $\angle GFD$ .  Therefore:
$\angle B + \angle D + \angle GFD = 180$ $\angle B + \angle D + 100 = 180$ $\angle B + \angle D = 80^\circ$ , and the answer is thus $\boxed{80}$

Answer: D 80
Problem: The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]
$\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

Solution: To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension.  Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$ , which is the easiest way to avoid fractions.  Labelling the right midpoint as $M$ , and the bottom midpoint as $N$ , we know that $DN = NC = 6$ , and $BM = MC = 3$
$[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$
$[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$
$[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$
$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$
$[\triangle AMN] = 72 - 18 - 9 - 18$
$[\triangle AMN] = 27$ , and the answer is $\boxed{27}$

Answer: B 27
Problem: The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]
$\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

Solution: The above answer is fast, but satisfying, and assumes that the area of $\triangle AMN$ is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label $AB = CD = l$ and $BC = DA = h$
Labelling $m$ and $n$ as the right and lower midpoints respectively, and redoing all the work above, we get:
$[\triangle ABN] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}$
$[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}$
$[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}$
$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$
$[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}$
$[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27$ , and the answer is $\boxed{27}$

Answer: B 27
Problem: The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]
$\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

Solution: Let's assume, for simplicity, that the sides of the rectangle are $9$ and $8.$ The area of the 3 triangles would then be $8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,$ $4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,$ $4\cdot 9\cdot\frac{1}{2} = 18.$ Adding these up, we get $45$ , and subtracting that from $72$ , we get $27$ , so the answer is $\boxed{27}$

Answer: B 27
Problem: What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?
[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); }  label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SSE); label("$10$",2*dir(150),ESE); label("$9$",2*dir(180),E); label("$8$",2*dir(210),ENE); label("$7$",2*dir(240),NNE); label("$6$",2*dir(270),N); label("$5$",2*dir(300),NNW); label("$4$",2*dir(330),WNW); [/asy]
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90$

Solution: At $10:00$ , the hour hand will be on the $10$ while the minute hand on the $12$
This makes them $\frac{1}{6}$ th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{60}$

Answer: C 60
Problem: What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?
[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); }  label("$3$",2*dir(0),W); label("$2$",2*dir(30),WSW); label("$1$",2*dir(60),SSW); label("$12$",2*dir(90),S); label("$11$",2*dir(120),SSE); label("$10$",2*dir(150),ESE); label("$9$",2*dir(180),E); label("$8$",2*dir(210),ENE); label("$7$",2*dir(240),NNE); label("$6$",2*dir(270),N); label("$5$",2*dir(300),NNW); label("$4$",2*dir(330),WNW); [/asy]
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90$

Solution: We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are $12$ numbers, the distance between one number will be $360\div 12=30$ .
If the time is $10:00$ , then the hour hand will be on $10$ , and the minute hand will be on, $12$ , making them $2$ numbers apart, so they will be $60$ degrees apart, or $\boxed{60}$

Answer: C 60
Problem: The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?
[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9));  label(rotate(30)*"Bjorn",(12,6.75),SE); label(rotate(37)*"Alberto",(11,8.25),NW);  label("$0$",(0,0),S); label("$1$",(4,0),S); label("$2$",(8,0),S); label("$3$",(12,0),S); label("$4$",(16,0),S); label("$5$",(20,0),S); label("$0$",(0,0),W); label("$15$",(0,3),W); label("$30$",(0,6),W); label("$45$",(0,9),W); label("$60$",(0,12),W); label("$75$",(0,15),W);  label("H",(6,-2),S); label("O",(8,-2),S); label("U",(10,-2),S); label("R",(12,-2),S); label("S",(14,-2),S);  label("M",(-4,11),N); label("I",(-4,9),N); label("L",(-4,7),N); label("E",(-4,5),N); label("S",(-4,3),N); [/asy]
$\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution: After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60.  Thus the answer is $60-45=15$ $\boxed{15}$

Answer: A 15
Problem: The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?
[asy] for (int a = 0; a < 6; ++a) { for (int b = 0; b < 6; ++b) { dot((4*a,3*b)); } } draw((0,0)--(20,0)--(20,15)--(0,15)--cycle); draw((0,0)--(16,12)); draw((0,0)--(16,9));  label(rotate(30)*"Bjorn",(12,6.75),SE); label(rotate(37)*"Alberto",(11,8.25),NW);  label("$0$",(0,0),S); label("$1$",(4,0),S); label("$2$",(8,0),S); label("$3$",(12,0),S); label("$4$",(16,0),S); label("$5$",(20,0),S); label("$0$",(0,0),W); label("$15$",(0,3),W); label("$30$",(0,6),W); label("$45$",(0,9),W); label("$60$",(0,12),W); label("$75$",(0,15),W);  label("H",(6,-2),S); label("O",(8,-2),S); label("U",(10,-2),S); label("R",(12,-2),S); label("S",(14,-2),S);  label("M",(-4,11),N); label("I",(-4,9),N); label("L",(-4,7),N); label("E",(-4,5),N); label("S",(-4,3),N); [/asy]
$\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution: We see that each dot is $15$ units away from the nearest one above it. So the answer is $\boxed{15}$

Answer: A 15
Problem: A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?
$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

Solution: We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$ . To get the greatest area while keeping a perimeter of $160$ , the sides should all be $40$ . that means an area of $1600$ . Right now, the area is $20 \times 60$ which is $1200$ $1600-1200=400$ which is $\boxed{400}$

Answer: D 400
Problem: The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this   service center?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130$

Solution: There are $160-40=120$ miles between the third and tenth exits, so the service center is at milepost $40+(3/4)(120) = 40+90=\boxed{130}$

Answer: E 130
Problem: Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N); label("B",(3.1,-.2),S); label("C",(.6,-.2),S); [/asy]
$\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450$

Solution: Plants shared by two beds have been counted
twice, so the total is $500 + 450 + 350 - 50 - 100 = \boxed{1150}$

Answer: C 1150
Problem: Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N); label("B",(3.1,-.2),S); label("C",(.6,-.2),S); [/asy]
$\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450$

Solution: Bed A has $350$ plants it doesn't
share with B or C. Bed B has $400$ plants it doesn't
share with A or C. And C has $250$ it doesn't share
with A or B. The total is $350 + 400 + 250 + 50 + 100 = \boxed{1150}$ plants.

Answer: C 1150
Problem: Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$

Solution: The largest sum occurs when $13$ is placed in the center. This sum is $13 + 10 + 1 = 13 + 7 + 4 = \boxed{24}$ . Note: Two other common sums, $18$ and $21$ , are also possible.

Answer: D 24
Problem: Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$

Solution: Since the horizontal sum equals the vertical sum, twice this sum will be the sum
of the five numbers plus the number in the center. When the center number is $13$ , the sum is the largest, \[[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{24}\] The other
four numbers are divided into two pairs with equal sums.

Answer: D 24
Problem: The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$

Solution: The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$ . The percentage of games lost is just $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{27}$

Answer: B 27
Problem: The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$

Solution: The Won/Lost ratio is 11/4 so, for some number $N$ , the team won $11N$ games and lost $4N$ games. Thus, the team played $15N$ games and the fraction of games lost is $\dfrac{4N}{15N}=\dfrac{4}{15}\approx 0.27=\boxed{27}.$ -Clara Garza

Answer:  27
Problem: The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?
$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30$

Solution: First, find the total amount of the girl's ages and add it to the total amount of the boy's ages. It equals $(20)(15)+(15)(16)=540$ . The total amount of everyone's ages can be found from the average age, $17\cdot40=680$ . Then you do $680-540=140$ to find the sum of the adult's ages. The average age of an adult is divided among the five of them, $140\div5=\boxed{28}$

Answer: C 28
Problem: In trapezoid $ABCD$ , the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is
[asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8));  label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$16$",(8,0),S); label("$3$",(4,1.5),E); [/asy]
$\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48$

Solution: [asy] draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); draw((4,3)--(4,0),dashed); draw((12,3)--(12,0),dashed); draw((3.2,0)--(3.2,.8)--(4,.8));  label("$A$",(0,0),SW); label("$B$",(4,3),NW); label("$C$",(12,3),NE); label("$D$",(16,0),SE); label("$8$",(8,3),N); label("$8$",(8,0),S); label("$3$",(4,1.5),E); label("$4$",(2,0),S); label("$4$",(14,0),S); label("$5$",(0,0)--(4,3),NW); label("$5$",(12,3)--(16,0),NE); [/asy]
There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$ . The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of $5$ , because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$ $5$ $16$ , and $5$ . Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = \boxed{34}$ units.

Answer: D 34
Problem: Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$

Solution: There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \cdot 3 \cdot 4 = 60$ license plates.
Adding $2$ letters to the start gives $7\cdot 3 \cdot 4 = 84$ plates.
Adding $2$ letters to the middle gives $5 \cdot 5 \cdot 4 = 100$ plates.
Adding $2$ letters to the end gives $5 \cdot 3 \cdot 6 = 90$ plates.
Adding a letter to the start and middle gives $6 \cdot 4 \cdot 4 = 96$ plates.
Adding a letter to the start and end gives $6 \cdot 3 \cdot 5 = 90$ plates.
Adding a letter to the middle and end gives $5 \cdot 4 \cdot 5 = 100$ plates.
You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\boxed{40}$

Answer: D 40
Problem: Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}.
When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$

Solution: Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters.
In general, when three numbers have the same fixed sum, their product will be maximal when they are as close together as possible.  This is a 3D analogue of the fact that a rectangle with fixed perimeter maximizes its area when the sides are equal (ie when it becomes a square).  In this case, no matter where you add the letters, there will be $5 + 3 + 4 + 2 = 14$ letters in total.  If you divide them as evenly as possible among the three groups, you get $5, 5, 4$ , which is a possible situation.
As before, the answer is $5 \cdot 5 \cdot 4 - 5 \cdot 3 \cdot 4 = 40$ , and the correct choice is $\boxed{40}$

Answer: D 40