Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What number should be removed from the list \[1,2,3,4,5,6,7,8,9,10,11\] so that the average of the remaining numbers is $6.1$
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution: Similar to the first solution, the current total is $66$ .  Since there are $11$ numbers on the list, taking $1$ number away will leave $10$ numbers.  If those $10$ numbers have an average of $6.1$ , then those $10$ numbers must have a sum of $10 \times 6.1 = 61$ .  Thus, the number that was removed must be $66 - 61 = 5$ , and the answer is $\boxed{5}$

Answer: B 5
Problem: In the fall of 1996, a total of 800 students participated in an annual school clean-up day.  The organizers of the event expect that in each of the years 1997, 1998, and 1999, participation will increase by 50% over the previous year.  The number of participants the organizers will expect in the fall of 1999 is
$\text{(A)}\ 1200 \qquad \text{(B)}\ 1500 \qquad \text{(C)}\ 2000 \qquad \text{(D)}\ 2400 \qquad \text{(E)}\ 2700$

Solution: If the participation increases by $50\%$ , then it is the same as saying participation is multipled by a factor of $100\% + 50\% = 1 + 0.5 = 1.5$
In 1997, participation will be $800 \cdot 1.5 = 1200$
In 1998, participation will be $1200 \cdot 1.5 = 1800$
In 1999, participation will be $1800 \cdot 1.5 = 2700$ , giving an answer of $\boxed{2700}$

Answer: E 2700
Problem: In the fall of 1996, a total of 800 students participated in an annual school clean-up day.  The organizers of the event expect that in each of the years 1997, 1998, and 1999, participation will increase by 50% over the previous year.  The number of participants the organizers will expect in the fall of 1999 is
$\text{(A)}\ 1200 \qquad \text{(B)}\ 1500 \qquad \text{(C)}\ 2000 \qquad \text{(D)}\ 2400 \qquad \text{(E)}\ 2700$

Solution: Since the percentage increase is the same each year, this is an example of exponential growth with a base of $1.5$ .  In three years, there will be $1.5^3 = \frac{27}{8}$ times as many participants.  Multiplying this by the $800$ current participants, there are $2700$ participants, and the answer is $\boxed{2700}$

Answer: E 2700
Problem: Six different digits from the set \[\{ 1,2,3,4,5,6,7,8,9\}\] are placed in the squares in the figure shown so that the sum of the entries in the vertical column is 23 and the sum of the entries in the horizontal row is 12.
The sum of the six digits used is
[asy] unitsize(18); draw((0,0)--(1,0)--(1,1)--(4,1)--(4,2)--(1,2)--(1,3)--(0,3)--cycle); draw((0,1)--(1,1)--(1,2)--(0,2)); draw((2,1)--(2,2)); draw((3,1)--(3,2)); label("$23$",(0.5,0),S); label("$12$",(4,1.5),E); [/asy]
$\text{(A)}\ 27 \qquad \text{(B)}\ 29 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 35$

Solution: Looking at the vertical column, the three numbers sum to $23$ .  If we make the numbers on either end $9$ and $8$ in some order, the middle number will be $6$ .  This is the minimum for the intersection.
Looking at the horizontal row, the four numbers sum to $12$ .  If we minimize the three numbers on the right to $123$ , the first number has a maximum value of $6$ .  This is the maximum for the intersection
Thus, the minimum of the intersection is $6$ , and the maximum of the intersection is $6$ .  This means the intersection must be $6$ , and the other numbers must be $9$ and $8$ in the column, and $123$ in the row.  The sum of all the numbers is $12 + 23 - 6 = 29$ , and the answer is $\boxed{29}$

Answer: B 29
Problem: The remainder when the product $1492\cdot 1776\cdot 1812\cdot 1996$ is divided by 5 is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

Solution: To determine a remainder when a number is divided by $5$ , you only need to look at the last digit.  If the last digit is $0$ or $5$ , the remainder is $0$ .  If the last digit is $1$ or $6$ , the remainder is $1$ , and so on.
To determine the last digit of $1492\cdot 1776\cdot 1812\cdot 1996$ , you only need to look at the last digit of each number in the product.  Thus, we compute $2\cdot 6\cdot 2\cdot 6 = 12^2 = 144$ .  The last digit of the number $1492\cdot 1776\cdot 1812\cdot 1996$ is also $4$ , and thus the remainder when the number is divided by $5$ is also $4$ , which gives an answer of $\boxed{4}$

Answer: E 4
Problem: The remainder when the product $1492\cdot 1776\cdot 1812\cdot 1996$ is divided by 5 is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

Solution: $1492\cdot 1776\cdot 1812\cdot 1996 \mod 5$
$\equiv 2\cdot 1\cdot 2\cdot 1 \mod 5$
$=4$ , which is option $\boxed{4}$

Answer: E 4
Problem: $1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$
$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$

Solution: Put the numbers in groups of $4$
$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996)$
The first group has a sum of $0$
The second group increases the two positive numbers on the end by $1$ , and decreases the two negative numbers in the middle by $1$ .  Thus, the second group also has a sum of $0$
Continuing the pattern, every group has a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\boxed{0}$

Answer: C 0
Problem: $1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$
$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$

Solution: Let any term of the series be $t_n$ . Realize that at every $n\equiv0 \pmod4$ , the sum of the series is 0. For $t_{1996}$ we know $1996\equiv0 \pmod4$ so the solution is $\boxed{0}$

Answer: C 0
Problem: Figure $OPQR$ is a square.  Point $O$ is the origin, and point $Q$ has coordinates (2,2).  What are the coordinates for $T$ so that the area of triangle $PQT$ equals the area of square $OPQR$
[asy] pair O,P,Q,R,T; O = (0,0); P = (2,0); Q = (2,2); R = (0,2); T = (-4,0); draw((-5,0)--(3,0)); draw((0,-1)--(0,3)); draw(P--Q--R); draw((-0.2,-0.8)--(0,-1)--(0.2,-0.8)); draw((-0.2,2.8)--(0,3)--(0.2,2.8)); draw((-4.8,-0.2)--(-5,0)--(-4.8,0.2)); draw((2.8,-0.2)--(3,0)--(2.8,0.2)); draw(Q--T);  label("$O$",O,SW); label("$P$",P,S); label("$Q$",Q,NE); label("$R$",R,W); label("$T$",T,S); [/asy]
$\text{(A)}\ (-6,0) \qquad \text{(B)}\ (-4,0) \qquad \text{(C)}\ (-2,0) \qquad \text{(D)}\ (2,0) \qquad \text{(E)}\ (4,0)$

Solution: The area of $\square OPQR$ is $2^2 = 4$
The area of $\triangle PQT$ is $\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ$
If we set the areas equal, the area of $\triangle PQT$ is $4$ .  Also, note that $PQ=2$ .  Plugging those in, we get:
$4 = \frac{1}{2} \cdot PT \cdot 2$
$PT = 4$
If $PT = 4$ , and $PO = 2$ , then $OT = 2$ , and $T$ must be $2$ units to the left of the origin.  This would be $(-2,0)$ , giving answer $\boxed{2,0}$

Answer: C 2,0
Problem: Ana's monthly salary was $$2000$ in May.  In June she received a 20% raise.  In July she received a 20% pay cut.  After the two changes in June and July, Ana's monthly salary was
$\text{(A)}\ 1920\text{ dollars} \qquad \text{(B)}\ 1980\text{ dollars} \qquad \text{(C)}\ 2000\text{ dollars} \qquad \text{(D)}\ 2020\text{ dollars} \qquad \text{(E)}\ 2040\text{ dollars}$

Solution: Notice that a 20% raise translates to Ana's monthly salary multiplied by $(100 + 10)\% = 120\%$ , and a 20% pay cut translates to her salary multiplied by $100\% - 20\% = 80\%$ , so
\[2000 \cdot 120\% = 2400\] \[2400 \cdot 80\% = 1920 \rightarrow \boxed{1920}\]

Answer: A 1920
Problem: The pie charts below indicate the percent of students who prefer golf, bowling, or tennis at East Junior High School and West Middle School.  The total number of students at East is 2000 and at West, 2500.  In the two schools combined, the percent of students who prefer tennis is
[asy] unitsize(18); draw(circle((0,0),4)); draw(circle((9,0),4)); draw((-4,0)--(0,0)--4*dir(352.8)); draw((0,0)--4*dir(100.8)); draw((5,0)--(9,0)--(4*dir(324)+(9,0))); draw((9,0)--(4*dir(50.4)+(9,0)));  label("$48\%$",(0,-1),S); label("bowling",(0,-2),S); label("$30\%$",(1.5,1.5),N); label("golf",(1.5,0.5),N); label("$22\%$",(-2,1.5),N); label("tennis",(-2,0.5),N);  label("$40\%$",(8.5,-1),S); label("tennis",(8.5,-2),S); label("$24\%$",(10.5,0.5),E); label("golf",(10.5,-0.5),E); label("$36\%$",(7.8,1.7),N); label("bowling",(7.8,0.7),N);  label("$\textbf{East JHS}$",(0,-4),S); label("$\textbf{2000 students}$",(0,-5),S); label("$\textbf{West MS}$",(9,-4),S); label("$\textbf{2500 students}$",(9,-5),S); [/asy]
$\text{(A)}\ 30\% \qquad \text{(B)}\ 31\% \qquad \text{(C)}\ 32\% \qquad \text{(D)}\ 33\% \qquad \text{(E)}\ 34\%$

Solution: In the first school, $2000 \cdot 22\% = 2000 \cdot 0.22 = 440$ students prefer tennis.
In the second school, $2500 \cdot 40\% = 2500 \cdot 0.40 = 1000$ students prefer tennis.
In total, $440 + 1000 = 1440$ students prefer tennis out of a total of $2000 + 2500 = 4500$ students
This means $\frac{1440}{4500}\cdot 100\% = \frac{32}{100} \cdot 100\% = 32\%$ of the students in both schools prefer tennis, giving answer $\boxed{32}$

Answer: C 32
Problem: Suppose there is a special key on a calculator that replaces the number $x$ currently displayed with the number given by the formula $1/(1-x)$ .  For example, if the calculator is displaying 2 and the special key is pressed, then the calculator will display -1 since $1/(1-2)=-1$ .  Now suppose that the calculator is displaying 5.  After the special key is pressed 100 times in a row, the calculator will display
$\text{(A)}\ -0.25 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 0.8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 5$

Solution: We look for a pattern, hoping this sequence either settles down to one number, or that it forms a cycle that repeats.
After $1$ press, the calculator displays $\frac{1}{1 - 5} = -\frac{1}{4}$
After $2$ presses, the calculator displays $\frac{1}{1 - (-\frac{1}{4})} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$
After $3$ presses, the calculator displays $\frac{1}{1 - \frac{4}{5}} = \frac{1}{\frac{1}{5}} = 5$
Thus, every three presses, the display will be $5$ .  On press $3\cdot 33 = 99$ , the display will be $5$ .  One more press will give $-\frac{1}{4}$ , which is answer $\boxed{0.25}$

Answer: A 0.25
Problem: How many subsets containing three different numbers can be selected from the set \[\{ 89,95,99,132, 166,173 \}\] so that the sum of the three numbers is even?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$

Solution: To have an even sum with three numbers, we must add either $E+O+O$ , or $E + E + E$ , where $O$ represents an odd number, and $E$ represents an even number.
Since there are not three even numbers in the given set, $E+E+E$ is impossible.  Thus, we must choose two odd numbers, and one even number.
There are $2$ choices for the even number.
There are $4$ choices for the first odd number.  There are $3$ choices for the last odd number.  But the order in which we pick these 2 numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of $2$ .   Thus, we have $\frac{4\cdot 3}{2} = 6$ choices for a pair of odd numbers.
In total, there are $2$ choices for an even number, and $6$ choices for the odd numbers, giving a total of $2\cdot 6 = 12$ possible choices for a 3-element set that has an even sum.  This is option $\boxed{12}$

Answer: D 12
Problem: The manager of a company planned to distribute a $$50$ bonus to each employee from the company fund, but the fund contained $$5$ less than what was needed.  Instead the manager gave each employee a $$45$ bonus and kept the remaining $$95$ in the company fund.  The amount of money in the company fund before any bonuses were paid was
$\text{(A)}\ 945\text{ dollars} \qquad \text{(B)}\ 950\text{ dollars} \qquad \text{(C)}\ 955\text{ dollars} \qquad \text{(D)}\ 990\text{ dollars} \qquad \text{(E)}\ 995\text{ dollars}$

Solution: Let $p$ be the number of people in the company, and $f$ be the amount of money in the fund.
The first sentence states that $50p = f + 5$
The second sentence states that $45p = f - 95$
Subtracing the second equation from the first, we get $5p = 100$ , leading to $p = 20$
Plugging that number into the first equation gives $50\cdot 20 = f + 5$ , leading to $f = 995$ , which is answer $\boxed{995}$

Answer: E 995
Problem: The manager of a company planned to distribute a $$50$ bonus to each employee from the company fund, but the fund contained $$5$ less than what was needed.  Instead the manager gave each employee a $$45$ bonus and kept the remaining $$95$ in the company fund.  The amount of money in the company fund before any bonuses were paid was
$\text{(A)}\ 945\text{ dollars} \qquad \text{(B)}\ 950\text{ dollars} \qquad \text{(C)}\ 955\text{ dollars} \qquad \text{(D)}\ 990\text{ dollars} \qquad \text{(E)}\ 995\text{ dollars}$

Solution: Since the company must employ a whole number of employees, the amount of money in the fund must be $5$ dollars less than a multiple of $50$ .  Only options $A$ and $E$ satisfy that requirement.
Additionally, the number must be $95$ more than a multiple of $45$ .  Since $45 \cdot 20 = 900$ , the only number that is $95$ more than a multiple of $45$ out of options $A$ and $E$ is option $\boxed{995}$

Answer: E 995
Problem: The measure of angle $ABC$ is $50^\circ$ $\overline{AD}$ bisects angle $BAC$ , and $\overline{DC}$ bisects angle $BCA$ .  The measure of angle $ADC$ is
[asy] pair A,B,C,D; A = (0,0); B = (9,10); C = (10,0); D = (6.66,3); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--cycle); draw(A--D--C);  label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,N); label("$50^\circ $",(9.4,8.8),SW); [/asy]
$\text{(A)}\ 90^\circ \qquad \text{(B)}\ 100^\circ \qquad \text{(C)}\ 115^\circ \qquad \text{(D)}\ 122.5^\circ \qquad \text{(E)}\ 125^\circ$

Solution: Let $\angle CAD = \angle BAD = x$ , and let $\angle ACD = \angle BCD = y$
From $\triangle ABC$ , we know that $50 + 2x + 2y = 180$ , leading to $x + y = 65$
From $\triangle ADC$ , we know that $x + y + \angle D = 180$ .  Plugging in $x + y = 65$ , we get $\angle D = 180 - 65 = 115$ , which is answer $\boxed{115}$

Answer: C 115
Problem: Walter has exactly one penny, one nickel, one dime and one quarter in his pocket.  What percent of one dollar is in his pocket?
$\text{(A)}\ 4\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 41\% \qquad \text{(E)}\ 59\%$

Solution: Walter has $1 + 5 + 10 + 25 = 41$ cents in his pocket.  There are $100$ cents in a dollar.  Therefore, Walter has $\frac{41}{100}\cdot 100\% = 41\%$ of a dollar in his pocket, and the right answer is $\boxed{41}$

Answer: D 41
Problem: Jose is $4$ years younger than Zack.  Zack is $3$ years older than Inez.  Inez is $15$ years old.  How old is Jose?
$\text{(A)}\ 8 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22$

Solution: Zack is $3$ years older than Inez, who is $15$ .  Therefore, Zack is $15 + 3 = 18$ years old.
Jose is $4$ years younger than Zack, who is $18$ .  Therefore, Jose is $18 - 4 = 14$ years old, and the answer is $\boxed{14}$

Answer: C 14
Problem: A teacher tells the class,
Ben thinks of $6$ , and gives his answer to Sue.  What should Sue's answer be?
$\text{(A)}\ 18 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 30$

Solution: Ben adds $1$ to $6$ , getting $1 + 6 = 7$ .  Then Ben doubles it, to get $7\cdot 2 = 14$ .  He gives this number to Sue.  Then Sue subtracts $1$ from the number, to get $14 - 1 = 13$ .  Finally, Sue doubles the result to get $13\cdot 2 = 26$ , and the answer is $\boxed{26}$

Answer: C 26
Problem: Find the smallest whole number that is larger than the sum
\[2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4}+5\dfrac{1}{5}.\]
$\text{(A)}\ 14 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 17 \qquad \text{(E)}\ 18$

Solution: Adding the whole numbers gives $2 + 3 + 4 + 5 = 14$
Adding the fractions gives $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{30}{60} + \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{77}{60}$ .  This will create one more whole, and a fraction that is less than $1$ .  Thus, the smallest whole number that is less than $15$ plus some fractional part is $16$ , and the answer is $\boxed{16}$

Answer: C 16
Problem: Find the smallest whole number that is larger than the sum
\[2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4}+5\dfrac{1}{5}.\]
$\text{(A)}\ 14 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 17 \qquad \text{(E)}\ 18$

Solution: Convert the fractional parts to decimals, and approximate the answer. $2.5 + 3.33 + 4.25 + 5.2 = 15.28$ , and the answer is $16$ , which is $\boxed{16}$

Answer: C 16
Problem: Find the smallest whole number that is larger than the sum
\[2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4}+5\dfrac{1}{5}.\]
$\text{(A)}\ 14 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 17 \qquad \text{(E)}\ 18$

Solution: Clearly, break the mixed numbers into $2+3+4+5+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
This is the same as $14+(\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{5})$ . Notice that clearly, $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ , and $1>\frac{1}{4}+\frac{1}{5}>\frac{1}{6}$ (clearly, $\frac{1}{4}>\frac{1}{6}$ ). This means that the fractional parts is between 1 and 2, so when that is added to 14, the original expression is between 15, and 16. So the least integer that is bigger than this sum is clearly 16. So select $\boxed{16}$

Answer: C 16
Problem: Figures $I$ $II$ , and $III$ are squares.  The perimeter of $I$ is $12$ and the perimeter of $II$ is $24$ .  The perimeter of $III$ is
[asy] draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle); draw((9,0)--(9,9)); draw((9,6)--(12,6)); label("$III$",(4.5,4),N); label("$II$",(12,2.5),N); label("$I$",(10.5,6.75),N); [/asy]
$\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(D)}\ 81$

Solution: Since the perimeter of $I$ $12$ , each side is $\frac{12}{4} = 3$
Since the perimeter of $II$ is $24$ , each side is $\frac{24}{4} = 6$
The side of $III$ is equal to the sum of the sides of $I$ and $II$ .  Therefore, the side of $III$ is $3 + 6 = 9$
Since $III$ is also a square, it has an perimeter of $9\cdot 4 = 36$ , and the answer is $\boxed{36}$

Answer: C 36
Problem: Figures $I$ $II$ , and $III$ are squares.  The perimeter of $I$ is $12$ and the perimeter of $II$ is $24$ .  The perimeter of $III$ is
[asy] draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle); draw((9,0)--(9,9)); draw((9,6)--(12,6)); label("$III$",(4.5,4),N); label("$II$",(12,2.5),N); label("$I$",(10.5,6.75),N); [/asy]
$\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(D)}\ 81$

Solution: Let a side of $I$ equal $x$ , and let a side of $II$ equal $y$ . The perimeter of $I$ is $4x$ , and the perimeter of $II$ is $4y$ . One side of $III$ has length $x+y$ , so the perimeter is $4x+4y$ , which just so happens to be the sum of the perimeters of $I$ and $II$ , giving us $12+24=36$ , or answer $\boxed{36}$

Answer: C 36
Problem: An American traveling in Italy wishes to exchange American money (dollars) for Italian money (lire).  If 3000 lire = 1.60, how much lire will the traveler receive in exchange for 1.00?
$\text{(A)}\ 180 \qquad \text{(B)}\ 480 \qquad \text{(C)}\ 1800 \qquad \text{(D)}\ 1875 \qquad \text{(E)}\ 4875$

Solution: $\frac{3000 \text{ lire}}{1.6 \text{ dollars}} = \frac{x \text{ lire}}{1 \text{ dollar}}$
$x = 1875$ , and the answer is $\boxed{1875}$

Answer: D 1875
Problem: Three congruent circles with centers $P$ $Q$ , and $R$ are tangent to the sides of rectangle $ABCD$ as shown.  The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R$ .  The area of the rectangle is
[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,W); label("$R$",R,W); [/asy]
$\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

Solution: If circle $Q$ has diameter $4$ , then so do congruent circles $P$ and $R$ .  Draw a diameter through $P$ parallel to $AD$ .  The diameter will be congruent to $AD$ , and thus $AD = 4$ , which is the height of the rectangle.
Draw a horizontal line $PQR$ that extends to the sides of the rectangle.  This line is $2$ diameters long, so it has length $4\cdot 2 = 8$ .  It is parallel and congruent to $AB$ , so the width of the rectangle is $8$
Thus, the area of the rectangle is $lw = 4\cdot 8 = 32$ , and the answer is $\boxed{32}$

Answer: C 32
Problem: Three congruent circles with centers $P$ $Q$ , and $R$ are tangent to the sides of rectangle $ABCD$ as shown.  The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R$ .  The area of the rectangle is
[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,W); label("$R$",R,W); [/asy]
$\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

Solution: The area of circle $P$ and circle $R$ almost fill the rectangle.  Circle $P$ has radius $\frac{4}{2} = 2$ , and so does circle $R$ .  Thus, the sum of their areas is $\pi \cdot 2^2 + \pi \cdot 2^2 = 8\pi$ .  Since $\pi \approx 3.14$ , the area of the circles is just over $24$ .  The area of the rectangle is greater than the area of the circles. $64$ would be too high, as the two circles appear to take up much more than half the area of the rectangle.  But $24$ is too low, as it is less than the area of the two circles.  Thus, the only reasonable answer is $\boxed{32}$ , which is $32$

Answer: C 32
Problem: lucky year is one in which at least one date, when written in the form month/day/year, has the following property: The product of the month times the day equals the last two digits of the year .  For example, 1956 is a lucky year because it has the date 7/8/56 and $7\times 8 = 56$ .  Which of the following is NOT a lucky year?
$\text{(A)}\ 1990 \qquad \text{(B)}\ 1991 \qquad \text{(C)}\ 1992 \qquad \text{(D)}\ 1993 \qquad \text{(E)}\ 1994$

Solution: We examine only the factors of $90, 91, 92, 93,$ and $94$ that are less than $13$ , because for a year to be lucky, it must have at least one factor between $1$ and $12$ to represent the month.
$90$ has factors of $1, 2, 3, 5, 6, 9,$ and $10$ .  Dividing $90$ by the last number $10$ , gives $9$ .  Thus, $10/9/90$ is a valid date, and $1990$ is a lucky year.
$91$ has factors of $1$ and $7$ .  Dividing $91$ by $7$ gives $13$ , and $7/13/91$ is a valid date.  Thus, $1991$ is a lucky year.
$92$ has factors of $1, 2$ and $4$ .  Dividing $92$ by $4$ gives $23$ , and $4/23/92$ is a valid date.  Thus, $1992$ is a lucky year.
$93$ has factors of $1$ and $3$ .  Dividing $93$ by $3$ gives $31$ , and $3/31/93$ is a valid date.  Thus, $1993$ is a lucky year.
$94$ has factors of $1$ and $2$ .  Dividing $94$ by $2$ gives $47$ , and no month has $47$ days.  Thus, $1994$ is not a lucky year, and the answer is $\boxed{1994}$

Answer: E 1994
Problem: A team won $40$ of its first $50$ games.  How many of the remaining $40$ games must this team win so it will have won exactly $70 \%$ of its games for the season?
$\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution: Noting that 70% is the same as $\frac{70}{100}=\frac{7}{10}$ , and that, when x is the amount of wins in the last 40 games, the fraction of games won is $\frac{40+x}{50+40}=\frac{40+x}{90}$ , all we have to do is set them equal: \[\frac{40+x}{90}=\frac{7}{10}\] \[40+x=63\] \[x=\boxed{23}\]

Answer: B 23