Navvye/Olympiad-Problems-CSV huggingface dataset:
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: Alternatively we can note that they will play a total of $40+50=90$ games and must win $0.7(90)=63$ games. Since they won $40$ games already they need $63-40=\boxed{23}$ more games.

Answer: B 23
Problem: What is the $100^\text{th}$ digit to the right of the decimal point in the decimal form of $4/37$
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution: $\frac{4}{37}=\frac{12}{111}=\frac{108}{999}=0.108108108...$
Since this repeats every three digits, digit number x = digit number (x mod 3), and the 100th digit = (100 mod 3)th digit = 1st digit = $\boxed{1}$

Answer: B 1
Problem: Students from three middle schools worked on a summer project.
The total amount paid for the students' work was 744.  Assuming each student received the same amount for a day's work, how much did the students from Balboa school earn altogether?
$\text{(A)}\ 9.00\text{ dollars} \qquad \text{(B)}\ 48.38\text{ dollars} \qquad \text{(C)}\ 180.00\text{ dollars} \qquad \text{(D)}\ 193.50\text{ dollars} \qquad \text{(E)}\ 258.00\text{ dollars}$

Solution: Altogether, the summer project totaled $(7)(3)+(4)(5)+(5)(9)=21+20+45=86$ days of work for a single student. This equals $744/86=9$ dollars per day per student. The students from Balboa school earned $9(4)(5)=\boxed{180.00}$

Answer: C 180.00
Problem: The table below gives the percent of students in each grade at Annville and Cleona elementary schools:
\[\begin{tabular}{rccccccc}&\textbf{\underline{K}}&\textbf{\underline{1}}&\textbf{\underline{2}}&\textbf{\underline{3}}&\textbf{\underline{4}}&\textbf{\underline{5}}&\textbf{\underline{6}}\\ \textbf{Annville:}& 16\% & 15\% & 15\% & 14\% & 13\% & 16\% & 11\%\\ \textbf{Cleona:}& 12\% & 15\% & 14\% & 13\% & 15\% & 14\% & 17\%\end{tabular}\]
Annville has 100 students and Cleona has 200 students.  In the two schools combined, what percent of the students are in grade 6?
$\text{(A)}\ 12\% \qquad \text{(B)}\ 13\% \qquad \text{(C)}\ 14\% \qquad \text{(D)}\ 15\% \qquad \text{(E)}\ 28\%$

Solution: By the tables, Annville has $11$ 6th graders and Cleona has $34$ . Together they have $45$ 6th graders and $300$ total students, so the percent is $\frac{45}{300}=\frac{15}{100}= \boxed{15}$

Answer: D 15
Problem: The number $6545$ can be written as a product of a pair of positive two-digit numbers.  What is the sum of this pair of numbers?
$\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222$

Solution: The prime factorization of $6545$ is $5\cdot7\cdot11\cdot17  =385\cdot17$ , which contains a three digit number, but we want 6545 to be expressed as ab x cd. Now we do trial and error: \[5\cdot7=35 \text{,   } 11\cdot17=187 \text{  X}\] \[5\cdot11=55 \text{,   } 7\cdot17=119 \text{  X}\] \[5\cdot17=85 \text{,   } 7\cdot11=77 \text{ }\surd\] \[85+77= \boxed{162}\]

Answer: A 162
Problem: How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?
$\text{(A)}\ 1120 \qquad \text{(B)}\ 1400 \qquad \text{(C)}\ 1800 \qquad \text{(D)}\ 2025 \qquad \text{(E)}\ 2500$

Solution: Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are $5*5*8*7= \boxed{1400}$ numbers.

Answer: B 1400
Problem: Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes $5$ hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 9\qquad\text{(D)}\ 10\qquad\text{(E)}\ 11$

Solution: Say you are on the Houston-bound bus that left at 12:30 in the afternoon, looking out the window to see how many buses you pass. At 12:45 pm, the Dallas bus that left at 8:00 am is 4:45 away (Note - $a:b$ $a$ is for hrs. and $b$ is for min.) from Dallas, and therefore 15 minutes from Houston. Your bus is also 15 minutes from Houston, so you are delighted at the first bus you have passed.
At 1:15 pm, the 9:00 am Dallas bus meets you again, being 4:15 away from Dallas and therefore 45 minutes from Houston. After a while you might notice that the buses meet you in 30 minute intervals after 12:45 pm; indeed, the bus that left an hour later than the bus most recently has traveled 30 minutes closer, and so have you, for a total of an hour closer, and when you passed the most recent bus, the bus after it had indeed left an hour later. You decide to predict how many buses you pass by finding all valid times that are 30 minute intervals from 12:45. They are
12:45, 1:15, 1:45, 2:15, 2:45, 3:15, 3:45, 4:15, 4:45, 5:15, and 5:30, your arrival time.
Hence, you will pass $\boxed{10}$ Dallas-bound buses.

Answer: D 10
Problem: $\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=$
$\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$

Solution: $1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45$
$\frac{45+55}{10} = \dfrac{100}{10} = \boxed{10}$

Answer: D 10
Problem: Given that $\text{1 mile} = \text{8 furlongs}$ and $\text{1 furlong} = \text{40 rods}$ , the number of rods in one mile is
$\text{(A)}\ 5 \qquad \text{(B)}\ 320 \qquad \text{(C)}\ 660 \qquad \text{(D)}\ 1760 \qquad \text{(E)}\ 5280$

Solution: \[(1\ \text{mile}) \left( \frac{8\ \text{furlongs}}{1\ \text{mile}} \right) \left( \frac{40\ \text{rods}}{1\ \text{furlong}} \right) = \boxed{320}\]

Answer: B 320
Problem: The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$ . Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\boxed{0}$

Answer: A 0
Problem: The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: We can easily compute the product of the first 6 positive integers: $(1*2*3*4*5*6)=6!=720$ Therefore the units digit must be $\boxed{0}$

Answer: A 0
Problem: If $\angle A = 60^\circ$ $\angle E = 40^\circ$ and $\angle C = 30^\circ$ , then $\angle BDC =$
[asy] pair A,B,C,D,EE; A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5); draw(A--C--D); draw(B--EE--A); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",EE,N); [/asy]
$\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ$

Solution: The sum of the angles in a triangle is $180^\circ$ . We can find $\angle ABE = 80^\circ$ , so $\angle CBD = 180-80=100^\circ$
\[\angle BDC = 180-100-30=\boxed{50}\]

Answer: B 50
Problem: For how many three-digit whole numbers does the sum of the digits equal $25$
$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Solution: Because $8+8+8=24$ , it follows that one of the digits must be a $9$ . The other two digits them have a sum of $25-9=16$ . The groups of digits that produce a sum of $25$ are $799, 889$ and can be arranged as follows
\[799,979,997,889,898,988\]
The number of configurations is $\boxed{6}$

Answer: C 6
Problem: A shopper buys a $100$ dollar coat on sale for $20\%$ off.  An additional $5$ dollars are taken off the sale price by using a discount coupon.  A sales tax of $8\%$ is paid on the final selling price.  The total amount the shopper pays for the coat is
$\text{(A)}\ \text{81.00 dollars} \qquad \text{(B)}\ \text{81.40 dollars} \qquad \text{(C)}\ \text{82.00 dollars} \qquad \text{(D)}\ \text{82.08 dollars} \qquad \text{(E)}\ \text{82.40 dollars}$

Solution: After the $20\%$ sale, the coat costs $100(0.8)=80$ dollars. Then $5$ dollars are taken off for a cost of $80-5=75$ . Adding on the sales tax, the final amount is $(75)(1.08)=\boxed{81.00}$

Answer: A 81.00
Problem: For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution: We should list all the positive divisors of $36$ and count them. By trial and error, the divisors of $36$ are found to be $1,2,3,4,6,9,12,18,36$ , for a total of $9$ . However, $1$ and $2$ can't be equal to $N+2$ for a POSITIVE integer N, so the number of possibilities is $\boxed{7}$

Answer: A 7
Problem: For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution: To find the number of positive divisors of $36$ , first prime factorize $36$ to get $2^2*3^2$ . Then add $1$ to the power of both $2$ and $3$ to get $3$ . Multiply $3*3$ to get $9$ . Since the problem is asking only for positive integer values of N, subtract $2$ from $9$ (since $2-2$ and $1-2$ result in integers that are not positive) to get $\boxed{7}$

Answer: A 7
Problem: Last summer $100$ students attended basketball camp.  Of those attending, $52$ were boys and $48$ were girls.  Also, $40$ students were from Jonas Middle School and $60$ were from Clay Middle School.  Twenty of the girls were from Jonas Middle School.  How many of the boys were from Clay Middle School?
$\text{(A)}\ 20 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 52$

Solution: Make a table with the information given.
\[\begin{tabular}{c|ccc} & \text{Jonas} & \text{Clay} & \text{total} \\ \hline \text{boys} & & & 52 \\ \text{girls} & 20 & & 48 \\ \text{total} & 40 & 60 & 100  \end{tabular}\]
Because the first two columns must add up to the third column, and the same with the rows, the rest of the empty boxes can be filled in.
\[\begin{tabular}{c|ccc} & \text{Jonas} & \text{Clay} & \text{total} \\ \hline \text{boys} & 20 & 32 & 52 \\ \text{girls} & 20 & 28 & 48 \\ \text{total} & 40 & 60 & 100  \end{tabular}\]
The number of boys from Clay is $\boxed{32}$

Answer: B 32
Problem: Two children at a time can play pairball.  For $90$ minutes, with only two children playing at time, five children take turns so that each one plays the same amount of time.  The number of minutes each child plays is
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36$

Solution: There are $2 \times 90 = 180$ minutes of total playing time. Divided equally among the five children, each child gets $180/5 = \boxed{36}$ minutes.

Answer: E 36
Problem: The perimeter of one square is $3$ times the perimeter of another square.  The area of the larger square is how many times the area of the smaller square?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$

Solution: Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$ . If the perimeter of one is $3$ times the other's, then $a=3b$ . The area of the larger square over the area of the smaller square is
\[\frac{a^2}{b^2} = \frac{(3b)^2}{b^2} = \frac{9b^2}{b^2} = \boxed{9}\]

Answer: E 9
Problem: Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour.  If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number of hours it will take her to shovel it clean is closest to
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 12$

Solution: Her driveway has $(4)(10)(3)=120$ cubic yards of snow. After the first hour she would have $120-20=100$ cubic yards, then $100-19=81$ $81-18=63$ $63-17=46$ $46-16=30$ $30-15=15$ , and $15-14=1$ cubic yard after the seventh hour. It will take her a little more than seven hours to shovel it clean, which is closest to $\boxed{7}$

Answer: D 7
Problem: Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters.  Another square, $ABCD$ , has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles.  The area of the square $ABCD$ is
[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); [/asy]
$\text{(A)}\ 16 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 64$

Solution: The radius of each semicircle is $2$ , half the sidelength of the square. The line straight down the middle of square $ABCD$ is the sum of two radii and the length of the smaller square, which is equivalent to its side length. The area of $ABCD$ is $(4+2+2)^2 = \boxed{64}$

Answer: E 64
Problem: A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs.  The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
$\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$

Solution: If a person gets three gumballs of each of the three colors, that is, $9$ gumballs, then the $10^{\text{th}}$ gumball must be the fourth one for one of the colors. Therefore, the person must buy $\boxed{10}$ gumballs.

Answer: C 10
Problem: $2$ by $2$ square is divided into four $1$ by $1$ squares.  Each of the small squares is to be painted either green or red.  In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square?  There may be as few as zero or as many as four small green squares.
$\text{(A)}\  4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16$

Solution: If a green square cannot share its top or right side with a red square, then a red square can not share its bottom or left side with a green square. Let us split this up into several cases.
Case 1: There are no green squares. This can be done in $1$ way.
Case 2: There is one green square and three red squares. This can only be done when the green square's top and right edges are against the edge, so there is $1$ way.
Case 3: There are two green squares and two red squares. This happens when the two green squares are in the two top squares or two right squares, so there are $2$ ways.
Case 4: There are three green squares and one red square. Similar to case 2, this happens when the red square's left and bottom edges are against the edge, so there is $1$ way.
Case 5: There are four green squares and zero red squares. $1$ way.
\[1+1+2+1+1 = \boxed{6}\]

Answer: B 6
Problem: Find the sum of the digits in the answer to
$\underbrace{9999\cdots 99}_{94\text{ nines}} \times \underbrace{4444\cdots 44}_{94\text{ fours}}$
where a string of $94$ nines is multiplied by a string of $94$ fours.
$\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072$

Solution: Notice that:
$9 \cdot 4 = 36$ and $3+6 = 9 = 9 \cdot 1$
$99 \cdot 44 = 4356$ and $4+5+3+6 = 18 = 9 \cdot 2$
So the sum of the digits of $x$ 9s times $x$ 4s is simply $x \cdot 9$ (Try to find the proof that it works for all values of $x$ ~MATHWIZARD10).
Therefore the answer is $94 \cdot 9 = \boxed{846.}$

Answer: A 846.
Problem: Find the sum of the digits in the answer to
$\underbrace{9999\cdots 99}_{94\text{ nines}} \times \underbrace{4444\cdots 44}_{94\text{ fours}}$
where a string of $94$ nines is multiplied by a string of $94$ fours.
$\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072$

Solution: \[\underbrace{9999\cdots 99}_{94\text{ nines}} \cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = (10^{94}-1)\cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{94\text{ fours}} \underbrace{000\cdots 0}_{94\text{ zeros}} - \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{93\text{ fours}} 3 \underbrace{555\cdots 5}_{93\text{ zeros}}6\]
\[4 \cdot 93 + 3 + 5 \cdot 93 + 6 = 9 \cdot 94 = \boxed{846}\]

Answer: A 846
Problem: When the fraction $\dfrac{49}{84}$ is expressed in simplest form, then the sum of the numerator and the denominator will be
$\text{(A)}\ 11 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 133$

Solution: \begin{align*} \dfrac{49}{84} &= \dfrac{7^2}{2^2\cdot 3\cdot 7} \\ &= \dfrac{7}{2^2\cdot 3} \\ &= \dfrac{7}{12}. \end{align*}
The sum of the numerator and denominator is $7+12=19\rightarrow \boxed{19}$

Answer: C 19
Problem: A can of soup can feed $3$ adults or $5$ children.  If there are $5$ cans of soup and $15$ children are fed, then how many adults would the remaining soup feed?
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Solution: A can of soup will feed $5$ children so $15$ children are feed by $3$ cans of soup. Therefore, there are $5-3=2$ cans for adults, so $3 \times 2 =\boxed{6}$ adults are fed.

Answer: B 6
Problem: To control her blood pressure, Jill's grandmother takes one half of a pill every other day.  If one supply of medicine contains $60$ pills, then the supply of medicine would last approximately
$\text{(A)}\ 1\text{ month} \qquad \text{(B)}\ 4\text{ months} \qquad \text{(C)}\ 6\text{ months} \qquad \text{(D)}\ 8\text{ months} \qquad \text{(E)}\ 1\text{ year}$

Solution: If Jill's grandmother takes one half of a pill every other day, she takes a pill every $4$ days. Since she has $60$ pills, the supply will last $60 \times 4=240$ days which is about $\boxed{8}$

Answer: D 8
Problem: Consider the operation $*$ defined by the following table:
\[\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}\]
For example, $3*2=1$ .  Then $(2*4)*(1*3)=$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: Using the chart, $(2*4)=3$ and $(1*3)=3$ . Therefore, $(2*4)*(1*3)=3*3=\boxed{4}$

Answer: D 4
Problem: Consider the operation $*$ defined by the following table:
\[\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}\]
For example, $3*2=1$ .  Then $(2*4)*(1*3)=$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: By the chart, we can see that the " $*$ " operation is actually multiplication modulo $5$ . Thus, we can do $(2*4)*(1*3)\rightarrow(2\cdot4)\cdot(1\cdot3)=8\cdot3=24\rightarrow\boxed{4}$

Answer: D 4
Problem: Consider this histogram of the scores for $81$ students taking a test:
[asy] unitsize(12); draw((0,0)--(26,0)); draw((1,1)--(25,1)); draw((3,2)--(25,2)); draw((5,3)--(23,3)); draw((5,4)--(21,4)); draw((7,5)--(21,5)); draw((9,6)--(21,6)); draw((11,7)--(19,7)); draw((11,8)--(19,8)); draw((11,9)--(19,9)); draw((11,10)--(19,10)); draw((13,11)--(19,11)); draw((13,12)--(19,12)); draw((13,13)--(17,13)); draw((13,14)--(17,14)); draw((15,15)--(17,15)); draw((15,16)--(17,16));  draw((1,0)--(1,1)); draw((3,0)--(3,2)); draw((5,0)--(5,4)); draw((7,0)--(7,5)); draw((9,0)--(9,6)); draw((11,0)--(11,10)); draw((13,0)--(13,14)); draw((15,0)--(15,16)); draw((17,0)--(17,16)); draw((19,0)--(19,12)); draw((21,0)--(21,6)); draw((23,0)--(23,3)); draw((25,0)--(25,2));  for (int a = 1; a < 13; ++a) {     draw((2*a,-.25)--(2*a,.25)); }  label("$40$",(2,-.25),S); label("$45$",(4,-.25),S); label("$50$",(6,-.25),S); label("$55$",(8,-.25),S); label("$60$",(10,-.25),S); label("$65$",(12,-.25),S); label("$70$",(14,-.25),S); label("$75$",(16,-.25),S); label("$80$",(18,-.25),S); label("$85$",(20,-.25),S); label("$90$",(22,-.25),S); label("$95$",(24,-.25),S);  label("$1$",(2,1),N); label("$2$",(4,2),N); label("$4$",(6,4),N); label("$5$",(8,5),N); label("$6$",(10,6),N); label("$10$",(12,10),N); label("$14$",(14,14),N); label("$16$",(16,16),N); label("$12$",(18,12),N); label("$6$",(20,6),N); label("$3$",(22,3),N); label("$2$",(24,2),N);  label("Number",(4,8),N); label("of Students",(4,7),N);  label("$\textbf{STUDENT TEST SCORES}$",(14,18),N); [/asy]
The median is in the interval labeled
$\text{(A)}\ 60 \qquad \text{(B)}\ 65 \qquad \text{(C)}\ 70 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 80$

Solution: Since $81$ students took the test, the median is the score of the $41^{st}$ student. The five rightmost intervals include $2+3+6+12+16=39$ students, so the $41^{st}$ one must lie in the next interval, which is $\boxed{70}$

Answer: C 70
Problem: If each of the three operation signs, $+$ $\text{--}$ $\times$ , is used exactly ONCE in one of the blanks in the expression
\[5\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}4\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}6\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}3\]
then the value of the result could equal
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 19$

Solution: There are a reasonable number of ways to place the operation signs, so guess and check to find that $5-4+6 \times 3 = \boxed{19}$

Answer: E 19
Problem: The word " HELP " in block letters is painted in black with strokes $1$ unit wide on a $5$ by $15$ rectangular white sign with dimensions as shown.  The area of the white portion of the sign, in square units, is
[asy] unitsize(12); fill((0,0)--(0,5)--(1,5)--(1,3)--(2,3)--(2,5)--(3,5)--(3,0)--(2,0)--(2,2)--(1,2)--(1,0)--cycle,black); fill((4,0)--(4,5)--(7,5)--(7,4)--(5,4)--(5,3)--(7,3)--(7,2)--(5,2)--(5,1)--(7,1)--(7,0)--cycle,black); fill((8,0)--(8,5)--(9,5)--(9,1)--(11,1)--(11,0)--cycle,black); fill((12,0)--(12,5)--(15,5)--(15,2)--(13,2)--(13,0)--cycle,black); fill((13,3)--(14,3)--(14,4)--(13,4)--cycle,white); draw((0,0)--(15,0)--(15,5)--(0,5)--cycle);  [/asy]
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 34 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 38$

Solution: Count the number of black squares in each letter. H has 11, E has 11, L has 7, and P has 10, giving the number of black squares to be $11+11+7+10=39$ . The total number of squares is $(15)(5)=75$ and the number of white squares is $75-39=\boxed{36}$

Answer: D 36
Problem: The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,2,3$ .  Then $A+B=$
\[\begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
The last square in the top row cannot be either 1 or 3, so it must be 2.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
The other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of $1+3=\boxed{4}$

Answer: C 4