Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: $\diamondsuit$ and $\Delta$ are whole numbers and $\diamondsuit \times \Delta =36$ .  The largest possible value of $\diamondsuit + \Delta$ is
$\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20\ \qquad \text{(E)}\ 37$

Solution: Since it doesn't take too long, we can just make a table with all the possible values of the sum \[\begin{array}{|c|c|c|} \multicolumn{3}{}{} \\ \hline \diamondsuit & \Delta & \diamondsuit + \Delta \\ \hline 36 & 1 & 37 \\ \hline 18 & 2 & 20 \\ \hline 12 & 3 & 15 \\ \hline 9 & 4 & 13 \\ \hline 6 & 6 & 12 \\ \hline \end{array}\]
Clearly $37\rightarrow \boxed{37}$ is the largest.

Answer: E 37
Problem: Placing no more than one $\text{X}$ in each small square , what is the greatest number of $\text{X}$ 's that can be put on the grid shown without getting three $\text{X}$ 's in a row vertically, horizontally, or diagonally?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
[asy] for(int a=0; a<4; ++a)  {   draw((a,0)--(a,3));  } for(int b=0; b<4; ++b)  {   draw((0,b)--(3,b));  } [/asy]

Solution: By the Pigeonhole Principle , if there are at least $7$ $\text{X}$ 's, then there will be some row with $3$ $\text{X}$ 's.  We can put in $6$ by leaving out the three boxes in one of the main diagonals.
$\rightarrow \boxed{6}$

Answer: E 6
Problem: The average weight of $6$ boys is $150$ pounds and the average weight of $4$ girls is $120$ pounds.  The average weight of the $10$ children is
$\text{(A)}\ 135\text{ pounds} \qquad \text{(B)}\ 137\text{ pounds} \qquad \text{(C)}\ 138\text{ pounds} \qquad \text{(D)}\ 140\text{ pounds} \qquad \text{(E)}\ 141\text{ pounds}$

Solution: Let the $6$ boys have total weight $S_B$ and let the $4$ girls have total weight $S_G$ .  We are given
\begin{align*} \frac{S_B}{6} &= 150 \\ \frac{S_G}{4} &= 120  \end{align*}
We want the average of the $10$ children, which is \[\frac{S_B+S_G}{10}\] From the first two equations , we can determine that $S_B=900$ and $S_G=480$ , so $S_B+S_G=1380$ .  Therefore, the average we desire is \[\frac{1380}{10}=138 \rightarrow \boxed{138}\]

Answer: C 138
Problem: What is the $100\text{th}$ number in the arithmetic sequence $1,5,9,13,17,21,25,...$
$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$

Solution: To get from the $1^\text{st}$ term of an arithmetic sequence to the $100^\text{th}$ term, we must add the common difference $99$ times.  The first term is $1$ and the common difference is $5-1=9-5=13-9=\cdots = 4$ , so the $100^\text{th}$ term is \[1+4(99)=397 \rightarrow \boxed{397}\]

Answer: A 397
Problem: What is the $100\text{th}$ number in the arithmetic sequence $1,5,9,13,17,21,25,...$
$\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405$

Solution: Alternatively you could create an equation for the arithmetic sequence: $a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3$
$a_{100}=4(100)-3=397$ , or $\boxed{397}$

Answer: A 397
Problem: The glass gauge on a cylindrical coffee maker shows that there are $45$ cups left when the coffee maker is $36\%$ full.  How many cups of coffee does it hold when it is full?
$\text{(A)}\ 80 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 130 \qquad \text{(E)}\ 262$
[asy] draw((5,0)..(0,-1.3)..(-5,0)); draw((5,0)--(5,10)); draw((-5,0)--(-5,10)); draw(ellipse((0,10),5,1.3)); draw(circle((.3,1.3),.4)); draw((-.1,1.7)--(-.1,7.9)--(.7,7.9)--(.7,1.7)--cycle); fill((-.1,1.7)--(-.1,4)--(.7,4)--(.7,1.7)--cycle,black); draw((-2,11.3)--(2,11.3)..(2.6,11.9)..(2,12.2)--(-2,12.2)..(-2.6,11.9)..cycle); [/asy]

Solution: Let the amount of coffee the maker will hold when full be $x$ .  Then, \[.36x=45 \Rightarrow x=125 \rightarrow \boxed{125}\]

Answer: C 125
Problem: A fifth number, $n$ , is added to the set $\{ 3,6,9,10 \}$ to make the mean of the set of five numbers equal to its median .  The number of possible values of $n$ is
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ \text{more than }4$

Solution: The possible medians after $n$ is added are $6$ $n$ , or $9$ .  Now we use casework
Case 1: The median is $6$
In this case, $n<6$ and \[\frac{3+n+6+9+10}{5}=6 \Rightarrow n=2\] so this case contributes $1$
Case 2: The median is $n$
We have $6<n<9$ and \[\frac{3+6+n+9+10}{5}=n \Rightarrow n=7\] so this case also contributes $1$
Case 3: The median is $9$
We have $9<n$ and \[\frac{3+6+9+n+10}{5}=9 \Rightarrow 17\] so this case adds $1$
In all there are $3\rightarrow \boxed{3}$ possible values of $n$

Answer: C 3
Problem: Maria buys computer disks at a price of $4$ for $$5$ and sells them at a price of $3$ for $$5$ .  How many computer disks must she sell in order to make a profit of $$100$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 1200$

Solution: This is the equivalent of saying she buys $12$ for $$15$ and sells $12$ for $$20$ , so for every dozen disks she sells, she profits $$5$
She needs to profit $$100$ , so she needs to sell $\frac{100}{5}=20$ dozen disks, which is $240\rightarrow \boxed{240}$

Answer: D 240
Problem: $.4+.02+.006=$
$\text{(A)}\ .012 \qquad \text{(B)}\ .066 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .24 \qquad \text{(E)} .426$

Solution: $.4+.02+.006 = .400 + .020 + .006 = .426\rightarrow \boxed{.426}$

Answer: E .426
Problem: $\frac{2}{25}=$
$\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5$

Solution: $\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{.08}$

Answer: B .08
Problem: $2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$

Solution: Find that \[(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180\] Which gives us \begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus \boxed{1800}

Answer: E 1800
Problem: $2(81+83+85+87+89+91+93+95+97+99)=$
$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$

Solution: $2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ $83+97$ $85+95$ $87+93$ $89+91$ $180$ .  Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$ .  But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$ , which is $\boxed{1800}$

Answer: E 1800
Problem: Martians measure angles in clerts.  There are $500$ clerts in a full circle.  How many clerts are there in a right angle?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 250$

Solution: The right angle is $1/4$ of the circle, hence it contains $500/4=125\rightarrow \boxed{125}$ clerts.

Answer: C 125
Problem: The smallest product one could obtain by multiplying two numbers in the set $\{ -7,-5,-1,1,3 \}$ is
$\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\ -15 \qquad \text{(D)}\ -1 \qquad \text{(E)}\ 3$

Solution: To get the smallest possible product, we want to multiply the smallest negative number by the largest positive number.  These are $-7$ and $3$ , respectively, and their product is $-21$ , which is $\boxed{21}$

Answer: B 21
Problem: The large cube shown is made up of $27$ identical sized smaller cubes.  For each face of the large cube, the opposite face is shaded the same way.  The total number of smaller cubes that must have at least one face shaded is
$\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24$
[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy]

Solution: Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares.
There are $10$ painted unit squares on the half of the cube shown, so there are $10\cdot 2=20$ unit cubes with at least one face painted, thus our answer is $\boxed{20}$

Answer: C 20
Problem: The large cube shown is made up of $27$ identical sized smaller cubes.  For each face of the large cube, the opposite face is shaded the same way.  The total number of smaller cubes that must have at least one face shaded is
$\text{(A)}\ 10 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 22 \qquad \text{(E)}\ 24$
[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy]

Solution: Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus $27 - 7 = 20$ our answer is $\boxed{20}$

Answer: C 20
Problem: When finding the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$ , the least common denominator used is
$\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040$

Solution: We want the least common multiple of $2,3,4,5,6,7$ , which is $420$ , or choice $\boxed{420}$

Answer: C 420
Problem: $4(299)+3(299)+2(299)+298=$
$\text{(A)}\ 2889 \qquad \text{(B)}\ 2989 \qquad \text{(C)}\ 2991 \qquad \text{(D)}\ 2999 \qquad \text{(E)}\ 3009$

Solution: We can make use of the distributive property as follows: \begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\ &= (4+3+2+1)(299)-1 \\ &= 10(299)-1 \\ &= 2989 \\ \end{align*}
$\boxed{2989}$

Answer: B 2989
Problem: A computer can do $10,000$ additions per second.  How many additions can it do in one hour?
$\text{(A)}\ 6\text{ million} \qquad \text{(B)}\ 36\text{ million} \qquad \text{(C)}\ 60\text{ million} \qquad \text{(D)}\ 216\text{ million} \qquad \text{(E)}\ 360\text{ million}$

Solution: There are $3600$ seconds per hour, so we have \begin{align*} \frac{3600\text{ seconds}}{\text{hour}}\cdot \frac{10,000\text{ additions}}{\text{second}} &= \frac{36,000,000\text{ additions}}{\text{hour}} \\ &= 36\text{ million additions per hour}  \end{align*}
$\boxed{36}$

Answer: B 36
Problem: The sale ad read: "Buy three tires at the regular price and get the fourth tire for 3 dollars." Sam paid 240 dollars for a set of four tires at the sale.  What was the regular price of one tire?
$\text{(A)}\ 59.25\text{ dollars} \qquad \text{(B)}\ 60\text{ dollars} \qquad \text{(C)}\ 70\text{ dollars} \qquad \text{(D)}\ 79\text{ dollars} \qquad \text{(E)}\ 80\text{ dollars}$

Solution: Let the regular price of one tire be $x$ .  We have \begin{align*} 3x+3=240 &\Rightarrow 3x=237 \\ &\Rightarrow x=79  \end{align*}
$\boxed{79}$ Good Job!

Answer: D 79
Problem: Joyce made $12$ of her first $30$ shots in the first three games of this basketball game, so her seasonal shooting average was $40\%$ .  In her next game, she took $10$ shots and raised her seasonal shooting average to $50\%$ .  How many of these $10$ shots did she make?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: After the fourth game, she took $40$ shots, $50\%$ of which she made, so she made $40\times .5=20$ shots.  Twelve of them were made in the first three games, so in the last game she made $20-12=8$ shots.
$\boxed{8}$

Answer: E 8
Problem: Half the people in a room left.  One third of those remaining started to dance.  There were then $12$ people who were not dancing.  The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$

Solution: Let the original number of people in the room be $x$ .  Half of them left, so $\frac{x}{2}$ of them are left in the room.
After that, one third of this group is dancing, so $\frac{x}{2}-\frac{1}{3}\left( \frac{x}{2}\right) =\frac{x}{3}$ people are not dancing.
This is given to be $12$ , so \[\frac{x}{3}=12\Rightarrow x=36\]
$\boxed{36}$

Answer: C 36
Problem: Half the people in a room left.  One third of those remaining started to dance.  There were then $12$ people who were not dancing.  The original number of people in the room was
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 72$

Solution: First note that of the $\frac{1}{2}$ people remaining in the room, $\frac{2}{3}$ are not dancing. Therefore $\frac{1}{2}\cdot\frac{2}{3}= \frac{1}{3}$ of the original amount of people in the room is $12$ . The answer is $\boxed{36}$

Answer: C 36
Problem: "If a whole number $n$ is not prime, then the whole number $n-2$ is not prime." A value of $n$ which shows this statement to be false is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 23$

Solution: To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime.  Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.
Now we just check the statement for $n=9,12,16$ .  If $n=12$ or $n=16$ , then $n-2$ is $10$ or $14$ , which aren't prime.  However, $n=9$ makes $n-2=7$ , which is prime, so $n=9$ proves the statement false.
Therefore, the answer is $\boxed{9}$ , 9.

Answer: A 9
Problem: Suppose $n^{*}$ means $\frac{1}{n}$ , the reciprocal of $n$ .  For example, $5^{*}=\frac{1}{5}$ .  How many of the following statements are true?
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

Solution: We can just test all of these statements: \begin{align*} 3^*+6^* &= \frac{1}{3}+\frac{1}{6} \\ &= \frac{1}{2} \neq 9^* \\ 6^*-4^* &= \frac{1}{6}-\frac{1}{4} \\ &= \frac{-1}{12} \neq 2^* \\ 2^*\cdot 6^* &= \frac{1}{2}\cdot \frac{1}{6} \\ &= \frac{1}{12} = 12^* \\ 10^* \div 2^* &= \frac{1}{10}\div \frac{1}{2} \\ &= \frac{1}{5} = 5^*  \end{align*}
The last two statements are true and the first two aren't, so $\boxed{2}$

Answer: C 2
Problem: Assume the adjoining chart shows the $1980$ U.S. population, in millions, for each region by ethnic group.  To the nearest percent , what percent of the U.S. Black population lived in the South?
\[\begin{tabular}[t]{c|cccc}  & NE & MW & South & West \\ \hline White & 42 & 52 & 57 & 35 \\ Black & 5 & 5 & 15 & 2 \\ Asian & 1 & 1 & 1 & 3 \\ Other & 1 & 1 & 2 & 4  \end{tabular}\]
$\text{(A)}\ 20\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 56\% \qquad \text{(E)}\ 80\%$

Solution: There are $5+5+15+2=27$ million Blacks living in the U.S.  Out of these, $15$ of them live in the South, so the percentage is $\frac{15}{27}\approx \frac{60}{108}\approx 56\%$
$\boxed{56}$

Answer: D 56
Problem: A multiple choice examination consists of $20$ questions.  The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question.  John's score on the examination is $48$ .  What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$

Solution: Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank.  We are given that \begin{align} c+w+b &= 20 \\ 5c-2w &= 48  \end{align}
Adding equation $(2)$ to double equation $(1)$ , we get \[7c+2b=88\]
Since we want to maximize the value of $c$ , we try to find the largest multiple of $7$ less than $88$ .  This is $84=7\times 12$ , so let $c=12$ .  Then we have \[7(12)+2b=88\Rightarrow b=2\]
Finally, we have $w=20-12-2=6$ .  We want $c$ , so the answer is $12$ , or $\boxed{12}$

Answer: D 12
Problem: A multiple choice examination consists of $20$ questions.  The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question.  John's score on the examination is $48$ .  What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$

Solution: If John answered 16 questions correctly, then he answered at most 4 questions incorrectly, giving him at least $16 \cdot 5 - 4 \cdot 2 = 72$ points. Therefore, John did not answer 16 questions correctly. If he answered 12 questions correctly and 6 questions incorrectly (leaving 2 questions unanswered), then he scored $12 \cdot 5 - 6 \cdot 2 = 48$ points. As all other options are less than 12, we conclude that 12 is the most questions John could have answered correctly, and the answer is $\boxed{12}$

Answer: D 12
Problem: The smallest sum one could get by adding three different numbers from the set $\{ 7,25,-1,12,-3 \}$ is
$\text{(A)}\ -3 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 21$

Solution: To find the smallest sum, we just have to find the smallest 3 numbers and add them together.
Obviously, the numbers are $-3, -1, 7$ , and adding them gets us $3$
$\boxed{3}$

Answer: C 3
Problem: The product $(1.8)(40.3+.07)$ is closest to
$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$

Solution: Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ is the closest to $72$ , so the answer is $\boxed{74}$

Answer: C 74
Problem: How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution: No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.
Clearly it must be true that for any positive integers $a$ $b$ , and $c$ with $a>b>c$ \[\sqrt{a}>\sqrt{b}>\sqrt{c}\]
If we let $a=9$ $b=8$ , and $c=4$ , then we get \[\sqrt{9}>\sqrt{8}>\sqrt{4}\] \[3>\sqrt{8}>2\]
Therefore, the smallest whole number between $\sqrt{8}$ and $\sqrt{80}$ is $3$
Similarly, if we let $a=81$ $b=80$ , and $c=64$ , we get \[\sqrt{81}>\sqrt{80}>\sqrt{64}\] \[9>\sqrt{80}>8\]
So $8$ is the largest whole number between $\sqrt{8}$ and $\sqrt{80}$
So we know that we just have to find the number of integers from 3 to 8 inclusive. If we subtract 2 from every number in this set (which doesn't change the number of integers in the set at all), we find that now all we need to do is find the number of integers there are from 1 to 6, which is obviously 6.
$\boxed{6}$

Answer: B 6
Problem: In the product shown, $\text{B}$ is a digit.  The value of $\text{B}$ is
\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution: Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.
So, $2 \times \text{B}$ has a units digit of $6$ , so $\text{B}$ is either $3$ or $8$ . If $\text{B}=3$ , then the product is $32\times 73$ , which is clearly too small, so $\text{B}=8$
$\boxed{8}$

Answer: E 8
Problem: Using only the paths and the directions shown, how many different routes are there from $\text{M}$ to $\text{N}$
[asy] draw((0,0)--(3,0),MidArrow); draw((3,0)--(6,0),MidArrow); draw(6*dir(60)--3*dir(60),MidArrow); draw(3*dir(60)--(0,0),MidArrow); draw(3*dir(60)--(3,0),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(6,0),MidArrow); draw(6*dir(60)--5.1961524227066318805823390245176*dir(30),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--3*dir(60),MidArrow); draw(5.1961524227066318805823390245176*dir(30)--(3,0),MidArrow); label("M",6*dir(60),N); label("N",(6,0),SE); label("A",3*dir(60),NW); label("B",5.1961524227066318805823390245176*dir(30),NE); label("C",(3,0),S); label("D",(0,0),SW); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: There is 1 way to get from C to N. There is only one way to get from D to N, which is DCN.
Since A can only go to C or D, which each only have 1 way to get to N each, there are $1+1=2$ ways to get from A to N.
Since B can only go to A, C or N, and A only has 2 ways to get to N, C only has 1 way and to get from B to N is only 1 way, there are $2+1+1=4$ ways to get from B to N.
M can only go to either B or A, A has 2 ways and B has 4 ways, so M has $4+2=6$ ways to get to N.
6 is $\boxed{6}$

Answer: E 6
Problem: If $\text{A}*\text{B}$ means $\frac{\text{A}+\text{B}}{2}$ , then $(3*5)*8$ is
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16\qquad \text{(E)}\ 30$

Solution: We just plug in and evaluate: \begin{align*} (3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\ &= 4*8 \\ &= \frac{4+8}{2} \\ &= 6 \\ \end{align*}
$\boxed{6}$

Answer: A 6
Problem: The perimeter of the polygon shown is
$\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48$
$\text{(E)}\ \text{cannot be determined from the information given}$

Solution: You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines!
[asy] unitsize(12); draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((8,3)--(8,0)--(2.7,0),dashed); [/asy]
So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get $\boxed{28}$

Answer: C 28