Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A straight concrete sidewalk is to be $3$ feet wide, $60$ feet long, and $3$ inches thick.  How many cubic yards of concrete must a contractor order for the sidewalk if concrete must be ordered in a whole number of cubic yards?
$\text{(A)}\ 2 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ \text{more than 20}$

Solution: This is a $1$ yard by $20$ yard by $1/12$ yard sidewalk, so its volume in yards is \[1\times 20\times \frac{1}{12} = 1.\overline{6}.\] Since concrete must be ordered in a whole number of cubic yards, we need $2\rightarrow \boxed{2}$

Answer: A 2
Problem: Each corner of a rectangular prism is cut off.  Two (of the eight) cuts are shown.  How many edges does the new figure have?
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); draw((2,0)--(3,1.8)--(4,1)--cycle,linewidth(1)); draw((2,3)--(4,4)--(3,2)--cycle,linewidth(1)); [/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 48$
Assume that the planes cutting the prism do not intersect anywhere in or on the prism.

Solution: In addition to the original $12$ edges, each original vertex contributes $3$ new edges.
There are $8$ original vertices, so there are $12+3\times 8=36$ edges in the new figure $\rightarrow \boxed{36}$

Answer: C 36
Problem: There are $120$ seats in a row.  What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119$

Solution: Let $p$ be a person seated and $o$ is an empty seat
The pattern of seating that results in the fewest occupied seats is $\text{opoopoopoo...po}$ .
We can group the seats in 3s like this: $\text{opo opo opo ... opo}.$
There are a total of $40=\boxed{40}$ groups

Answer: B 40
Problem: The annual incomes of $1,000$ families range from $8200$ dollars to $98,000$ dollars.  In error, the largest income was entered on the computer as $980,000$ dollars.  The difference between the mean of the incorrect data and the mean of the actual data is
$\text{(A)}\ \text{882 dollars} \qquad \text{(B)}\ \text{980 dollars} \qquad \text{(C)}\ \text{1078 dollars} \qquad \text{(D)}\ \text{482,000 dollars} \qquad \text{(E)}\ \text{882,000 dollars}$

Solution: Let $S$ be the sum of all the incomes but the largest one.  For the actual data, the mean is $\frac{S+98000}{1000}$ , and for the incorrect data the mean is $\frac{S+980000}{1000}$ .  The difference is $882, or \rightarrow \boxed{882}$

Answer: A 882
Problem: Several students are seated at a large circular table.  They pass around a bag containing $100$ pieces of candy.  Each person receives the bag, takes one piece of candy and then passes the bag to the next person.  If Chris takes the first and last piece of candy, then the number of students at the table could be
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25$

Solution: If this is the case, then if there were only $99$ pieces of candy, the bag would have gone around the table a whole number of times.  Thus, the number of students is a divisor of $99$ .  The only choice that satisfies this is choice $\boxed{11}$

Answer: B 11
Problem: Three $\Delta$ 's and a $\diamondsuit$ will balance nine $\bullet$ 's.  One $\Delta$ will balance a $\diamondsuit$ and a $\bullet$
[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$\Delta \hspace{2 mm}\Delta \hspace{2 mm}\Delta \hspace{2 mm}\diamondsuit $",(-12,6.5),S); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm} \bullet $",(12,5.2),N); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet $",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label("$\Delta $",(34,6.5),S);  label("$\bullet \hspace{2 mm}\diamondsuit $",(54,6.5),S); [/asy]
How many $\bullet$ 's will balance the two $\diamondsuit$ 's in this balance?
[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label("$?$",(-12,8.5),S); label("$\diamondsuit \hspace{2 mm}\diamondsuit $",(12,4.5),S); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: For simplicity, suppose $\Delta = a$ $\diamondsuit = b$ and $\bullet = c$ .  Then, \[3a+b=9c\] \[a=b+c\] and we want to know what $2b$ is in terms of $c$ .  Substituting the second equation into the first, we have \[4b=6c\Rightarrow 2b=3c\]
Thus, we need $3$ $\bullet$ 's $\rightarrow \boxed{3}$

Answer: C 3
Problem: How many different patterns can be made by shading exactly two of the nine squares?  Patterns that can be matched by flips and/or turns are not considered different.  For example, the patterns shown below are not considered different.
[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,gray); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle,gray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1)); draw((2,0)--(2,3),linewidth(1)); draw((0,1)--(3,1),linewidth(1)); draw((1,0)--(1,3),linewidth(1)); draw((0,2)--(3,2),linewidth(1)); fill((6,0)--(8,0)--(8,1)--(6,1)--cycle,gray); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle,linewidth(1)); draw((8,0)--(8,3),linewidth(1)); draw((6,1)--(9,1),linewidth(1)); draw((7,0)--(7,3),linewidth(1)); draw((6,2)--(9,2),linewidth(1)); fill((14,1)--(15,1)--(15,3)--(14,3)--cycle,gray); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle,linewidth(1)); draw((14,0)--(14,3),linewidth(1)); draw((12,1)--(15,1),linewidth(1)); draw((13,0)--(13,3),linewidth(1)); draw((12,2)--(15,2),linewidth(1)); fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle,linewidth(1)); draw((20,0)--(20,3),linewidth(1)); draw((18,1)--(21,1),linewidth(1)); draw((19,0)--(19,3),linewidth(1)); draw((18,2)--(21,2),linewidth(1)); [/asy]
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$

Solution: We break this into cases.
Case 1: At least one square is a vertex: WLOG , suppose one of them is in the upper-left corner.  Then, consider the diagonal through that square.  The two squares on that diagonal could be the second square, or the second square is on one side of the diagonal.
The square is reflectionally symmetric about this diagonal, so we only consider the squares on one side, giving another three possibilities.
In this case, there are $2+3=5$ distinct squares.
Case 2: At least one square is on an edge, but no square is on a vertex: There are clearly two edge-edge combinations and one edge-center combination, so this case has $3$ squares.
In total, there are $3+5=8$ distinct squares $\rightarrow \boxed{8}$

Answer: C 8
Problem: $(1+11+21+31+41)+(9+19+29+39+49)=$
$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$

Solution: We make use of the associative and commutative properties of addition to rearrange the sum as \begin{align*} (1+49)+(11+39)+(21+29)+(31+19)+(41+9) &= 50+50+50+50+50 \\ &= 250 \Longrightarrow \boxed{250}

Answer: E 250
Problem: $(1+11+21+31+41)+(9+19+29+39+49)=$
$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$

Solution: Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:
\[(1+9)+(11+19)+(21+29)+(31+39)+(41+49),\] which gives us \[10+30+50+70+90 = 250 = \boxed{250}.\]

Answer: E 250
Problem: $\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}=$
$\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246$

Solution: \begin{align*} \frac{2}{10}+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\ &= \frac{246}{1000} \\ &= .246 \rightarrow \boxed{.246}

Answer: D .246
Problem: Which of the following numbers is the largest?
$\text{(A)}\ .99 \qquad \text{(B)}\ .9099 \qquad \text{(C)}\ .9 \qquad \text{(D)}\ .909 \qquad \text{(E)}\ .9009$

Solution: We have $.99>.9099>.909>.9009>.9$ , so choice $\boxed{.99}$ is the largest.

Answer: A .99
Problem: Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$
$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution: $401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \[\frac{400}{.2}=2000\rightarrow \boxed{2000}\]

Answer: E 2000
Problem: $-15+9\times (6\div 3) =$
$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$

Solution: We use the order of operations here to get
\begin{align*} -15+9\times (6\div 3) &= -15+9\times 2 \\ &= -15+18 \\ &= 3 \rightarrow \boxed{3}

Answer: D 3
Problem: If the markings on the number line are equally spaced, what is the number $\text{y}$
[asy] draw((-4,0)--(26,0),Arrows); for(int a=0; a<6; ++a)  {   draw((4a,-1)--(4a,1));  } label("0",(0,-1),S); label("20",(20,-1),S); label("y",(12,-1),S); [/asy]
$\text{(A)}\ 3 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 16$

Solution: Five steps are taken to get from $0$ to $20$ .  Each step is of equal size, so each step is $4$ .  Three steps are taken from $0$ to $y$ , so $y=3\times 4=12\rightarrow \boxed{12}$

Answer: C 12
Problem: If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$

Solution: We have \begin{align*} 20(25)+10(10) &= 10(25)+n(10) \\ 600 &= 250+10n \\ 35 &= n \implies \boxed{35}

Answer: D 35
Problem: $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$

Solution: We use the distributive property to get \[3\times 4+2\times 4+2\times 3 = 26 \rightarrow \boxed{26}\]

Answer: E 26
Problem: $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$

Solution: Since $\frac12+\frac13+\frac14 > \frac12+\frac14+\frac14 = 1$ , we have \[(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) > 2\times 3\times 4 \times 1 = 24\] The only answer choice greater than $24$ is $\boxed{26}$

Answer: E 26
Problem: $(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$

Solution: We can just bash it out, getting $24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \Longrightarrow \boxed{26}$ -fn106068

Answer: E 26
Problem: What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?
$\text{(A)}\ 50^\circ \qquad \text{(B)}\ 120^\circ \qquad \text{(C)}\ 135^\circ \qquad \text{(D)}\ 150^\circ \qquad \text{(E)}\ 165^\circ$

Solution: The smaller angle makes up $5/12$ of the circle which is the clock.  A circle is $360^\circ$ , so the measure of the smaller angle is \[\frac{5}{12}\cdot 360^\circ = 150^\circ \rightarrow \boxed{150}\]

Answer: D 150
Problem: The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is
[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution: Let $[ABC]$ denote the area of figure $ABC$
Clearly, $[BEDC]=[ABCD]-[ABE]$ .  Using basic area formulas,
Since $AE+ED=BC=10$ and $ED=6$ $AE=4$ and the area of $\triangle ABE$ is $4(4)=16$
Finally, we have $[BEDC]=80-16=64\rightarrow \boxed{64}$

Answer: D 64
Problem: In how many ways can $47$ be written as the sum of two primes
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$

Solution: For $47$ to be written as the sum of two integers , one must be odd and the other must be even.  There is only one even prime, namely $2$ , so one of the numbers must be $2$ , making the other $45$
However, $45$ is not prime, so there are no ways to write $47$ as the sum of two primes $\rightarrow \boxed{0}$

Answer: A 0
Problem: The number $\text{N}$ is between $9$ and $17$ .  The average of $6$ $10$ , and $\text{N}$ could be
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution: We know that $9<N<17$ and we wish to bound $\frac{6+10+N}{3}=\frac{16+N}{3}$
From what we know, we can deduce that $25<N+16<33$ , and thus \[8.\overline{3}<\frac{N+16}{3}<11\]
The only answer choice that falls in this range is choice $\boxed{10}$

Answer: B 10
Problem: The graph below shows the total accumulated dollars (in millions) spent by the Surf City government during $1988$ .  For example, about $.5$ million had been spent by the beginning of February and approximately $2$ million by the end of April.  Approximately how many millions of dollars were spent during the summer months of June, July, and August?
[asy] unitsize(18); for (int a=1; a<13; ++a)  {   draw((a,0)--(a,.5));  } for (int b=1; b<6; ++b)  {   draw((-.5,2b)--(0,2b));  } draw((0,0)--(0,12)); draw((0,0)--(14,0)); draw((0,0)--(1,.9)--(2,1.9)--(3,2.6)--(4,4.3)--(5,4.5)--(6,5.7)--(7,8.2)--(8,9.4)--(9,9.8)--(10,10.1)--(11,10.2)--(12,10.5)); label("J",(.5,0),S); label("F",(1.5,0),S); label("M",(2.5,0),S); label("A",(3.5,0),S); label("M",(4.5,0),S); label("J",(5.5,0),S); label("J",(6.5,0),S); label("A",(7.5,0),S); label("S",(8.5,0),S); label("O",(9.5,0),S); label("N",(10.5,0),S); label("D",(11.5,0),S); label("month F=February",(16,0),S); label("$1$",(-.6,2),W); label("$2$",(-.6,4),W); label("$3$",(-.6,6),W); label("$4$",(-.6,8),W); label("$5$",(-.6,10),W); label("dollars in millions",(0,11.9),N); [/asy]
$\text{(A)}\ 1.5 \qquad \text{(B)}\ 2.5 \qquad \text{(C)}\ 3.5 \qquad \text{(D)}\ 4.5 \qquad \text{(E)}\ 5.5$

Solution: Since we want to know how much money is spent in June, July and August, we need the difference between the amount of money spent by the beginning of June and the amount of money spent by the end of August.
We estimate these to be about $2.2$ million and $4.8$ million, respectively.  The difference is \[4.8-2.2=2.6\approx 2.5 \rightarrow \boxed{2.5}\]

Answer: B 2.5
Problem: The figure may be folded along the lines shown to form a number cube.  Three number faces come together at each corner of the cube.  What is the largest sum of three numbers whose faces come together at a corner?
[asy] draw((0,0)--(0,1)--(1,1)--(1,2)--(2,2)--(2,1)--(4,1)--(4,0)--(2,0)--(2,-1)--(1,-1)--(1,0)--cycle); draw((1,0)--(1,1)--(2,1)--(2,0)--cycle); draw((3,1)--(3,0)); label("$1$",(1.5,1.25),N); label("$2$",(1.5,.25),N); label("$3$",(1.5,-.75),N); label("$4$",(2.5,.25),N); label("$5$",(3.5,.25),N); label("$6$",(.5,.25),N); [/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution: It is clear that $6$ $5$ , and $4$ will not come together to get a sum of $15$
The faces $6$ $5$ , and $3$ come together at a common vertex, making the maximal sum $6+5+3=14\rightarrow \boxed{14}$

Answer: D 14
Problem: Jack had a bag of $128$ apples.  He sold $25\%$ of them to Jill.  Next he sold $25\%$ of those remaining to June.  Of those apples still in his bag, he gave the shiniest one to his teacher.  How many apples did Jack have then?
$\text{(A)}\ 7 \qquad \text{(B)}\ 63 \qquad \text{(C)}\ 65 \qquad \text{(D)}\ 71 \qquad \text{(E)}\ 111$

Solution: First he gives $128\times .25 = 32$ apples to Jill, so he has $128-32=96$ apples left.  Then he gives $96\times .25 = 24$ apples to June, so he has $96-24=72$ left.
Finally, he gives one to the teacher, leaving $71\rightarrow \boxed{71}$

Answer: D 71
Problem: The letters $\text{A}$ $\text{J}$ $\text{H}$ $\text{S}$ $\text{M}$ $\text{E}$ and the digits $1$ $9$ $8$ $9$ are "cycled" separately as follows and put together in a numbered list:
\[\begin{tabular}[t]{lccc}  & & AJHSME & 1989 \\  & & & \\ 1. & & JHSMEA & 9891 \\ 2. & & HSMEAJ & 8919 \\ 3. & & SMEAJH & 9198 \\  & & ........ &  \end{tabular}\]
What is the number of the line on which $\text{AJHSME 1989}$ will appear for the first time?
$\text{(A)}\ 6 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 24$

Solution: Every $4\text{th}$ line has $1989$ as part of it and every $6\text{th}$ line has $\text{AJHSME}$ as part of it.  In order for both to be part of line $n$ $n$ must be a multiple of $4$ and $6$ , the least of which is $\text{lcm}(4,6)=12\rightarrow \boxed{12}$

Answer: C 12
Problem: An artist has $14$ cubes, each with an edge of $1$ meter.  She stands them on the ground to form a sculpture as shown.  She then paints the exposed surface of the sculpture.  How many square meters does she paint?
$\text{(A)}\ 21 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 33 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 42$
[asy] draw((0,0)--(2.35,-.15)--(2.44,.81)--(.09,.96)--cycle); draw((.783333333,-.05)--(.873333333,.91)--(1.135,1.135)); draw((1.566666667,-.1)--(1.656666667,.86)--(1.89,1.1)); draw((2.35,-.15)--(4.3,1.5)--(4.39,2.46)--(2.44,.81)); draw((3,.4)--(3.09,1.36)--(2.61,1.4)); draw((3.65,.95)--(3.74,1.91)--(3.29,1.94)); draw((.09,.96)--(.76,1.49)--(.71,1.17)--(2.2,1.1)--(3.6,2.2)--(3.62,2.52)--(4.39,2.46)); draw((.76,1.49)--(.82,1.96)--(2.28,1.89)--(2.2,1.1)); draw((2.28,1.89)--(3.68,2.99)--(3.62,2.52)); draw((1.455,1.135)--(1.55,1.925)--(1.89,2.26)); draw((2.5,2.48)--(2.98,2.44)--(2.9,1.65)); draw((.82,1.96)--(1.55,2.6)--(1.51,2.3)--(2.2,2.26)--(2.9,2.8)--(2.93,3.05)--(3.68,2.99)); draw((1.55,2.6)--(1.59,3.09)--(2.28,3.05)--(2.2,2.26)); draw((2.28,3.05)--(2.98,3.59)--(2.93,3.05)); draw((1.59,3.09)--(2.29,3.63)--(2.98,3.59)); [/asy]

Solution: We can consider the contributions of the sides of the three layers and the tops of the layers separately.
Layer $n$ (counting from the top starting at $1$ ) has $4$ side faces each with $n$ unit squares, so the sides of the pyramid contribute $4+8+12=24$ for the surface area.
The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a $3\times 3$ square, hence this contributes $9$ for the surface area.
You do not have to count the side underneath, since it is not exposed.
Thus, the artist paints $24+9=33 \rightarrow \boxed{33}$ square meters.

Answer: C 33
Problem: The diagram shows part of a scale of a measuring device.  The arrow indicates an approximate reading of
[asy] draw((-3,0)..(0,3)..(3,0)); draw((-3.5,0)--(-2.5,0)); draw((0,2.5)--(0,3.5)); draw((2.5,0)--(3.5,0)); draw((1.8,1.8)--(2.5,2.5)); draw((-1.8,1.8)--(-2.5,2.5)); draw((0,0)--3*dir(120),EndArrow); label("$10$",(-2.6,0),E); label("$11$",(2.6,0),W); [/asy]
$\text{(A)}\ 10.05 \qquad \text{(B)}\ 10.15 \qquad \text{(C)}\ 10.25 \qquad \text{(D)}\ 10.3 \qquad \text{(E)}\ 10.6$

Solution: Clearly the arrow marks a value between $10.25$ and $10.5$ , so only $\text{C}$ and $\text{D}$ are possible.
Looking, we see that the arrow is closer to $10.3$ , so $\boxed{10.3}$

Answer: D 10.3
Problem: The product $8\times .25\times 2\times .125 =$
$\text{(A)}\ \frac18 \qquad \text{(B)}\ \frac14 \qquad \text{(C)}\ \frac12 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: Converting the decimals to fractions , this is \begin{align*} 8\times \frac{1}{4} \times 2\times \frac{1}{8} &= \frac{8\times 2}{4\times 8} \\ &= \frac{16}{32} \\ &= \frac{1}{2} \rightarrow \boxed{12}

Answer: C 12
Problem: $\frac{1}{10}+\frac{2}{20}+\frac{3}{30} =$
$\text{(A)}\ .1 \qquad \text{(B)}\ .123 \qquad \text{(C)}\ .2 \qquad \text{(D)}\ .3 \qquad \text{(E)}\ .6$

Solution: Each of the fractions simplify to $\frac{1}{10}$ , so this sum is \begin{align*} \frac{1}{10}+\frac{1}{10}+\frac{1}{10} &= \frac{3}{10} \\ &= .3 \rightarrow \boxed{.3}

Answer: D .3
Problem: The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$
[asy] unitsize(12); //Force a white background in middle even when transparent fill((3,1)--(12,1)--(12,4)--(3,4)--cycle,white); //Black Squares, Gray Border (blends better than white) for(int a=0; a<7; ++a)  {   filldraw((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black,gray);  } for(int b=7; b<15; ++b)  {   filldraw((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black,gray);  } for(int c=1; c<7; ++c)  {   filldraw((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black,gray);  } filldraw((6,4)--(7,4)--(7,5)--(6,5)--cycle,black,gray); filldraw((7,5)--(8,5)--(8,6)--(7,6)--cycle,black,gray); filldraw((8,4)--(9,4)--(9,5)--(8,5)--cycle,black,gray); //White Squares, Black Border filldraw((7,4)--(8,4)--(8,5)--(7,5)--cycle,white,black); for(int a=0; a<7; ++a)  {   filldraw((2a+1,0)--(2a+2,0)--(2a+2,1)--(2a+1,1)--cycle,white,black);  } for(int b=9; b<15; ++b)  {   filldraw((b-1,14-b)--(b,14-b)--(b,15-b)--(b-1,15-b)--cycle,white,black);  } for(int c=1; c<7; ++c)  {   filldraw((c+1,c)--(c+2,c)--(c+2,c+1)--(c+1,c+1)--cycle,white,black);  } label("same",(6.3,2.45),N); label("pattern here",(7.5,1.4),N); [/asy]

Solution: It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are $8$ rows, there are $8$ more black squares than the white squares. $8\rightarrow \boxed{8}$

Answer: B 8
Problem: If $\angle \text{CBD}$ is a right angle , then this protractor indicates that the measure of $\angle \text{ABC}$ is approximately
[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D--B); label("O",(-2.5,0),W); label("A",A,W); label("B",B,S); label("C",C,W); label("D",D,E); label("0",(-1.8,0),W); label("20",(-1.7,.5),NW); label("160",(1.6,.5),NE); label("180",(1.7,0),E); [/asy]
$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ$

Solution: We have that $20^{\circ}+\angle ABD +\angle CBD=160^{\circ}$ , or $\angle ABD +\angle CBD=140^{\circ}$ . Since $\angle CBD$ is a right angle, we have $\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \boxed{50}$

Answer: C 50
Problem: Suppose the estimated $20$ billion dollar cost to send a person to the planet Mars is shared equally by the $250$ million people in the U.S. Then each person's share is
$\text{(A)}\ 40\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 80\text{ dollars} \qquad \text{(D)}\ 100\text{ dollars} \qquad \text{(E)}\ 125\text{ dollars}$

Solution: We want the cost per person, which is \begin{align*} \frac{20\text{ billion}}{250\text{ million}} &= \frac{20000\text{ million}}{250\text{ million}} \\ &= 80 \rightarrow \boxed{80}

Answer: C 80
Problem: If rose bushes are spaced about $1$ foot apart, approximately how many bushes are needed to surround a circular patio whose radius is $12$ feet?
$\text{(A)}\ 12 \qquad \text{(B)}\ 38 \qquad \text{(C)}\ 48 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 450$

Solution: The circumference of the patio is $2\cdot 12\cdot \pi =24\pi \approx 75.398$ .  Since the bushes are spaced $1$ foot apart, about $75\rightarrow \boxed{75}$ are needed.

Answer: D 75