Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A bar graph shows the number of hamburgers sold by a fast food chain each season.  However, the bar indicating the number sold during the winter is covered by a smudge.  If exactly $25\%$ of the chain's hamburgers are sold in the fall, how many million hamburgers are sold in the winter?
[asy] size(250);  void bargraph(real X, real Y, real ymin, real ymax, real ystep, real tickwidth,  string yformat, Label LX, Label LY, Label[] LLX, real[] height,pen p=nullpen) { draw((0,0)--(0,Y),EndArrow); draw((0,0)--(X,0),EndArrow); label(LX,(X,0),plain.SE,fontsize(9)); label(LY,(0,Y),plain.NW,fontsize(9)); real yscale=Y/(ymax+ystep);  for(real y=ymin; y<ymax; y+=ystep) { draw((-tickwidth,yscale*y)--(0,yscale*y)); label(format(yformat,y),(-tickwidth,yscale*y),plain.W,fontsize(9)); }  int n=LLX.length; real xscale=X/(2*n+2); for(int i=0;i<n;++i) { real x=xscale*(2*i+1); path P=(x,0)--(x,height[i]*yscale)--(x+xscale,height[i]*yscale)--(x+xscale,0)--cycle; fill(P,p); draw(P); label(LLX[i],(x+xscale/2),plain.S,fontsize(10)); } for(int i=0;i<n;++i) draw((0,height[i]*yscale)--(X,height[i]*yscale),dashed); }  string yf="%#.1f"; Label[] LX={"Spring","Summer","Fall","Winter"}; for(int i=0;i<LX.length;++i) LX[i]=rotate(90)*LX[i]; real[] H={4.5,5,4,4};  bargraph(60,50,1,5.1,0.5,2,yf,"season","hamburgers (millions)",LX,H,yellow); fill(ellipse((45,30),7,10),brown); [/asy]
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 3.5 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 4.5$

Solution: What we want to find is the number of hamburgers sold in the winter. Since we don't know what it is, let's call it $x$ . From the graph, we know that in Spring, 4.5 million hamburgers were sold, in the Summer was 5 million and in the Fall was 4 million. We know that the number of hamburgers sold in Fall is exactly $\frac{1}{4}$ of the total number of hamburgers sold, so we can say that...
$4 \times \text{Fall} = \text{Spring} + \text{Winter} + \text{Fall} + \text{Summer}$
$4 \times 4 = 4.5 + 4 + x + 5$
$16 = x + 13.5$
$2.5 = x$
The answer is 2.5, or $\boxed{2.5}$

Answer: A 2.5
Problem: A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side.  Fence posts are to be placed every $12$ meters along the fence including the two posts where the fence meets the rock wall.  What is the fewest number of posts required to fence an area $36$ m by $60$ m?
[asy] unitsize(12); draw((0,0)--(16,12)); draw((10.66666,8)--(6.66666,13.33333)--(1.33333,9.33333)--(5.33333,4)); label("WALL",(7,4),SE); [/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution: Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$ m by $60$ m rectangle, so the $60$ m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$ m. We can start by counting $60\div12+1$ on the $60$ m fence (since we also count the $0$ m post). Next, we have the two $36$ m fences. There are a total of $36\div12+1-1$ fences on that side since the $0$ m and $60$ m fence posts are also part of the $36$ m fences. So we have $6+2\cdot3=12$ minimum fence posts needed to box a $36$ m by $60$ m grazing area, or answer $\boxed{12}$

Answer: B 12
Problem: At the beginning of a trip, the mileage odometer read $56,200$ miles.  The driver filled the gas tank with $6$ gallons of gasoline.  During the trip, the driver filled his tank again with $12$ gallons of gasoline when the odometer read $56,560$ .  At the end of the trip, the driver filled his tank again with $20$ gallons of gasoline.  The odometer read $57,060$ .  To the nearest tenth, what was the car's average miles-per-gallon for the entire trip?
$\text{(A)}\ 22.5 \qquad \text{(B)}\ 22.6 \qquad \text{(C)}\ 24.0 \qquad \text{(D)}\ 26.9 \qquad \text{(E)}\ 27.5$

Solution: The first six gallons are irrelevant. We start with the odometer at $56,200$ miles, and a full gas tank. The total gas consumed by the car during the trip is equal to the total gas the driver had to buy to make the tank full again, i.e., $12+20=32$ gallons. The distance covered is $57,060 - 56,200 = 860$ miles. Hence the average MPG ratio is $860 / 32 \approx 26.9 \rightarrow \boxed{26.9}$

Answer: D 26.9
Problem: The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to
$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$

Solution: \[\frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}\] Which is closest to $3\rightarrow\boxed{3}$

Answer: D 3
Problem: Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined.  How many of the resulting figures can be folded into a topless cubical box?
[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.
In total, there are $6\rightarrow\boxed{6}$ good possibilities.

Answer: E 6
Problem: Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined.  How many of the resulting figures can be folded into a topless cubical box?
[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: Fold the four squares into the four sides of a cube. Then, there are six edges "open" (for lack of better term). For each open edge, we can add a square/side, so the answer is $6\rightarrow\boxed{6}$

Answer: E 6
Problem: The large circle has diameter $\text{AC}$ .  The two small circles have their centers on $\text{AC}$ and just touch at $\text{O}$ , the center of the large circle.  If each small circle has radius $1$ , what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
[asy] pair A=(-2,0), O=origin, C=(2,0); path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), M=X..Y..Z..cycle; filldraw(M, black, black); draw(reflect(A,C)*M); draw(A--C, dashed);  label("A",A,W); label("C",C,E); label("O",O,SE); dot((-1,0)); dot(O); dot((1,0)); label("$1$",(-.5,0),N); label("$1$",(1.5,0),N); [/asy]
$\text{(A)}\ \text{between }\frac{1}{2}\text{ and 1} \qquad \text{(B)}\ 1 \qquad \text{(C)}\ \text{between 1 and }\frac{3}{2}$
$\text{(D)}\ \text{between }\frac{3}{2}\text{ and 2} \qquad \text{(E)}\ \text{cannot be determined from the information given}$

Solution: The small circle has radius $1$ , thus its area is $\pi$
The large circle has radius $2$ , thus its area is $4\pi$
The area of the semicircle above $AC$ is then $2\pi$
The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\pi$ . This means that the area of the shaded part is $2\pi-\pi=\pi$ . 
This is equal to the area of a small circle, hence the correct answer is $\boxed{1}$

Answer: B 1
Problem: [katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]

Solution: By the associative property , we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{1}[/katex]

Answer: A 1
Problem: [katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]

Solution: Notice that the $9 \times 11$ in the denominator of the first fraction cancels with the same term in the second fraction, the $7$ s in the numerator and denominator of the second fraction cancel, and the $3 \times 5$ in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with $\boxed{1}$

Answer: A 1
Problem: $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

Solution: To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have:
[mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{945}[/mathjax].

Answer: B 945
Problem: $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

Solution: We can express each of the terms as a difference from $100$ and then add the negatives using $\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ &= 1000 - 55\\ &= \boxed{945}

Answer: B 945
Problem: $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

Solution: Instead of breaking the sum then rearranging, we can rearrange directly: \begin{align*} 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= \boxed{945}

Answer: B 945
Problem: $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

Solution: The finite arithmetic sequence formula states that the sum in the sequence is equal to $\frac{n}{2}\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.
Applying the formula, we have: \[\frac{10}{2}\cdot(90+99)=\boxed{945}\]

Answer: B 945
Problem: $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

Solution: The expression is equal to the sum of integers from $1$ to $99$ minus the sum of integers from $1$ to $89$ , so it is equal to $\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{945}$

Answer: B 945
Problem: $\frac{10^7}{5\times 10^4}=$
$\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution: We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: \[\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}\]
We know that $10^3 = 10 \times 10 \times 10$ , so
\begin{align*} \frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\ &= 2\times 10\times 10 \\ &= 200 \\ \end{align*}
So the answer is $\boxed{200}$

Answer: D 200
Problem: The area of polygon $ABCDEF$ , in square units, is
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Solution: [asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(6,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of.
If we continue segment $\overline{FE}$ until it reaches the right side at $G$ , we create two rectangles - one on the top and one on the bottom.
We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\times5 = 30$ . For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.
Note that $GC+GB=9$ , and $GB=AF=5$ , so we must have \[GC+5=9\Rightarrow GC=4\]
The area of the bottom rectangle is then \[(DC)(GC)=4\times 4=16\]
Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$
$\boxed{46}$

Answer: C 46
Problem: The area of polygon $ABCDEF$ , in square units, is
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Solution: [asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(0,0),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
Let $\langle ABCDEF \rangle$ be the area of polygon $ABCDEF$ . Also, let $G$ be the intersection of $DC$ and $AF$ when both are extended.
Clearly, \[\langle ABCDEF \rangle = \langle ABCG \rangle - \langle GFED \rangle\]
Since $AB=6$ and $BC=9$ $\langle ABCG \rangle =6\times 9=54$
To compute the area of $GFED$ , note that \[AB=GD+DC\] \[BC=GF+FA\]
We know that $AB=6$ $DC=4$ $BC=9$ , and $FA=5$ , so \[6=GD+4\Rightarrow GD=2\] \[9=GF+5\Rightarrow GF=4\]
Thus $\langle GFED \rangle = 4\times 2=8$
Finally, we have \begin{align*} \langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\ &= 54-8 \\ &= 46 \\ \end{align*}
This is answer choice $\boxed{46}$

Answer: C 46
Problem: A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?
$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$

Solution: We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.
Let's say that $500\text{ sheets}=5\text{ cm}\Rightarrow \frac{500 \text{ sheets}}{5 \text{ cm}} = 1$ . So by multiplying $7.5 \text{ cm}$ by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets
\begin{align*} \frac{7.5 \times 500}{5} &= 7.5 \times 100 \\ &= 750 \text{ sheets} \\ \end{align*}
$750$ is $\boxed{750}$

Answer: D 750
Problem: A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?
$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$

Solution: We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness.
Our proportion is \[\frac{5}{500}=\frac{7.5}{x}\] where $x$ is the number we are looking for.
Next, we cross-multiply to get $5x=500 \times 7.5$ so $x=750$ which is $\boxed{750}$

Answer: D 750
Problem: A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$

Solution: The best way to solve this problem is to find patterns and to utilize them to our advantage. For example, we can't really do anything without knowing how many squares there are in the 37th row. But who wants to continue the diagram for 37 rows? And what if the problem said 100,000th row? It'll still be possible - but not if your method is to continue the diagram...
So hopefully there's a pattern. We find a pattern by noticing what is changing from row 1 to row 2. Basically, for the next row, we are just adding 2 squares (1 on each side) to the number of squares we had in the previous row.  So each time we're adding 2. So how can we find $N$ , if $N$ is the number of squares in the $a^\text{th}$ row of this diagram? We can't just say that $N = 1 + 2a$ , because it doesn't work for the first row. But since 1 is the first term, we have to EXCLUDE the first term, meaning that we must subtract 1 from a. Thus, $N = 1 + 2\times(a - 1) = 2a - 1$ . So in the 37th row we will have $2 \times 37 - 1 = 74 - 1 = 73$
You may now be thinking - aha, we're finished. But we're only half finished. We still need to find how many black squares there are in these 73 squares. Well let's see - they alternate white-black-white-black... but we can't divide by two - there aren't exactly as many white squares as black squares... there's always 1 more white square... aha! If we subtract 1 from the number of squares (1 white square), we will have exactly 2 times the number of black squares.
Thus, the number of black squares is $\frac{73 - 1}{2} = \frac{72}{2} = 36$
36 is $\boxed{36}$

Answer: C 36
Problem: A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$

Solution: Note that each row adds one black and one white square to the end of the previous one, and then shifts the new row over a little.  Since the first row had no black squares, the number of black squares in any row is one less than the row number.
Now that this has been established, we just have $37-1=36\rightarrow \boxed{36}$

Answer: C 36
Problem: A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$

Solution: We can easily spot the pattern. In the first row, there are $0$ black squares. In the second row, there is $1$ black square. So the pattern is that in order to find the number of black squares in a row, you need to subtract $1$ from the row number. Therefore, the number of black squares in the $37\text{th}$ row we subtract 1. $37-1=36\rightarrow \boxed{36}$

Answer: C 36
Problem: If $a = - 2$ , the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is
$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$

Solution: Since all the numbers are small, we can just evaluate the set to be \[\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}\]
The largest number is $6$ , which corresponds to $-3a$
$\boxed{3}$

Answer: A 3
Problem: How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$

Solution: This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.
If you ever learned about complementary counting , this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we count how many that DON'T?
So let's find the number of numbers. Obviously, we'd start by subtracting 100 from 400, getting us 300, but we're not done. Since just subtracting includes the number 400, we must subtract one (because 400 isn't allowed - it says between), getting us 299.
So how many numbers are there that DON'T have a 2? Well, we have 2 possibilities for the hundreds digit (1, 3, note that 2 is not allowed), 9 possibilities for the tens digit (1, 3, 4, 5, ... , 9, 0), and 9 possibilities for the ones digit. $2 \times 9 \times 9 = 162$ . However, one of the numbers we counted is $100$ , which isn't allowed, so there are $162-1=161$ numbers without a 2.
Since there are 299 numbers in total and 161 that DON'T have any 2's, $299 - 161 = 138$ numbers WILL have at least one two.
$\boxed{138}$

Answer: C 138
Problem: How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$

Solution: As in the previous solution, get rid of the $400$ to make things easier. This gives us the numbers from $100$ to $399$
Let $A$ be the event that the first digit is $2$ $B$ be the event that the second digit is $2$ , and $C$ be the event that the third digit is $2$ . Then, PIE says that our answer will be $|A|+|B|+|C|-|A\cup B|-|A\cup C|-|B\cup C|+|A\cup B\cup C|$
We have that $|A|$ is just $1\times10\times10=100$ $|B|=3\times1\times10=30$ , and $|C|=3\times10\times1=30$
Next, $|A\cup B|$ is just having something in the form $22\_$ , so there are $10$ ways. Similarly, $|A\cup C|$ means that we have $2\_2$ , so there are again $10$ ways. Finally, $|B\cup C|$ means that our number is like $\_22$ , so there are $3$ ways.
Finally, $|A\cup B\cup C|$ counts the number of three digit numbers with all three digits $2$ , which there is only $1$ of: $222$
Putting this together, our answer will be $100+30+30-10-10-3+1=\boxed{138}$

Answer: C 138