Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The area of the smallest square that will contain a circle of radius 4 is
$\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

Solution: Draw a square circumscribed around the circle.  (Alternately, the circle is inscribed in the square.)  If the circle has radius $4$ , it has diameter $8$ .  Two of the diameters of the circle will run parallel to the sides of the square.  Thus, the smallest square that contains it has side length $8$ , and area $8\times8=64$ $\boxed{64}$

Answer: D 64
Problem: Walter gets up at 6:30 a.m., catches the school bus at 7:30 a.m., has 6 classes that last 50 minutes each, has 30 minutes for lunch, and has 2 hours additional time at school.  He takes the bus home and arrives at 4:00 p.m.  How many minutes has he spent on the bus?
$\text{(A)}\ 30 \qquad \text{(B)}\ 60 \qquad \text{(C)}\ 75 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 120$

Solution: There are $4\frac{1}{2}$ hours from 7:30 a.m. to noon, and $4$ hours from noon to 4 p.m., meaning Walter was away from home for a total of $8\frac{1}{2}$ hours, or $8.5\times 60 = 510$ minutes.
Tallying up the times he was not on the bus, he has $6\times 50 = 300$ minutes in classes, $30$ minutes at lunch, and $2\times 60 = 120$ minutes of "additional time".  That is a total of $300 + 30 + 120 = 450$ minutes that Walter was away from home, but not on the bus.
Therefore, Walter spent $510 - 450 = 60$ minutes on the bus, and the answer is $\boxed{60}$

Answer: B 60
Problem: $\angle 1 + \angle 2 = 180^\circ$
$\angle 3 = \angle 4$
Find $\angle 4.$
[asy] pair H,I,J,K,L; H = (0,0); I = 10*dir(70); J = I + 10*dir(290); K = J + 5*dir(110); L = J + 5*dir(0); draw(H--I--J--cycle); draw(K--L--J); draw(arc((0,0),dir(70),(1,0),CW)); label("$70^\circ$",dir(35),NE); draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S); label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE); label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE); [/asy] $\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ$

Solution: Using the left triangle, we have:
$\angle 1 + 70 + 40 = 180$
$\angle 1 = 180 - 110$
$\angle 1 = 70$
Using the given fact that $\angle 1 + \angle 2 = 180$ , we have $\angle 2 = 180 - 70 = 110$
Finally, using the right triangle, and the fact that $\angle 3 = \angle 4$ , we have:
$\angle 2 + \angle 3 + \angle 4 = 180$
$110 + \angle 3 + \angle 4 = 180$
$110 + \angle 4+ \angle 4 = 180$
$2\angle 4 = 70$
$\angle 4 = 35$
Thus, the answer is $\boxed{35}$

Answer: D 35
Problem: Three bags of jelly beans contain 26, 28, and 30 beans.  The ratios of yellow beans to all beans in each of these bags are $50\%$ $25\%$ , and $20\%$ , respectively.  All three bags of candy are dumped into one bowl.  Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?
$\text{(A)}\ 31\% \qquad \text{(B)}\ 32\% \qquad \text{(C)}\ 33\% \qquad \text{(D)}\ 35\% \qquad \text{(E)}\ 95\%$

Solution: In bag $A$ , there are $26$ jellybeans, and $50\%$ are yellow.  That means there are $26\times 50\% = 26\times 0.50 = 13$ yellow jelly beans in this bag.
In bag $B$ , there are $28$ jellybeans, and $25\%$ are yellow.  That means there are $28\times 25\% = 28\times 0.25 = 7$ yellow jelly beans in this bag.
In bag $C$ , there are $30$ jellybeans, and $20\%$ are yellow.  That means there are $30\times 20\% = 30\times 0.20 = 6$ yellow jelly beans in this bag.
In all three bags, there are $13 + 7 + 6 = 26$ yellow jelly beans in total , and $26 + 28 + 30 = 84$ jelly beans of all types in total.
Thus, $\frac{26}{84} = \frac{26}{84}\cdot 100\% = \frac{13 }{ 42} \cdot 100\% = 30.9\%$ of all jellybeans are yellow.  Thus, the correct answer is $\boxed{31}$

Answer: A 31
Problem: There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8.  What is the difference between the largest and smallest integers in the set?
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution: When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum to 4. 2 and 2 does not work because that would result in two modes. However, 1 and 3 does, and so our answer is 8-1=7. $\boxed{7}$

Answer: D 7
Problem: A cube has eight vertices (corners) and twelve edges.  A segment, such as $x$ , which joins two vertices not joined by an edge is called a diagonal.  Segment $y$ is also a diagonal.  How many diagonals does a cube have?
[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution: On each face, there are $2$ diagonals like $x$ .  There are $6$ faces on a cube.  Thus, there are $2\times 6 = 12$ diagonals that are "x-like".
Every "y-like" diagonal must connect the bottom of the cube to the top of the cube.  Thus, for each of the $4$ bottom vertices of the cube, there is a different "y-like" diagonal.  So there are $4$ "y-like" diagonals.
This gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer $\boxed{16}$

Answer: E 16
Problem: A cube has eight vertices (corners) and twelve edges.  A segment, such as $x$ , which joins two vertices not joined by an edge is called a diagonal.  Segment $y$ is also a diagonal.  How many diagonals does a cube have?
[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution: There are $8$ vertices on a cube(A, B, C, D, E, F, G, H).  If you pick one vertice, it will have 7 line segments associated with it, so you will have $\frac{8\cdot7}{2} = 28$ segments within the cube.  The division by $2$ is necessary because you counted both the segment from $A$ to $B$ and the segment from $B$ to $A$
But not all $28$ of these segments are diagonals.  Some are edges.  There are $4$ edges on the top, $4$ edges on the bottom, and $4$ edges that connect the top to the bottom.  So there are $12$ edges total, meaning that there are $28 - 12 = 16$ segments that are not edges.  All of these segments are diagonals, and thus the answer is $\boxed{16}$

Answer: E 16
Problem: A cube has eight vertices (corners) and twelve edges.  A segment, such as $x$ , which joins two vertices not joined by an edge is called a diagonal.  Segment $y$ is also a diagonal.  How many diagonals does a cube have?
[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution: Consider picking one point on the corner of the cube.  That point has $3$ "x-like" diagonals , and $1$ "y-like" diagonal that ends on the opposite vertex.  Thus, each vertex has $3 + 1 = 4$ diagonals associated with it.  There are $8$ vertices on the cube, giving a total of $8 \cdot 4 = 32$ diagonals.
However, each diagonal was counted as both a "starting point" and an "ending point".  So there are really $\frac{32}{2} = 16$ diagonals, giving an answer of $\boxed{16}$

Answer: E 16
Problem: At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for $$5$ .  This week they are on sale at 5 boxes for $$4$ .  The percent decrease in the price per box during the sale was closest to
$\text{(A)}\ 30\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 65\%$

Solution: Last week, each box was $\frac{5}{4} = 1.25$
This week, each box is $\frac{4}{5} = 0.80$
Percent decrease is given by $\frac{X_{old} - X_{new}}{X_{old}} \cdot 100\%$
This, the percent decrease is $\frac{1.25 - 0.8}{1.25}\cdot 100\% = \frac{45}{125} \cdot 100\% = 36\%$ , which is closest to $\boxed{35}$

Answer: B 35
Problem: If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9$ , what is the sum of $a$ and $b$
$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$

Solution: Notice that the numerator of the first fraction cancels out the denominator of the second fraction, and the numerator of the second fraction cancels out the denominator of the third fraction, and so on.
The only numbers left will be $a$ in the numerator from the last fraction and $2$ in the denominator from the first fraction.  (The $b$ will cancel with the numerator of the preceeding number.)  Thus, $\frac{a}{2} = 9$ , and $a=18$
Since the numerator is always one more than the denominator, $b = a-1 = 17$ , and $a+ b = 18 + 17 = 35$ , giving an answer of $\boxed{35}$

Answer: D 35
Problem: If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9$ , what is the sum of $a$ and $b$
$\text{(A)}\ 11 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 37$

Solution: Find a pattern.  If $a=4$ , then the expression is just the first two terms, which is $2$
If $a=5$ , then the expression is the first three terms, giving $2.5$
If $a=6$ , the expression is the first four terms, giving $3$
If $a=7$ , the expression will be the first five terms, giving $3.5$
Conjecture that the expression is always going to equal $\frac{a}{2}$ , and thus when $a=18$ , the expression will be $9$ , as desired.
As above, when $a=18$ $b=17$ , and the sum is $35$ , or $\boxed{35}$

Answer: D 35
Problem: Each corner cube is removed from this $3\text{ cm}\times 3\text{ cm}\times 3\text{ cm}$ cube.  The surface area of the remaining figure is
[asy] draw((2.7,3.99)--(0,3)--(0,0)); draw((3.7,3.99)--(1,3)--(1,0)); draw((4.7,3.99)--(2,3)--(2,0)); draw((5.7,3.99)--(3,3)--(3,0));  draw((0,0)--(3,0)--(5.7,0.99)); draw((0,1)--(3,1)--(5.7,1.99)); draw((0,2)--(3,2)--(5.7,2.99)); draw((0,3)--(3,3)--(5.7,3.99));  draw((0,3)--(3,3)--(3,0)); draw((0.9,3.33)--(3.9,3.33)--(3.9,0.33)); draw((1.8,3.66)--(4.8,3.66)--(4.8,0.66)); draw((2.7,3.99)--(5.7,3.99)--(5.7,0.99)); [/asy]
$\text{(A)}\ 19\text{ sq.cm} \qquad \text{(B)}\ 24\text{ sq.cm} \qquad \text{(C)}\ 30\text{ sq.cm} \qquad \text{(D)}\ 54\text{ sq.cm} \qquad \text{(E)}\ 72\text{ sq.cm}$

Solution: The original cube has $6$ square surfaces that each have an area of $3^2 = 9$ , for a toal surface area of $6\cdot 9 = 54$
Since no two corner cubes touch, we can examine the effect of removing each corner cube individually.
Each corner cube contribues $3$ faces each of surface area $1$ to the big cube, so the surface area is decreased by $3$ when the cube is removed.
However, when the cube is removed, $3$ faces on the 3x3x3 cube will be revealed, increasing the surface area by $3$
Thus, the surface area does not change with the removal of a corner cube, and it remains $54$ , which is answer $\boxed{54}$

Answer: D 54
Problem: A two-inch cube $(2\times 2\times 2)$ of silver weighs 3 pounds and is worth 200 dollars. How much is a three-inch cube of silver worth?
$\textbf{(A) }\text{300 dollars} \qquad \textbf{(B) }\text{375 dollars} \qquad \textbf{(C) }\text{450 dollars} \qquad \textbf{(D) }\text{560 dollars}\qquad \textbf{(E) }\text{675 dollars}$

Solution: The two-inch cube has a volume of $8$ cubic inches, and the three-inch cube has a volume of $27$ cubic inches. Thus, the three-inch cube has a weight that is $\frac{27}{8}$ times that of the two-inch cube. Then its value is $\frac{27}{8} \cdot 200 = \boxed{675}$

Answer: E 675
Problem: There are positive integers that have these properties:
The product of the digits of the largest integer with both properties is
$\text{(A)}\ 7 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 60$

Solution: Five-digit numbers will have a minimum of $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ as the sum of their squares if the five digits are distinct and non-zero.  If there is a zero, it will be forced to the left by rule #2.
No digit will be greater than $7$ , as $8^2 = 64$
Trying four digit numbers $WXYZ$ , we have $w^2 + x^2 + y^2 + z^2 = 50$ with $0 < w < x < y < z < 8$
$z=7$ will not work, since the other digits must be at least $1^2 + 2^2 + 3^2 = 14$ , and the sum of the squares would be over $50$
$z=6$ will give $w^2 + x^2 + y^2 = 14$ $(w,x,y) = (1,2,3)$ will work, giving the number $1236$ .  No other number with $z=6$ will work, as $w, x,$ and $y$ would have to be greater.
$z=5$ will give $w^2 + x^2 + y^2 = 25$ $y=4$ forces $x=3$ and $w=0$ , which has a leading zero, and then we have 345 which is a 3-digit number.
$z=4$ can only give the number $1234$ , which does not satisfy the condition of the problem.
Thus, the number in question is $1236$ , and the product of the digits is $36$ , giving $\boxed{36}$ as the answer.

Answer: C 36
Problem: All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together.  What is the rightmost digit (the units digit) of the product?
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: All the tens digits of the product will be irrelevant to finding the units digit.  Thus, we are searching for the units digit of $(2\cdot 4\cdot 6 \cdot 8) \cdot (2 \cdot 4 \cdot 6 \cdot 8) \cdot (2\cdot 4\cdot 6 \cdot 8) \cdot ...$
There will be $10$ groups of $4$ numbers.  The number now can be rewritten as $(2\cdot 4 \cdot 6 \cdot 8)^{10}$
Multiplying, we get $384^{10}$
Again, we can disregard the tens and hundreds digit of $384$ , since we only want the units digit of the number, leaving $4^{10}$
Now, we try to find a pattern to the units digit of $4^n$ .  To compute this quickly, we once again discard all tens digits and higher.
$4^1 = 4$
$4^2 = 4\cdot 4 = 1\underline{6}$ , discard the $1$
$4^3 = 1 \cdot 4 = \underline{4}$
$4^4 = 4 \cdot 4 = 1\underline{6}$ , discard the $1$
$4^5 = 1 \cdot 4 = \underline{4}$
Those equalities are, in reality, congruences $\mod {10}$
Thus, the pattern of the units digits is $\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}$ .  The cycle repeats so that term $n$ is the same as term $n+2$ .  The tenth number in the cycle is $6$ , giving an answer of $\boxed{6}$

Answer: D 6
Problem: All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together.  What is the rightmost digit (the units digit) of the product?
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: Again, the value we seek is equal to $(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$ .
We can use equivalence to simplify.
$(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$
$\equiv (2\cdot 4\cdot (-4)\cdot (-2))^{10} \mod 10$
$\equiv 64^{10} \mod 10$
$\equiv 4^{10} \mod 10$
$\equiv 2^{20} \mod 10$
$\equiv 8^6\cdot 2^2 \mod 10$
$\equiv (-2)^6\cdot 2^2 \mod 10$
$\equiv 2^8 \mod 10$
$\equiv 256 \mod 10$
$\equiv 6 \mod 10$
Thus our answer is $\boxed{6}$

Answer: D 6
Problem: How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: The factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18,$ and $36$
The multiples of $4$ up to $36$ are $4, 8, 12, 16, 20, 24, 28, 32$ and $36$
Only $4, 12$ and $36$ appear on both lists, so the answer is $3$ , which is option $\boxed{3}$

Answer: B 3
Problem: How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: $36 = 4^1 \cdot 3^2$ . All possible factors of $36$ will be here, except for ones divisible by $2$ and not by $4$ $(1+1)\cdot (2+1) = 6$ . Subtract factors not divisible by $4$ , which are $1$ $3^1$ , and $3^2$ $6-3=3$ , which is $\boxed{3}$

Answer: B 3
Problem: How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution: Divide $36$ by $4$ , and the remaining factors, when multiplied by $4$ , will be factors of $36$
$36 \div 4 = 9$ , which has $3$ factors, giving us option $\boxed{3}$

Answer: B 3
Problem: The 64 whole numbers from 1 through 64 are written, one per square, on a checkerboard (an 8 by 8 array of 64 squares).  The first 8 numbers are written in order across the first row, the next 8 across the second row, and so on.  After all 64 numbers are written, the sum of the numbers in the four corners will be
$\text{(A)}\ 130 \qquad \text{(B)}\ 131 \qquad \text{(C)}\ 132 \qquad \text{(D)}\ 133 \qquad \text{(E)}\ 134$

Solution: Obviously $1$ is in the top left corner, $8$ is in the top right corner, and $64$ is in the bottom right corner. To find the bottom left corner, subtract $7$ from $64$ which is $57$ . Adding the results gives $1+8+57+64=130$ which is answer $\boxed{130}$

Answer: A 130
Problem: What is the smallest result that can be obtained from the following process?
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 56$

Solution: Since we want the smallest possible result, and we are only adding and multiplying positive numbers over $1$ , we can "prune" the set to the three smallest numbers $\{3,5,7\}$ .  Using bigger numbers will create bigger sums and bigger products.
From there, compute the $3$ ways you can do the two operations:
$(3+5)7 = 8\cdot 7 = 56$
$(3 + 7)5 = 10\cdot 5 = 50$
$(7+5)3 = 12\cdot 3 = 36$
The smallest number is 36, giving an answer of $\boxed{36}$

Answer: C 36
Problem: Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month.  If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution: Call this month "Month 0".  Make a table of the fish that Brent and Gretel have each month.
$\text{Month / Brent / Gretel}$
$\text{0 / 4 / 128}$
$\text{1 / 16 / 256}$
$\text{2 / 64 / 512}$
$\text{3 / 256 / 1024}$
$\text{4 / 1024 / 2048}$
$\text{5 / 4096 / 4096}$
You could create a similar table without doing all of the calculations by converting all the goldfish into powers of $2$ .  In this table, you could increase Brent's goldfish by two powers of $2$ , while increasing Gretel's fish by one power of $2$
Either way, in $5$ months they will have the same number of fish, giving an answer of $\boxed{5}$

Answer: B 5
Problem: Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month.  If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution: Create two equations.
Brent starts with $4$ goldfish and they quadruple each month, giving us $y=4\cdot 4^{x}$ or $y=2^{2x+2}$
Gretel starts with $128$ goldfish and they double each month, giving us $y=128 \cdot 2^{x}$ or $y=2^{x+7}$
Using substitution, $2^{2x+2}=2^{x+7}$ . Thus $2x+2=x+7$ giving us $x=5$ , which is $\boxed{5}$

Answer: B 5
Problem: Points $A$ and $B$ are 10 units apart.  Points $B$ and $C$ are 4 units apart.  Points $C$ and $D$ are 3 units apart.  If $A$ and $D$ are as close as possible, then the number of units between them is
$\text{(A)}\ 0 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 11 \qquad \text{(E)}\ 17$

Solution: If $AB = 10$ and $BC=4$ , then $(10 - 4) \le AC \le (10 + 4)$ by the triangle inequality .  In the triangle inequality, the equality is only reached when the "triangle" $ABC$ is really a degenerate triangle, and $ABC$ are collinear.
Simplifying, this means the smallest value $AC$ can be is $6$
Applying the triangle inequality on $ACD$ with $AC = 6$ and $CD = 3$ , we know that $6 - 3 \le AD \le 6 + 3$ when $AC$ is minimized.  If $AC$ were larger, then $AD$ could be larger, but we want the smallest $AD$ possible, and not the largest.  Thus, $AD$ must be at least $3$ , but cannot be smaller than $3$ .  Therefore, $\boxed{3}$ is the answer.

Answer: B 3
Problem: If 5 times a number is 2, then 100 times the reciprocal of the number is
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 250 \qquad \text{(E)}\ 500$

Solution: If $5$ times a number is $2$ , then $5x = 2 \rightarrow x = \frac{2}{5}$ , and the number is $\frac{2}{5}$
If the number is $\frac{2}{5}$ , then $100$ times its reciprocal is $100$ times $\frac{5}{2}$ , which is $250$ , giving an answer of $\boxed{250}$

Answer: D 250
Problem: When Walter drove up to the gasoline pump, he noticed that his gasoline tank was 1/8 full.  He purchased 7.5 gallons of gasoline for 10 dollars.  With this additional gasoline, his gasoline tank was then 5/8 full.  The number of gallons of gasoline his tank holds when it is full is $\text{(A)}\ 8.75 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11.5 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 22.5$

Solution: The tank started at $\frac{1}{8}$ full, and ended at $\frac{5}{8}$ full.  Therefore, Walter filled $\frac{5}{8} - \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$ of the tank.
If Walter fills half the tank with $7.5$ gallons, then Walter can fill two halves of the tank (or a whole tank) with $7.5 \times 2 = 15$ gallons, giving an answer of $\boxed{15}$

Answer: D 15
Problem: Let $x$ be the number \[0.\underbrace{0000...0000}_{1996\text{ zeros}}1,\] where there are 1996 zeros after the decimal point.  Which of the following expressions represents the largest number?
$\text{(A)}\ 3+x \qquad \text{(B)}\ 3-x \qquad \text{(C)}\ 3\cdot x \qquad \text{(D)}\ 3/x \qquad \text{(E)}\ x/3$

Solution: Estimate each of the options.
$A$ will give a number that is just over $3$
$B$ will give a number that is just under $3$ .  This eliminates $B$ , because $A$ is bigger.
$C$ will give a number that is barely over $0$ , since it is three times a tiny number.  This eliminates $C$ , because $A$ is bigger.
$D$ will give a huge number. $\frac{1}{x}$ will get very, very large in magnitude when $x$ gets close to zero.  You can see this by examining the sequence $x=0.1$ , which gives $10$ as the reciprocal, $x=0.01$ , which gives $100$ as the reciprocal, and $x=0.001$ , which gives $1000$ as the reciprocal.  Thus, $D$ will be huge, and this eliminates $A$
$E$ will give a small number, since you're dividing a tiny number into thirds.  This eliminates $E$ , and gives $\boxed{3}$ as the answer.

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

Solution: Adding all of the numbers gives us $\frac{11\cdot12}{2}=66$ as the current total. Since there are $11$ numbers, the current average is $\frac{66}{11}=6$ . We need to take away a number from the total and then divide the result by $10$ because there will only be $10$ numbers left to give an average of $6.1$ . Setting up the equation:
$\frac{66-x}{10}=6.1$
$66 - x = 61$
$x = 5$
Thus, the answer is $\boxed{5}$

Answer: B 5