Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Dots are spaced one unit apart, horizontally and vertically.  The number of square units enclosed by the polygon is
[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution: Notice that the extra triangle on the top with area $1$ can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area $1$ . This creates a $2*3$ rectangle, with a area of $6$ . The answer is $\boxed{6}$ ~sakshamsethi

Answer: B 6
Problem: A child's wading pool contains 200 gallons of water.  If water evaporates at the rate of 0.5 gallons per day and no other water is added or removed, how many gallons of water will be in the pool after 30 days?
$\text{(A)}\ 140 \qquad \text{(B)}\ 170 \qquad \text{(C)}\ 185 \qquad \text{(D)}\ 198.5 \qquad \text{(E)}\ 199.85$

Solution: $30$ days multiplied by $0.5$ gallons a day results in $15$ gallons of water loss.
The remaining water is $200-15=185=\boxed{185}$

Answer: C 185
Problem: For a sale, a store owner reduces the price of a $$10$ scarf by $20\%$ .  Later the price is lowered again, this time by one-half the reduced price.  The price is now
$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}$

Solution: $100\%-20\%=80\%$
$10\times80\%=10\times0.8$
$10\times0.8=8$
$\frac{8}{2}=4=\boxed{4.00}$

Answer: C 4.00
Problem: For a sale, a store owner reduces the price of a $$10$ scarf by $20\%$ .  Later the price is lowered again, this time by one-half the reduced price.  The price is now
$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}$

Solution: The first discount has percentage 20, which is then discounted again for half of the already discounted price.
$100-20=80$
$\frac{80}{2}=40$
$40\%\times10=10\times0.4=4=\boxed{4.00}$

Answer: C 4.00
Problem: Each of the letters $\text{W}$ $\text{X}$ $\text{Y}$ , and $\text{Z}$ represents a different integer in the set $\{ 1,2,3,4\}$ , but not necessarily in that order.  If $\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1$ , then the sum of $\text{W}$ and $\text{Y}$ is
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

Solution: There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\{ 1,2,3,4\}$
$1$ has factor $1$
$2$ has factors $1$ and $2$
$3$ has factors $1$ and $3$
$4$ has factors $1$ $2$ , and $4$
From here, we note that even though all numbers have the factor $1$ , only $4$ has another factor other than $1$ in the set (ie. $2$
We could therefore have one fraction be $\frac{4}{2}$ and another $\frac{3}{1}$
The sum of the numerators is $4+3=7=\boxed{7}$

Answer: E 7
Problem: Harry has 3 sisters and 5 brothers.  His sister Harriet has $\text{S}$ sisters and $\text{B}$ brothers.  What is the product of $\text{S}$ and $\text{B}$
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$

Solution: Harry has 3 sisters and 5 brothers. His sister, being a girl, would have 1 less sister and 1 more brother.
$S = 3-1=2$
$B = 5+1=6$
$S\cdot B = 2\times6=12=\boxed{12}$

Answer: C 12
Problem: $2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) + 4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)=$
$\text{(A)}\ 45 \qquad \text{(B)}\ 49 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$

Solution: Taking the first product, we have
$\left(1-\frac{1}{2}\right)=\frac{1}{2}$
$\frac{1}{2}\times2=1$
Looking at the second, we get
$\left(1-\frac{1}{3}\right)=\frac{2}{3}$
$\frac{2}{3}\times3=2$
We seem to be going up by $1$
Just to check,
$1-\frac{1}{n}=\frac{n-1}{n}$
$\frac{n-1}{n}\times n=n-1$
Now that we have discovered the pattern, we have to find the last term.
$1-\frac{1}{10}=\frac{9}{10}$
$\frac{9}{10}\times10=9$
The sum of all numbers from $1$ to $9$ is
$\frac{9\cdot10}{2}=45=\boxed{45}$

Answer: A 45
Problem: An Annville Junior High School, $30\%$ of the students in the Math Club are in the Science Club, and $80\%$ of the students in the Science Club are in the Math Club.  There are 15 students in the Science Club.  How many students are in the Math Club?
$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 40$

Solution: If $80\%$ of the people in the science club of $15$ people are in the Math Club, $\frac{4}{5}\times15=12$ people are in the both the Math Club and the Science Club.
These $12$ people make up $30\%$ of the Math Club.
Setting up a proportion:
$12\cdot 1.00 =0.30\cdot x$
$\frac{12}{0.3} = 40=x$
There are $40=\boxed{40}$ people in the Math Club.

Answer: E 40
Problem: Estimate the year in which the population of Nisos will be approximately 6,000.
$\text{(A)}\ 2050 \qquad \text{(B)}\ 2075 \qquad \text{(C)}\ 2100 \qquad \text{(D)}\ 2125 \qquad \text{(E)}\ 2150$

Solution: We could triple the population every $25$ years and make a chart:
Year: 2000
Population: 200
Year: 2025
Population: 600
Year: 2050
Population: 1800
Year: 2075
Population: 5400
Year: 2100
Population: 16200
The closest year is 2075, or $\boxed{2075}$

Answer: B 2075
Problem: Estimate the year in which the population of Nisos will be approximately 6,000.
$\text{(A)}\ 2050 \qquad \text{(B)}\ 2075 \qquad \text{(C)}\ 2100 \qquad \text{(D)}\ 2125 \qquad \text{(E)}\ 2150$

Solution: We could find out how many periods of 25 years we need to triple by dividing our total from our present number.
$\frac{6000}{200}=30$
The power of $3$ that $30$ is closest to is $27=3^3$
Therefore, after $3$ periods, we will be closest to $6000$
$\boxed{2075}$

Answer: B 2075
Problem: In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
$\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}$

Solution: We can divide the total area by how much will be occupied per person:
$\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}$ can stay on the island at its maximum capacity.
We can divide 16600 by the current population in $1998$ which is 200 to see by what factor the population increases:
$\frac{16600}{200}=83$ -fold increase in population.
Thus, the population increases by a factor $83$ .  This is very close to $3 \times 3 \times 3 \times 3 = 81$ , and so there are about $4$ triplings of the island's population.
It takes $4\times25=100=\boxed{100}$ years to triple the island's population four times in succession.

Answer: C 100
Problem: In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
$\text{(A)}\ 50\text{ yrs.} \qquad \text{(B)}\ 75\text{ yrs.} \qquad \text{(C)}\ 100\text{ yrs.} \qquad \text{(D)}\ 125\text{ yrs.} \qquad \text{(E)}\ 150\text{ yrs.}$

Solution: We can continue the pattern, and because the pattern increases numbers rapidly, it won't be hard. $\frac{24900 \text{ acres}}{1.5 \text{ acres per person}}=16600 \text{ people}$ can live on the island at its maximum capacity.
$1998: 200 \text{ people}$ $2023: 600 \text{ people}$ $2048: 1800 \text{ people}$ $2073: 5400 \text{ people}$ $2098: 16200 \text{ people}$ After the year $2098$ , it will not be possible for the next increase to occur, because when $16200$ is tripled, it is way more than the maximum capacity. Thus, the answer is $2098-1998 = 100$ , or $\boxed{100}$

Answer: C 100
Problem: $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box.  How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution: Each small cube would have dimensions $1\times 1\times 1$ making each cube a unit cube.
If there are $16$ cubes per face and there are $5$ faces we are counting, we have $16\times 5= 80$ cubes.
Some cubes are on account of overlap between different faces.
We could reduce this number by subtracting the overlap areas, which could mean subtracting 4 cubes from each side and 12 from the bottom.
$80-(4\times 4+12)=80-28=52=\boxed{52}$

Answer: B 52
Problem: $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box.  How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution: You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".
There are $4$ cubes in the center of the top face that do not fit the description.  Remove those.
Once you remove those cubes on top, you must go down one level and remove the cubes in the same position on the second layer.  Thus, $4$ more cubes are removed.
Finally, you repeat this on the third layer, for $4$ more cubes.
Once you do the top three layers, you will be on the bottom layer, and you don't remove any more cubes.
This means you removed $4 + 4 + 4 = 12$ cubes of the $4 \times 4 \times 4 = 64$ cubes.
Thus, $64 - 12 = 52$ cubes remain, and the answer is $\boxed{52}$

Answer: B 52
Problem: $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box.  How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution: We can use casework to solve this.
Case $1$ : The cubes on the sides
There are four sides of a cube which means $64$ cubes, but there will be an overlap of $16$ cubes ( $4$ sides $\times$ $4$ cubes on each side).
So there are $16 \times 4-16=64-16=48$ cubes on the sides.
Case $2$ : The cubes on the bottom
There are $16$ cubes on the bottom, but by counting the cubes on the sides, $12$ of the cubes are already accounted for. $16-12=4$ cubes on the bottom that aren't already counted.
$48+4=\boxed{52}$ cubes touching at least one side. Since there is no top, we don't need to count the top face cubes.

Answer: B 52
Problem: Terri produces a sequence of positive integers by following three rules.  She starts with a positive integer, then applies the appropriate rule to the result, and continues in this fashion.
Rule 1: If the integer is less than 10, multiply it by 9.
Rule 2: If the integer is even and greater than 9, divide it by 2.
Rule 3: If the integer is odd and greater than 9, subtract 5 from it.
A sample sequence: $23, 18, 9, 81, 76, \ldots .$
Find the $98^\text{th}$ term of the sequence that begins $98, 49, \ldots .$
$\text{(A)}\ 6 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 54$

Solution: We could start by looking for a pattern.
$98, 49, 44, 22, 11, 6, 54, 27, 22, 11, 6, \ldots .$
From here, we see that we have a pattern of $22, 11, 6, 54, 27, \ldots .$ after $98, 49, 44$
Our problem is now really
Find the $95^\text{th}$ term of the sequence that goes $22, 11, 6, 54, 27, 22, 11, 6, 54, 27, 22, \ldots .$
There are 5 terms in each repetition of the pattern, and $95\equiv0\pmod{5}$ , so the answer is $\boxed{27}$

Answer: D 27
Problem: A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?
[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy]
$\text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120$

Solution: The numbers that are shaded are the triangular numbers, which are numbers in the form $\frac{(n)(n+1)}{2}$ for positive integers.  They can also be generated by starting with $1$ , and adding $1, 2, 3, 4...$ as in the description of the problem.
Squares that have the same remainder after being divided by $8$ will be in the same column.  Thus, we want to find when the last remainder, from $0$ to $7$ , is found.
So, instead of adding $1, 2, 3, 4, 5, 6, 7, 8, 9, 10...$ , we can effectively either add $1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3...$ or subtract $7, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5...$ if we are only concerned about remainders when divided by $8$ .  We will pick the number that keeps the terms on the list between $1$ and $8$ .  We get:
$1$
$1 + 2 = 3$
$3 + 3 = 6$
$6 - 4 = 2$
$2 + 5 = 7$
$7 - 2 = 5$
$5 - 1 = 4$
$4 + 0 = 4$
$4 + 1 = 5$
$5 + 2 = 7$
$7 - 5 = 2$
$2 + 4 = 6$
$6 - 3 = 3$
$3 - 2 = 1$
$1 - 1 = 0$
Finally, a term with $0$ is found, and checking, all numbers $1$ through $7$ are also on the right side of the list.  This means the last term in our sequence is the first time that column $8$ is shaded.  There are $15$ terms in the sequence, leading to an answer of $\frac{15\cdot 16}{2} = 120$ , which is choice $\boxed{120}$

Answer: E 120
Problem: A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?
[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy]
$\text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120$

Solution: Note that the triangular numbers up to $120$ are $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120$ .  When you divide each of those numbers by $8$ , all remainders must be present.  We first search for number(s) that are evenly divisible by $8$ ; if two such numbers exist, we search for numbers that leave a remainder of $1$ , etc.
Quickly scanning the list, only $6, 10, 28, 36, 66, 78$ and $120$ are even.  That smaller list doesn't have any multiples of $8$ until it hits $120$ .  So $\boxed{120,}$ must be the answer.

Answer: E 120,
Problem: A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?
[asy] unitsize(20); for(int a = 0; a < 10; ++a) { draw((0,a)--(8,a)); } for (int b = 0; b < 9; ++b) { draw((b,0)--(b,9)); } draw((0,0)--(0,-.5)); draw((1,0)--(1,-1.5)); draw((.5,-1)--(1.5,-1)); draw((2,0)--(2,-.5)); draw((4,0)--(4,-.5)); draw((5,0)--(5,-1.5)); draw((4.5,-1)--(5.5,-1)); draw((6,0)--(6,-.5)); draw((8,0)--(8,-.5)); fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black); fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black); fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black); label("$2$",(1.5,8.2),N); label("$4$",(3.5,8.2),N); label("$5$",(4.5,8.2),N); label("$7$",(6.5,8.2),N); label("$8$",(7.5,8.2),N); label("$9$",(0.5,7.2),N); label("$11$",(2.5,7.2),N); label("$12$",(3.5,7.2),N); label("$13$",(4.5,7.2),N); label("$14$",(5.5,7.2),N); label("$16$",(7.5,7.2),N); [/asy]
$\text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120$

Solution: The numbers shaded are triangular numbers of the form $\frac{(n)(n+1)}{2}$ .  For this number to be divisible by $8$ , the numerator must be divisible by $16$ .  Since only one of $n$ and $n+1$ can be even, only one of them can have factors of $2$ .  Therefore, the first time the whole expression is divisible by $8$ is when either $n+1=16$ or when $n=16$ .  This gives $n=15$ as the first time $\frac{n(n+1)}{2}$ is divisible by $8$ , which gives $120$ .  No other triangular number lower than that is divisible by $8$ , and thus the $8^{th}$ column on the checkerboard won't be filled until then.  That gives $\boxed{120}$ as the right answer.

Answer: E 120
Problem: Three generous friends, each with some money, redistribute the money as followed:
Amy gives enough money to Jan and Toy to double each amount has.
Jan then gives enough to Amy and Toy to double their amounts. 
Finally, Toy gives enough to Amy and Jan to double their amounts.
If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 216\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 288$

Solution: If Toy had $36$ dollars at the beginning, then after Amy doubles his money, he has $36 \times 2 = 72$ dollars after the first step.
Then Jan doubles his money, and Toy has $72 \times 2 = 144$ dollars after the second step.
Then Toy doubles whatever Amy and Jan have.  Since Toy ended up with $36$ , he spent $144 - 36 = 108$ to double their money.  Therefore, just before this third step, Amy and Jan must have had $108$ dollars in total.  And, just before this step, Toy had $144$ dollars.  Altogether, the three had $144 + 108 = 252$ dollars, and the correct answer is $\boxed{252}$

Answer: D 252
Problem: $\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000} =$
$\text{(A)}\ 0.0026 \qquad \text{(B)}\ 0.0197 \qquad \text{(C)}\ 0.1997 \qquad \text{(D)}\ 0.26 \qquad \text{(E)}\ 1.997$

Solution: Convert each fraction to a decimal.
$\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000}$
$0.1+0.09+0.009+0.0007=0.1997$
$\boxed{0.1997}$

Answer: C 0.1997
Problem: Ahn chooses a two-digit integer, subtracts it from 200, and doubles the result.  What is the largest number Ahn can get?
$\text{(A)}\ 200 \qquad \text{(B)}\ 202 \qquad \text{(C)}\ 220 \qquad \text{(D)}\ 380 \qquad \text{(E)}\ 398$

Solution: The smallest two-digit integer he can subtract from $200$ is $10$ .  This will give the largest result for that first operation, and doubling it will keep it as the largest number possible.
\[200-10=190\] \[190\times2=380\]
$\boxed{380}$

Answer: D 380
Problem: Which of the following numbers is the largest?
$\text{(A)}\ 0.97 \qquad \text{(B)}\ 0.979 \qquad \text{(C)}\ 0.9709 \qquad \text{(D)}\ 0.907 \qquad \text{(E)}\ 0.9089$

Solution: Using the process of elimination:
Tenths digit: 
All tenths digits are equal, at $9$
Hundreths digit: $A$ $B$ , and $C$ all have the same hundreths digit of $7$ , and it is greater than the hundredths of $D$ or $E$ (which is $0$ ).  Eliminate both $D$ and $E$
Thousandths digit: $B$ has the largest thousandths digit of the remaining answers, and is the correct answer. $A$ has an "invisible" thousandths digit of $0$ , while $C$ also has a thousdandths digit of $0$
$\boxed{0.979}$

Answer: B 0.979
Problem: Julie is preparing a speech for her class.  Her speech must last between one-half hour and three-quarters of an hour.  The ideal rate of speech is 150 words per minute.  If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?
$\text{(A)}\ 2250 \qquad \text{(B)}\ 3000 \qquad \text{(C)}\ 4200 \qquad \text{(D)}\ 4350 \qquad \text{(E)}\ 5650$

Solution: Since there are $60$ minutes in an hour, we need a number of words between $(\frac{1}{2}\times 60)\times150=4500$ and $(\frac{3}{4}\times60)\times150=6750$ .  The only such choice is $\boxed{5650}$

Answer: E 5650
Problem: There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10.  The sum of these two multiples of 7 is
$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$

Solution: Writing out all two digit numbers that have a digital sum of $10$ , you get $19, 28, 37, 46, 55, 64, 73, 82,$ and $91$ .  The two numbers on that list that are divisible by $7$ are $28$ and $91$ .  Their sum is $28+91=119$ , choice $\boxed{119}$

Answer: A 119
Problem: There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10.  The sum of these two multiples of 7 is
$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$

Solution: Writing out all the two digit multiples of $7$ , you get $14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,$ and $91$ .  Again you find $28$ and $91$ have a digital sum of $10$ , giving answer $\boxed{119}$

Answer: A 119
Problem: In the number $74982.1035$ the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?
$\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000$

Solution: The digit $9$ is $5$ places to the left of the digit $3$ . Thus, it has a place value that is $10^5 = 100,000$ times greater. $\boxed{100,000}$

Answer: C 100,000
Problem: In the number $74982.1035$ the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?
$\text{(A)}\ 1,000 \qquad \text{(B)}\ 10,000 \qquad \text{(C)}\ 100,000 \qquad \text{(D)}\ 1,000,000 \qquad \text{(E)}\ 10,000,000$

Solution: The digit $9$ is in the $100$ s place.  The digit $3$ is in the $\frac{1}{1000}$ ths place.  The ratio of these two numbers is $\frac{100}{\frac{1}{1000}} = 100\times1000 = 100,000$ , and the answer is $\boxed{100,000}$

Answer: C 100,000