Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: $1,000,000,000,000-777,777,777,777=$
$\text{(A)}\ 222,222,222,222 \qquad  \text{(B)}\ 222,222,222,223 \qquad  \text{(C)}\ 233,333,333,333 \qquad \\ \text{(D)}\ 322,222,222,223 \qquad  \text{(E)}\ 333,333,333,333$

Solution: \begin{align*} 1,000,000,000,000-777,777,777,777 &= 999,999,999,999-777,777,777,777+1 \\ &= 222,222,222,222+1 \\ &= 222,222,222,223 \rightarrow \boxed{222,222,222,223}

Answer: B 222,222,222,223
Problem: $\frac{16+8}{4-2}=$
$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

Solution: \begin{align*} \frac{16+8}{4-2} &= \frac{24}{2} \\ &= 12\rightarrow \boxed{12}

Answer: C 12
Problem: If $991+993+995+997+999=5000-N$ , then $N=$
$\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25$

Solution: \begin{align*} 991+993+995+997+999=5000-N \\ &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\ &\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\ &\Rightarrow 5000-25=5000-N \\ &\Rightarrow N=25\rightarrow \boxed{25}

Answer: E 25
Problem: Which number in the array below is both the largest in its column and the smallest in its row?  (Columns go up and down, rows go right and left.) \[\begin{tabular}[t]{ccccc} 10 & 6 & 4 & 3 & 2 \\ 11 & 7 & 14 & 10 & 8 \\ 8 & 3 & 4 & 5 & 9 \\ 13 & 4 & 15 & 12 & 1 \\ 8 & 2 & 5 & 9 & 3 \end{tabular}\]
$\text{(A)}\ 1 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 15$

Solution: The largest numbers in the first, second, third, fourth and fifth columns are $13,7,15,12,9$ respectively.  Of these, only $7$ is the smallest in its row $\rightarrow \boxed{7}$

Answer: C 7
Problem: The value of $\frac{(487,000)(12,027,300)+(9,621,001)(487,000)}{(19,367)(.05)}$ is closest to
$\text{(A)}\ 10,000,000 \qquad \text{(B)}\ 100,000,000 \qquad \text{(C)}\ 1,000,000,000 \qquad \text{(D)}\ 10,000,000,000 \qquad \text{(E)}\ 100,000,000,000$

Solution: We can make the approximations \begin{align*} 487,000 &\approx 500,000 \\  12,027,300 &\approx 12,000,000 \\ 9,621,001 &\approx 10,000,000 \\ 19,367 &\approx 20,000. \end{align*}
Using these instead of the original numbers for an estimate, we have \begin{align*} \frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &= \frac{500,000\times 22,000,000}{1000} \\ &= 500,000 \times 22,000 \\ &= 1.1\times 10^{10} \\ &\approx 10,000,000,000 \rightarrow \boxed{10,000,000,000}

Answer: D 10,000,000,000
Problem: What is the largest quotient that can be formed using two numbers chosen from the set $\{ -24, -3, -2, 1, 2, 8 \}$
$\text{(A)}\ -24 \qquad \text{(B)}\ -3 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$

Solution: Let the two chosen numbers be $a$ and $b$ .  To maximize the quotient, we first have either $a,b>0$ or $a,b<0$ , and from there we maximize $|a|$ and minimize $|b|$
For the case $a,b<0$ , we have $a=-24$ and $b=-2$ , which gives us $(-24)/(-2)=12$ .  For the case $a,b>0$ , we have $a=8$ and $b=1$ , which gives us $8/1=8$
Since $12>8$ , our answer is $\boxed{12}$

Answer: D 12
Problem: How many whole numbers from $1$ through $46$ are divisible by either $3$ or $5$ or both?
$\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \text{(E)}\ 27$

Solution: There are $\left\lfloor \frac{46}{3}\right\rfloor =15$ numbers divisible by $3$ $\left\lfloor\frac{46}{5}\right\rfloor =9$ numbers divisible by $5$ , so at first we have $15+9=24$ numbers that are divisible by $3$ or $5$ , except we counted the multiples of $\text{LCM}(3,5)=15$ twice, once for $3$ and once for $5$
There are $\left\lfloor \frac{46}{15}\right\rfloor =3$ numbers divisible by $15$ , so there are $24-3=21$ numbers divisible by $3$ or $5$ $\rightarrow \boxed{21}$

Answer: B 21
Problem: The area in square units of the region enclosed by parallelogram $ABCD$ is
[asy] unitsize(24); pair A,B,C,D; A=(-1,0); B=(0,2); C=(4,2); D=(3,0);  draw(A--B--C--D); draw((0,-1)--(0,3)); draw((-2,0)--(6,0)); draw((-.25,2.75)--(0,3)--(.25,2.75)); draw((5.75,.25)--(6,0)--(5.75,-.25)); dot(origin); dot(A); dot(B); dot(C); dot(D); label("$y$",(0,3),N); label("$x$",(6,0),E); label("$(0,0)$",origin,SE); label("$D (3,0)$",D,SE); label("$C (4,2)$",C,NE); label("$A$",A,SW); label("$B$",B,NW); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$

Solution: The base is $\overline{BC}=4$ . The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is $4\cdot 2=8\Rightarrow \boxed{8}$

Answer: B 8
Problem: There are several sets of three different numbers whose sum is $15$ which can be chosen from $\{ 1,2,3,4,5,6,7,8,9 \}$ .  How many of these sets contain a $5$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

Solution: Let the three-element set be $\{ a,b,c \}$ and suppose that $a=5$
We need $b+c=10$ and $b\neq c$ .  This gives us four solutions, so there are $4$ sets with a $5$ also with the desired properties $\rightarrow \boxed{4}$

Answer: B 4
Problem: If $\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}$ , then $N=$
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1991 \qquad \text{(E)}\ 1992$

Solution: Note that for all integers $n\neq 0$ \[\frac{(n-1)+n+(n+1)}{n}=3.\] Thus, we must have $N=1991\rightarrow \boxed{1991}$

Answer: D 1991
Problem: If $\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}$ , then $N=$
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1991 \qquad \text{(E)}\ 1992$

Solution: As we know that $\frac{1990+1991+1992}{n}$ has to be some multiple of $\frac{2+3+4}{3}$ , then we know that the first equation is $995$ (1990/2) times bigger than the second one(in my solution), so the bottom must be $3\cdot995=\boxed{1991}$ -fn106068

Answer: D 1991
Problem: How many zeros are at the end of the product \[25\times 25\times 25\times 25\times 25\times 25\times 25\times 8\times 8\times 8?\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution: \begin{align*} (5^2)^7\times (2^3)^3 &=  5^{14}\times 2^{9} \\ &= 5^5\times 10^9  \end{align*} Since $5^5$ doesn't end in a $0$ , we can conclude that the product ends in $9$ zeroes $\rightarrow \boxed{9}$

Answer: C 9
Problem: Several students are competing in a series of three races.  A student earns $5$ points for winning a race, $3$ points for finishing second and $1$ point for finishing third.  There are no ties.  What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$

Solution: There are two ways for a student to get $11$ $5+5+1$ and $5+3+3$ .  Clearly if someone gets one of these combinations someone else could get the other, so we are not guaranteed the most points with $11$
There is only one way to get $13$ points: $5+5+3$ .  In this case, the largest score another person could get is $5+3+3=11$ , so having $13$ points guarantees having more points than any other person $\rightarrow \boxed{13}$

Answer: D 13
Problem: Several students are competing in a series of three races.  A student earns $5$ points for winning a race, $3$ points for finishing second and $1$ point for finishing third.  There are no ties.  What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$

Solution: If someone gets $11$ points, the possible combinations are $5,5,1$ and $5,3,3$ If he gets $5,3,3$ then someone else can be $5,3,3$ which would not guarantee victory.
If we have 13 points, the only way to make this is $5,5,3$ . There is no way to get any number of points higher than this, so the answer is $\boxed{13}$ .---stjwyl

Answer: D 13
Problem: The $16$ squares on a piece of paper are numbered as shown in the diagram.  While lying on a table, the paper is folded in half four times in the following sequence:
(1) fold the top half over the bottom half
(2) fold the bottom half over the top half
(3) fold the right half over the left half
(4) fold the left half over the right half.
Which numbered square is on top after step $4$
[asy] unitsize(18); for(int a=0; a<5; ++a)  {   draw((a,0)--(a,4));  } for(int b=0; b<5; ++b)  {   draw((0,b)--(4,b));  } label("$1$",(0.5,3.1),N); label("$2$",(1.5,3.1),N); label("$3$",(2.5,3.1),N); label("$4$",(3.5,3.1),N); label("$5$",(0.5,2.1),N); label("$6$",(1.5,2.1),N); label("$7$",(2.5,2.1),N); label("$8$",(3.5,2.1),N); label("$9$",(0.5,1.1),N); label("$10$",(1.5,1.1),N); label("$11$",(2.5,1.1),N); label("$12$",(3.5,1.1),N); label("$13$",(0.5,0.1),N); label("$14$",(1.5,0.1),N); label("$15$",(2.5,0.1),N); label("$16$",(3.5,0.1),N); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution: Suppose we undo each of the four folds, considering just the top square until we completely unfold the paper. $x$ will be marked in the square if the face that shows after all the folds is face up, $y$ if that face is facing down.
Step 0: [asy] unitsize(18); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); label("$x$",(0.5,0.1),N); [/asy] Step 1: [asy] unitsize(18); draw((0,0)--(2,0)--(2,1)--(0,1)--cycle); draw((1,0)--(1,1)); label("$y$",(0.5,0.1),N); [/asy] Step 2: [asy] unitsize(18); draw((0,0)--(4,0)--(4,1)--(0,1)--cycle);  draw((1,0)--(1,1)); draw((2,0)--(2,1)); draw((3,0)--(3,1)); label("$y$",(0.5,0.1),N); [/asy] Step 3: [asy] unitsize(18); draw((0,0)--(4,0)--(4,2)--(0,2)--cycle); draw((0,1)--(4,1)); draw((1,0)--(1,2)); draw((2,0)--(2,2)); draw((3,0)--(3,2)); label("$y$",(0.5,1.1),N); [/asy] Step 4: [asy] unitsize(18); for(int a=0; a<5; ++a)  {   draw((a,0)--(a,4));  } for(int b=0; b<5; ++b)  {   draw((0,b)--(4,b));  } label("$y$",(0.5,1.1),N); [/asy] The marked square is in the same spot as the number $9\rightarrow \boxed{9}$

Answer: B 9
Problem: An auditorium with $20$ rows of seats has $10$ seats in the first row.  Each successive row has one more seat than the previous row.  If students taking an exam are permitted to sit in any row, but not next to another student in that row, then the maximum number of students that can be seated for an exam is
$\text{(A)}\ 150 \qquad \text{(B)}\ 180 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 460$

Solution: We first note that if a row has $n$ seats, then the maximum number of students that can be seated in that row is $\left\lceil \frac{n}{2} \right\rceil$ , where $\lceil x \rceil$ is the smallest integer greater than or equal to $x$ .  If a row has $2k$ seats, clearly we can only fit $k$ students in that row.  If a row has $2k+1$ seats, we can fit $k+1$ students by putting students at the ends and then alternating between skipping a seat and putting a student in.
For each row with $10+k$ seats, there is a corresponding row with $29-k$ seats.  The sum of the maximum number of students for these rows is \[\left\lceil \frac{10+k}{2}\right\rceil +\left\lceil \frac{29-k}{2} \right\rceil = 20.\] There are $20/2=10$ pairs of rows, so the maximum number of students for the exam is $20\times 10=200\rightarrow \boxed{200}$

Answer: C 200
Problem: The average (arithmetic mean) of $10$ different positive whole numbers is $10$ .  The largest possible value of any of these numbers is
$\text{(A)}\ 10 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 55 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 91$

Solution: If the average of the numbers is $10$ , then their sum is $10\times 10=100$
To maximize the largest number of the ten, we minimize the other nine.  Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$ .  Their sum is $45$
The sum of nine of the numbers is $45$ , and the sum of all ten is $100$ so the last number must be $100-45=55\rightarrow \boxed{55}$

Answer: C 55
Problem: In the addition problem, each digit has been replaced by a letter.  If different letters represent different digits then what is the value of $C$
[asy] unitsize(18); draw((-1,0)--(3,0)); draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4)); label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1),N); label("$A$",(1.5,1.1),N); label("$B$",(2.5,1.1),N); label("$A$",(2.5,0.1),N); label("$3$",(0.5,-.1),S); label("$0$",(1.5,-.1),S); label("$0$",(2.5,-.1),S); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$

Solution: From this we have \[111A+11B+C=300.\] Clearly, $A<3$ .  Since $B,C\leq 9$ \[111A > 201 \Rightarrow A\geq 2.\] Thus, $A=2$ and $11B+C=78$ .  From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{1}$

Answer: A 1
Problem: In the addition problem, each digit has been replaced by a letter.  If different letters represent different digits then what is the value of $C$
[asy] unitsize(18); draw((-1,0)--(3,0)); draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4)); label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1),N); label("$A$",(1.5,1.1),N); label("$B$",(2.5,1.1),N); label("$A$",(2.5,0.1),N); label("$3$",(0.5,-.1),S); label("$0$",(1.5,-.1),S); label("$0$",(2.5,-.1),S); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$

Solution: Using logic, $a+b+c= 10$ , therefore $b+a+1$ (from the carry over) $= 10$ .
So $b+a=9$ $A+1=3$
Thus, $A=2$ and $11B+C=78$ .  From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{1}$

Answer: A 1
Problem: For every $3^\circ$ rise in temperature, the volume of a certain gas expands by $4$ cubic centimeters.  If the volume of the gas is $24$ cubic centimeters when the temperature is $32^\circ$ , what was the volume of the gas in cubic centimeters when the temperature was $20^\circ$
$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 40$

Solution: We know that \[T=32-3k\Rightarrow V=24-4k.\] Setting $k=4$ , we get the volume when the temperature is $20^\circ$ , which is $24-4(4)=8\rightarrow \boxed{8}$

Answer: A 8
Problem: The Pythagoras High School band has $100$ female and $80$ male members.  The Pythagoras High School orchestra has $80$ female and $100$ male members.  There are $60$ females who are members in both band and orchestra.  Altogether, there are $230$ students who are in either band or orchestra or both.  The number of males in the band who are NOT in the orchestra is
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 70$

Solution: There are $100+80-60=120$ females in either band or orchestra, so there are $230-120=110$ males in either band or orchestra.  Suppose $x$ males are in both band and orchestra. \[80+100-x=110\Rightarrow x=70.\] Thus, the number of males in band but not orchestra is $80-70=10\rightarrow \boxed{10}$

Answer: A 10
Problem: A cube of edge $3$ cm is cut into $N$ smaller cubes, not all the same size.  If the edge of each of the smaller cubes is a whole number of centimeters, then $N=$
$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

Solution: If none of the cubes have edge length $2$ , then all of the cubes have edge length $1$ , meaning they all are the same size, a contradiction.
It is clearly impossible to split a cube of edge $3$ into two or more cubes of edge $2$ with extra unit cubes, so there is one $2\times 2\times 2$ cube and $3^3-2^3=19$ unit cubes.
The total number of cubes, $N$ , is $1+19=20\rightarrow \boxed{20}$

Answer: E 20
Problem: What is the smallest sum of two $3$ -digit numbers that can be obtained by placing each of the six digits $4,5,6,7,8,9$ in one of the six boxes in this addition problem?
[asy] unitsize(12); draw((0,0)--(10,0)); draw((-1.5,1.5)--(-1.5,2.5)); draw((-1,2)--(-2,2)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); draw((1,4)--(3,4)--(3,6)--(1,6)--cycle); draw((4,1)--(6,1)--(6,3)--(4,3)--cycle); draw((4,4)--(6,4)--(6,6)--(4,6)--cycle); draw((7,1)--(9,1)--(9,3)--(7,3)--cycle); draw((7,4)--(9,4)--(9,6)--(7,6)--cycle); [/asy]
$\text{(A)}\ 947 \qquad \text{(B)}\ 1037 \qquad \text{(C)}\ 1047 \qquad \text{(D)}\ 1056 \qquad \text{(E)}\ 1245$

Solution: Let the two three-digit numbers be $\overline{abc}$ and $\overline{def}$ .  Their sum is equal to $100(a+d)+10(b+e)+(c+f)$
To minimize this, we need to minimize the contribution of the $100$ factor, so we let $a=4$ and $d=5$ .  Similarly, we let $b=6$ $e=7$ , and then $c=8$ and $f=9$ .  The sum is \[100(9)+10(13)+(17)=1047 \rightarrow \boxed{1047}\]

Answer: C 1047
Problem: Which digit of $.12345$ , when changed to $9$ , gives the largest number?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: When dealing with positive decimals, the leftmost digits affect the change in value more.  Thus, to get the largest number, we change the $1$ to a $9 \rightarrow \boxed{1}$

Answer: A 1
Problem: Which of the following could not be the units digit [ones digit] of the square of a whole number?
$\text{(A)}\ 1 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: We see that $1^2=1$ $2^2=4$ $5^2=25$ , and $4^2=16$ , so already we know that either $\text{E}$ is the answer or the problem has some issues.
For integers, only the units digit affects the units digit of the final result, so we only need to test the squares of the integers from $0$ through $9$ inclusive.  Testing shows that $8$ is unachievable, so the answer is $\boxed{8}$

Answer: E 8
Problem: Which of the following is closest to the product $(.48017)(.48017)(.48017)$
$\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110$

Solution: Clearly, \[.4<.48017<.5\] Since the function $f(x)=x^3$ is strictly increasing, we can say that \[.4^3<.48017^3<.5^3\] from which it follows that $\text{A}$ is much too small and $\text{C}$ is much too large, so $\boxed{0.110}$ is the answer.

Answer: B 0.110
Problem: Which of the following is closest to the product $(.48017)(.48017)(.48017)$
$\text{(A)}\ 0.011 \qquad \text{(B)}\ 0.110 \qquad \text{(C)}\ 1.10 \qquad \text{(D)}\ 11.0 \qquad \text{(E)}\ 110$

Solution: Since $0.48017$ is quite close to $0.5$ , or $\dfrac{1}{2}$ , we can look for the answer choice that is just below $(\dfrac{1}{2})^3=\dfrac{1}{8}=0.125$ , which would be $\boxed{0.110}$

Answer: B 0.110
Problem: When three different numbers from the set $\{ -3, -2, -1, 4, 5 \}$ are multiplied, the largest possible product is
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$

Solution: First we try for a positive product, meaning we either pick three positive numbers or one positive number and two negative numbers.
It is clearly impossible to pick three positive numbers.  If we try the second case, we want to pick the numbers with the largest absolute values, so we choose $5$ $-3$ and $-2$ .  Their product is $30\rightarrow \boxed{30}$

Answer: C 30
Problem: A dress originally priced at $80$ dollars was put on sale for $25\%$ off.  If $10\%$ tax was added to the sale price, then the total selling price (in dollars) of the dress was
$\text{(A)}\ \text{45 dollars} \qquad \text{(B)}\ \text{52 dollars} \qquad \text{(C)}\ \text{54 dollars} \qquad \text{(D)}\ \text{66 dollars} \qquad \text{(E)}\ \text{68 dollars}$

Solution: After the price reduction, the sale price is $80-.25\times 80 = 60$ dollars.  The tax makes the final price $60+.1\times 60 = 66$ dollars $\rightarrow \boxed{66}$

Answer: D 66
Problem: The numbers on the faces of this cube are consecutive whole numbers.  The sum of the two numbers on each of the three pairs of opposite faces are equal.  The sum of the six numbers on this cube is
[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)); label("$15$",(1.5,1.2),N); label("$11$",(4,2.3),N); label("$14$",(2.5,3.7),N); [/asy]
$\text{(A)}\ 75 \qquad \text{(B)}\ 76 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 81$

Solution: The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$
In the second case, the common sum would be $(10+11+12+13+14+15)/3=25$ , so $11$ must be paired with $14$ , which
it isn't.
Thus, the only possibility is the first case and the sum of the six numbers is $81\rightarrow \boxed{81}$

Answer: E 81
Problem: There are twenty-four $4$ -digit numbers that use each of the four digits $2$ $4$ $5$ , and $7$ exactly once.  Listed in numerical order from smallest to largest, the number in the $17\text{th}$ position in the list is
$\text{(A)}\ 4527 \qquad \text{(B)}\ 5724 \qquad \text{(C)}\ 5742 \qquad \text{(D)}\ 7245 \qquad \text{(E)}\ 7524$

Solution: For each choice of the thousands digit, there are $6$ numbers with that as the thousands digit.  Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need $5$ more numbers.
We can just list from here: $5247,5274,5427,5472,5724 \rightarrow \boxed{5724}$

Answer: B 5724
Problem: One proposal for new postage rates for a letter was $30$ cents for the first ounce and $22$ cents for each additional ounce (or fraction of an ounce).  The postage for a letter weighing $4.5$ ounces was
$\text{(A)}\ \text{96 cents} \qquad \text{(B)}\ \text{1.07 dollars} \qquad \text{(C)}\ \text{1.18 dollars} \qquad \text{(D)}\ \text{1.20 dollars} \qquad \text{(E)}\ \text{1.40 dollars}$

Solution: After the first ounce, there are $3.5$ ounces left.  Since each additional ounce or fraction of an ounce adds $22$ cents to the total cost, we need to add $4\times 22$ to the cost for the first ounce.
So, the total price is $30+4\times 22=118$ cents. The answer is choice $\boxed{1.18}$

Answer: C 1.18
Problem: A bag contains only blue balls and green balls.  There are $6$ blue balls.  If the probability of drawing a blue ball at random from this bag is $\frac{1}{4}$ , then the number of green balls in the bag is
$\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36$

Solution: The total number of balls in the bag must be $4\times 6=24$ , so there are $24-6=18$ green balls $\rightarrow \boxed{18}$

Answer: B 18
Problem: The area of this figure is $100\text{ cm}^2$ .  Its perimeter is
[asy] draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); draw((1,2)--(1,1)--(2,1)--(2,0),dashed); [/asy]
$\text{(A)}\ \text{20 cm} \qquad \text{(B)}\ \text{25 cm} \qquad \text{(C)}\ \text{30 cm} \qquad \text{(D)}\ \text{40 cm} \qquad \text{(E)}\ \text{50 cm}$

Solution: Since the area of the whole figure is $100$ , each square has an area of $25$ and the side length is $5$
There are $10$ sides of this length, so the perimeter is $10(5)=50\rightarrow \boxed{50}$

Answer: E 50
Problem: $1990-1980+1970-1960+\cdots -20+10 =$
$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$

Solution: In the middle, we have $\cdots + 1010-1000+990 -\cdots$
If we match up the back with the front, and then do the same for the rest, we get pairs with $2000$ and $-2000$ , so these will cancel out.  In the middle, we have $2000-1000$ which doesn't cancel, but gives us $1000 \rightarrow \boxed{1000}$

Answer: D 1000
Problem: $1990-1980+1970-1960+\cdots -20+10 =$
$\text{(A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990$

Solution: We can see that there are $199$ terms in total. We can also see that the first $198$ numbers form groups of two that add to $10$ each. Dividing to see how many pairs we have, $198$ $2$ $99$ groups of ten, or $990$ . However, we have to remember to add the 199th term ( $10$ ), so we get $990$ $10$ $1000$ , which gives us $\boxed{1000}$

Answer: D 1000