Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow
tint and $45\%$ water. Five liters of yellow tint are added to
the original mixture. What is the percent of yellow tint
in the new mixture?
$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50$

Solution: Since $30\%$ of the original $30$ liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are $9+5=14$ liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This gives $40\%$ of yellow tint in the new mixture, which is $\boxed{40}$

Answer: C 40
Problem: The product of the two $99$ -digit numbers
$303,030,303,...,030,303$ and $505,050,505,...,050,505$
has thousands digit $A$ and units digit $B$ . What is the sum of $A$ and $B$
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$

Solution: We can first make a small example to find out $A$ and $B$ . So,
$303\times505=153015$
The ones digit plus thousands digit is $5+3=8$
Note that the ones and thousands digits are, added together, $8$ . (and so on...) So the answer is $\boxed{8}$ This is a direct multiplication way.

Answer: D 8
Problem: Pick two consecutive positive integers whose sum is less than $100$ . Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$

Solution: Let the smaller of the two numbers be $x$ . Then, the problem states that $(x+1)+x<100$ $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$ $2x+1$ is obviously odd, so only answer choices C and E need to be considered.
$2x+1=131$ contradicts the fact that $2x+1<100$ , so the answer is $\boxed{79}$

Answer: C 79
Problem: Pick two consecutive positive integers whose sum is less than $100$ . Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$

Solution: Since for two consecutive numbers $a$ and $b$ , the difference between their squares are $a^2-b^2=(a+b)(a-b)$ , which equals to $a+b$ , because $a$ and $b$ are consecutive. And because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and that the sum of $a$ and $b$ is less than 100, you can eliminate all answers expect for $\boxed{79}$

Answer: C 79
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. From the information given in the problem, we can say that $\frac{x}{y}=0.45.$ Next, the Unicorns win 6 more games and lose 2 more, for a total of $6+2=8$ games played during district play. We are told that they end the season having won half of their games, or $0.5$ of their games. We can write another equation: $\frac{x+6}{y+8}=0.5.$ This gives us a system of equations: $\frac{x}{y}=0.45$ and $\frac{x+6}{y+8}=0.5.$ We first multiply both sides of the first equation by $y$ to get $x=0.45y.$ Then, we multiply both sides of the second equation by $(y+8)$ to get $x+6=0.5(y+8).$ Applying the Distributive Property gives yields $x+6=0.5y+4.$ Now we substitute $0.45y$ for $x$ to get $0.45y+6=0.5y+4.$ Solving gives us $y=40.$ Since the problem asks for the total number of games, we add on the last 8 games to get the solution $\boxed{48}$

Answer: A 48
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: Simplifying 45% to $\frac{9}{20}$ , we see that the numbers of games before district play are a multiple of 20. After that the Aces played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is $\boxed{48}$ , which is 20(2)+8.

Answer:  48
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: First we simplify $45$ % to $\frac{9}{20}$ . Ratio of won to total is $\frac{9}{20}$ , but ratio of total number won to total number played is $\frac{9x}{20x}$ for some $x$ . After they won 6 more games and lost 2 more games the number of games they won is $9x+6$ , and the total number of games is $20x+8$ . Turning it into a fraction we get $\frac{9x+6}{20x+8}=\frac{1}{2}$ , so solving for $x$ we get $x=2.$ Plugging in 2 for $x$ we get $20(2)+8=\boxed{48}$

Answer:  48
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: Because 45% can be simplified to $9/20$ , and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was $8$ games, so in total there must be $20x+8$ . The only answer in this format is $\boxed{48}$

Answer: A 48
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: Let $n$ be the number of pre-district games. Therefore, we can write the percentage of total games won as a weighted average, namely $.45(n)+.75(8)=(n+8)(.5)$ . Solving this equation for $n$ gives $40$ , but since the problem asked for all games, the answer is $n+8=40+8=\boxed{48}$

Answer: A 48
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: Let the number of games they won be $x$ and the number of games they lost be $y$ . We are told that $\frac{x}{x+y}=\frac{9}{20}$ , which can be manipulated to $20x=9x+9y$ which simplifies down to $11x=9y$ . Then, after district games, we are told $\frac{x+6}{x+y+8}=\frac12$ , which can be changed into $2x+12=x+y+8$ which simplifies down to $x+4=y$ . Then we can solve for $x$ using substitution: \[11x=9x+36\] \[2x=36\] \[x=18\] Now that we know $x=18$ , we can figure out that $y=22$ $18+22=40$ . Now we need to add on the district games: $40+8=\boxed{48}$

Answer: A 48
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: Let $x$ be the total number of games before the district play. The Unicorns have $0.45x$ wins, therefore the rest $0.55x$ are losses. But after the district play, they won 6 and lost 2 more games. We can solve for x by forming the equation $0.45x+6=0.55x+2$ . Subtracting $2$ and $0.45x$ from both sides gives us $0.10x=4$ , and from here we multiply both sides by 10 to get $x=40$ . We are not finished yet as the problem is asking for the total games (which includes the games after the district play), so we add 8 to our value of x to get our answer which is $\boxed{48}$

Answer: A 48
Problem: Before the district play, the Unicorns had won $45$ % of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution: We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played $48$ games in total. Then, after district play, they would have won $24$ games. Now, consider the situation before district play. The Unicorns would have won $18$ games out of $40$ . Converting to a percentage, $18/40 = 45$ %. Thus, the answer is $\boxed{48}$

Answer: A 48
Problem: A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$

Solution: The shortest segments would be perpendicular to the square. The lemming went $x$ meters horizontally and $y$ meters vertically. No matter how much it went, the lemming would have been $x$ and $y$ meters from the sides and $10-x$ and $10-y$ meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: $\frac {\cancel{x}+10-\cancel{x}+\cancel{y}+10-\cancel{y}}{4} =$ $\boxed{5}$

Answer: C 5
Problem: A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$

Solution: For any point in the square, the sum of its distance to the left edge and right edge is equal to $10$ , and the sum of its distance to the up edge and down edge is also equal to $10$ . Thus, the answer is $\boxed{5}$ , and the moving progress is misguide at all.

Answer: C 5
Problem: What is the area of the shaded pinwheel shown in the $5 \times 5$ grid?
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$

Solution: The area of the square around the pinwheel is 25. The area of the pinwheel is equal to $\text{the square } - \text{ the white space.}$ Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is $25-(15+4)$ which is $\boxed{6}$

Answer: B 6
Problem: What is the area of the shaded pinwheel shown in the $5 \times 5$ grid?
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$

Solution: We'd like to use Pick's Theorem on one of the kites, except it doesn't immediately apply since there is a single vertex (in the middle of the diagram) of each kite that does not lie on a lattice point.
We can remedy this be pretending the figure is twice as big:
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } for(i=0; i<5; i=i+1) { draw((i+.5,0)--(i+.5,5), rgb(0.6,0.6,0.6)); draw((0,i+.5)--(5,i+.5), rgb(0.6,0.6,0.6)); } [/asy]
Now we can safely use Pick's Theorem:
\[A=\frac{b}{2}+i-1=\frac{6}{2}+4-1=6\]
However since we scaled the figure's dimensions by $2$ , we scaled its area by $4$ (since the area of similar shapes scales quadratically with the scaling factor).  Therefore the area of each kite is $\frac{6}{4}$ and the area of all four kites combined is $\boxed{6}$

Answer: B 6
Problem: What is the area of the shaded pinwheel shown in the $5 \times 5$ grid?
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$

Solution: The area of any kite (concave OR convex) with diagonals $p$ $q$ is $\frac{1}{2}pq$ .  Let $p$ be the smaller diagonal and $q$ be the longer diagonal.  Then by Pythagorean Theorem $p=\sqrt{2}$ . Similarly, $q$ is $\sqrt{2}$ less than half of the diagonal of the $5 \times 5$ grid, or $q=\frac{5\sqrt{2}}{2}-\sqrt{2}=\frac{3\sqrt{2}}{2}$ .  Therefore the area of the four kites is just:
\[A=4\cdot\frac{1}{2}pq=4\cdot\frac{1}{2}\cdot\sqrt{2}\cdot\frac{3\sqrt{2}}{2}=\boxed{6}\]

Answer: B 6
Problem: Mindy made three purchases for $\textdollar 1.98$ dollars, $\textdollar 5.04$ dollars, and $\textdollar 9.89$ dollars. What was her total, to the nearest dollar?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$

Solution: The three prices round to $\textdollar 2$ $\textdollar 5$ , and $\textdollar 10$ , which have a sum of $\boxed{17}$

Answer: D 17
Problem: On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 26$

Solution: As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly, $\boxed{13}$

Answer: C 13
Problem: Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]
$\textbf{(A)}\ 15 \qquad  \textbf{(B)}\ 20 \qquad  \textbf{(C)}\ 24 \qquad  \textbf{(D)}\ 30 \qquad  \textbf{(E)}\ 40$

Solution: Drawing segments $AC$ and $BD$ , the number of triangles outside square $ABCD$ is the same as the number of triangles inside the square. Thus areas must be equal so the area of $ABCD$ is half the area of the larger square which is $\frac{60}{2}=\boxed{30}$

Answer: D 30
Problem: Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]
$\textbf{(A)}\ 15 \qquad  \textbf{(B)}\ 20 \qquad  \textbf{(C)}\ 24 \qquad  \textbf{(D)}\ 30 \qquad  \textbf{(E)}\ 40$

Solution: If the side length of the larger square is $x$ , the side length of the smaller square is $\frac{\sqrt{2} \cdot x}{2}$ . Therefore the area of the smaller square is $\frac{x^2}{2}$ , half of the larger square's area, $x^2$
Thus, the area of the smaller square in the picture is $\frac{60}{2}=\boxed{30}$

Answer: D 30
Problem: The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
[asy] size(150); draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));[/asy]
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution: If the two rectangles were seperate, the perimeter would be $2(2(2+4)=24$ . It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \times 2 = \boxed{20}$

Answer: C 20
Problem: The table shows some of the results of a survey by radiostation KACL. What percentage of the males surveyed listen to the station?
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & ? & 26 & ?\\ \hline Females & 58 & ? & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
$\textbf{(A)}\ 39\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 55\qquad\textbf{(E)}\ 75$

Solution: Filling out the chart, it becomes
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & 78 & 26 & 104\\ \hline Females & 58 & 38 & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
Thus, the percentage of males surveyed that listen to the station is $100 \cdot \frac{78}{104} \%= \boxed{75}$

Answer: E 75
Problem: What is the product of $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1002\qquad\textbf{(C)}\ 1003\qquad\textbf{(D)}\ 2005\qquad\textbf{(E)}\ 2006$

Solution: The numerator in each fraction cancels out with the denominator of the next fraction. There are only two numbers that didn't cancel: $\frac{2006}{2}=\boxed{1003}$

Answer: C 1003
Problem: How many two-digit numbers have digits whose sum is a perfect square?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$

Solution: There is $1$ integer whose digits sum to $1$ $10$
There are $4$ integers whose digits sum to $4$ $13, 22, 31, \text{and } 40$
There are $9$ integers whose digits sum to $9$ $18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90$
There are $3$ integers whose digits sum to $16$ $79, 88, \text{and } 97$
Two digits cannot sum to $25$ or any greater square since the greatest sum of digits of a two-digit number is $9 + 9 = 18$
Thus, the answer is $1 + 4 + 9 + 3 = \boxed{17}$

Answer: C 17
Problem: Antonette gets $70 \%$ on a 10-problem test, $80 \%$ on a 20-problem test and $90 \%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 77\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 83\qquad\textbf{(E)}\ 87$

Solution: $70 \% \cdot 10=7$
$80 \% \cdot 20=16$
$90 \% \cdot 30=27$
Adding them up gets $7+16+27=50$ . The overall percentage correct would be $\frac{50}{60}=\frac{5}{6}=5 \cdot 16.\overline{6}=83.\overline{3} \approx \boxed{83}$

Answer: D 83
Problem: Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same $760$ -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in $45$ seconds and Chandra reads a page in $30$ seconds.
If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?
$\textbf{(A)}\ 7,600\qquad\textbf{(B)}\ 11,400\qquad\textbf{(C)}\ 12,500\qquad\textbf{(D)}\ 15,200\qquad\textbf{(E)}\ 22,800$

Solution: The information is the same for Problems 14,15, and 16. Therefore, we shall only use the information we need. All we need for this problem is that there's 760 pages, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds. A lot of people will find how long it takes Bob to read the book, how long it takes Chandra to read the book, and then find the seconds. However, if we just set up the expression, we can find an easier way.
$760\cdot45 - 760\cdot30 = 760(45-30) = 760(15) = \boxed{11,400}$

Answer: B 11,400
Problem: Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$

Solution: Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve.
Let $x$ be the number of pages that Chandra reads.
$30x = 45(760-x)$ Distribute the $45$
$30x = 45(760) - 45x$ Add $45x$ to both sides
$75x = 45(760)$ Divide both sides by $15$ to make it easier to solve
$5x = 3(760)$ Divide both sides by $5$
$x = 3(152) = \boxed{456}$

Answer: C 456
Problem: Problems 14, 15 and 16 involve Mrs. Reed's English assignment.
A Novel Assignment
The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$

Solution: Chandra and Bob read at a rate of $30:45$ seconds per page, respectively. Simplifying that gets us Bob reads $2$ pages for every $3$ pages that Chandra reads. Therefore Chandra should read $\frac{3}{2+3}=\frac{3}{5}$ of the book. $\frac{3}{5}\cdot760$ $\boxed{456}$

Answer: C 456