Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\tfrac{1}{2}$ hours late. Had the accident happened $90$ miles farther along the line, it would have arrived only $3$ hours late. The length of the trip in miles was:
$\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550$

Solution: Let the speed of the train be $x$ miles per hour, and let $D$ miles be the total distance of the trip, where $x$ and $D$ are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled $D-x$ miles after the crash. In other words, the train travelled for $\frac{(D-x)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x}{3x/4}\text{hr}$ after the crash. So, not including the half hour detention after the crash, the entire trip took \[\left(1+\frac{D-x}{3x/4}\right)\text{hours}.\] Now let us consider the alternative trip, where the accident happened $90$ miles farther along the line. Originally, the accident happened $x\frac{\text{mi}}{\text{hr}}\cdot(1\text{ hr})=x\text{ mi}$ down the line, but in this situation it happens $x+90$ miles from the line. Therefore the accident happens $\frac{(x+90)\text{ mi}}{x\frac{\text{mi}}{\text{hr}}}=\frac{x+90}{x}\text{hr}$ into the trip. The length of the remaining part of the trip is now $D-x-90$ , so the remaining trip takes $\frac{(D-x-90)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x-90}{3x/4}\text{hr}$ . So, not including the half hour detention after the crash, the entire trip took \[\left(\frac{x+90}{x}+\frac{D-x-90}{3x/4}\right)\text{hours}.\] We are given that the trip that actually happened resulted in being $3.5$ hours late, while the alternative trip would have resulted in being only $3$ hours late. Therefore the first trip took exactly $\frac{1}{2}$ hour more: \[1+\frac{D-x}{3x/4}=\frac{x+90}{x}+\frac{D-x-90}{3x/4}+\frac{1}{2}.\] After simplification and cancellation, we get the equation \[0=\frac{90}{x}-\frac{90}{3x/4}+\frac{1}{2}.\] Therefore $(90/x)+(1/2)=(120/x)$ , so $1/2=30/x$ . We solve for $x$ and have the original speed of the train: $x=60$
We must now solve for the length of the actual trip. If the train had gone $60\frac{\text{mi}}{\text{hr}}$ for the entire trip, then it would have taken $\frac{D}{60}\text{hr}$ , and the train would have been on time. But the trip actually took $1+\frac{D-60}{45}$ hours, not including the half-hour detention, and the train was $3.5$ hours late. Therefore \[\frac{D}{60}+3.5=1+\frac{D-60}{45}+0.5\Rightarrow 3D+630=180+4D-240+90\Rightarrow D=600.\] The length of the trip in miles was $\boxed{600}$

Answer: C 600
Problem: A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$ , he must sell:
$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\  \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$

Solution: The boy buys $3$ oranges for $10$ cents or $1$ orange for $\frac{10}{3}$ cents.  He sells them at $\frac{20}{5}=4$ cents each.  
That means for every orange he sells, he makes a profit of $4-\frac{10}{3}=\frac{2}{3}$ cents.
To make a profit of $100$ cents, he needs to sell $\frac{100}{\frac{2}{3}}=\boxed{150}$

Answer: B 150
Problem: A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$ , he must sell:
$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\  \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$

Solution: The boy buys $3$ oranges for $10$ cents. He sells them at $5$ for $20$ cents. So, he buys $15$ for $50$ cents and sells them at $15$ for $60$ cents, so he makes $10$ cents of profit on every $15$ oranges. To make $100$ cents of profit, he needs to sell $15 \cdot \frac{100}{10} = \boxed{150}$ oranges.

Answer:  150
Problem: The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution: Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\frac{9}{2}$ ounces of alcohol. We want $\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\frac{9}{2}$ , to be $30\%$ , or $0.3$ , of the total amount of shaving lotion, $9+N$ ). Solving, we find that \[9=0.6(9+N)\implies9=5.4+0.6N\implies3.6=0.6N\implies6=N.\] So, the total amount of water we need to add is $\boxed{6}$

Answer: D 6
Problem: The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution: The concentration of alcohol after adding $n$ ounces of water is $\frac{4.5}{9+n}$
To get a solution of 30% alcohol, we solve $\frac{4.5}{9+n}=\frac{3}{10}$
$45=27+3n$
$18=3n$
$6=n \implies \boxed{6}$

Answer: D 6
Problem: The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:
$\textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi\qquad\textbf{(E)}\ \text{none of these}$

Solution: We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\frac{5280}{6\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\boxed{880}$

Answer: C 880
Problem: A man has part of $ $4500$ invested at $4$ % and the rest at $6$ %. If his annual return on each investment is the same, the average rate of interest which he realizes of the $4500 is:
$\textbf{(A)}\ 5\% \qquad \textbf{(B)}\ 4.8\% \qquad \textbf{(C)}\ 5.2\% \qquad \textbf{(D)}\ 4.6\% \qquad \textbf{(E)}\ \text{none of these}$

Solution: You are trying to find $\frac{2(0.06x)}{4500}$ , where $x$ is the principle for one investment. To find $x$ , solve $0.04(4500-x) = 0.06x$ $X$ will come out to be $1800$ . Then, plug in x into the first equation, $\frac{2(0.06)(1800)}{4500}$ , to get $0.048$ . Finally, convert that to a percentage and you get $\boxed{4.8}$

Answer: B 4.8
Problem: The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to 
the nearest ten-thousandth of a square inch is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 1$

Solution: There are 5 significant digits, $1$ $1$ $0$ $2$ , and $5$ . The answer is $\boxed{5}$

Answer: D 5
Problem: The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click. 
The speed of the train in miles per hour is approximately the number of clicks heard in:
$\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad \textbf{(D)}\ 5\text{ minutes}\\ \textbf{(E)}\ \text{none of these}$

Solution: We assume that the clicks are heard at the head of the train. Then if the train's speed in miles per hour is $x$ , we can convert it to clicks per minute: \[\frac{x\text{ mile}}{\text{hr}}\cdot\left(\frac{1\text{ hr}}{60\text{ min}}\right)\cdot\left(\frac{5280\text{ ft}}{1\text{ mile}}\right)\cdot\left(\frac{1\text{ click}}{30\text{ ft}}\right)=\frac{5280x}{1800}\cdot\frac{\text{click}}{\text{min}}.\] Therefore every minute, on average, $5280x/1800$ clicks are heard. The number of clicks heard in $y$ minutes is $y\cdot 5280x/1800$ , so the number of clicks heard in $1800/5280$ minutes is equal to $x$ . In other words, the speed of the train in miles per hour is equal to the number of clicks heard in $1800/5280$ minutes, which is approximately one-third of a minute, or $\boxed{20}$

Answer: A 20
Problem: If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$ , then the diameter of the circumscribed circle is:
$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$

Solution: By the Extended Law of Sines, the diameter, or twice the circumradius $R$ , is given by \[2R=\frac{12\text{ inches}}{\sin30^\circ}=\boxed{24}.\]

Answer: C 24
Problem: Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$ . The obtained value, $m$ , is an exact divisor of:
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$

Solution: Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$ , where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$ . The only number that fits the first equation is $y=6$ , so $m=-18$ . The only choice that is a multiple of 18 is $\boxed{36}$

Answer: C 36
Problem: If $f(a)=a-2$ and $F(a,b)=b^2+a$ , then $F(3,f(4))$ is:
$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$

Solution: We find $f(4)=(4)-2=2$ , so $F(3,f(4))=F(3,2)=(2)^2+3=\boxed{7}$

Answer: C 7
Problem: The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$

Solution: \[a^x=b\] \[b^y=a\] \[{a^x}^y=a\] \[xy=1\] \[\log_a b\log_b a=1\] As a result, the answer should be $\boxed{1}$

Answer: A 1
Problem: The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$

Solution: Apply the change of base formula to $\log_a b$ and $\log_b a$ . For simplicity, let us convert to base-10 log.
By change of base, the expression becomes $\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{1}$

Answer: A 1
Problem: The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches. 
The length of the common internal tangent is:
$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

Solution: [asy] size(400); draw((0,0)--(41,0)); draw((0,0)--(45/41,200/41)--(1645/41,-160/41)); draw((0,0)--(1600/41,-360/41)--(41,0)); draw(circle((0,0),5)); draw(circle((41,0),4)); label("$A$",(0,0),W); label("$B$",(41,0),E); label("$C$",(45/41,200/41),N); label("$D$",(1645/41,-160/41),SE); label("$E$",(1600/41,-360/41),E); [/asy]
Let $A$ be the center of the circle with radius $5$ , and $B$ be the center of the circle with radius $4$ . Let $\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$ . Extend $\overline{BD}$ past $D$ to point $E$ such that $\overline{BE}\perp\overline{AE}$ . Since $\overline{AC}\perp\overline{CD}$ and $\overline{BD}\perp\overline{CD}$ $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$
Since the centers of the two circles are $41$ inches apart, $AB=41$ . Also, $BE=4+5=9$ . Using the Pythagorean Theorem on right triangle $ABE$ $CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40$ . The length of the common internal tangent is $\boxed{40}$

Answer: E 40
Problem: The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:
$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}$

Solution: Since both lengths are positive, the AM-GM Inequality is satisfied. The correct relationship between $a$ and $b$ is $\boxed{2}$

Answer: E 2
Problem: If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude, 
then the ratio of the smaller base to the larger base is:
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5}$

Solution: [asy] draw((0,0)--(1,3)--(4,3)--(5,0)--cycle); draw((0,0)--(4,3)); draw((4,3)--(4,0)); draw((3.8,0)--(3.8,0.2)--(4,0.2)); label("$A$",(0,0),W); label("$B$",(1,3),NW); label("$C$",(4,3),NE); label("$D$",(5,0),E); label("$E$",(4,0),S); label("1",(2,1.5),NW); [/asy]
Let $a$ be the length of the smaller base of isosceles trapezoid $ABCD$ , and $1$ be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is $\frac a1=a$ . Let point $E$ be the foot of the altitude from $C$ to $\overline{AD}$
Since the larger base of the isosceles trapezoid equals a diagonal, $AC=AD=1$ . Since the smaller base equals the altitude, $BC=CE=a$ . Since the trapezoid is isosceles, $DE=\frac{1-a}{2}$ , so $AE = 1-\frac{1-a}{2} = \frac{a+1}{2}$ . Using the Pythagorean Theorem on right triangle $ACE$ , we have \[a^2+\left(\frac{a+1}{2}\right)^2=1\] Multiplying both sides by $4$ gives \[4a^2+(a+1)^2=4\] Expanding the squared binomial and rearranging gives \[5a^2+2a-3=0\] This can be factored into $(5a-3)(a+1)=0$ . Since a must be positive, $5a+3=0$ , so $a=\frac 35$ . The ratio of the smaller base to the larger base is $\boxed{35}$

Answer: D 35
Problem: One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$ , then the length of the shortest side is
$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units}$

Solution: Let the triangle have side lengths $14, 6+x,$ and $8+x$ . The area of this triangle can be computed two ways. We have $A = rs$ , and $A = \sqrt{s(s-a)(s-b)(s-c)}$ , where $s = 14+x$ is the semiperimeter. Therefore, $4(14+x)=\sqrt{(14+x)(x)(8)(6)}$ . Solving gives $x = 7$ as the only valid solution. This triangle has sides $13,14$ and $15$ , so the shortest side is $\boxed{13}$

Answer: B 13
Problem: One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$ , then the length of the shortest side is
$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units}$

Solution: Label the tangent points on $BC, CA, AB$ as $D, E, F$ respectively. Let $AF=AE=6$ $BF=BD=8$ , and $CE=CD=x.$ The problem is a matter of solving for $x$ . To this, we use the fact that if $A,B,C$ are the angles of a triangle, then $\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{B}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{A}{2}} = 1.$ We know that $\tan{\frac{A}{2}} = \frac{2}{3}$ $\tan{\frac{B}{2}} = \frac{1}{2}$ , and $\tan{\frac{C}{2}} = \frac{4}{x},$ so we have the equation $\frac{1}{2}\cdot \frac{2}{3} + \frac{1}{2}\cdot \frac{4}{x} + \frac{4}{x}\cdot \frac{2}{3} = 1.$ Solving this equation yields $x=7$ , so the shortest side has length $\boxed{13}$

Answer: B 13
Problem: Two high school classes took the same test. One class of $20$ students made an average grade of $80\%$ ; the other class of $30$ students made an average grade of $70\%$ . The average grade for all students in both classes is:
$\textbf{(A)}\ 75\%\qquad \textbf{(B)}\ 74\%\qquad \textbf{(C)}\ 72\%\qquad \textbf{(D)}\ 77\%\qquad \textbf{(E)\ }\text{none of these}$

Solution: The desired average can be found by dividing the total number of points earned by the total number of students. There are $20\cdot 80+30\cdot 70=3700$ points earned and $20+30=50$ students. Thus, our answer is $\frac{3700}{50}$ , or $\boxed{74}$

Answer: B 74
Problem: The points $(6,12)$ and $(0,-6)$ are connected by a straight line. Another point on this line is:
$\textbf{(A) \ }(3,3)  \qquad \textbf{(B) \ }(2,1) \qquad \textbf{(C) \ }(7,16) \qquad \textbf{(D) \ }(-1,-4) \qquad \textbf{(E) \ }(-3,-8)$

Solution: The slope of this line is $\frac{y_2-y_1}{x_2-x_1}=\frac{12+6}{6-0}=3$ . Hence, its equation is $y=3x-6$ . The only given point which satisfies these conditions is $\boxed{3,3}$

Answer: A 3,3
Problem: The difference of the roots of $x^2-7x-9=0$ is:
$\textbf{(A) \ }+7  \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85}$

Solution: Denote the $2$ roots of this quadratic as $r_1$ and $r_2$ . Note that $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$ . By Vieta's Formula's $r_1+r_2=7$ , and $r_1r_2=-9$ . Thus, $r_1-r_2=\sqrt{49+4\cdot 9}=\boxed{85}$

Answer: E 85
Problem: Two equal circles in the same plane cannot have the following number of common tangents.
$\textbf{(A) \ }1  \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these}$

Solution: Two congruent coplanar circles will either be tangent to one another (resulting in $3$ common tangents), intersect one another (resulting in $2$ common tangents), or be separate from one another (resulting in $4$ common tangents).
Having only $\boxed{1}$ common tangent is impossible, unless the circles are non-congruent and internally tangent.

Answer: A 1
Problem: The function $x^2+px+q$ with $p$ and $q$ greater than zero has its minimum value when:
$\textbf{(A) \ }x=-p  \qquad \textbf{(B) \ }x=\frac{p}{2} \qquad \textbf{(C) \ }x=-2p \qquad \textbf{(D) \ }x=\frac{p^2}{4q} \qquad$
$\textbf{(E) \ }x=\frac{-p}{2}$

Solution: The minimum value of this parabola is found at its turning point, on the line $\boxed{2}$ .
Indeed, the turning point of any function of the form $ax^2+bx+c$ has an x-coordinate of $\frac{-b}{2a}$ . This can be seen at the average of the quadratic's two roots (whose sum is $\frac{-b}{a}$ ) or (using calculus) as the value of its derivative set equal to $0$

Answer: E 2
Problem: A house and store were sold for $\textdollar 12,000$ each. The house was sold at a loss of $20\%$ of the cost, and the store at a gain of $20\%$ of the cost. The entire transaction resulted in:
$\textbf{(A) \ }\text{no loss or gain}  \qquad \textbf{(B) \ }\text{loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \ }\text{none of these}$

Solution: Denote the original price of the house and the store as $h$ and $s$ , respectively. It is given that $\frac{4h}{5}=\textdollar 12,000$ , and that $\frac{6s}{5}=\textdollar 12,000$ . Thus, $h=\textdollar 15,000$ $s=\textdollar10,000$ , and $h+s=\textdollar25,000$ . This value is $\textdollar1000$ higher than the current price of the property, $2\cdot \textdollar12,000$ . Hence, the transaction resulted in a $\boxed{1000}$

Answer: B 1000
Problem: A merchant bought some goods at a discount of $20\%$ of the list price. He wants to mark them at such a price that he can give a discount of $20\%$ of the marked price and still make a profit of $20\%$ of the selling price. The per cent of the list price at which he should mark them is:
$\textbf{(A) \ }20  \qquad \textbf{(B) \ }100 \qquad \textbf{(C) \ }125 \qquad \textbf{(D) \ }80 \qquad \textbf{(E) \ }120$

Solution: Let $C$ represent the cost of the goods, and let $L$ $S$ , and $M$ represent the list, selling, and marked prices of the goods, respectively. Hence, we have three equations, which we need to manipulate in order to relate $M$ and $L$
$C=\frac{4}{5}L$
$S=C+\frac{1}{5}S$
$S=\frac{4}{5}M$
We find that $M=\frac{5}{4}L$ . Hence, the percent of the list price which should be marked is $\boxed{125}$

Answer: C 125
Problem: $\log p+\log q=\log(p+q)$ only if:
$\textbf{(A) \ }p=q=\text{zero}  \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad$
$\textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1}$

Solution: $\log p+\log q=\log(p+q)\implies \log pq=\log(p+q)\implies pq=p+q\implies \boxed{1}$

Answer: D 1
Problem: A powderman set a fuse for a blast to take place in $30$ seconds. He ran away at a rate of $8$ yards per second. Sound travels at the rate of $1080$ feet per second. When the powderman heard the blast, he had run approximately: $\textbf{(A)}\ \text{200 yd.}\qquad\textbf{(B)}\ \text{352 yd.}\qquad\textbf{(C)}\ \text{300 yd.}\qquad\textbf{(D)}\ \text{245 yd.}\qquad\textbf{(E)}\ \text{512 yd.}$

Solution: Let $p(t)=24t$ be the number of feet the powderman is from the blast at $t$ seconds after the fuse is lit, and let $q(t)=1080t-32400$ be the number of feet the sound has traveled. We want to solve for $p(t)=q(t)$ \[24t=1080t-32400\] \[1056t=32400\] \[t=\frac{32400}{1056}\] \[t=\frac{675}{22}=30.6\overline{81}\] The number of yards the powderman is from the blast at time $t$ is $\frac{24t}3=8t$ , so the answer is $8(30.6\overline{81})$ , which is about $245$ yards. $\boxed{245}$

Answer: D 245
Problem: If $\left(r+\frac1r\right)^2=3$ , then $r^3+\frac1{r^3}$ equals
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6$

Solution: We know $r+\frac1r=\sqrt3$ . Cubing this gives $r^3+3r+\frac3r+\frac1{r^3}=3\sqrt3$ . But $3r+\frac3r=3\left(r+\frac1r\right)=3\sqrt3$ , so subtracting this from the first equation gives $r^3+\frac1{r^3}=\boxed{0}$ . 
(Actually, $r+\frac1r$ could have been equal to $-\sqrt3$ instead of $\sqrt3$ , but this would have led to the same answer. Also, this answer implies that $r^6=-1$ , which means that $r$ is a complex number.)

Answer: C 0
Problem: The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
$\textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3$

Solution: If the radius of the circle is $r$ , then the perimeter of the first triangle is $3\left(\frac{2r}{\sqrt3}\right)=2r\sqrt3$ , and the perimeter of the second is $3r\sqrt3$ . So the ratio is $\boxed{23}$

Answer: E 23
Problem: If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by
$\textbf{(A) \ } 9-k  \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1$

Solution: Let $n = 10a+b$ . The problem states that $10a+b=k(a+b)$ . We want to find $x$ , where $10b+a=x(a+b)$ . Adding these two equations gives $11(a+b) = (k+x)(a+b)$ . Because $a+b \neq 0$ , we have $11 = k + x$ , or $x = \boxed{11}$

Answer: C 11
Problem: The percent that $M$ is greater than $N$ is:
$(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}$

Solution: $M-N$ is the amount by which $M$ is greater than $N$ . We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\boxed{100}$

Answer: B 100
Problem: A barn with a roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high.  It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720$

Solution: The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$
$((65+50)2)2+130=590 \Rightarrow \boxed{590}$

Answer: D 590