Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Mr. Jones sold two pipes at $\textdollar{ 1.20}$ each. Based on the cost, his profit on one was
$20$ % and his loss on the other was $20$ %. 
On the sale of the pipes, he:
$\textbf{(A)}\ \text{broke even}\qquad \textbf{(B)}\ \text{lost }4\text{ cents} \qquad\textbf{(C)}\ \text{gained }4\text{ cents}\qquad \\ \textbf{(D)}\ \text{lost }10\text{ cents}\qquad \textbf{(E)}\ \text{gained }10\text{ cents}$

Solution: For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or $\frac{6}{5}$ of its original value. This tells us that the original price was $\frac{5}{6}\cdot1.20 = $1.00$
For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or $\frac{4}{5}$ of its original value. This tells us that the original price was $\frac{5}{4}\cdot1.20 = $1.50$
Thus, his total cost was $$2.50$ and his total revenue was $$2.40$
Therefore, he $\boxed{10}$

Answer: D 10
Problem: A man has $\textdollar{10,000 }$ to invest. He invests $\textdollar{4000}$ at 5% and $\textdollar{3500}$ at 4%. 
In order to have a yearly income of $\textdollar{500}$ , he must invest the remainder at:
$\textbf{(A)}\ 6\%\qquad\textbf{(B)}\ 6.1\%\qquad\textbf{(C)}\ 6.2\%\qquad\textbf{(D)}\ 6.3\%\qquad\textbf{(E)}\ 6.4\%$

Solution: The man currently earns $4000 \cdot \frac{5}{1000} + 3500 \cdot \frac{4}{1000} = 340$ dollars. So, we need to find the value of $x$ such that \[2500 \cdot \frac{x}{1000} = 160.\] Solving, we get $x = \boxed{6.4}$

Answer: E 6.4
Problem: A nickel is placed on a table. The number of nickels which can be placed around it, each tangent to it and to two others is:
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

Solution: Arranging the nickels in a hexagonal fashion, we see that only $\boxed{6}$ nickels can be placed around the central nickel.

Answer: C 6
Problem: In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$

Solution: Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation: \[4x+2y=14+2(x+y)\] \[4x+2y=14+2x+2y\] \[2x=14\] \[x=\boxed{7}\]

Answer: B 7
Problem: When you simplify $\left[ \sqrt [3]{\sqrt [6]{a^9}} \right]^4\left[ \sqrt [6]{\sqrt [3]{a^9}} \right]^4$ , the result is:
$\textbf{(A)}\ a^{16} \qquad\textbf{(B)}\ a^{12} \qquad\textbf{(C)}\ a^8 \qquad\textbf{(D)}\ a^4 \qquad\textbf{(E)}\ a^2$

Solution: This simplifies to \[(a^{\frac{9}{6}/3})^4 \cdot (a^{\frac{9}{3}/6})^4 = (a^{\frac{1}{2}})^4 \cdot (a^{\frac{1}{2}})^4 = (a^2)(a^2) = \boxed{4}.$

Answer: D 4
Problem: Given two positive integers $x$ and $y$ with $x < y$ . The percent that $x$ is less than $y$ is:
$\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x)\qquad \textbf{(E)}\ 100(x - y)$

Solution: Suppose that $x$ is $p$ percent less than $y$ . Then $x = \frac{100 - p}{100}y$ , so that $y - x = \frac{p}{100}y$ . Solving for $p$ , we get $p = \frac{100(y-x)}{y}$ , or $\boxed{100}$

Answer: C 100
Problem: The sum of three numbers is $98$ . The ratio of the first to the second is $\frac {2}{3}$ , 
and the ratio of the second to the third is $\frac {5}{8}$ . The second number is:
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$

Solution: Let the $3$ numbers be $a,$ $b,$ and $c$ . We see that \[a+b+c = 98\] and \[\frac{a}{b} = \frac{2}{3} \Rrightarrow 3a = 2b\] \[\frac{b}{c} = \frac{5}{8} \Rrightarrow 8b = 5c\] Writing $a$ and $c$ in terms of $b$ we have $a = \frac{2}{3} b$ and $c = \frac{8}{5} b$ . 
Substituting in the sum, we have \[\frac{2}{3} b + b + \frac{8}{5} b = 98\] \[\frac{49}{15} b = 98\] \[b = 98 \cdot \frac{15}{49} \Rrightarrow b = 30\] $\boxed{30}$

Answer: C 30
Problem: The sum of all numbers of the form $2k + 1$ , where $k$ takes on integral values from $1$ to $n$ is:
$\textbf{(A)}\ n^2\qquad\textbf{(B)}\ n(n+1)\qquad\textbf{(C)}\ n(n+2)\qquad\textbf{(D)}\ (n+1)^2\qquad\textbf{(E)}\ (n+1)(n+2)$

Solution: The sum of the odd integers $2k-1$ from $1$ to $n$ is $n^2$ . However, in this problem, the sum is instead $2k+1$ , starting at $3$ rather than $1$ . To rewrite this, we note that $2k-1$ is $2$ less than $2k+1$ for every $k$ we add, so for $n$ $k$ 's, we subtract $2n$ , giving us $n^2+2n$ ,which factors as $n(n+2) \implies \boxed{2}$

Answer: C 2
Problem: If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$

Solution: Let the angle be $\theta$ and the sides around it be $a$ and $b$ . 
The area of the triangle can be written as \[A =\frac{a \cdot b \cdot \sin(\theta)}{2}\] The doubled sides have length $2a$ and $2b$ , while the angle is still $\theta$ . Thus, the area is \[\frac{2a \cdot 2b \cdot \sin(\theta)}{2}\] \[\Rrightarrow  \frac{4ab \sin \theta}{2} = 4A\] \[\boxed{4}\]

Answer: C 4
Problem: Mr. J left his entire estate to his wife, his daughter, his son, and the cook. 
His daughter and son got half the estate, sharing in the ratio of $4$ to $3$ . 
His wife got twice as much as the son. If the cook received a bequest of $\textdollar{500}$ , then the entire estate was:
$\textbf{(A)}\ \textdollar{3500}\qquad \textbf{(B)}\ \textdollar{5500}\qquad \textbf{(C)}\ \textdollar{6500}\qquad \textbf{(D)}\ \textdollar{7000}\qquad \textbf{(E)}\ \textdollar{7500}$

Solution: The wife, daughter, son, and cook received estates in the ratio $6:4:3:1.$ The estate is worth $6+4+3+1 = 14$ units of $$500.,$ which is $\boxed{7000}$

Answer: D 7000
Problem: In our number system the base is ten. If the base were changed to four you would count as follows: $1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be:
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$

Solution: The $20^{\text{th}}$ number will be the value of $20_{10}$ in base $4$ . Thus, we see \[20_{10} = (1) \cdot 4^2 + (1) \cdot 4^1 + 0 \cdot 4^0\] \[= 110_{4}\]
$\boxed{110}$

Answer: E 110
Problem: If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$

Solution: Suppose $n$ is even. So, we have $n^2(n+1)(n-1).$ Out of these three numbers, at least one of them is going to be a multiple of 3. $n^2$ is also a multiple of 4. Therefore, this expression is always divisible by $\boxed{12}.$

Answer: A 12
Problem: The number of scalene triangles having all sides of integral lengths, and perimeter less than $13$ is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$

Solution: We can write all possible triangles adding up to 12 or less \[(2, 4, 5)=11\] \[(3, 4, 5)=12\] \[(2, 3, 4)=9\]
This leaves $\boxed{3}$ scalene triangles.

Answer: C 3
Problem: Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40^{\circ}$ ; then $\angle AOB$ equals:
$\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}$

Solution: First, from triangle $ABO$ $\angle AOB = 180^\circ - \angle BAO - \angle ABO$ . Note that $AO$ bisects $\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\angle BAO = \angle BAT/2$ . Similarly, $\angle ABO = \angle ABR/2$
Also, $\angle BAT = 180^\circ - \angle BAP$ , and $\angle ABR = 180^\circ - \angle ABP$ . Hence,
$\angle AOB = 180^\circ - \angle BAO - \angle ABO =  180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} =  180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}=   \frac{\angle BAP + \angle ABP}{2}.$
Finally, from triangle $ABP$ $\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ$ , so \[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70}.\]

Answer:  70
Problem: A circle is inscribed in a triangle with sides $8, 15$ , and $17$ . The radius of the circle is:
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7$

Solution: We know that $A = sr$ , where $A$ is the triangle's area, $s$ its semiperimeter, and $r$ its inradius. Since this particular triangle is a right triangle (which we can verify by the Pythagorean theorem), the area is half of $8*15 = 120$ , and the semiperimeter is half of $8 + 15 + 17 = 40$ . Therefore, the inradius is $\frac{120}{40} = 3$ , so our answer is $\boxed{3}$ and we are done.

Answer: D 3
Problem: The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$

Solution: First, square both sides.  This gives us
\[\sqrt{5x-1}^2+2\cdot\sqrt{5x-1}\cdot\sqrt{x-1}+\sqrt{x-1}^2=4 \Longrightarrow 5x-1+2\cdot\sqrt{(5x-1)\cdot(x-1)}+x-1=4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}+6x-2=4\] Then, adding $-6x$ to both sides gives us
\[2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4\] After that, adding $2$ to both sides will give us
\[2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6\] Next, we divide both sides by 2 which gives us
\[\frac{2\cdot\sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2} \Longrightarrow \sqrt{5x^2-6x+1}=-3x+3\] Finally, solving the equation, we get
\[5x^2-6x+1=(-3x+3)^2 \Longrightarrow 5x^2-6x+1=9x^2-18x+9\] \[\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9)\] \[\Longrightarrow -4x^2+12x-8=0 \Longrightarrow -4(x-1)(x-2)=0\] \[\Longrightarrow x-1=0 \text{or}\ x-2=0 \Longrightarrow x=1 \text{or}\ x=2\] Plugging 1 and 2 into the original equation, $\sqrt{5x-1}+\sqrt{x-1}=2$ , we see that when $x=1$
\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot1-1}+\sqrt{1-1}=2 \Longrightarrow \sqrt4+\sqrt0=2 \Longrightarrow 2+0=2 \Longrightarrow 2=2\] the equation is true. On the other hand, we note that when $x=2$
\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot2-1}+\sqrt{2-1}=2 \Longrightarrow \sqrt9+\sqrt1=2 \Longrightarrow 3+1=2 \Longrightarrow 4=0\] the equation is false. 
Therefore the answer is $\boxed{1}$

Answer: D 1
Problem: The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$

Solution: Let us test the answer choices, for it is in this case simpler and quicker. $x=0$ and $x=2/3$ obviously doesn't work, since square roots of negative numbers are not real. $x=2$ doesn't work either, because the first term is already bigger than 2, and the second term cannot be negative. By the process of elimination, the answer is $\boxed{1}$

Answer: D 1
Problem: If $\log x-5 \log 3=-2$ , then $x$ equals:
$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$

Solution: $\log x-5 \log 3=-2$
$\log x- \log 3^5=-2$
$\log x- \log 243 =-2$
$\log x / 243 = -2$
$x/243 = 10^{-2}$
$x=\frac{243}{100}$ $x=\boxed{2.43}$

Answer: C 2.43
Problem: In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:
$\textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents}$

Solution: If the clerk mistook $x$ nickels as quarters, then every mistake inflates the total by $20$ cents. In order to correct this, we have to subtract $20$ cents $x$ times, for a total of $20x$ cents. We can do the same for the $x$ dimes that were turned into pennies (or cents). This exchange would increase the total value of the coins by $9x$
As a total, you get $-20x + 9x = -11x$ . In order to correct the amount, the clerk should $\boxed{11}$

Answer:  11
Problem: Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:
$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$
Solution

Solution: Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.
$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$ , which gives the following roots $+3$ and $-3$
$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$ , which in turn leads to $3x^2-2x=0$ . The roots here are $0$ and $\frac{2}{3}$
$\sqrt{x^2-7}=\sqrt{x-1}$ , when squared, also turns into a quadratic equation: $x^2 - x - 6 = 0$ . Binomial factoring gives us the roots $-2$ and $3$
We can clearly see that, between all of the equations, there is $\boxed{3}$

Answer: B 3
Problem: Four positive integers are given. Select any three of these integers, find their arithmetic average, 
and add this result to the fourth integer. Thus the numbers $29, 23, 21$ , and $17$ are obtained. One of the original integers is:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17$

Solution: Define numbers $a, b, c,$ and $d$ to be the four numbers. In order to satisfy the following conditions, the system of equation should be constructed. (It doesn't matter which variable is which.) \[\frac{a+b+c}{3}+d=29\] \[\frac{a+b+d}{3}+c=23\] \[\frac{a+c+d}{3}+b=21\] \[\frac{b+c+d}{3}+a=17\] Adding all of the equations together, we get: $2(a+b+c+d)=90$ . This means that $a+b+c+d=45$
We can determine that $\frac{a+b+c}{3}+d+16=a+b+c+d$ . This, with some algebra, means that $\frac{1}{3}(a+b+c)=8$ $d$ must be $\boxed{21}$

Answer: B 21
Problem: The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$ , and $y=\frac{2}{3}$ intersect in:
$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$

Solution: We first convert each of the lines into slope-intercept form ( $y = mx + b$ ):
$2x+3y-6=0 ==> 3y = -2x + 6 ==> y = -\frac{2}{3}x + 2$
$4x - 3y - 6 = 0 ==> 4x - 6 = 3y ==> y = \frac{4}{3}x - 2$
$x = 2$ stays as is.
$y = \frac{2}{3}$ stays as is
We can graph the four lines here: [1]
When we do that, the answer turns out to be $\boxed{1}$

Answer: B 1
Problem: The expressions $a+bc$ and $(a+b)(a+c)$ are:
$\textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0$

Solution: Using the FOIL method, we see that $(a+b)(a+c) = a^2 + ab + ac + bc.$ We want to solve \[a + bc = a^2 + ab + ac + bc\] \[a = a^2 + ab + ac\] \[a((a + b + c) - 1) = 0\] These expressions are $\boxed{1.}$

Answer: C 1.
Problem: In order to pass $B$ going $40$ mph on a two-lane highway, $A$ , going $50$ mph, must gain $30$ feet. 
Meantime, $C, 210$ feet from $A$ , is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds, 
then, in order to pass safely, $A$ must increase his speed by:
$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$

Solution: Let $V_A, V_B, V_C$ be $A, B, C$ 's velocity, respectively. We want to pass $B$ before we collide with $C$ . Since $A$ and $B$ are going in the same direction and $V_A>V_B$ $A$ will pass $B$ in $\frac{30\mathrm{ft}}{V_A-V_B}$ time. Since $A$ and $C$ are going in opposite directions, their relative velocity is $V_A+V_C$ , so the amount of time before $A$ will collide with $C$ is given by $\frac{210\mathrm{ft}}{V_A+V_C}$ . We want to pass $B$ before we collide with $C$ , so $V_A$ must satisfy the inequality $\frac{30\mathrm{ft}}{V_A-V_B}<\frac{210\mathrm{ft}}{V_A+V_C}$ . We can eliminate all the units, simplifying the inequality to $\frac{1}{V_A-V_B} < \frac{7}{V_A+V_C}$ . Solving this and substituting our known values of $V_B$ and $V_C$ yields $330<6V_A$ , so $A$ must increase his speed by $\boxed{5}$ miles per hour.

Answer: C 5
Problem: A point $P$ is outside a circle and is $13$ inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$ so that the external segment of the secant $PQ$ is $9$ inches and $QR$ is $7$ inches. The radius of the circle is:
$\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7"$

Solution: Using the Secant-Secant Power Theorem, you can get $9(16)=(13-r)(13+r)$ , where $r$ is the radius of the given circle. Solving the equation, you get a quadratic: $r^2-25$ . A radius cannot be negative so the answer is $\boxed{5}$

Answer: C 5
Problem: The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$ , 
since division by zero is not allowed. For other values of $x$
$\textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1.$

Solution: $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)} = \frac{2x^2-2x-4}{(x+1)(x-2)}$
This can be factored as $\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \implies \frac{(2)(x+1)(x-2)}{(x+1)(x-2)}$ , which cancels out to $2 \implies \boxed{2}$

Answer: B 2
Problem: $A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days. 
The number of days required for A to do the job alone is:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$

Solution: Let $A$ do $r_A$ of the job per day, $B$ do $r_B$ of the job per day, and $C$ do $r_C$ of the job per day. These three quantities have unit $\frac{\text{job}}{\text{day}}$ . Therefore our three conditions give us the three equations: \begin{align*} (2\text{ days})(r_A+r_B)&=1\text{ job},\nonumber\\ (4\text{ days})(r_B+r_C)&=1\text{ job},\nonumber\\ (2.4\text{ days})(r_C+r_A)&=1\text{ job}.\nonumber \end{align*} We divide the three equations by the required constant so that the coefficients of the variables become 1: \begin{align*} r_A+r_B&=\frac{1}{2}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_B+r_C&=\frac{1}{4}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_C+r_A&=\frac{5}{12}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align*} If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side $r_A+r_B+r_C$ , so if we subtract $r_B+r_C$ (the value of which we know) from both equations, we obtain the value of $r_A$ , which is what we wish to determine anyways. So we add these three equations and divide by two: \[r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.\] Hence: \begin{align*} r_A &= (r_A+r_B+r_C)-(r_B+r_C)\nonumber\\ &=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}-\frac{1}{4}\cdot\frac{\text{job}}{\text{day}}\nonumber\\ &=\frac{1}{3}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align*} This shows that $A$ does one third of the job per day. Therefore, if $A$ were to do the entire job himself, he would require $\boxed{3}$ days.

Answer: B 3
Problem: In $\triangle ABC$ $AB=AC$ $\angle A=40^\circ$ . Point $O$ is within the triangle with $\angle OBC \cong \angle OCA$ . 
The number of degrees in $\angle BOC$ is:
$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}$

Solution: Since $\triangle ABC$ is an isosceles triangle, $\angle ABC = \angle ACB = 70^{\circ}$ . Let $\angle OBC = \angle OCA = x$ . Since $\angle ACB = 70$ $\angle OCB = 70 - x$ . The angle of $\triangle OBC$ add up to $180$ , so $\angle BOC = 180 - (x + 70 - x) = \boxed{110}$

Answer: A 110
Problem: A bank charges $\textdollar{6}$ for a loan of $\textdollar{120}$ . The borrower receives $\textdollar{114}$ and repays the loan in $12$ easy installments of $\textdollar{10}$ a month. The interest rate is approximately:
$\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%$

Solution: The borrower pays $\textdollar{120}$ in a single year for a loan of $\textdollar{114}$ . This means that the bank charges an interest of $\textdollar{6}$ for a loan of $\textdollar{114}$ over a single year, so that the annual interest rate is $100*\frac{\textdollar{6}}{\textdollar{114}} = \frac{100}{19} \approx 5$ percentage points. Therefore, our answer is $\boxed{5}$ and we are done.

Answer: A 5
Problem: If $\log 2 = .3010$ and $\log 3 = .4771$ , the value of $x$ when $3^{x+3} = 135$ is approximately
$\textbf{(A) \ } 5  \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63$

Solution: Taking the logarithm in base $3$ of both sides, we get $x+3 = \log_3 135$ . Using the property $\log ab = \log a + \log b$ , we get $x+3 = \log_3 5 + \log_3 3^3$ , or $x = \log_3 5$ . Converting into base $10$ gives $x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}$ . Now, plugging in the values yeilds $\boxed{1.47}$

Answer: B 1.47
Problem: The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$

Solution: To begin, let's notice that the inscribed circle of the right triangle is its incircle , and that the radius of the incircle is the right triangle's inradius . In this case, the hypotenuse is 10, and the inradius is 1. The formula for the inradius of a right triangle is $r = (a+b-c)/2$ , where $r$ is the length of the inradius of the triangle, $c$ is the length hyptotenuse the the right triangle, and $a$ and $b$ are the lengths of the legs of the right triangle. From this, we can plug in values to notice that:
\begin{align*} r & = \frac{a+b-c}{2}\\\\ 1 & = \frac{a+b-10}{2}\\\\ 2 & = a+b-10\\\\ 12 &= a+b\\\\ \end{align*}
From this, we arrive at $a+b+c = 12+10 = 22$ . The answer is clearly $\boxed{22}$

Answer: B 22
Problem: A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$ . He was born in:
$\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806$

Solution: If a man born in the 19th century was $x$ years of in the year $x^2$ , it implies that the year the man was born was $x^2-x$ . So, if the man was born in the first half of the 19th century, it means that $x^2-x < 1850$ . Noticing that $40^2 - 40 = 1560$ and $50^2-50 = 2450$ , we see that $40 < x < 50$ . We can guess values until we hit a solution. $43^2-43 = 1806$ , so we see that the man had to have been $43$ years old in the year $1849=43^2$ , so the answer is $\boxed{1806}$

Answer: E 1806