Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Mr. $A$ owns a home worth $ $10,000$ .  He sells it to Mr. $B$ at a $10\%$ profit based on the worth of the house. Mr. $B$ sells the house back to Mr. $A$ at a $10\%$ loss.  Then:
$\mathrm{(A) \ A\ comes\ out\ even  } \qquad$ $\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}$ $\qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ deal } \qquad$ $\mathrm{(D) \ A\ loses\ 900\ on\ the\ deal }$ $\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal }$

Solution: Mr. $A$ sells his home for $(1 + 10$ $)$ $\cdot$ $10,000$ dollars $=$ $1.1$ $\cdot$ $10,000$ dollars $=$ $11,000$ dollars to Mr. $B$ . Then, Mr. $B$ sells it at a price of $(1-10$ $)$ $\cdot$ $11,000$ dollars $=$ $0.9$ $\cdot$ $11,000$ dollars $=$ $9,900$ dollars, thus $11,000 - 9,900$ $=$ $\boxed{1100}$

Answer: B 1100
Problem: An equilateral triangle is drawn with a side of length $a$ . A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:
$\textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$

Solution: The perimeter of the first triangle is $3a$ . The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing $3a+\frac{3a}{2}+\frac{3a}{4}+\cdots$
The starting term is $3a$ , and the common ratio is $1/2$ . Therefore, the sum is $\frac{3a}{1-\frac{1}{2}}=\boxed{6}$

Answer: D 6
Problem: The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$ , is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$

Solution: Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3.  Therefore $6$ must divide the given expression.
Plugging in $n=2$ yields $6$ .  So the largest possibility is $6$
Clearly the answer is $\boxed{6}$

Answer: E 6
Problem: The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$ , is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$

Solution: In general, $r!$ $n(n+1)(n+2)...(n+r-1)$ were $r$ and $n$ are integers. So here $3!$ $n^3$ $n$ always for any integer $n$ .Hence,the correct answer is $6$ $\boxed{6}$

Answer: E 6
Problem: If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution: The two poles formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\frac1{\frac1{20}+\frac1{80}}$ , or $\frac1{\frac1{16}}=\boxed{16}$

Answer: C 16
Problem: If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution: The two lines can be represented as $y=\frac{-x}{5}+20$ and $y=\frac{4x}{5}$ .
Solving the system,
$\frac{-x}{5}+20=\frac{4x}{5}$
$20=x.$
So the lines meet at an $x$ -coordinate of 20.
Solving for the height they meet,
\[y=\frac{4\cdot 20}{5}\] \[y=\boxed{16}.\]

Answer: C 16
Problem: If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$ , then $AC+BC$ must be
$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$

Solution: Because $AB$ is the diameter of the semi-circle, it follows that $\angle C = 90$ . Now we can try to eliminate all the solutions except for one by giving counterexamples.
$\textbf{(A):}$ Set point $C$ anywhere on the perimeter of the semicircle except on $AB$ . By triangle inequality, $AC+BC>AB$ , so $\textbf{(A)}$ is wrong. $\textbf{(B):}$ Set point $C$ on the perimeter of the semicircle infinitesimally close to $AB$ , and so $AC+BC$ almost equals $AB$ , therefore $\textbf{(B)}$ is wrong. $\textbf{(C):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$ , it follows that $\textbf{(C)}$ is wrong. $\textbf{(E):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$ , it follows that $\textbf{(E)}$ is wrong.
Therefore, the only possible case is $\boxed{2}$

Answer: D 2
Problem: A number which when divided by $10$ leaves a remainder of $9$ , when divided by $9$ leaves a remainder of $8$ , by $8$ leaves a remainder of $7$ , etc., down to where, when divided by $2$ , it leaves a remainder of $1$ , is:
$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$

Solution: If we add $1$ to the number, it becomes divisible by $10, 9, 8, \cdots, 2, 1$ . The LCM of $1$ through $10$ is $2520$ , therefore the number we want to find is $2520-1=\boxed{2519}$

Answer: D 2519
Problem: $\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:
$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$ $\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$

Solution: First, note that we can pull the exponents out of every factor, since they are all squared. This results in $\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}$ Now, multiplying the numerators together gives $\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1}{x^3-1}\right)^{4}$ ,
which simplifies to $\boxed{1}$

Answer: C 1
Problem: If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$ , where $a, b, c$ are other than zero, then $x$ equals:
$\textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc}$ $\textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab}$

Solution: Note that $\frac{1}{a}=\frac{1}{x}+\frac{1}{y}$ $\frac{1}{b}=\frac{1}{x}+\frac{1}{z}$ , and $\frac{1}{c}=\frac{1}{y}+\frac{1}{z}$ . Therefore
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\]
Therefore
\[\frac{1}{x}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}-\frac{1}{2c}\]
A little algebraic manipulation yields that
\[x=\boxed{2}\]

Answer: E 2
Problem: If you are given $\log 8\approx .9031$ and $\log 9\approx .9542$ , then the only logarithm that cannot be found without the use of tables is:
$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$

Solution: While $\log 17 = \log(8 + 9)$ , we cannot easily deal with the logarithm of a sum. Furthermore, $17$ is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefore $\boxed{17}$

Answer: A 17
Problem: Let $R=gS-4$ . When $S=8$ $R=16$ . When $S=10$ $R$ is equal to:
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these}$

Solution: Our first procedure is to find the value of $g$ . With the given variables' values, we can see that $8g-4=16$ so $g=\frac{20}{8}=\frac{5}{2}$
With that, we can replace $g$ with $\frac{5}{2}$ . When $S=10$ , we can see that $10\times\frac{5}{2}-4=\frac{50}{2}-4=25-4=\boxed{21}$

Answer: D 21
Problem: If five geometric means are inserted between $8$ and $5832$ , the fifth term in the geometric series:
$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$

Solution: We can let the common ratio of the geometric sequence be $r$ $5832$ is given to be the seventh term in the geometric sequence as there are five terms between it and $a_1$ if we consider $a_1=8$ .
By the formula for each term in a geometric sequence, we find that $a_n=a_1r^{n-1}$ or $(5382)=(8)r^6$ We divide by eight to find: $r^6=729$ $r=\pm 3$
Because $a_2$ will not be between $8$ and $5832$ if $r=-3$ we can discard it as an extraneous solution. We find $r=3$ and $a_5=a_1r^4=(8)(3)^4=\boxed{648}$

Answer: A 648
Problem: If the radius of a circle is increased $100\%$ , the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$

Solution: Increasing by $100\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$
Since the area of the circle is given by the formula $\pi r^2,$ the area of the new circle is $\pi (2r)^2 = 4\pi r^2.$ The area is quadrupled, or increased by $\boxed{300}$

Answer: C 300
Problem: The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$

Solution: The area of a triangle is $\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\frac12 \cdot 2r \cdot r = \boxed{2}$

Answer: A 2
Problem: The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution: Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{5}$

Answer: B 5
Problem: Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\\textbf{(E)}\ \text{Only 1 is true}$

Solution: The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \frac{a^x}{a^y}$ , so $\boxed{1}$

Answer: E 1
Problem: When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$

Solution: Using synthetic division, we get that the remainder is $\boxed{2}$

Answer: D 2
Problem: When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$

Solution: By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{2.}$

Answer: D 2.
Problem: When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$

Solution: Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)$ , so $x^{13} - 1$ is divisible by $x-1$ , meaning $(x^{13} - 1) + 2$ leaves a remainder of $\boxed{2.}$

Answer: D 2.
Problem: The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$ $8\text{ in}^{2}$ , and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these}$

Solution: If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:
\[ab=12, bc=8, ac=6.\]
The volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.
\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\\ a^2b^2c^2 &= 576\\ abc &= \boxed{24}

Answer: B 24
Problem: Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$

Solution: Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\boxed{28}$

Answer: D 28
Problem: A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays  $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:
$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textbf{(D)} \ \ 45.83\qquad\textbf{(E)} \ \ 177.08$

Solution: $12\frac{1}{2}\%$ is the same as $\frac{1}{8}$ , so the man sets one eighth of each month's rent aside, so he only gains $\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\%$ , or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was for the whole year, so he collected $\frac{875}{12}$ dollars each month as rent. This is only $\frac{7}{8}$ of the monthly rent, so the monthly rent in dollars is $\frac{875}{12}\cdot \frac{8}{7}=\boxed{83.33}$

Answer: B 83.33
Problem: The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$

Solution: $x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4$
$3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark$
There is $\boxed{1}$

Answer: E 1
Problem: The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$

Solution: We can create symmetry in the equation: \[x+\sqrt{x-2} = 4\] \[x-2+\sqrt{x-2} = 2.\] Let $y = \sqrt{x-2}$ , then we have \[y^2+y-2 = 0\] \[(y+2)(y-1) = 0\] The two roots are $\sqrt{x-2} = -2, 1$
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \Rightarrow \boxed{1}$

Answer: E 1
Problem: The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$

Solution: $\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$ $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$ $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\boxed{5}$

Answer: D 5
Problem: The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$

Solution: $\log_{5}\frac{(125)(625)}{25}$ can be also represented as $\log_{5}\frac{(5^3)(5^4)}{5^2}= \log_{5}\frac{(5^7)}{5^2}= \log_{5} 5^5$ which can be solved to get $\boxed{5}$

Answer: D 5
Problem: If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$

Solution: We have $b=\log_{10}{10^b}$ . Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$ . Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$ , the left side becomes $\log_{10}{\dfrac{10^b}{n}}$ . Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$ $m=\boxed{10}$

Answer: E 10
Problem: If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$

Solution: adding $\log_{10} n$ to both sides: \[\log_{10} m + \log_{10} n=b\] using the logarithm property: $\log_a {b} + \log_a {c}=\log_a{bc}$ \[\log_{10} {mn}=b\] rewriting in exponential notation: \[10^b=mn\] \[m=\boxed{10}\] ~Vndom

Answer: E 10
Problem: A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37\text{ mph}$

Solution: The car takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}$ to get from $A$ to $B$ . Also, it takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}$ to get from $B$ to $A$ . Therefore, the average speed is $\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2}{7}\text{ mph}$ , which is closest to $\boxed{34}$

Answer: B 34
Problem: Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$ $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph} \qquad \textbf{(C)}\ 12\text{ mph} \qquad \textbf{(D)}\ 16\text{ mph} \qquad \textbf{(E)}\ 20\text{ mph}$

Solution: Let the speed of boy $A$ be $a$ , and the speed of boy $B$ be $b$ . Notice that $A$ travels $4$ miles per hour slower than boy $B$ , so we can replace $b$ with $a+4$
Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to make an equation, where we set the time to be equal: \[\frac{48}{a}=\frac{72}{a+4}\] Cross-multiplying gives $48a+192=72a$ . Isolating the variable $a$ , we get the equation $24a=192$ , so $a=\boxed{8}$

Answer: B 8
Problem: From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{None of these}$

Solution: Let us represent the number of boys $b$ , and the number of girls $g$
From the first sentence, we get that $2(g-15)=b$
From the second sentence, we get $5(b-45)=g-15$
Expanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$
Substituting $b$ for $2g-30$ , we get $5(2g-30)=g+210$ . Solving for $g$ , we get $g = \boxed{40}$

Answer: A 40
Problem: $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$

Solution: By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{8}$

Answer: D 8
Problem: $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$

Solution: We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.
\[x^2 + 7^2 = 25^2\] \[x^2 = 625 - 49\] \[x^2 = 576\] \[x = 24\]
Since the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$ . The base of the ladder has moved so the new base is say $(7+y)$ . The hypotenuse remains the same at 25ft. So,
\[20^2 + (7+y)^2 = 25^2\] \[400 + 49 + y^2 + 14y = 625\] \[y^2 + 14y - 176 = 0\] \[y^2 + 22y - 8y - 176\] \[x(y+22) - 8(y+22)\] \[(y-8)(y+22)\]
Disregarding the negative solution to equation the solution to the problem is $\boxed{8}$

Answer: D 8
Problem: The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$

Solution: It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter $6 in$ \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]
Cross sectional area of pipe with diameter $1 in$
\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}\]
So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
\[\frac{9\pi}{\frac{\pi}{4}} = 36\]
So the answer is $\boxed{36}$

Answer: D 36
Problem: In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$

Solution: The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$ . Therefore the inradius is $\boxed{4}$

Answer: B 4
Problem: In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$

Solution: Since this is a right triangle, we have \[\frac{a+b-c}{2}=\boxed{4}\]

Answer:  4
Problem: A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)}\ 120\% \qquad \textbf{(D)}\ 80\% \qquad \textbf{(E)}\ 75\%$

Solution: Without loss of generality, we can set the list price equal to $100$ . The merchant buys the goods for $100*.75=75$ . Let $x$ be the marked price.
We then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\boxed{125}$

Answer:  125