Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution: Let $n=5m+k$ for some natural numbers $m$ $k$ such that $k\in\{0,1,2,3,4\}$ . Notice that $n<5^3=125$ . Thus \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor+\left\lfloor\frac{2n}{125}\right\rfloor\] For smaller $n$ , we temporarily let $\left\lfloor\frac{2n}{125}\right\rfloor=0$ \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{2(5m+k)}{5}\right\rfloor+\left\lfloor\frac{2(5m+k)}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{10m+2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[3m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=2m+\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] To minimize $n$ , we let $\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{10m+2k}{25}\right\rfloor=0$ , then \[m=\left\lfloor\frac{2k}{5}\right\rfloor\] Since $k<5$ $m>0$ , the only integral value of $m$ is $1$ , from which we have $k=3,4\Longrightarrow n=8,9$
Now we let $\left\lfloor\frac{5m+k}{25}\right\rfloor=0$ and $\left\lfloor\frac{10m+2k}{25}\right\rfloor=1$ , then \[m=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] Since $k<5$ $10m>15\Longrightarrow m\ge2$
If $m>2$ , then \[m>\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] which is a contradiction.
Thus $m=2\Longrightarrow\left\lfloor\frac{2k}{5}\right\rfloor=1\Longrightarrow n=13,14$
Finally, the sum of the four smallest possible $n=8+9+13+14=44$ and $4+4=8$ $\boxed{8}$

Answer: B 8
Problem: Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution: We first note that the number of 0's in $n!$ is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:
$\textbf{CASE ONE: } 5\leq 2n < 25.$
The only way we can fulfill the requirements is if $\lfloor{\dfrac{n}{5}}\rfloor = 1$ and $\lfloor{\dfrac{2n}{5}}\rfloor=3$ which means that $5\leq n <10$ and $15\leq 2n < 20$ . The only way this works is if $n = 8 \text{ or } 9.$
$\textbf{CASE TWO: } 25 \leq 2n$
Since we want the smallest values of $n$ , we first try it when $2n<30.$ Thus $(2n)!$ has 6 zeros, which implies that $n!$ must have 2. The only way to do this while maintaining our restrictions for $2n$ is if $n = 13 \text{ or } 14.$
So the sum of the four values is $8+9+13+14=44$ so the digit is sum is $\boxed{8}.$

Answer: B 8
Problem: Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution: We will use trial and error to determine the answer to this problem. If $n = 5$ , then $n!$ has $1$ zero, and $2n!$ will have $2$ zeros. But $3 \cdot 1 \neq 2$ so $n = 5$ does not work. Similarly $n = 6, 7$ do not work either. But $n = 8$ works because $8!$ has $1$ zero and $16!$ has $3$ zeros. Note that $n = 9$ also works because $9!$ has $1$ zero and $18!$ has $3$ zeros. After performing trial and error several times, we find that $n = 10, 11, 12$ do not work but $n = 13, 14$ do work. Therefore, the four smallest values of $n$ are $8, 9, 13, 14$ . Therefore adding them together gives $44$ and our answer is $\boxed{8}.$

Answer: B 8
Problem: A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$

Solution: We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b, ac \le bc$ , from the first equation we get $abc \le 6bc$ . Thus $a\le 6$ . From the second equation we see that $a > 2$ . Thus $a\in \{3, 4, 5, 6\}$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.

Answer: B 10
Problem: A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$

Solution: The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields
\[2(ab+bc+ca)=abc.\]
Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]
First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\geqslant3$
Also note that $c \geq b \geq a > 0$ , hence $\frac{1}{a} \geq \frac{1}{b}  \geq \frac{1}{c}$ .
Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$ , so $a \leq 6$
So we have $a=3, 4, 5$ or $6$
Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ . From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$ , we have $b \leq \frac{2}{k}$ . Thus $\frac{1}{k}<b \leq \frac{2}{k}$
When $a=3$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.
When $a=4$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.
When $a=5$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$
When $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.

Answer: B 10
Problem: Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $\textdollar 2.50$ each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $\textdollar 0.75$ for her to make. In dollars, what is her profit for the day?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution: She first sells one-half of her $48$ loaves, or $\frac{48}{2}=24$ loaves. Each loaf sells for $\textdollar 2.50$ , so her total earnings in the morning is equal to \[24\cdot \textdollar 2.50 = \textdollar 60\]
This leaves 24 loaves left, and Bridget will sell $\dfrac{2}{3}\times 24=16$ of them for a price of $\textdollar\frac{2.50}{2}=\textdollar 1.25$ . Thus, her total earnings for the afternoon is \[16\cdot \textdollar 1.25  = \textdollar 20\]
Finally, Bridget will sell the remaining $24-16=8$ loaves for a dollar each. This is a total of $\textdollar 1\cdot 8 = \textdollar 8$
The total amount of money she makes is equal to $60+20+8=\textdollar 88$
However, since Bridget spends $\textdollar 0.75$ making each loaf of bread, the total cost to make the bread is equal to $\textdollar 0.75\cdot48=\textdollar 36$
Her total profit is the amount of money she spent subtracted from the amount of money she made, which is \[88-36=52\implies\boxed{52}\]

Answer: E 52
Problem: problem_id
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object

Solution: Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house.
Case 1: $\text{Y}$ is the $3^\text{rd}$ house.
The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$
Case 2: $\text{Y}$ is the last house.
There are two possible arrangements:
$\text{B}-\text{O}-\text{R}-\text{Y}$
$\text{O}-\text{B}-\text{R}-\text{Y}$
The answer is $1+2=\boxed{3}$

Answer: B 3
Problem: problem_id
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object

Solution: There are $4!=24$ arrangements without restrictions. There are $3!\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\text{BY}$ ], $\text{O}$ , and $\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.
By symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, so our answer is $12\cdot\frac{1}{2}\cdot\frac{1}{2}= \boxed{3}$

Answer: B 3
Problem: problem_id
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object

Solution: To start with, the blue house is either the first or second house.
If the blue house is the first, then the orange must follow, leading to $2$ cases: $\text{B-O-R-Y}$ and $\text{B-O-Y-R}$
If the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\text{O-B-R-Y}$ .
Therefore, our answer is $\boxed{3}$

Answer: B 3
Problem: problem_id
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object

Solution: Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean
is $87$ and the median is $90$
Thus, the solution is \[90-87=3\implies\boxed{3}\]

Answer: C 3
Problem: problem_id
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object

Solution: The percentage who scored $100$ points is $100\%-(10\%+35\%+30\%)=100\%-75\%=25\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\%$ .The lower $10\%+35\%=45\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\%$ is $100$ points and the lower $45\%$ is $70$ or $80$ ).The mean is $10\%\cdot70+35\%\cdot80+30\%\cdot90+25\%\cdot100=7+28+27+25=87$ .
So, our solution is $90-87=3\Rightarrow\boxed{3}$ ~sosiaops

Answer: C 3
Problem: problem_id
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object

Solution: The $\le 80$ -point scores make up $10\%+35\% = 45\% < 50\%$ of the scores, but the $\le 90$ -point scores make up $45\%+30\% = 75\% > 50\%$ of the scores, so the median is $90$
$10\%$ of scores were $70-90 = -20$ more than the median, $35\%$ were $-10$ more, $100\%-75\% = 25\%$ were $10$ more, and the rest were equal. This means that the mean score is $10\%\cdot(-20)+35\%\cdot(-10)+25\%\cdot10 = -2 + (-3.5) + 2.5 = -3$ more than the median, so their difference is $\left|-3\right| = \boxed{3}$ .
~ emerald_block

Answer: C 3
Problem: Nonzero real numbers $x$ $y$ $a$ , and $b$ satisfy $x < a$ and $y < b$ . How many of the following inequalities must be true?
$\textbf{(I)}\ x+y < a+b\qquad$
$\textbf{(II)}\ x-y < a-b\qquad$
$\textbf{(III)}\ xy < ab\qquad$
$\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Let us denote $a = x + k$ where $k > 0$ and $b = y + l$ where $l > 0$ . We can write that $x + y < x + y + k + l \implies x + y < a + b$
It is important to note that $1$ counterexample fully disproves a claim. Let's try substituting $x=-3,y=-4,a=1,b=4$
$\textbf{(II)}$ states that $x-y<a-b \implies -3 - (-4) < 1 - 4  \implies 1 < -3$ .Therefore, $\textbf{(II)}$ is false.
$\textbf{(III)}$ states that $xy<ab \implies (-3) \cdot (-4) < 1 \cdot 4 \implies 12 < 4$ . Therefore, $\textbf{(III)}$ is false.
$\textbf{(IV)}$ states that $\frac{x}{y}<\frac{a}{b} \implies \frac{-3}{-4} < \frac{1}{4} \implies 0.75 < 0.25$ . Therefore, $\textbf{(IV)}$ is false.
One of our four inequalities is true, hence, our answer is $\boxed{1}$

Answer: B 1
Problem: The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: We find that the area of the triangle is $\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}$ . By the Pythagorean Theorem , we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.
Let $h$ be the third height of the triangle. We have $\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{3}$

Answer: C 3
Problem: The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Notice that we now have a 30-60-90 triangle, with the angle between sides $2\sqrt{3}$ and $4\sqrt{3}$ equal to $60^{\circ}$ . Dropping an altitude from the right angle to the hypotenuse, we see that our desired height is $\boxed{3}$ (We can also check from the other side).

Answer: C 3
Problem: problem_id
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
Name: Text, dtype: object

Solution: Let $a=1$ . Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$ . Our next set starting with $3$ is $\{3,4,5,6,7\}$ . Our average is $25\div 5=5$
Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{4}$

Answer: B 4
Problem: problem_id
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
Name: Text, dtype: object

Solution: We are given that \[b=\frac{a+a+1+a+2+a+3+a+4}{5}\] \[\implies b =a+2\]
We are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is \[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]
Thus, the answer is $\boxed{4}$

Answer: B 4
Problem: problem_id
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
Name: Text, dtype: object

Solution: We know from experience that the average of $5$ consecutive numbers is the $3^\text{rd}$ one or the $1^\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\boxed{4}$

Answer:  4
Problem: problem_id
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
Name: Text, dtype: object

Solution: The list of numbers is $\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}$ so $b=a+2$ . The new list is $\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}$ and the average is $a+4 \Longrightarrow \boxed{4}$

Answer: B 4
Problem: problem_id
bdf1157d287c23856113e8dc3c9bdb32    A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32    A customer who intends to purchase an applianc...
Name: Text, dtype: object

Solution: Let the listed price be $x$ . Since all the answer choices are above $\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:
Coupon 1: $x\times10\%=.1x$
Coupon 2: $20$
Coupon 3: $18\%\times(x-100)=.18x-18$
For coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\implies x>200$ and $.1x>.18x-18\implies.08x<18\implies x<225$
The only choice that satisfies such conditions is $\boxed{219.95}$

Answer: C 219.95
Problem: problem_id
bdf1157d287c23856113e8dc3c9bdb32    A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32    A customer who intends to purchase an applianc...
Name: Text, dtype: object

Solution: For coupon $1$ to be the most effective, we want 10% of the price to be greater than 20. This clearly occurs if the value is over 200. For coupon 1 to be more effective than coupon 3, we want to minimize the value over 200, so $\boxed{219.95}$ is the smallest number over 200.

Answer: C 219.95
Problem: The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution: [asy]//Needs refining (hmm I think it's fine --bestwillcui1) size(12cm); fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); for(int i=-2;i<=8;i+=1)   draw((i,-12)--(i,12),grey); for(int j=-12;j<=12;j+=1)   draw((-2,j)--(8,j),grey); draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis dot((0,0)); dot((6,8)); draw((-2,10.66667)--(8,7.33333),Arrows); draw((7.33333,12)--(-0.66667,-12),Arrows); draw((6,8)--(0,8)); draw((6,8)--(0,0)); draw(rightanglemark((0,10),(6,8),(0,-10),20)); label("$A$",(6,8),NE); label("$a$", (0,5),W); label("$a$",(0,-5),W); label("$a$",(3,4),NW); label("$P$",(0,10),SW); label("$Q$",(0,-10),NW); // wanted to import graph and use xaxis/yaxis but w/e label("$x$",(9,0),E); label("$y$",(0,13),N); [/asy] Note that if the $y$ -intercepts have a sum of $0$ , the distance from the origin to each of the intercepts must be the same. Call this distance $a$ . Since the $\angle PAQ = 90^\circ$ , the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is $\sqrt{6^2+8^2} = 10$ , this means $a=10$ , and the length of the hypotenuse is $2a = 20$ . Since the $x$ -coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{60}$

Answer: D 60
Problem: The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution: We can let the two lines be \[y=mx+b\] \[y=-\frac{1}{m}x-b\] This is because the lines are perpendicular, hence the $m$ and $-\frac{1}{m}$ , and the sum of the y-intercepts is equal to 0, hence the $b, -b$
Since both lines contain the point $(6,8)$ , we can plug this into the two equations to obtain \[8=6m+b\] and \[8=-6\frac{1}{m}-b\]
Adding the two equations gives \[16=6m+\frac{-6}{m}\] Multiplying by $m$ gives \[16m=6m^2-6\] \[\implies 6m^2-16m-6=0\] \[\implies 3m^2-8m-3=0\] Factoring gives \[(3m+1)(m-3)=0\]
Plugging $m=3$ into one of our original equations, we obtain \[8=6(3)+b\] \[\implies b=8-6(3)=-10\]
Since $\bigtriangleup APQ$ has hypotenuse $2|b|=20$ and the altitude to the hypotenuse is equal to the the x-coordinate of point $A$ , or 6, the area of $\bigtriangleup APQ$ is equal to \[\frac{20\cdot6}{2}=\boxed{60}\]

Answer: D 60
Problem: The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution: Since their sum is $0$ , let the y intercepts be P $(0,a)$ and Q $(0,-a)$ . The slope of $AP$ is $\frac{8-a}{6}$ . The slope of AQ is $\frac{8+a}{6}$ . Since multiplying the slopes of perpendicular lines yields a product of $-1$ , we have $\frac{64-a^2}{36}=-1$ , which results in $a^2=100$ . We can use either the positive or negative solution because if we choose $10$ , then the other y-intercept is $-10$ ; but if we choose $-10$ , then the other y-intercept is $10$ . For simplicity, we choose that $a=10$ in this solution.
Now we have a triangle APQ with points A $(6,8)$ , P $(0,10)$ , and Q $(0,-10)$ . By the Pythagorean theorem, we have that $AP=\sqrt{6^2+2^2}=2\sqrt{10}$ , and that $AQ=\sqrt{6^2+18^2}=6\sqrt{10}$ $PQ$ is obviously $10-(-10)=20$ since they have the same $x$ coordinate. Now using Heron's formula, we have $\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{60}$

Answer: D 60
Problem: The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution: Using point-slope form, the first line has the equation \[y-8=m\left(x-6\right) \longrightarrow y=mx-6m+8\] The second line has the equation \[y-8=-\frac{1}{m}\left(x-6\right) \longrightarrow y=-\frac{x}{m}+\frac{6}{m}+8\] At the y-intercept, the value of the x-coordinate is $0$ , hence: the first equation is $y=-6m+8$ and the second is $y=\frac{6}{m}+8$ . Since the y-intercepts sum to $0$ , they are opposites, so: \[-6m+8=-\left(\frac{6}{m}+8\right)=-\frac{6}{m}-8\] \[6m-\frac{6}{m}=16\] Multiply both sides by m: $6m^{2}-6=16m \longrightarrow 3m^{2}-8m-3=0$ . The solution to this quadratic, using the quadratic formula, is: $\frac{8\pm\sqrt{64-4\left(3\right)\left(-3\right)}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}$ This yields $m=-\frac{1}{3}$ and $m=3$ . Plugging $m=3$ into the second equation, we get $y=\frac{6}{3}+8=10$ . Plugging $m=-\frac{1}{3}$ into the first equation, we get $y=-10$ So the base is $20$ and the height is $6$ , the area is $60 \Longrightarrow \boxed{60}$

Answer: D 60
Problem: problem_id
d52fac580a189f0980006e4b2e9b254b    David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b    David drives from his home to the airport to c...
Name: Text, dtype: object

Solution: Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.
Let $d$ be the distance still needed to travel after $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.
Simplifying gives $7d+525=10d$ , or $d=175$
Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{210}$ miles.

Answer: C 210