AMC8 2019

We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$. Since they can spend $30.00$ they have $3$ dollars left. Since soft drinks cost $1.00$ dollar each, they can buy 3 soft drinks, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 soft drinks, they bought a total of $9$ items. Therefore, the answer is $\boxed{\textbf{(D) }9}$.

We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$, so the bigger side is $10$, if we do $5 \cdot 2 = 10$. Now we get the sides of the big rectangle being $15$ and $10$, so the area is $\boxed{\textbf{(E)}\ 150}$. ~avamarora

We take a common denominator: \[\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.\] Since $2717<2805<2925$ it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$. Another approach to this problem is using the properties of one fraction being greater than another, also known as the butterfly method. That is, if $\frac{a}{b}>\frac{c}{d}$, then it must be true that $a * d$ is greater than $b * c$. Using this approach, we can check for at least two distinct pairs of fractions and find out the greater one of those two, logically giving us the expected answer of $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$. Remark: Instead of evaluating ad and bc to see which is bigger, difference of squares is sufficient. -Supernova77

A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$, each side is $\frac{52}{4}=13$. In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ = $12$ = $\overline{EC}$. Consider one of the $\overline{AB}$ = $13$, and $\overline{AE}$ = $12$. Using the Pythagorean theorem, we find that $\overline{BE}$ = $5$. You know the Pythagorean triple, (5, 12, 13). Thus the values of the two diagonals are $\overline{AC}$ = $24$ and $\overline{BD}$ = $10$. The area of a rhombus is = $\frac{d_1\cdot{d_2}}{2}$ = $\frac{24\cdot{10}}{2}$ = $120$ $\boxed{\textbf{(D)}\ 120}$

First, the tortoise walks at a constant rate, ruling out $(D)$. Second, when the hare is resting, the distance will stay the same, ruling out $(E)$ and $(C)$. Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out $(A)$. The answer $\boxed{\textbf{(B)}}$ is the only one left.

Lines of symmetry go through point $P$, and there are $8$ directions the lines could go, and there are $4$ dots at each direction.$\frac{4\times8}{80}=\boxed{\textbf{(C)} \frac{2}{5}}$.

We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257+x+y = 405,\] \[x+y = 148.\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \[x = 100,\] \[y=48.\] Now we know that the lowest score on the two other tests is $\boxed{\textbf{(A)}\ 48}$.

After Gilda gives $20$% of the marbles to Pedro, she has $80$% of the marbles left. If she then gives $10$% of what's left to Ebony, she has $(0.8*0.9)$ = $72$% of what she had at the beginning. Finally, she gives $25$% of what's left to her brother, so she has $(0.75*0.72)$ $\boxed{\textbf{(E)}\ 54}$ of what she had in the beginning left.

Using the formula for the volume of a cylinder $\pi r^2 h$, we get Alex, $108\pi$, and Felicia, $216\pi$. We can quickly notice that $\pi$ cancels out on both sides and that Alex's volume is $1/2$ of Felicia's leaving $1/2 = \boxed{\textbf{(B)}\ 1:2}$ as the answer.

On Monday, $20$ people come. On Tuesday, $26$ people come. On Wednesday, $16$ people come. On Thursday, $22$ people come. Finally, on Friday, $16$ people come. $20+26+16+22+16=100$, so the mean is $20$. The median is $(16, 16, 20, 22, 26)$ $20$. The coach figures out that actually $21$ people come on Wednesday. The new mean is $21$, while the new median is $(16, 20, 21, 22, 26)$ $21$. Also, the median increases by $1$ because now the median is $21$ instead of $20$. The median and mean both change, so the answer is $\boxed{\textbf{(B)}}$. Another way to compute the change in mean is to notice that the sum increased by $5$ with the correction. So, the average increased by $5/5 = 1$. Then, the median is computed the same way.

Let $x$ be the number of students taking both a math and a foreign language class. By P-I-E, we get $70 + 54 - x$ = $93$. Solving gives us $x = 31$. But we want the number of students taking only a math class, which is $70 - 31 = 39$. $\boxed{\textbf{(D)}\ 39}$

$B$ is on the top, and $R$ is on the side, and $G$ is on the right side. That means that (image $2$) $W$ is on the left side. From the third image, you know that $P$ must be on the bottom since $G$ is sideways. That leaves us with the back, so the back must be $A$. The front is opposite of the back, so the answer is $\boxed{\textbf{(A)}\ R}$.

Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$. Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$. Then, $N = 110$, and the sum of the digits of $N$ is $1+1+0 = \boxed{\textbf{(A) }2}$.

Let $\text{Day }1$ to $\text{Day\\ }2$ denote a day where one coupon is redeemed and the day when the second coupon is redeemed. If she starts on a $\text{Monday}$, she redeems her next coupon on $\text{Thursday}$. $\text{Thursday}$ to $\text{Sunday}$. Thus, $\textbf{(A)}\ \text{Monday}$ is incorrect. If she starts on a $\text{Tuesday}$, she redeems her next coupon on $\text{Friday}$. $\text{Friday}$ to $\text{Monday}$. $\text{Monday}$ to $\text{Thursday}$. $\text{Thursday}$ to $\text{Sunday}$. Thus $\textbf{(B)}\ \text{Tuesday}$ is incorrect. If she starts on a $\text{Wednesday}$, she redeems her next coupon on $\text{Saturday}$. $\text{Saturday}$ to $\text{Tuesday}$. $\text{Tuesday}$ to $\text{Friday}$. $\text{Friday}$ to $\text{Monday}$. $\text{Monday}$ to $\text{Thursday}$. And on $\text{Thursday}$, she redeems her last coupon. No Sunday occured; thus, $\boxed{\textbf{(C)}\ \text{Wednesday}}$ is correct. Checking for the other options, If she starts on a $\text{Thursday}$, she redeems her next coupon on $\text{Sunday}$. Thus, $\textbf{(D)}\ \text{Thursday}$ is incorrect. If she starts on a $\text{Friday}$, she redeems her next coupon on $\text{Monday}$. $\text{Monday}$ to $\text{Thursday}$. $\text{Thursday}$ to $\text{Sunday}$. Checking for the other options gave us negative results; thus, the answer is $\boxed{\textbf{(C)}\ \text{Wednesday}}$.

The number of people wearing caps and sunglasses is $\frac{2}{5}\cdot35=14$. So then, 14 people out of the 50 people wearing sunglasses also have caps. $\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}$

The only option that is easily divisible by $55$ is $110$, which gives 2 hours of travel. And, the formula is $\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$. And, $\text{Average Speed}$ = $\frac{\text{Total Distance}}{\text{Total Time}}$. Thus, $\frac{125}{50} = \frac{5}{2}$. Both are equal and thus our answer is $\boxed{\textbf{(D)}\ 110}.$

We rewrite: \[\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}\] The middle terms cancel, leaving us with \[\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}\]

We have $2$ dice with $2$ evens and $4$ odds on each die. For the sum to be even, the 2 rolls must be $2$ odds or $2$ evens. Ways to roll $2$ odds (Case $1$): The total number of ways to obtain $2$ odds on 2 rolls is $4*4=16$, as there are $4$ possible odds on the first roll and $4$ possible odds on the second roll. Ways to roll $2$ evens (Case $2$): Similarly, we have $2*2=4$ ways to obtain 2 evens. Probability is $\frac{20}{36}=\frac{5}{9}$, or $\framebox{C}$.

Each team wins once and loses once to get the highest points. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. Against the other 3 teams, there are 6 games. If each team wins each of those six games, they get $3\cdot6=18$ points This brings a total of $18+6=\boxed {\textbf {(C) }24}$ points. Note that this can be easily seen to be the best case as follows. Let $x_A$ be the number of points $A$ gets, $x_B$ be the number of points $B$ gets, and $x_C$ be the number of points $C$ gets. Since $x_A = x_B = x_C$, to maximize $x_A$, we can just maximize $x_A + x_B + x_C$. But in each match, if one team wins then the total sum increases by $3$ points, whereas if they tie, the total sum increases by $2$ points. So, it is best if there are the fewest ties possible.

We have that $(x^2-5)^2 = 16$ if and only if $x^2-5 = \pm 4$. If $x^2-5 = 4$, then $x^2 = 9 \implies x = \pm 3$, giving 2 solutions. If $x^2-5 = -4$, then $x^2 = 1 \implies x = \pm 1$, giving 2 more solutions. All four of these solutions work, so the answer is $\boxed{\textbf{(D) }4}$. Further, the equation is a quartic in $x$, so by the Fundamental Theorem of Algebra, there can be at most four real solutions.

First, we need to find the coordinates where the graphs intersect. We want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$ $y=5$, and $y=1+x$ intersect at $(4,5)$, we call this line x. Doing the same thing, we get $x=-4.$ Thus, $y=5$. Also, $y=5$ and $y=1-x$ intersect at $(-4,5)$, and we call this line y. It's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$ $y=1-x$ and $y=1+x$ intersect at $(0,1)$, we call this line z. Using the Shoelace Theorem we get: \[\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}\] $=$ So, our answer is $\boxed{\textbf{(E)}\ 16.}$ We might also see that the lines $y$ and $x$ are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by $-1$ to get the other. As the base is horizontal, this is an isosceles triangle with base 8, as the intersection points have a distance of 8. The height is $5-1=4,$ so $\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}$ Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!

Suppose the fraction of discount is $x$. That means $(1-x)(1+x)=0.84$; so, $1-x^{2}=0.84$, and $(x^{2})=0.16$, procuring $x=0.4$. Therefore, the price was increased and decreased by $40$%, or $\boxed{\textbf{(E)}\ 40}$.

Given the information above, we start with the equation $\frac{t}{4}+\frac{2t}{7} + 15 + x = t$, where $t$ is the total number of points scored and $x\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \frac{13}{28}t$, or $28x+28\cdot 15=13t$. Since $t$ is necessarily divisible by 28, let $t=28u$ where $u \ge 0$ and divide by 28 to obtain $x + 15 = 13u$. Then, it is easy to see $u=2$ ($t=56$) is the only candidate remaining, giving $x=\boxed{\textbf{(B)} 11}$.

We use the line-segment ratios to infer area ratios and height ratios. Areas: $AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$. $BE:BD = 1:2 \text{ (midpoint)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$. Heights: Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$. $AD:DC = 1:2 \implies CD:CA = 2:3 \implies \text{height } h_D$ from $\overline{BC}$ to $D$ is $\frac{2}{3}h_A$. $BE:BD = 1:2 \text{ (midpoint)} \implies \text{height } h_E$ from $\overline{BC}$ to $E$ is $\frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$. Conclusion: $\frac{[EBF]} {[ABF]} = \frac{[EBF]} {[EBF] + [ABE]} = \frac{[EBF]} {[EBF]+60}$, and also $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$. So, $\frac{[EBF]} {[EBF] + 60} = \frac{1}{3}$, and thus, $[EBF] = \boxed{\textbf{(B) }30}$

Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2). This method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$. In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \choose 2$. This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$. We can also do this because the apples are indistinguishable. $20 \choose 2$ is $190$, therefore the answer is $\boxed{\textbf{(C) }190}$.