## AMC 8 2019

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 1 point for each correct answer. There is no penalty for wrong answers.
- No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted.
- Figures are not necessarily drawn to scale.
- You will have
**40 minutes**working time to complete the test.

Ike and Mike go into a sandwich shop with a total of $\$30.00$ to spend. Sandwiches cost $\$4.50$ each and soft drinks cost $\$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy?

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

$\textbf{D}$

As $30\div4.5=6\dfrac23$, Ike and Mike can buy at most 6 sandwiches, which costs $6\times4.5=27$ dollars. The rest $3$ dollars can afford $3$ soft drinks. Counting both sandwiches and soft drinks gives a total of $9$ items.

Three identical rectangles are put together to form rectangle $ABCD$, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$?

$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

$\textbf{E}$

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. Given that the short side is $5$, the long side is $2\times5=10$. Now we get the sides of the big rectangles being $15$ and $10$, so the area is $150$.

Which of the following is the correct order of the fractions $\dfrac{15}{11},\dfrac{19}{15},$ and $\dfrac{17}{13},$ from least to greatest?

$\textbf{(A) }\dfrac{15}{11}< \dfrac{17}{13}< \dfrac{19}{15} \qquad\newline$ $\textbf{(B) }\dfrac{15}{11}< \dfrac{19}{15}<\dfrac{17}{13} \qquad\newline$ $\textbf{(C) }\dfrac{17}{13}<\dfrac{19}{15}<\dfrac{15}{11} \qquad\newline$ $\textbf{(D) } \dfrac{19}{15}<\dfrac{15}{11}<\dfrac{17}{13} \qquad\newline$ $\textbf{(E) } \dfrac{19}{15}<\dfrac{17}{13}<\dfrac{15}{11}$

$\textbf{E}$

The given fractions can be rewrote as $\dfrac{15}{11}=1\dfrac{4}{11},\dfrac{19}{15}=1\dfrac{4}{15},$ and $\dfrac{17}{13}=1\dfrac{4}{17}$. Since a positive fraction is greater with a smaller denominator, we have $\dfrac{4}{15}<\dfrac{4}{13}<\dfrac{4}{11}$. Hence, $1\dfrac{4}{15}<1\dfrac{4}{13}<1\dfrac{4}{11}\Rightarrow\dfrac{19}{15}<\dfrac{17}{13}<\dfrac{15}{11}$.

Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$?

$\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$

$\textbf{D}$

A rhombus has sides of equal length. Given that the perimeter of the rhombus is $52$, each side is $52/4=13$. In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}=\overline{EC}=12$. Consider one of the right triangles $\triangle ABE$, $\overline{AB}=13$, and $\overline{AE}=12$. Using the Pythagorean theorem, we find that $\overline{BE}=5$ (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 $\dfrac12\overline{AC}\cdot\overline{BD}=\dfrac12\times24\times10=120$.

A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance $d$ traveled by the two animals over time $t$ from start to finish?

$\textbf{B}$

The tortoise walks at a constant rate, so his distance vs. time graph is a straight line (D is incorrect). When the hare is at rest, the graph is a horizontal line (C and E are incorrect). Finally, the tortoise wins the race, meaning that the time of the straight line is lesser (A is incorrect). Hence, the answer is B.

There are $81$ grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point $P$ is in the center of the square. Given that point $Q$ is randomly chosen among the other $80$ points, what is the probability that the line $PQ$ is a line of symmetry for the square?

$\textbf{(A) }\dfrac{1}{5}\qquad\textbf{(B) }\dfrac{1}{4} \qquad\textbf{(C) }\dfrac{2}{5} \qquad\textbf{(D) }\dfrac{9}{20} \qquad\textbf{(E) }\dfrac{1}{2}$

$\textbf{C}$

There are $8$ directions the lines could go, with $4$ dots at each direction. So the probability is $\dfrac{4\times8}{80}=\dfrac{2}{5}$.

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$, $94$, and $87$. In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?

$\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$

$\textbf{A}$

Given the average score, the lowest score achieved in one of the rest two tests only when another score is $100$. Assuming the lowest score is $x$, the average score is \[\dfrac{76+94+87+x+100}{5} = 81\Rightarrow x=48\]

Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?

$\textbf{(A) }20\qquad\textbf{(B) }33\dfrac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$

$\textbf{E}$

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 $80\%\times90\%=72\%$ of what she had at the beginning. Finally, she gives $25\%$ of what's left to Jimmy, so she has $72\%\times75\%=54\%$ of what she had in the beginning left.

Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume of one of Alex's cans to the volume one of Felicia's cans?

$\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1$

$\textbf{B}$

The volume of Alex's can is $$V_a=\pi r^2h=\pi\times3^2\times12=108\pi$$ and the volume of Felicia's can is $$V_f=\pi r^2h=\pi\times6^2\times6=216\pi$$ So the ratio is $$\dfrac{V_a}{V_f}=\dfrac{108\pi}{216\pi}=\dfrac12$$

The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually $21$ participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?

$\textbf{(A) }$ The mean increases by $1$ and the median does not change.$\newline$

$\textbf{(B) }$ The mean increases by $1$ and the median increases by $1$.$\newline$

$\textbf{(C) }$ The mean increases by $1$ and the median increases by $5$.$\newline$

$\textbf{(D) }$ The mean increases by $5$ and the median increases by $1$.$\newline$

$\textbf{(E) }$ The mean increases by $5$ and the median increases by $5$.

$\textbf{B}$

The numbers of students from Monday to Friday are $20, 26, 16, 22$ and $16$. So the mean is $(20+26+16+22+16)/5=20$, and the median is $20$. When the data of Wednesday is corrected, the mean is $(20+26+21+22+16)/5=21$, and the median is $21$. So both the mean and the median are increased by $1$.

The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eight graders taking a foreign language class. How many eight graders take only a math class and not a foreign language class?

$\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$

$\textbf{D}$

The number of students taking both math and foreign language class is $70+54-93=31$. So the number of students taking only the math class is $70-31=39$.

The faces of a cube are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face?

$\textbf{(A) }\textrm{red}\qquad\textbf{(B) }\textrm{white}\qquad\textbf{(C) }\textrm{green}\qquad\textbf{(D) }\textrm{brown}\qquad\textbf{(E) }\textrm{purple}$

$\textbf{A}$

Figure 2 can be achieved when figure 1 rotates $90^\circ$ counterclockwise in the horizontal direction, so $W$ is on the opposite face of $G$. Similarly, figure 3 can be achieved when figure 1 rotates $90^\circ$ clockwise in the vertical direction, so $B$ is on the opposite face of $P$. The only possible choice on the opposite face of $A$ is $R$.

A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

$\textbf{A}$

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 whether 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 $1+1+0=2$.

Isabella has $6$ coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every $10$ days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the $6$ dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?

$\textbf{(A) }\textrm{Monday}\qquad\textbf{(B) }\textrm{Tuesday}\qquad\textbf{(C) }\textrm{Wednesday}\qquad\textbf{(D) }\textrm{Thursday}\qquad\textbf{(E) }\textrm{Friday}$

$\textbf{C}$

Supposing Isabella redeems the last coupon on Sunday, then $10$ days before Sunday is Thursday, then $10$ days before Thursday is Monday $\ldots$ Repeating the sequence for 5 times gives the first day when Isabella redeems her first coupon $$\text{Saturday}\rightarrow\text{Tuesday}\rightarrow\text{Friday}\rightarrow\text{Monday}\rightarrow\text{Thursday}\rightarrow\text{Sunday}$$ Given that Pete's Sweet Treats is closed on Sunday, if Isabella redeems her first coupon on any day above, she will redeem a coupon on Sunday finally. So Isabella can not redeem her first coupon on any day above, which leaves Wednesday the only possible choice.

On a beach $50$ people are wearing sunglasses and $35$ people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is $\dfrac{2}{5}$. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?

$\textbf{(A) }\dfrac{14}{85}\qquad\textbf{(B) }\dfrac{7}{25}\qquad\textbf{(C) }\dfrac{2}{5}\qquad\textbf{(D) }\dfrac{4}{7}\qquad\textbf{(E) }\dfrac{7}{10}$

$\textbf{B}$

The number of people wearing both sunglasses and caps is $35\times\dfrac25=14$. So the probability we are looking for is $\dfrac{14}{50}=\dfrac{7}{25}$.

Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip?

$\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

$\textbf{D}$

The average speed $50$ miles per hour of the entire trip is the total distance divided by the total time. For part 1 of the trip, the distance $15$ miles, and the time is $15/30=0.5$ hour. For part 2 of the trip, supposing the distance is $x$ miles, then the time is $x/55$ hours. The total distance is $x+15$, and the total time is $0.5+x/55$, so the average speed of the entire trip can be expressed as $$\dfrac{x+15}{0.5+x/55}=50\Rightarrow x=110$$

What is the value of the product\[\left(\frac{1\times3}{2\times2}\right)\left(\frac{2\times4}{3\times3}\right)\left(\frac{3\times5}{4\times4}\right)\cdots\left(\frac{97\times99}{98\times98}\right)\left(\frac{98\times100}{99\times99}\right)?\]

$\textbf{(A) }\dfrac{1}{2}\qquad\textbf{(B) }\dfrac{50}{99}\qquad\textbf{(C) }\dfrac{9800}{9801}\qquad\textbf{(D) }\dfrac{100}{99}\qquad\textbf{(E) }50$

$\textbf{B}$

We rewrite:\[\frac{1}{2}\times\left(\frac{3\times2}{2\times3}\right)\left(\frac{4\times3}{3\times4}\right)\cdots\left(\frac{99\times98}{98\times99}\right)\times\frac{100}{99}\]The middle terms cancel, leaving us with \[\frac12\times\frac{100}{99}=\frac{50}{99}\]

The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number?

$\textbf{(A) }\dfrac{4}{9}\qquad\textbf{(B) }\dfrac{1}{2}\qquad\textbf{(C) }\dfrac{5}{9}\qquad\textbf{(D) }\dfrac{3}{5}\qquad\textbf{(E) }\dfrac{2}{3}$

$\textbf{C}$

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.$\newline$

Ways to roll $2$ odds (Case $1$): The total number of ways to obtain $2$ odds on 2 rolls is $4\times4=16$, as there are $4$ possible odds on the first roll and $4$ possible odds on the second roll.$\newline$

Ways to roll $2$ evens (Case $2$): Similarly, we have $2\times2=4$ ways to obtain 2 evens.$\newline$

The probability we are looking for is $\dfrac{4\times4+2\times2}{6\times6}=\dfrac{5}{9}$.

In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$

$\textbf{C}$

We can name the top three teams as $A$, $B$, and $C$. To get the greatest possible points, they must beat the other three teams $D$, $E$, $F$. Therefore, $A$, $B$, $C$ must each obtain $(3+3+3)=9$ points. However, they play against each team twice, for a total of $18$ points against $D$, $E$, and $F$. For games between $A$, $B$, and $C$, each team will play 4 times. To get a equal point, each team will tie 4 times (get 4 points), or win 2 times while lose 2 times (get 6 point). Apparently the second choice gains the most points, with a total points of $18+6=24$.

How many different real numbers $x$ satisfy the equation\[(x^{2}-5)^{2}=16?\]

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8$

$\textbf{D}$

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

What is the area of the triangle formed by the lines $y=5$, $y=1+x$, and $y=1-x$?

$\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16$

$\textbf{E}$

Simply by graphing the lines, we can see that the height of the triangle is 4, and the base is 8. So the area of the triangle is $\dfrac12\times4\times8=16$.

A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased?

$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

$\textbf{E}$

Supposing the percent increased and decreased is $x$, then we get $$(1+x)(1-x)=0.84\Rightarrow1-x^2=0.84\Rightarrow x^2=0.16\Rightarrow x=0.4$$ The percentage is $40\%$.

After Euclid High School's last basketball game, it was determined that $\dfrac{1}{4}$ of the team's points were scored by Alexa and $\dfrac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?

$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$

$\textbf{B}$

Supposing the team's points is $x$, then the other 7 team members get $$x-\dfrac14x-\dfrac27x-15=\dfrac{13}{28}x-15$$ points. The number of score must be an integer, so $x$ is a multiple of $28$. Given that none of the other $7$ team members scored more than $2$ points, we have $$0\leq\dfrac{13}{28}x-15\leq14\Rightarrow31\dfrac{4}{13}\leq x\leq62\dfrac{6}{13}$$ Since $x$ is a multiple of 28, we get $x=56$. The other 7 team members get a total points of $$\dfrac{13}{28}\times56-15=11$$

In triangle $ABC$, point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$. Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $BC$ and line $AE$. Given that the area of $\triangle ABC$ is $360$, what is the area of $\triangle EBF$?

$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

$\textbf{B}$

For $\triangle ABD$ and $\triangle BCD$, $AD:DC=1:2$, so the area of $\triangle ABD$ is $$S_{\triangle ABD}=\dfrac13S_{\triangle ABC}=120$$ The height from $D$ to line $BC$ is $\dfrac23$ of the height from $A$ to $BC$. Since $E$ is the midpoint of $\overline{BD}$, we have $$S_{\triangle ABE}=\dfrac12S_{\triangle ABD}=60$$The height from $E$ to $BC$ is half the height from $D$ to $BC$, hence $\dfrac13$ the height from $A$ to $BC$. Supposing the area of $\triangle BEF$ is $x$, then the area of $\triangle ABF$ is $x+60$. Since $\triangle BEF$ and $\triangle ABF$ share the same base $BF$, and the height from $E$ to $BF$ is $\dfrac13$ of the height from $A$ to $BF$, the ratio of area is $$\dfrac{S_{\triangle BEF}}{S_{\triangle ABF}}=\dfrac{x}{x+60}=\dfrac13\Rightarrow x=30$$

Alice has 24 apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?

$\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$

$\textbf{C}$

Let $a$, $b$, and $c$ represent the number of apples they get. We have $$a+b+c=24$$ Given that each of them has at least two apples. To convert the restricted condition into “each of them has at least 1 apple”, we can name $a'=a-1$, $b'=b-1$, and $c'=c-1$. Now $$a'+b'+c'=21$$ Supposing the 21 apples are in a line like $$\square\ \square\ \square\cdots\square\ \square\ \square$$ We need to insert two boards between apples to separate these apples into 3 parts like $$\square\ \square\ |\ \square\cdots\square\ \square\ |\ \square$$ Each part represents the number $a'$, $b'$, and $c'$, which is no less than one. There are 20 places between apples to insert, and the two boards can not be inserted into the same place. The number of ways of choosing is $\dbinom{20}{2}=190$.

So the number of ways of sharing apples we are looking for is 190.