AMC 8 2020
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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.
Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need?
$\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad$
$\textbf{E}$
The amount of sugar is twice the amount of juice, so he need $2\times3=6$ cups of sugar.$\newline$
The amount of water is 4 times the amount of sugar, so he need $4\times6=24$ cups of water.
Four friends do yardwork for their neighbors over the weekend, earning $\$15, \$20, \$25,$ and $\$40,$ respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned $\$40$ give to the others?
$\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25$
$\textbf{C}$
They earn $\$15+\$20+\$25+\$40\$=100$ in total. So each of them will get $\dfrac{\$100}{4}=\$25$ since they decide to split their earnings equally. For the guy who earned $\$40$, he need to give $\$40-\$25=\$15$ to others.
Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest?
$\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$
$\textbf{D}$
The area of the garden is $6\times8=48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, there are a total of $48\times4=192$ strawberry plants, each of which produces $10$ strawberries on average. Accordingly, she can expect to harvest $192\times10=1920$ strawberries.
Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$
$\textbf{B}$
By observation of the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=37$ dots.
Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?
$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$
$\textbf{C}$
The pitcher is three fourths full, or 75$\%$ full, of juice. Since juice was equally poured, each cup received $75\%\div5=15\%$ of the total capacity.
Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?
$\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}$
$\textbf{A}$
Write the order of the cars as $\square\square\square\square\square$, where the left end of the row represents the back of the train and the right end represents the front. Call the people $A$, $D$, $K$, $M$, and $S$ respectively. The first condition gives $M\square\square\square\square$, so we try $MAS\square\square$, $M\square AS\square$, and $M\square\square AS$. In the first case, as $D$ sat in front of $A$, we must have $MASDK$ or $MASKD$, both of which do not comply with the last condition. In the second case, we obtain $MKASD$, which works, while the third case is obviously impossible, since it results in there being no way for $D$ to sit in front of $A$. It follows that, with the only possible arrangement being $MKASD$, the person sitting in the middle car is $\text{Aaron}$.
How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.)
$\textbf{(A) }\textrm{9} \qquad \textbf{(B) }\textrm{10} \qquad \textbf{(C) }\textrm{15} \qquad \textbf{(D) }\textrm{21}\qquad \textbf{(E) }\textrm{28}$
$\textbf{C}$
Firstly, observe that the second digit of such a number cannot be $1$ or $2$, because the digits must be distinct and increasing (since the first number is already known as $2$). The second digit also cannot be $4$ as the number must be less than $2400$, so it must be $3$. It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$. That can be done in $\dbinom{6}{2} = \dfrac{6 \times 5}{2 \times 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $15$.
Ricardo has $2020$ coins, some of which are pennies ($1$-cent coins) and the rest of which are nickels ($5$-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?
$\textbf{(A) }\textrm{8062} \qquad \textbf{(B) }\textrm{8068} \qquad \textbf{(C) }\textrm{8072} \qquad \textbf{(D) }\textrm{8076}\qquad \textbf{(E) }\textrm{8082}$
$\textbf{C}$
Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is $2019$, giving a total of $(2019\times 5 + 1)$ cents. Analogously, the amount of money he has will be least when he has the greatest number of pennies. As he must have at least one nickel, the greatest number of pennies he can have is also $2019$, giving him a total of $(2019\times 1 + 5)$ cents. Hence, the required difference is $$(2019\times 5 + 1)-(2019\times 1 + 5)=2019\times 4-4=2018\times4=8072$$
Akash's birthday cake is in the form of a $4 \times 4 \times 4$ inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into $64$ smaller cubes, each measuring $1 \times 1 \times 1$ inch, as shown below. How many small pieces will have icing on exactly two sides?
$\textbf{(A) }\textrm{12} \qquad \textbf{(B) }\textrm{16} \qquad \textbf{(C) }\textrm{18} \qquad \textbf{(D) }\textrm{20}\qquad \textbf{(E) }\textrm{24}$
$\textbf{D}$
Noticing that only the corner cubes on the bottom face (4 corner cubes) and the center-of-edge cubes on the non-bottom faces (2 center-of-edge cubes on each of the 8 edges) have icing on exactly two sides. So the total number is $4+2\times8=20$.
Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$
$\textbf{C}$
Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$, respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other and take them as a whole, now we have $3$ elements to arrange, which have $3!=6$ ways. However, the $S$ and $T$ can exchange their order in the $S-T$ system, so the real number of ways to arrange the $3$ element is $2\cdot3!=12$. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements.
After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
$\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$
$\textbf{E}$
Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \dfrac{\text{distance}}{\text{time}}$, her speed is $\dfrac{6}{1/6} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph. So the difference of speeds is $36-12 = 24\ \text{mph}$.
For a positive integer $n$, the factorial notation $n!$ represents the product of the integers from $n$ to $1$. (For example, $6! = 6 \times 5 \times 4 \times3 \times 2 \times 1$.) What value of $N$ satisfies the following equation?\[5! \times 9! = 12 \times N!\]
$\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14$
$\textbf{A}$
We have $5!\times9!=5\times4\times3\times2\times9!=12\times N!$, where $4\times3$ can be canceled by the $12$ on the right side of the equation. So $5\times2\times9!=10!=N!$. Thus $N=10$.
Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$
$\textbf{B}$
After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=36+x$ total socks. This means the probability of drawing a purple sock is $\dfrac{18+x}{36+x}$, so we obtain\[\dfrac{18+x}{36+x}=\dfrac{3}{5}\] By solving the equation we get $x=9$.
There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities?
$\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000$
$\textbf{D}$
We can see that the dotted line is exactly halfway between $4{,}500$ and $5{,}000$, so it is at $4{,}750$. As this is the average population of all $20$ cities, the total population is simply $4{,}750 \times 20 = 95{,}000$.
Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \dfrac13 \qquad \textbf{(E) }300$
$\textbf{C}$
According to the information we have $15\%x=20\%y$, so the ratio of $\dfrac{y}{x}=\dfrac{15\%}{20\%}=\dfrac{15}{20}=0.75=75\%$.
Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by B?
$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$
$\textbf{E}$
The sum of each line is $A+B+C$, $B+D$, $A+F+E$, $B+F$ and $C+D+E$. By adding them together we get $$2A+3B+2C+2D+2E+2F=2(A+B+C+D+E+F)+B=47$$ Since $A,B,C,D,E,F$ represents different digits from $1$ to $6$, we have $A+B+C+D+E+F=1+2+3+4+5+6=21$. So $$2(A+B+C+D+E+F)+B=2\times21+B=47\rightarrow B=5$$
How many factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1, 2, 3, 4, 6,$ and $12.$)
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$
$\textbf{B}$
We prime factorize $2020$ as $2^2\times 5\times 101$. Recalling the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number which has one factor is $1$. For a number to have exactly two factors, it must be prime, and the only prime factors of $2020$ are $2$, $5$, and $101$. For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of $2020$ is $4$. Thus, there are $5$ factors of $2020$ which themselves have $1$, $2$, or $3$ factors (namely $1$, $2$, $4$, $5$, and $101$), so the number of factors of $2020$ that have more than $3$ factors is $12-5=7$.
Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
$\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$
$\textbf{A}$
Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so the radius $OC = 17$. By symmetry, $O$ is in fact the midpoint of $DA$, so $OD=OA=16/2=8$. By the Pythagorean theorem in right-angled triangle $ODC$, we have that $CD=\sqrt{17^2-8^2}=15$. Accordingly, the area of $ABCD$ is $16\cdot 15=240$.
A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$
$\textbf{B}$
A number is divisible by $15$ means that it is divisible by $3$ and $5$. The latter means the last digit must be either $5$ or $0$, and the former means the sum of the digits must be divisible by $3$. If the last digit is $0$, the first digit would be $0$ (because the digits alternate), which is not possible. Hence the last digit must be $5$, and the number is of the form $5\square 5\square 5$. If the unknown digit is $x$, we deduce $$5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}$$ So $x$ must be a multiple of $3$. It can be $0$, $3$, $6$, or $9$. So there are $4$ options: $50505$, $53535$, $56565$, and $59595$.
A scientist walking through a forest recorded as integers the heights of $5$ trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
$\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$
$\textbf{B}$
We are given that all tree heights are integers. Since Tree 2 has a height of $11$ meters, we can deduce that Trees 1 and 3 both have a height of $22$ meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of $11$ and $22$, $44$ and $88$, or $44$ and $22$. Checking each of these, in the first case, the average is $17.6$ meters, which doesn't end in $.2$ as the problem requires. Therefore, we consider the other cases. With $44$ and $88$, the average is $37.4$ meters, which again does not end in $.2$; but with $44$ and $22$, the average is $24.2$ meters, which does. Consequently, the answer is $24.2$.
A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$-step paths are there from $P$ to $Q?$ (The figure shows a sample path.)
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$
$\textbf{A}$
Suppose we extend the chessboard infinitely with $2$ additional columns to the right. The red line shows the right-hand edge of the original board. The total number of paths from $P$ to $Q$, including invalid paths which cross over the red line, is then the number of paths which make $4$ steps up-and-right and $3$ steps up-and-left, which is $\dbinom{4+3}{3} = \dbinom{7}{3} = 35$.
We need to subtract the number of invalid paths, i.e. the number of paths that pass through $X$ or $Y$. To get to $X$, the marker has to make $3$ up-and-right steps, after which it can proceed to $Q$ with $3$ steps up-and-left and $1$ step up-and-right. Thus, the number of paths from $P$ to $Q$ that pass through $X$ is $1 \cdot \dbinom{3+1}{3} = 4$. Similarly, the number of paths that pass through $Y$ is $\dbinom{4+1}{1}\cdot 1 = 5$.
However, we have now double-counted the invalid paths which pass through both $X$ and $Y$; from the diagram, it is clear that there are only $2$ of these (as the marker can get from $X$ to $Y$ by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is $4+5-2=7$, and the number of valid paths from $P$ to $Q$ is $35-7 =28$.
When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.
For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$\[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\]When the same $6$-step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$\[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\]
$\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$
$\textbf{E}$
We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities of each stage:\[\{1,8,10,64\}\leftarrow\{4,5,32\}\leftarrow\{2,16\}\leftarrow\{1,8\}\leftarrow\{4\}\leftarrow\{2\}\leftarrow\{1\}\]where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$), but $16$ could come from $32$ or $5$ (as $\dfrac{32}{2} = 3 \times 5 + 1 = 16$, and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ at the end of the $6$-step process, and sum is $1+8+10+64=83$.
Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$
$\textbf{B}$
Without the restriction that each student receives at least one award, we could simply take each of the $5$ awards and choose one of the $3$ students to give it to, so that there would be $3^5=243$ ways to distribute the awards. We now need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are $3$ choices for which student that is, then $2^5 = 32$ ways of choosing a student to receive each of the awards, for a total of $3 \times 32 = 96$. However, if $2$ students both don't receive an award, then such a case would be counted twice among our $96$ above, so we need to add back in these cases. Of course, $2$ students both not receiving an award is equivalent to only $1$ student receiving all $5$ awards, so there are simply $3$ choices for which student that would be. Therefore, the total number of ways of distributing the awards is $243-96+3=150$.
A large square region is paved with $n^2$ gray square tiles, each measuring $s$ inches on a side. A border $d$ inches wide surrounds each tile. The figure below shows the case for $n=3$. When $n=24$ , the $576$ gray tiles cover $64\%$ of the area of the large square region. What is the ratio $\dfrac{d}{s}$ for this larger value of $n?$
$\textbf{(A) }\dfrac6{25} \qquad \textbf{(B) }\dfrac14 \qquad \textbf{(C) }\dfrac9{25} \qquad \textbf{(D) }\dfrac7{16} \qquad \textbf{(E) }\dfrac9{16}$
$\textbf{A}$
The area of the shaded region is $(24s)^2$. To find the area of the large square, we note that there is a $d$-inch border between each of the $23$ pairs of consecutive squares, as well as the 2 outer sides, for a total of $23+2 = 25$ times the length of the border, i.e. $25d$. Adding this to the total length of the consecutive squares, which is $24s$, the side length of the large square is $(24s+25d)$, yielding the equation $$\dfrac{(24s)^2}{(24s+25d)^2}=\dfrac{64}{100}$$ Taking the square root of both sides (and using the fact that lengths are non-negative) gives $$\dfrac{24s}{24s+25d}=\dfrac{8}{10} = \dfrac{4}{5}$$ and cross-multiplying now gives $$120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \dfrac{d}{s} = \dfrac{24}{100} =\dfrac{6}{25}$$
Rectangles $R_1$ and $R_2,$ and squares $S_1,S_2$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$
$\textbf{A}$
Let the side length of each square $S_k$ be $s_k$. The width of the 3322 can be expressed as $s_{1}+s_{2}+s_{3}=3322$. Similarly, the height $s_{1}-s_{2}+s_{3}=2020$. We subtract the second equation from the first to obtain $2s_{2}=1302$, and thus $s_{2}=651$.