AMC 8 2007
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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.
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$, $11$, $7$, $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$
$\textbf{D}$
The total number of hours for an average of 10 hours per week for 6 weeks is $6\times10=60$. So she must work $60-8-11-7-12-10=12$ hours for the final week.
$650$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?
$\textbf{(A)} \dfrac{2}{5} \qquad \textbf{(B)} \dfrac{1}{2} \qquad \textbf{(C)} \dfrac{5}{4} \qquad \textbf{(D)} \dfrac{5}{3} \qquad \textbf{(E)} \dfrac{5}{2}$
$\textbf{E}$
250 students prefer spaghetti while 100 students prefer manicotti. So the ratio is $\dfrac{250}{100}=\dfrac52$.
What is the sum of the two smallest prime factors of $250$?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$
$\textbf{C}$
The two smallest prime factors of 250 are 2 and 5. So the answer is $2+5=7$.
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$
$\textbf{D}$
Georgie has 6 ways to choose which window to enter, and then 5 ways to choose which window to leave. So the answer is $6\times5=30$.
Chandler wants to buy a $\$500$ mountain bike. For his birthday, his grandparents send him $\$50$, his aunt sends him $\$35$ and his cousin gives him $\$15$. He earns $\$16$ per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?
$\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 26 \qquad\textbf{(D)}\ 27 \qquad\textbf{(E)}\ 28$
$\textbf{B}$
He need to earn $\$500-\$50-\$35-\$15=\$400$ from the paper route. Since he earns $\$16$ per week, he needs to work for $400/16=25$ weeks.
The average cost of a long-distance call in the USA in $1985$ was $41$ cents per minute, and the average cost of a long-distance call in the USA in $2005$ was $7$ cents per minute. Find the approximate percent decrease in the cost per minute of a long- distance call.
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 34 \qquad\textbf{(D)}\ 41 \qquad\textbf{(E)}\ 80$
$\textbf{E}$
The percent decrease is $1-\dfrac{7}{41}=\dfrac{34}{41}\approx\dfrac{32}{40}=\dfrac45=80\%$.
The average age of $5$ people in a room is $30$ years. An $18$-year-old person leaves the room. What is the average age of the four remaining people?
$\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 26 \qquad\textbf{(C)}\ 29 \qquad\textbf{(D)}\ 33 \qquad\textbf{(E)}\ 36$
$\textbf{D}$
The total age of the 5 people is $5\times30=150$. When the 18-year-old person leaves, the total age of the rest 4 people is $150-18=132$. So the average age is $132/4=33$.
In trapezoid $ABCD$, $\overline{AD}$ is perpendicular to $\overline{DC}$, $AD = AB = 3$, and $DC = 6$. In addition, $E$ is on $\overline{DC}$, and $\overline{BE}$ is parallel to $\overline{AD}$. Find the area of $\triangle BEC$.
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$
$\textbf{B}$
The base of $\triangle BEC$ is $CE=3$, and the height is $BE=3$. So the area of $\triangle BEC$ is $\dfrac12\times3\times3=4.5$.
To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}$ cannot be determined
$\textbf{B}$
The number in the first row, last column must be a $3$ due to the fact that if a $3$ was in the first row, second column, there would be two 3s in that column. By the same reasoning, the number in the second row, last column has to be a $1$. Therefore, the number in the lower right-hand square is $2$.
For any positive integer $n$, define $\boxed{n}$ to be the sum of the positive factors of $n$. For example, $\boxed{6} = 1 + 2 + 3 + 6 = 12$. Find $\boxed{\boxed{11}}$ .
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 30$
$\textbf{D}$
\begin{align*} \boxed{\boxed{11}}&=\boxed{1+11} \\ &=\boxed{12} \\ &=1+2+3+4+6+12 \\ &=28 \end{align*}
Tiles $I, II, III$ and $IV$ are translated so one tile coincides with each of the rectangles $A, B, C$ and $D$. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $C$?
$\textbf{(A)}\ I \qquad \textbf{(B)}\ II \qquad \textbf{(C)}\ III \qquad \textbf{(D)}\ IV \qquad \textbf{(E)}$ cannot be determined
$\textbf{D}$
We first notice that tile III has a $0$ on the bottom and a $5$ on the right side. Since no other tile has a $0$ or a $5$, Tile III must be in rectangle $D$. Tile III also has a $1$ on the left, so Tile IV must be in Rectangle $C$.
A unit hexagram is composed of a regular hexagon of side length $1$ and its $6$ equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?
$\textbf{(A)}\ 1:1 \qquad \textbf{(B)}\ 6:5 \qquad \textbf{(C)}\ 3:2 \qquad \textbf{(D)}\ 2:1 \qquad \textbf{(E)}\ 3:1$
$\textbf{A}$
If the 6 equilateral triangular extensions grow inside but not outside, they will form a hexagon of the same size. So the ratio is $1:1$.
Sets $A$ and $B$, shown in the Venn diagram, have the same number of elements. Their union has $2007$ elements and their intersection has $1001$ elements. Find the number of elements in $A$.
$\textbf{(A)}\ 503 \qquad \textbf{(B)}\ 1006 \qquad \textbf{(C)}\ 1504 \qquad \textbf{(D)}\ 1507 \qquad \textbf{(E)}\ 1510$
$\textbf{C}$
The number of elements belong to $A$ but not belong to $B$ is $(2007-1001)/2=503$. So the total number of elements in $A$ is $503+1001=1504$.
The base of isosceles $\triangle ABC$ is $24$ and its area is $60$. What is the length of one of the congruent sides?
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 18$
$\textbf{C}$
The area of a triangle is $\dfrac{1}{2}bh$. We set the base equal to $24$, and the area equal to $60$, then we get the height of the triangle to be $5$. In this isosceles triangle, the height bisects the base. By using the Pythagorean Theorem, we get the length of the one of the congruent sides $\sqrt{5^2+12^2}=13$.
Let $a, b$ and $c$ be numbers with $0 < a < b < c$. Which of the following is impossible?
$\textbf{(A)} \ a + c < b \qquad \textbf{(B)} \ a \cdot b < c \qquad \textbf{(C)} \ a + b < c \qquad \textbf{(D)} \ a \cdot c < b \qquad \textbf{(E)}\dfrac{b}{c} = a$
$\textbf{A}$
According to the given rules, every number needs to be positive. Since $c$ is always greater than $b$, adding a positive number $a$ to $c$ will always make it greater than $b$. Therefore, the answer is A.
Amanda draws five circles with radii $1, 2, 3, 4$ and $5$. Then for each circle she plots the point $(C,A)$, where $C$ is its circumference and $A$ is its area. Which of the following could be her graph?
$\textbf{A}$
The circumference of a circle is obtained by simply multiplying the radius by $2\pi$. So the $C$-coordinate (in this case, it is the $x$-coordinate) will increase at a steady rate. The area, however, is obtained by squaring the radius and multiplying it by $\pi$. Since squares do not increase in an evenly spaced arithmetic sequence, the increase in the $A$-coordinates (aka the $y$-coordinates) will be much more significant. The answer is A.
A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow tint and $45\%$ water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?
$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 35 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 50$
$\textbf{C}$
There are $30\times30\%=9$ liters of yellow tint in the original mixture. After adding 5 liters of yellow tint, the percentage of yellow tint in the new mixture is $\dfrac{9+5}{30+5}=40\%$.
The product of the two $99$-digit numbers$$303,\!030,\!303,\!...,\!030,\!303 \text{ and } 505,\!050,\!505,\!...,\!050,\!505$$has thousands digit $A$ and units digit $B$. What is the sum of $A$ and $B$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$
$\textbf{D}$
Noticing that we only care about the thousands digit and units digit, the product can be simplified as $303\times505=153015$. So the thousands digit is 3, and the units digit is 5. Hence, the answer is 8.
Pick two consecutive positive integers whose sum is less than $100$. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 79 \qquad \textbf{(D)}\ 96 \qquad \textbf{(E)}\ 131$
$\textbf{C}$
Let the smaller of the two numbers be $x$. Then, the problem states that $(x+1)+x<100$. The difference of the squares is $$(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$$ $2x+1$ is obviously odd, so only answer choices C and E need to be considered. $2x+1=131$ contradicts the fact that $2x+1<100$, so the answer is C.
Before district play, the Unicorns had won $45\%$ of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 52 \qquad \textbf{(D)}\ 54 \qquad \textbf{(E)}\ 60$
$\textbf{A}$
Let $n$ be the number of pre-district games. So the Unicorns won $0.45n$ and lost $0.55n$ before district play. After district play, we have $$0.45n+6=0.55n+2\rightarrow n=40$$So the total number of game the Unicorns played in the season is $40+6+2=48$.
Two cards are dealt from a deck of four red cards labeled $A, B, C, D$ and four green cards labeled $A, B, C, D$. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
$\textbf{(A)} \dfrac{2}{7} \qquad \textbf{(B)} \dfrac{3}{8} \qquad \textbf{(C)} \dfrac{1}{2} \qquad \textbf{(D)} \dfrac{4}{7} \qquad \textbf{(E)} \dfrac{5}{8}$
$\textbf{D}$
We can choose 2 cards randomly from 8 cards, so the total number of results is $\dbinom{8}{2}=28$. To have the same color, we can choose 2 cards from 4 red cards, which has $\dbinom{4}{2}=6$ ways. Similarly, we can choose 2 cards from 4 green cards with $\dbinom{4}{2}=6$ ways. To have the same letter, there are 4 ways with pairs $(A,A)$, $(B,B)$, $(C,C)$ or $(D,D)$. So the probability of drawing a winning pair is $\dfrac{6+6+4}{28}=\dfrac47$.
A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90$ degree right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$
$\textbf{C}$
From any point in the square, the distance to the left side pluses the distance to the right side is the side length 10 meter, the distance to the top side pluses the distance to the bottom side is also the side length 10 meters. So the average of the distances between the lemming and each side of the square is $(10+10)/4=5$ meters, no matter where the lemming is.
What is the area of the shaded part shown in the $5\times5$ grid?
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$
$\textbf{B}$
The white part is composed of 4 unit squares and 4 congruent triangles. The area of each triangle is $\dfrac12\times3\times2.5=\dfrac{15}4$. So the area of the white part is $4\times1+4\times\dfrac{15}{4}=19$. Hence, the area of the shaded part is $5\times5-19=6$.
A bag contains four pieces of paper, each labeled with one of the digits "1, 2, 3" or "4", with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?
$\textbf{(A)} \dfrac{1}{4} \qquad \textbf{(B)} \dfrac{1}{3} \qquad \textbf{(C)} \dfrac{1}{2} \qquad \textbf{(D)} \dfrac{2}{3} \qquad \textbf{(E)} \dfrac{3}{4}$
$\textbf{C}$
The number of ways to choose three digits out of four is 4. The combinations of digits that give multiples of 3 are $(1,2,3)$ and $(2,3,4)$. Therefore, the probability is $\dfrac24=\dfrac12$.
On the dart board shown in the figure, the outer circle has radius $6$ and the inner circle has radius of 3. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?
$\textbf{(A)} \dfrac{17}{36} \qquad \textbf{(B)} \dfrac{35}{72} \qquad \textbf{(C)} \dfrac{1}{2} \qquad \textbf{(D)} \dfrac{37}{72} \qquad \textbf{(E)} \dfrac{19}{36}$
$\textbf{B}$
The area of the small circle is $9\pi$. So the area of each small sector is $3\pi$. The area of the large circle is $36\pi$. So the area of the large sector subtracts the area of the small sector is $12\pi-3\pi=9\pi$. Hence, the probability that a dart hits region 1 is $\dfrac{3\pi+9\pi+9\pi}{36\pi}=\dfrac7{12}$. The probability that a dart hits region 2 is $\dfrac{3\pi+3\pi+9\pi}{36\pi}=\dfrac{5}{12}$. To get an odd score finally, if the first dart hits region 1, the second dart must be in region 2. If the first dart hits region 2, the second dart must be in region 1. So the probability that the final score is odd is $$\dfrac{7}{12}\times\dfrac{5}{12}+\dfrac{5}{12}\times\dfrac{7}{12}=\dfrac{35}{72}$$