AMC 8 2002
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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.
A circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
$\textbf{D}$
The two lines can both intersect the circle twice, and intersect each other once, so $2+2+1= 5.$
How many different combinations of $\$5$ bills and $\$2$ bills can be used to make a total of $\$17$? Order does not matter in this problem.
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$
$\textbf{A}$
You cannot use more than 4 $\$5$ bills, but if you use 3 $\$5$ bills, you can add another $\$2$ bill to make a combination. You cannot use 2 $\$5$ bills since you have an odd number of dollars that need to be paid with $\$2$ bills. You can also use 1 $\$5$ bill and 6 $\$2$ bills to make another combination. There are no other possibilities, as making $\$17$ with 0 $\$5$ bills is impossible. So the answer is $2$.
What is the smallest possible average of four distinct positive even integers?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$
$\textbf{C}$
In order to get the smallest possible average, we want the 4 positive even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is $\dfrac{2+4+6+8}{4}=5$.
The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$
$\textbf{B}$
The palindrome right after 2002 is 2112. The product of the digits of 2112 is $4$.
Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?
$\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Wednesday} \qquad \text{(C)}\ \text{Friday} \qquad \text{(D)}\ \text{Saturday} \qquad \text{(E)}\ \text{Sunday}$
$\textbf{C}$
When 706 is divided by 7, the reminder is 6. 6 days after Saturday is Friday.
A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it?
$\text{(A)}\ \text{A} \qquad \text{(B)}\ \text{B} \qquad \text{(C)}\ \text{C} \qquad \text{(D)}\ \text{D} \qquad \text{(E)}\ \text{E}$
$\textbf{A}$
In the beginning the volume of water has a net gain of $20-18=2$ millimeters per minute. The volume of water in the birdbath increases at a constant rate until it reaches its maximum and starts overflowing to keep a constant volume. This is best represented by graph A.
The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?
$\text{(A)}\ 5 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$
$\textbf{E}$
From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\dfrac{5}{25}=20\%$.
$\textbf{Juan’s Old Stamping Grounds}$
Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and France, 6 cents each, Peru 4 cents each, and Spain 5 cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.)
How many of his European stamps were issued in the '80s?
$\text{(A)}\ 9 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 42$
$\textbf{D}$
France and Spain are European countries. The number of '80s stamps from France is $15$ and the number of '80s stamps from Spain is $9$. The total number of stamps is $15+9=24$.
His South American stamps issued before the '70s cost him
$\text{(A)}\ \$0.40 \qquad \text{(B)}\ \$1.06 \qquad \text{(C)}\ \$1.80 \qquad \text{(D)}\ \$2.38 \qquad \text{(E)}\ \$2.64$
$\textbf{B}$
Brazil 50s and 60s total 11 stamps cost 6 cents each. Peru 50s and 60s total 10 stamps cost 4 cents each. So the total cost is $11\times0.06+10\times0.04 =\$1.06$.
The average price of his '70s stamps is closest to
$\text{(A)}\ 3.5 \text{ cents} \qquad \text{(B)}\ 4 \text{ cents} \qquad \text{(C)}\ 4.5 \text{ cents} \qquad \text{(D)}\ 5 \text{ cents} \qquad \text{(E)}\ 5.4 \text{ cents}$
$\textbf{E}$
Juan has 12 Brazil 70s stamps of 6 cents each, 12 France 70s stamps of 6 cents each, 6 Peru 70s stamps of 4 cents each, and 13 Spain 70s stamps of 5 cents each. So the total number of 70s stamps is $12+12+6+13=43$. And the total costs of 70s stamps is $12\cdot6+12\cdot6+6\cdot4+13\cdot5=233$ cents. The average price is $233/43\approx5.4$ cents.
A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$
$\textbf{C}$
The seventh square has $7^2=49$ tiles. The sixth square has $6^2=36$ tiles. So the answer is $49-36=13$.
A board game spinner is divided into three regions labeled $A$, $B$ and $C$. The probability of the arrow stopping on region $A$ is $\dfrac{1}{3}$ and on region $B$ is $\dfrac{1}{2}$. The probability of the arrow stopping on region $C$ is
$\text{(A)}\ \dfrac{1}{12} \qquad \text{(B)}\ \dfrac{1}{6} \qquad \text{(C)}\ \dfrac{1}{5} \qquad \text{(D)}\ \dfrac{1}{3} \qquad \text{(E)}\ \dfrac{2}{5}$
$\textbf{B}$
Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is $1-\dfrac{1}{2}-\dfrac{1}{3}=\dfrac16$.
For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?
$\text{(A)}\ 250 \qquad \text{(B)}\ 500 \qquad \text{(C)}\ 625 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1000$
$\textbf{E}$
The volume of Carrie's box is $2\times2\times2=8$ times as large as Bert's box. So her box can carry $125\times8=1000$ jellybeans.
A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices. The total discount is
$\text{(A)}\ 35\% \qquad \text{(B)}\ 44\% \qquad \text{(C)}\ 50\% \qquad \text{(D)}\ 56\% \qquad \text{(E)}\ 60\%$
$\textbf{B}$
The final price is $(1-30\%)\times(1-20\%)=56\%$ of its original price, which means a $44\%$ discount.
Which of the following polygons has the largest area?
$\text{(A)} \text{A} \qquad \text{(B)}\ \text{B} \qquad \text{(C)}\ \text{C} \qquad \text{(D)}\ \text{D} \qquad \text{(E)}\ \text{E}$
$\textbf{E}$
Each polygon can be partitioned into unit squares and right triangles with side length $1$. The area of a square is 1, and the area of a right triangle is 0.5. Then we get the areas of each figures as 5, 5, 5, 4.5, and 5.5. Therefore, the polygon with the largest area is $E$.
Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?
$\text{(A)}\ X + Z = W + Y \qquad\newline$
$\text{(B)}\ W + X = Z \qquad \newline$
$\text{(C)}\ 3X + 4Y = 5Z\newline$
$\text{(D)}\ X +W = \dfrac{1}{2} (Y + Z) \qquad \newline$
$\text{(E)}\ X + Y = Z$
$\textbf{E}$
Since all triangles are right triangles, we can get the area of each triangle easily:
\begin{align*}
X&=\dfrac12(3)(3)=4.5\\
Y&=\dfrac12(4)(4)=8\\
Z&=\dfrac12(5)(5)=12.5\\
W&=\dfrac12(3)(4)=6
\end{align*}So $X+Y=Z$ is correct.
In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$
$\textbf{C}$
If she answered all question correctly, she will get 50 point. However, for each incorrect question, we need to subtract $5+2=7$ points from 50. Hence, we know she answered $(50-29)/7=3$ questions incorrectly, which means she answered 7 questions correctly.
Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each day for 3 days. How long would he have to skate the ninth day in order to average 85 minutes of skating each day for the entire time?
$\text{(A)}\ \text{1 hr} \qquad \text{(B)}\ \text{1 hr 10 min} \qquad \text{(C)}\ \text{1 hr 20 min} \qquad \text{(D)}\ \text{1 hr 40 min} \qquad \text{(E)}\ \text{2 hr}$
$\textbf{E}$
When the average time of 9 days is 85 minutes, he had skated $9\times85=765$ minutes in total. For the first 5 days he staked 75 minutes each day, and for the next 3 days he skated 90 minutes each day. So in total he staked $5\times75+3\times90=645$ minutes for the first 8 days. On the ninth day he had to skate for $765-645=120$ minutes, or 2 hours.
How many whole numbers between 99 and 999 contain exactly one 0?
$\text{(A)}\ 72 \qquad \text{(B)}\ 90 \qquad \text{(C)}\ 144 \qquad \text{(D)}\ 162 \qquad \text{(E)}\ 180$
$\textbf{D}$
The hundredth digit can not be 0. If the tenth digit is 0, then we have 9 choices for the hundred digit, and also 9 choices for the unit digit, which means $9\times9=81$ numbers in total. Similarly, if the unit digit is 0, we can also find 81 numbers. So the answer is $81+81=162$.
The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$. Altitude $\overline{XC}$ bisects $\overline{YZ}$. The area (in square inches) of the shaded region is
$\text{(A)}\ 1\dfrac{1}{2} \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 2\dfrac{1}{2} \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 3\dfrac{1}{2}$
$\textbf{D}$
The area of $\triangle XAB$ is $\dfrac14$ of $\triangle XYZ$. So the area of trapezoid $ABZY$ is $\dfrac34$ of $\triangle XYZ$. The shaded region is half of trapezoid $ABZY$. So the answer is $\dfrac12\times\dfrac34\times8=3$.
Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is
$\text{(A)}\ \dfrac{5}{16} \qquad \text{(B)}\ \dfrac{3}{8} \qquad \text{(C)}\ \dfrac{1}{2} \qquad \text{(D)}\ \dfrac{5}{8} \qquad \text{(E)}\ \dfrac{11}{16}$
$\textbf{E}$
Case 1: There are two heads and two tails. There are $\dbinom{4}{2} = 6$ ways to choose which two tosses are heads, and the other two must be tails.
Case 2: There are three heads, one tail. There are $\dbinom{4}{1} = 4$ ways to choose which of the four tosses is a tail.
Case 3: There are four heads and no tails. This can only happen in $1$ way.
There are a total of $2^4=16$ possible configurations, giving a probability of $\dfrac{6+4+1}{16} = \dfrac{11}{16}$.
Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom and sides.
$\text{(A)}\ 18 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36$
$\textbf{C}$
We can count the number of showing faces from each side. Noticing that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for $2\cdot(\text{front surface area} + \text{side surface area} + \text{top surface area})$. We find that this is $2\cdot(5 + 4 + 4) = 2\cdot13 = 26$.
A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?
$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{4}{9} \qquad \text{(C)}\ \dfrac{1}{2} \qquad \text{(D)}\ \dfrac{5}{9} \qquad \text{(E)}\ \dfrac{5}{8}$
$\textbf{B}$
The same pattern is repeated for every $6 \times 6$ tile. Looking closer, there is also symmetry of the top left $3 \times 3$ square, so the fraction of the entire floor in dark tiles is the same as the fraction in the square. Counting the tiles, there are $4$ dark tiles, and $9$ total tiles, giving a fraction of $\dfrac49$.
Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?
$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 70$
$\textbf{B}$
A pear gives $8/3$ ounces of juice per pear. An orange gives $8/2=4$ ounces of juice per orange. If the pear-orange juice blend used one pear and one orange each, the percentage of pear juice would be \[\dfrac{8/3}{8/3+4} = \dfrac{8}{8+12}=40\%\]
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
$\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{2}{5} \qquad \text{(E)}\ \dfrac{1}{2}$
$\textbf{B}$
Assume Moe, Loki, and Nick each give Ott $\$1$. Therefore, in the beginning Moe has $\$5$, Loki has $\$4$, and Nick has $\$3$. After everyone gives Ott some fraction of their money, the total money at the end situation will be the same as the original, which is $\$12$. Ott gets $\$ 3$. Thus, the answer is $\dfrac{3}{12}=\dfrac14$.