## AMC 8 2001

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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.

Casey's shop class is making a golf trophy. He has to paint $300$ dimples on a golf ball. If it takes him $2$ seconds to paint one dimple, how many minutes will he need to do his job?

$\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

$\textbf{D}$

It takes him $2\times300=600$ seconds to paint all the dimples, which is equivalent to $600/60=10$ minutes.

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

$\textbf{D}$

Let the numbers be $x$ and $y$. Then we have $x+y=11$ and $xy=24$. Since they are whole numbers, we can try the $(x,y)$ pairs from $(1,10)$ to $(5,6)$. The answer is $(3,8)$. So the larger number is 8.

Granny Smith has $\$63$. Elberta has $\$2$ more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?

$\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23$

$\textbf{E}$

Anjou has $\dfrac13\times63=21$ dollars. Elberta has $21+2=23$ dollars.

The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is

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

$\textbf{E}$

To make the smallest five-digit number, the higher order digits must be as small as possible, and the lower order digits as large as possible. Since it is an even number, the units digit should be 4. Then the tens digit should be 9.

On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is 1088 feet per second and one mile is 5280 feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning.

$\text{(A)}\ 1 \qquad \text{(B)}\ 1\dfrac{1}{2} \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 2\dfrac{1}{2} \qquad \text{(E)}\ 3$

$\textbf{C}$

During the $10$ seconds, the sound traveled $1088\times10=10880$ feet from the lightning to Snoopy. This is equivalent to $10880/5280\approx2$ miles.

Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is 60 feet. What is the distance in feet between the first and last trees?

$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 105 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 140$

$\textbf{B}$

There are $3$ spaces between the 1st and 4th trees, so each of these spaces has $60/3=20$ feet. Between the first and last trees there are $5$ spaces, so the distance between them is $20\times5=100$ feet.

$\textbf{Kites on Parade}$

Problems 7, 8 and 9 are about these kites.

To promote her school’s annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram below. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.

What is the number of square inches in the area of the small kite?

$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

$\textbf{A}$

The area of a kite is half the product of its diagonals. The diagonals have lengths of $6$ and $7$ inches, so the area is $\dfrac12\times6\times7=21$ square inches.

Genevieve puts bracing on her large kite in the form of cross-connecting opposite corners of the kite. How many inches of bracing material does she need?

$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 38 \qquad \text{(E)}\ 39$

$\textbf{E}$

Each diagonal of the large kite is $3$ times the length of the corresponding diagonal of the short kite since it was made with a grid $3$ times as long in each direction. The diagonals of the small kite are $6$ and $7$ inches, so the diagonals of the large kite are $18$ and $21$ inches, and the amount of bracing Genevieve needs is the sum of these lengths, which is $39$ inches.

The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?

$\text{(A)}\ 63 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 180 \qquad \text{(D)}\ 189 \qquad \text{(E)}\ 264$

$\textbf{D}$

The area of the large kite is $3\times3=9$ times as much as the area of the small kite. The area of the large grid is also 9 times as much as the area of the small gird. Hence, we get the area of the large kite $9\times21=189$ square inches, and the area of the large grid $9\times6\times7=378$ square inches. The area of the waste material is $378-189=189$ square inches.

A collector offers to buy state quarters for $2000\%$ of their face value. At that rate how much will Bryden get for his four state quarters?

$\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars}$

$\textbf{A}$

The face value of 4 state quarters is $\$1$. The collector will pay $\$1\times2000\%=\$20$ for that.

Points $A$, $B$, $C$ and $D$ have these coordinates: $A(3,2)$, $B(3,-2)$, $C(-3,-2)$ and $D(-3, 0)$. The area of quadrilateral $ABCD$ is

$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$

$\textbf{C}$

The area of trapezoid $\dfrac12\times(2+4)\times6=18$.

If $a\otimes b = \dfrac{a + b}{a - b}$, then $(6\otimes 4)\otimes 3 =$

$\text{(A)}\ 4 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 72$

$\textbf{A}$

$$6\otimes4=\dfrac{6+4}{6-4}=5$$ $$5\otimes3=\dfrac{5+3}{5-3}=4$$

Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 72$

$\textbf{D}$

The number of students who prefer cherry pie is $\dfrac12\times(36-12-8-6)=5$. So the degree in pie graph is $\dfrac{5}{36}\times360^\circ=50^\circ$.

Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?

- Meat: beef, chicken, pork
- Vegetables: baked beans, corn, potatoes, tomatoes
- Dessert: brownies, chocolate cake, chocolate pudding, ice cream

$\text{(A)}\ 4 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 72 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 144$

$\textbf{C}$

Tyler has $\dbinom31=3$ ways to choose meat, $\dbinom42=6$ ways to choose vegetables, and $\dbinom41=4$ ways to choose dessert. The total number of possible arrangements is $3\times6\times4=72$.

Homer began peeling a pile of 44 potatoes at the rate of 3 potatoes per minute. Four minutes later Christen joined him and peeled at the rate of 5 potatoes per minute. When they finished, how many potatoes had Christen peeled?

$\text{(A)}\ 20 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 33 \qquad \text{(E)}\ 40$

$\textbf{A}$

After the $4$ minutes of Homer peeling alone, he had peeled $4\times3=12$ potatoes. This means that there are $44-12=32$ potatoes left. Once Christen joins him, the two are peeling potatoes at a rate of $3+5=8$ potatoes per minute. So they finish peeling after another $32/8=4$ minutes. In these $4$ minutes, Christen peeled $4\times5=20$ potatoes.

A square piece of paper, 4 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?

$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{2} \qquad \text{(C)}\ \dfrac{3}{4} \qquad \text{(D)}\ \dfrac{4}{5} \qquad \text{(E)}\ \dfrac{5}{6}$

$\textbf{E}$

The smaller rectangles each have the same height as the original square, but have $\dfrac{1}{4}$ the length, since the paper is folded in half and then cut in half the same way. The larger rectangle has the same height as the original square but has $\dfrac{1}{2}$ the length. Therefore, the smaller rectangles have dimensions $4 \times 1$ and the larger rectangle has dimensions $4 \times 2$. The ratio of their perimeters is $\dfrac{2(4+1)}{2(4+2)}=\dfrac{5}{6}$.

For the game show $\textit{Who Wants To Be A Millionaire?}$, the dollar values of each question are shown in the following table (where K = 1000).

Between which two questions is the percent increase of the value the smallest?

$\text{(A)}\ \text{From 1 to 2} \qquad \text{(B)}\ \text{From 2 to 3} \qquad \text{(C)}\ \text{From 3 to 4} \qquad \text{(D)}\ \text{From 11 to 12} \qquad \text{(E)}\ \text{From 14 to 15}$

$\textbf{B}$

The value is doubled for most of the questions except $2$ to $3$, $3$ to $4$, and $11$ to $12$. The increase from $2$ to $3$ is $\dfrac{300-200}{200}=\dfrac12$. The increase from $3$ to $4$ is $\dfrac{500-300}{300}=\dfrac23$. The increase from $11$ to $12$ is $\dfrac{125-64}{64}\approx1$.

Therefore, the smallest percent increase is $\text{From 2 to 3}$.

Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?

$\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}$

$\textbf{D}$

If the product of the two numbers is a multiple of 5, at least one of the two numbers is 5. To find the probability that at least one of the two numbers is 5, we can consider another case when none of the two numbers is 5. In this case, the probability that none of the two numbers is 5 is $\dfrac56\times\dfrac56=\dfrac{25}{36}$. Therefore, the probability that at least one of the two numbers is 5 is $1-\dfrac{25}{36}=\dfrac{11}{36}$.

Car M traveled at a constant speed for a given time. This is shown by the dashed line. Car N traveled at twice the speed for the same distance. If Car N's speed and time are shown as solid line, which graph illustrates this?

$\textbf{D}$

Since car N has twice the speed, it must be twice as high on the speed axis. Also, since cars M and N travel at the same distance but car N has twice the speed, car N must take half the time. Therefore, line N must be half the size of line M. Since the speeds are constant, both lines are horizontal. Reviewing the graphs, we see that the only one satisfying these conditions is graph $\text{D}$.

Kaleana shows her test score to Quay, Marty and Shana, but the others keep theirs hidden. Quay thinks, "At least two of us have the same score." Marty thinks, "I didn't get the lowest score." Shana thinks, "I didn't get the highest score." List the scores from lowest to highest for Marty (M), Quay (Q) and Shana (S).

$\text{(A)}\ \text{S,Q,M} \qquad \text{(B)}\ \text{Q,M,S} \qquad \text{(C)}\ \text{Q,S,M} \qquad \text{(D)}\ \text{M,S,Q} \qquad \text{(E)}\ \text{S,M,Q}$

$\textbf{A}$

Since the only other score Quay knows is Kaleana's, and he knows that two of them have the same score, Quay and Kaleana must have the same score, $K=Q$. Marty knows that he didn't get the lowest score, and the only other score he knows is Kaleana's, so Marty must know that Kaleana has a lower score than him, $M>K$. Finally, Shana knows that she didn't get the highest score, and the only other score she knows is Kaleana's, so Shana must know that Kaleana has a higher score than her, $S<K$. Putting these together we have $S<Q<M$.

The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is

$\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$

$\textbf{D}$

To get the maximum possible value of the largest, the other values should be as small as possible. The sum of the five different positive integers is $15\times5=75$. The median is 18. The two numbers lower than the median should be 1 and 2. The number right after the median should be 19. Hence, the largest value is $75-1-2-18-19=35$.

On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is $\textbf{NOT}$ possible?

$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$

$\textbf{E}$

The highest possible score is $5\times20=100$ if you get every answer right. The second highest possible score is $5\times19+1=96$ if you get $19$ questions right and leave the remaining one blank. Therefore, no score between $96$ and $100$ is possible. The answer is 97.

Points $R$, $S$ and $T$ are vertices of an equilateral triangle, and points $X$, $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices?

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

$\textbf{D}$

There are 3 vertices and 3 midpoints. We have several ways to choose 3 of them to form a noncongruent triangle.

Case 1: We choose 3 vertices. Hence, we get $\triangle RST$.

Case 2: We choose 2 vertices, for example, $R$ and $S$. The rest 1 midpoint can not be the midpoint between the 2 chosen vertices. It must be one of the two other midpoints, $X$ or $Z$. By symmetry, no matter which one of the two other midpoints we choose, the result triangles are congruent. Hence, we get $\triangle RSX$.

Case 3: We choose 1 vertex, for example, $R$. The midpoints can be divided into 2 categories: adjacent point like $X$ or $Y$; and non-adjacent point like $Z$. We can choose 2 adjacent points, or 1 adjacent and 1 non-adjacent point. Hence, we have $\triangle RXY$ and $\triangle RXZ$.

Case 4: We choose 0 vertex and 3 midpoints. However, $\triangle XYZ$ is congruent with $\triangle RXY$.

In conclusion, we have 4 types of noncongruent triangles: $\triangle RST$, $\triangle RSX$, $\triangle RXY$ and $\triangle RXZ$.

Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?

$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$

$\textbf{B}$

We have 6 red triangles, 10 blue triangles and 16 white triangles in total. After subtracting 2 pairs of red triangles, 3 pairs of blue triangles and 2 red-white pairs, we have 0 red triangle, 4 blue triangles and 14 white triangles. The rest 4 blue triangles can not be paired to each other. So there will be 4 blue-white pairs. Now we have 10 white triangles left. Hence, there will be 5 pairs of white triangles.

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?

$\text{(A)}\ 5724 \qquad \text{(B)}\ 7245 \qquad \text{(C)}\ 7254 \qquad \text{(D)}\ 7425 \qquad \text{(E)}\ 7542$

$\textbf{D}$

The largest number made up of 2, 4, 5 and 7 is 7542, and the smallest number is 2457. So the larger number should be 2 time or 3 times as large as the smaller number.

Case 1: The larger number is twice the smaller number. Think about the units digit. If the units digit of the smaller number is 2, 4, 5, or 7, the units digit of the larger number is 4, 8, 0, or 4. Only 4 is possible for the units digit of the larger number. So the units digit of the smaller number must be 2 or 7. Now think about the thousands digit of the larger number. If the thousands digit of the larger number is 7, the thousands digit of the smaller number must be 3, which is impossible. Hence, the thousands digit of the larger number must be 5. So the larger number could be 5274 or 5724. Both 5274/2=2637 and 5724/2=2862 are impossible for the smaller number. Therefore, the larger number can not be twice of the smaller number.

Case 2: The larger number must be 3 times as large as the smaller number. Think about the units digit and the thousands digit in the same way. The thousands digit of the larger number must be 7. The units digit of the larger number should be 2 or 5. So the larger number could be 7452, 7542, 7245 or 7425. Dividing each of them by 3 and we found 7425 is the right answer.

For this question, we can choose the right answer in an easier way by dividing all of the choices by 2 or 3. Then we found the only possible choice is 7425/3=2475.