## AMC 8 1999

**If you like our free learning materials, please click the advertisement below. No paying, no donation, just a simple click. Thank you!!! 如果你喜欢我们的免费学习资料，请麻烦您点击下面的广告啦，为用爱发电的作者大大增加一点点广告收入，谢谢！！！**

**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.

$(6?3) + 4 - (2 - 1) = 5$. To make this statement true, the question mark between the 6 and the 3 should be replaced by

$\text{(A)} \div \qquad \text{(B)}\ \times \qquad \text{(C)} + \qquad \text{(D)}\ - \qquad \text{(E)}\ \text{None of these}$

$\textbf{A}$

Simplifying the given expression, we get $(6?3)=2.$ So the answer should be $\div$.

What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?

$\text{(A)}\ 30 \qquad \text{(B)}\ 45 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90$

$\textbf{C}$

At $10:00$, the hour hand will be on the $10$ while the minute hand on the $12$. The angle between them is $\dfrac16\times360^\circ=60^\circ$.

Which triplet of numbers has a sum NOT equal to 1?

$\text{(A)}\ (1/2,1/3,1/6) \qquad \text{(B)}\ (2,-2,1) \qquad \text{(C)}\ (0.1,0.3,0.6) \qquad \text{(D)}\ (1.1,-2.1,1.0) \qquad \text{(E)}\ (-3/2,-5/2,5)$

$\textbf{D}$

By adding each triplet, we can see that D gives us $0$, not $1$, as our sum.

The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?

$\text{(A)}\ 15 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 25 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

$\textbf{A}$

After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60. Thus the answer is $60-45=15$.

A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?

$\text{(A)}\ 100 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 300 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 500$

$\textbf{D}$

The perimeter of the $60\times20$ rectangle is $2\times(60+20)=160$ feet, and the area is $60\times20=1200\ \text{ft}^2$. If a square has the same perimeter, the side length should be $160/4=40$ feet, and the area is $40\times40=1600\ \text{ft}^2$. Hence, the answer is $1600-1200=400$.

Bo, Coe, Flo, Joe, and Moe have different amounts of money. Neither Joe nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Joe has more than Moe, but less than Bo. Who has the least amount of money?

$\text{(A)}\ \text{Bo} \qquad \text{(B)}\ \text{Coe} \qquad \text{(C)}\ \text{Flo} \qquad \text{(D)}\ \text{Joe} \qquad \text{(E)}\ \text{Moe}$

$\textbf{E}$

According to the given information, we have $$J<F$$ $$B<F$$ $$M<B$$ $$M<C$$ $$M<J<B$$ We can find it directly from the equations above that $M$ is less that $B$, $C$, and $J$. Since $J$ is less than $F$, $M$ is also less than $F$. Hence, Moe has the least amount of money.

The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center?

$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130$

$\textbf{E}$

The distance between the third and tenth exit is $160-40=120$ miles. Therefore, the distance between the third exit and the service center is $\dfrac34\times120=90$ miles. Hence, the milepost at the service center is $90+40=130$.

Six squares are colored, front and back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is

$\text{(A)}\ \text{B} \qquad \text{(B)}\ \text{G} \qquad \text{(C)}\ \text{O} \qquad \text{(D)}\ \text{R} \qquad \text{(E)}\ \text{Y}$

$\textbf{A}$

G is in opposite to O. Y is in opposite to R. The only possible face in opposite to W is B.

Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is

$\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450$

$\textbf{C}$

Plants shared by two beds have been counted twice, so the total is $500 + 450 + 350 - 50 - 100 = 1150$.

A complete cycle of a traffic light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly chosen time, what is the probability that the light will NOT be green?

$\text{(A)}\ \dfrac{1}{4} \qquad \text{(B)}\ \dfrac{1}{3} \qquad \text{(C)}\ \dfrac{5}{12} \qquad \text{(D)}\ \dfrac{1}{2} \qquad \text{(E)}\ \dfrac{7}{12}$

$\textbf{E}$

The probability of green is $\dfrac{25}{60} = \dfrac{5}{12}$, so the probability of not green is $1- \dfrac{5}{12} = \dfrac{7}{12}$.

Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is

$\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$

$\textbf{D}$

The sum of the horizontal row plus the sum of the vertical column equals to the sum of five numbers plus the number in the middle square. In order to get the largest horizontal or vertical sum, the number in the middle square should be as large as possible. Hence, we try 13 in the middle square. The rest 4 numbers can be separated as $1+10=4+7$. So the largest possible value for the horizontal or vertical sum is $1+10+13=24$.

The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$. To the nearest whole percent, what percent of its games did the team lose?

$\text{(A)}\ 24 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 73$

$\textbf{B}$

The percentage of loss is $\dfrac{4}{4+11}=\dfrac{4}{15}\approx27\%$.

The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults ?

$\text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 29 \qquad \text{(E)}\ 30$

$\textbf{C}$

The total number of ages of the 40 members is $17\times40=680$. The total number of ages of the 20 girls is $15\times20=300$. And the total number of ages of the 15 boys is $16\times15=240$. So the total number of ages of the 5 adults is $680-300-240=140$. Hence, the average of the adults is $140/5=28$.

In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is

$\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48$

$\textbf{D}$

By the Pythagorean Theorem, $AB=CD=5$. So the perimeter of $ABCD$ is $8+16+5+5=34$.

Bicycle license plates in Flatville each contain three letters. The first is chosen from the set $\{\text{C, H, L, P, R}\}$, the second from $\{\text{A, I, O}\}$, and the third from $\{\text{D, M, N, T}\}$.

When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters?

$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 60$

$\textbf{D}$

There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \times 3 \times 4 = 60$ license plates.

When adding 2 letters, we need to make the set length as even as possible in order to get the largest product. Hence, the set length should be 5, 5, and 4. The product is $5\times5\times4=100$.

Therefore, the answer is $100-60=40$.

Tori's mathematics test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered $70\%$ of the arithmetic, $40\%$ of the algebra, and $60\%$ of the geometry problems correctly, she did not pass the test because she got less than $60\%$ of the problems right. How many more problems would she have needed to answer correctly to earn a $60\%$ passing grade?

$\text{(A)}\ 1 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 11$

$\textbf{B}$

She answered $10\cdot70\%+30\cdot40\%+35\cdot60\%=40$ questions correctly. To get a $60\%$ passing grade, she needed to answer $75\cdot60\%=45$ questions correctly. Hence, she needed to answer $45-40=5$ more questions correctly to pass the exam.

Problems 17, 18, and 19 refer to the following:

At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie’s Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists these items: $1\dfrac{1}{2}$ cups flour, $2$ eggs, $3$ tablespoons butter, $\dfrac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.

Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15$

$\textbf{C}$

Since 108 students eat an average of two cookies apiece, they need at least $108\times2=216$ cookies. Considering that $216/15=14.4$, they need 15 pans of cookies. Each pan needs 2 eggs. So they need $15\times2=30$ eggs, or $30/6=5$ half-dozen of eggs.

They learn that a big concert is scheduled for the same night and attendance will be down $25\%$. How many recipes of cookies should they make for their smaller party?

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$

$\textbf{E}$

Now they need only $216\times75\%=162$ cookies. Since $162/15=10.8$, they need 11 recipes of cookies.

The drummer gets sick. The concert is cancelled. Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)

$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

$\textbf{B}$

Now they need to prepare 216 cookies again. Since $216/15=14.4$, they need 15 pans of cookies. Each pan need 3 tablespoons butter, so they need $15\times3=45$ tablespoons butter in total. Given that there are 8 tablespoons in a stick of butter, they need at least $45/8=5.625$ sticks of butter. To get a whole number, they need 6 sticks of butter.

Figure 1 is called a "stack map." The numbers tell how many cubes are stacked in each position. Fig. 2 shows these cubes, and Fig. 3 shows the view of the stacked cubes as seen from the front.

Which of the following is the front view for the stack map in Fig. 4?

$\textbf{B}$

In the front view, we only care about the height of each column. According to figure 4, the heights of the left, the middle, and the right column is 2, 3, and 4. Hence, the answer is B.

The degree measure of angle $A$ is

$\text{(A)}\ 20 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 45$

$\textbf{B}$

In $\triangle BCD$, $\angle B=180^\circ-100^\circ-40^\circ=40^\circ$. So $\angle A$ in $\triangle ABE$ is $\angle A=180^\circ-110^\circ-40^\circ=30^\circ$.

In a far-off land three fish can be traded for two loaves of bread and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?

$\text{(A)}\ \dfrac{3}{8} \qquad \text{(B)}\ \dfrac{1}{2} \qquad \text{(C)}\ \dfrac{3}{4} \qquad \text{(D)}\ 2\dfrac{2}{3} \qquad \text{(E)}\ 3\dfrac{1}{3}$

$\textbf{D}$

Three fish can be traded for two loaves of bread, and then traded for eight bags of rice. So each fish is worth $\dfrac83=2\dfrac{2}{3}$ bags of rice.

Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$?

$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$

$\textbf{C}$

The area of square $ABCD$ is $3\times3=9$. So the area of $\triangle BCM$ is $9\times\dfrac13=3$. The area of $\triangle BCM$ can be expressed as $$\dfrac12\cdot BC\cdot BM=3\rightarrow BM=2$$ Hence, we have $$CM=\sqrt{BM^2+BC^2}=\sqrt{2^2+3^2}=\sqrt{13}$$

When $1999^{2000}$ is divided by $5$, the remainder is

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

$\textbf{B}$

Since we only care about the units digit, the number can be simplified as $9^{2000}$. Now think about the pattern of the units digit of the power of 9. When the power is odd, the units digit is 9. When the power is even, the units digit is 1. So the units digit of $9^{2000}$ is 1. The reminder is 1 if the number is divided by 5.

Points $B$, $D$, and $J$ are midpoints of the sides of right triangle $ACG$. Points $K$, $E$, $I$ are midpoints of the sides of triangle $JDG$, etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$, then the total area of the shaded triangles is nearest

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

$\textbf{A}$

The area of the first shade triangle $BCD$ is $\dfrac12\times3\times3=\dfrac92$. The area of the second shade triangle $KDE$ is 1/4 the area of triangle $BCD$. We found that the area of the $n$th shade triangle is 1/4 of the $(n-1)$th shade triangle. Hence, the sum of 100 shade triangles is $$\dfrac92\left[1+\dfrac14+\left(\dfrac14\right)^2+\dots+\left(\dfrac14\right)^{99}\right]$$ To find the answer, let $$x=\left[1+\dfrac14+\left(\dfrac14\right)^2+\dots+\left(\dfrac14\right)^{99}\right]$$ then we have $$\dfrac14x=\left[\dfrac14+\left(\dfrac14\right)^2+\left(\dfrac14\right)^3+\dots+\left(\dfrac14\right)^{100}\right]$$ By subtracting the two equations, we get $$\dfrac34x=1-\left(\dfrac14\right)^{100}\approx1$$ So $x=\dfrac43$. The answer is $$\dfrac92\left[1+\dfrac14+\left(\dfrac14\right)^2+\dots+\left(\dfrac14\right)^{99}\right]=\dfrac92\cdot\dfrac43=6$$