1999 AMC 8 Exam Problems

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Used with permission of the Mathematical Association of America.

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1.

(6 ? 3)+4(21)=5.(6 \ ? \ 3) + 4 - (2 - 1) = 5. To make this statement true, the question mark between the 66 and the 33 should be replaced by

÷\div

×\times

++

-

None of these

Answer: A
Solution(s):

We have that 4(21)=41=3. 4 - (2 - 1) = 4 - 1 = 3.

Then (6 ? 3)+3=5 (6 \ ? \ 3) + 3 = 5 6 ? 3=2. 6 \ ? \ 3 = 2.

The only choice that works is ÷.\div.

Thus, A is the correct answer.

2.

What is the degree measure of the smaller angle formed by the hands of a clock at 1010 o'clock?

3030

4545

6060

7575

9090

Answer: C
Solution(s):

At 1010 o'clock, we have that the hour hand is at 1010 and the minute hand is at 12.12.

This means that the hands are a 16\frac{1}{6} of the entire circle apart. This is equal to 36016=60. 360^{\circ} \cdot \dfrac{1}{6} = 60^{\circ}.

Thus, C is the correct answer.

3.

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

(12,13,16)(\frac12,\frac13,\frac16)

(2,2,1)(2,-2,1)

(0.1,0.3,0.6)(0.1,0.3,0.6)

(1.1,2.1,1.0)(1.1,-2.1,1.0)

(32,52,5)(-\frac32,-\frac52,5)

Answer: D
Solution(s):

We have that A is the same as 36+26+16=1. \dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = 1.

B reduces to 22+1=1. 2 - 2 + 1 = 1.

C adds up to .1+.3+.6=1. .1 + .3 + .6 = 1.

D however adds to 1.12.1+1.0=0. 1.1 - 2.1 + 1.0 = 0.

To make sure, we check that E has a sum of 82+5=54=1. -\dfrac{8}{2} + 5 = 5 - 4 = 1.

Thus, D is the correct answer.

4.

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

1515

2020

2525

3030

3535

Answer: A
Solution(s):

Looking at the graph, we have that Alberto has biked 6060 miles and Bjorn 4545 miles after 44 hours.

The difference between the two distances is 6045=15 60 - 45 = 15 miles.

Thus, A is the correct answer.

5.

A rectangular garden 5050 feet long and 1010 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?

100100

200200

300300

400400

500500

Answer: D
Solution(s):

The current length of the fence is 2(50+10)=260=120 2(50 + 10) = 2 \cdot 60 = 120 feet. If all the sides become the same, then each side length is 120÷4=30 120 \div 4 = 30 feet. As such, the area of the square is then 302=900 30^2 = 900 square feet. Note that the original area of the garden is 5010=500 50 \cdot 10 = 500 square feet.

Therefore, the difference is 900500=400 900 - 500 = 400 square feet.

Thus, D is the correct answer.

6.

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

Bo

Coe

Flo

Jo

Moe

Answer: E
Solution(s):

Let B,C,F,J,B, C, F, J, and MM represent the amount of money that Bo, Coe, Flo, Jo, and Moe have respectively.

Then J<F and B<F. J \lt F \text{ and } B \lt F.

We also have B>M and C>M. B \gt M \text{ and } C \gt M.

Finally, we are given M<J<B. M \lt J \lt B.

From the first and second inequalities, we can get that F>M.F \gt M.

This shows that everyone has more money than Moe.

Thus, E is the correct answer.

7.

The third exit on a highway is located at milepost 4040 and the tenth exit is at milepost 160.160. 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?

9090

100100

110110

120120

130130

Answer: E
Solution(s):

There are 16040=120160 - 40 = 120 mileposts between the third and tenth exits.

This means 34\frac{3}{4} of the way is at the 40+34120=130 40 + \dfrac{3}{4} \cdot 120 = 130 milepost.

Thus, E is the correct answer.

8.

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

B

G

O

R

Y

Answer: A
Solution(s):

Consider the cube with the yellow face facing upwards.

Then we can fold the white, green, and orange faces down.

Since the blue face is attached to the green face, it will end up folding backwards.

This means that the blue face will end up facing opposite the white face.

Thus, A is the correct answer.

9.

Three flower beds overlap as shown. Bed A has 500500 plants, bed B has 450450 plants, and bed C has 350350 plants. Beds A and B share 5050 plants, while beds A and C share 100.100. The total number of plants is

850850

10001000

11501150

13001300

14501450

Answer: C
Solution(s):

Note there are 50+100=15050 + 100 = 150 plants that are in two beds and there are no plants in all three beds.

The total number of plants is then 500+450+350150=1150. 500 + 450 + 350 - 150 = 1150. We subtract to get rid of the plants that we counted twice.

Thus, C is the correct answer.

10.

A complete cycle of a traffic light takes 6060 seconds. During each cycle the light is green for 2525 seconds, yellow for 55 seconds, and red for 3030 seconds. At a randomly chosen time, what is the probability that the light will NOT be green?

14\dfrac{1}{4}

13\dfrac{1}{3}

512\dfrac{5}{12}

12\dfrac{1}{2}

712\dfrac{7}{12}

Answer: E
Solution(s):

During a given cycle, the light is not green for 30+5=3530 + 5 = 35 seconds.

Then the probability that it is not green is 3560=712. \dfrac{35}{60} = \dfrac{7}{12}.

Thus, E is the correct answer.

11.

Each of the five numbers 1,4,7,10,1, 4, 7, 10, and 1313 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

2020

2121

2222

2424

3030

Answer: D
Solution(s):

Let xx be the number in the middle. Then the sum of the horizontal and vertical sums is 1+4+7+10+13+x, 1 + 4 + 7 + 10 + 13 + x, since xx is counted twice. To maximize this, we want to maximize x,x, so we let x=13.x = 13.

Then, from the above, expression, we have that the largest possible value of the sum of the horizontal and vertical sums is 35+13=48. 35 + 13 = 48.

The sum of one direction is then 48÷2=24.48 \div 2 = 24.

Thus, D is the correct answer.

12.

The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is 11/4.11/4. To the nearest whole percent, what percent of its games did the team lose?

24%24\%

27%27\%

36%36\%

45%45\%

73%73\%

Answer: B
Solution(s):

The total number of games is 11+4=15,11 + 4 = 15, so the percent of games lost is 415100%=4203%27%. \dfrac{4}{15} \cdot 100 \% = \dfrac{4 \cdot 20}{3} \% \approx 27 \%.

Thus, B is the correct answer

13.

The average age of the 4040 members of a computer science camp is 1717 years. There are 2020 girls, 1515 boys, and 55 adults. If the average age of the girls is 1515 and the average age of the boys is 16,16, what is the average age of the adults?

2626

2727

2828

2929

3030

Answer: C
Solution(s):

The sum of the ages of everybody at the camp is 4017=680. 40 \cdot 17 = 680.

The sum of the ages of the girls is 2015=300 20 \cdot 15 = 300 and of the boys is 1516=240. 15 \cdot 16 = 240.

As such, we know that the ages of the adults must be 680300240=140. 680 - 300 - 240 = 140.

And therefore, the average age of the adults is then 140÷5=28. 140 \div 5 = 28.

Thus, C is the correct answer.

14.

In trapezoid ABCD,ABCD, the sides ABAB and CDCD are equal. The perimeter of ABCDABCD is

2727

3030

3232

3434

4848

Answer: D
Solution(s):

Let HH be where the altitude from BB to AD\overline{AD} intersects AD.\overline{AD}.

Then we have that AH=1682=4, AH = \dfrac{16 - 8}{2} = 4, since AB=CD.AB = CD.

We then have that AB=42+32=25=5. AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5.

Then the perimeter is 8+16+25=34. 8 + 16 + 2 \cdot 5 = 34.

Thus, D is the correct answer.

15.

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

2424

3030

3636

4040

6060

Answer: D
Solution(s):

There are currently 534=605 \cdot 3 \cdot 4 = 60 license plates that can be made.

If both letters are added to the first set, then there are 734=84 7 \cdot 3 \cdot 4 = 84 possible plates.

If they are both added to the second, there are 554=100 5 \cdot 5 \cdot 4 = 100 plates.

If they are added to the third, there are 536=90 5 \cdot 3 \cdot 6 = 90 choices.

If one is added to the first set and the other to the second set, there are 644=96 6 \cdot 4 \cdot 4 = 96 plates.

If the other is added to the third set, we get 635=90 6 \cdot 3 \cdot 5 = 90 possible plates.

Finally, if the letters are added to the second and third sets, there are 545=100 5 \cdot 4 \cdot 5 = 100 plates.

We see that 100100 is the greatest number of plates that we can achieve. This is an additional 10060=40100 - 60 = 40 plates.

Thus, D is the correct answer.

16.

Tori's mathematics test had 7575 problems: 1010 arithmetic, 3030 algebra, and 3535 geometry problems. Although she answered 70%70\% of the arithmetic, 40%40\% of the algebra, and 60%60\% of the geometry problems correctly, she did not pass the test because she got less than 60%60\% of the problems right.

How many more problems would she have needed to answer correctly to earn a 60%60\% passing grade?

11

55

77

99

1111

Answer: B
Solution(s):

She answered 10.7=7 10 \cdot .7 = 7 arithmetic questions correctly. She answered 30.4=12 30 \cdot .4 = 12 algebra ones correctly. She also got 35.6=2135 \cdot .6 = 21 geometry questions correctly.

This means she got a total of 7+12+21=40 7 + 12 + 21 = 40 questions correct.

As such, in order to get a 60%,60\%, Tori must have answered 75.6=45 75 \cdot .6 = 45 questions correctly. This means that she would have needed to answer an additional 4540=545 - 40 = 5 questions correctly.

Thus, B is the correct answer.

17.

Problems 17,18,17, 18, and 1919 refer to the following:

Cookies For a Crowd

At Central Middle School the 108108 students who take the AMC 8\to 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 1515 cookies, lists these items: 1121\dfrac{1}{2} cups of flour, 22 eggs, 33 tablespoons butter, 34\dfrac{3}{4} cups sugar, and 11 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.)

11

22

55

77

1515

Answer: C
Solution(s):

Since the students eat an average of 22 cookies each, they will eat a total of 1082=216108 \cdot 2 = 216 cookies.

Each recipe makes 1515 cookies, which means we need 21615=15 \left\lceil\dfrac{216}{15}\right\rceil = 15 full recipes to make enough cookies.

Each pan requires 22 eggs, which means we need 152=3015 \cdot 2 = 30 eggs. There are 66 eggs in a half-dozen, so we need 30÷6=530 \div 6 = 5 half-dozens.

Thus, C is the correct answer.

18.

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

66

88

99

1010

1111

Answer: E
Solution(s):

There will now only be 108(114)=10834=81 108 \cdot \left(1 - \dfrac{1}{4}\right) = 108 \cdot \dfrac{3}{4} = 81 people at the party. This means we need 81215=16215=11 \left\lceil \dfrac{81 \cdot 2}{15} \right\rceil = \left\lceil \dfrac{162}{15} \right\rceil = 11 recipes.

Thus, E is the correct answer.

19.

The drummer gets sick. The concert is cancelled. Walter and Gretel must make enough pans of cookies to supply 216216 cookies. There are 88 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)

55

66

77

88

99

Answer: B
Solution(s):

To make 216216 cookies, they have to make 21615=15 \left\lceil \dfrac{216}{15} \right\rceil = 15 pans. Since each pan requires 33 tablespoons of butter, all the pans will need 153=4515 \cdot 3 = 45 tablespoons.

They will then need 458=6 \left\lceil \dfrac{45}{8} \right\rceil = 6 sticks of butter.

Thus, B is the correct answer.

20.

Figure 11 is called a "stack map." The numbers tell how many cubes are stacked in each position. Fig. 22 shows these cubes, and Fig. 33 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?4?

Answer: B
Solution(s):

Note that the height of the column is the maximum height of the front and back columns.

In the left column, we have that the back column is taller with height 2.2.

In the middle column, we have that the front column is taller with height 3.3.

Finally, the right column has height 4.4.

Thus, B is the correct answer.

21.

The degree measure of angle AA is

2020

3030

3535

4040

4545

Answer: B
Solution(s):

Label the vertices as below.

We then have that ABC=180100=80 \angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ} by supplementary angles. Then we have CED=180110=70 \angle CED = 180^{\circ} - 110^{\circ} = 70^{\circ} again by supplementary angles. Using the sum of the interior angles of a triangle is 180,180^{\circ}, we get ECD=1807040 \angle ECD = 180^{\circ} - 70^{\circ} - 40^{\circ}=70. = 70^{\circ}. Then, using vertical angles, we have ACB=ECD=70. \angle ACB = \angle ECD = 70^{\circ}. Finally, using the sum of the interior angles of a triangle, we get A=1808070=30. \angle A = 180^{\circ} - 80^{\circ} - 70^{\circ} = 30^{\circ}.

Thus, B is the correct answer.

22.

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?

38\dfrac{3}{8}

12\dfrac{1}{2}

34\dfrac{3}{4}

2232\dfrac{2}{3}

3133\dfrac{1}{3}

Answer: D
Solution(s):

Let ff be the worth of a fish, ll for a loaf of bread, and rr for rice.

Then we are given 3f=2l and l=4r. 3f = 2l \text{ and } l = 4r.

Substituting, we get 3f=24r=8r 3f = 2 \cdot 4r = 8r f=83r. f = \dfrac{8}{3}r.

Thus, D is the correct answer.

23.

Square ABCDABCD has sides of length 3.3. Segments CMCM and CNCN divide the square's area into three equal parts. How long is segment CM?CM?

10\sqrt{10}

12\sqrt{12}

13\sqrt{13}

14\sqrt{14}

15\sqrt{15}

Answer: C
Solution(s):

The area of the square is 32=9,3^2 = 9, which means that the area of one region is 9÷3=3.9 \div 3 = 3.

This means the area of BMC\triangle BMC is 3,3, which means that 123BM=3 \dfrac{1}{2} \cdot 3 \cdot BM = 3 BM=2. BM = 2.

Since BMC\triangle BMC is right, we have CM=22+32=13. CM = \sqrt{2^2 + 3^2} = \sqrt{13}.

Thus, C is the correct answer.

24.

When 199920001999^{2000} is divided by 5,5, the remainder is

44

33

22

11

00

Answer: D
Solution(s):

Note that to find the remainder when divided by 5,5, we only care about the units digit.

This means we only have to observe how the powers of the units digit work, namely the powers of 9.9.

Then, looking at powers of 9,9, we see that the units digit alternates between 99 and 1:1: 9,81,729, 9, 81, 729, \cdots This means that 199920001999^{2000} ends in a 11 since the power is even.

The remainder when divided by 55 is then 1,1, since it is 11 more than a multiple of 10.10.

Thus, D is the correct answer.

25.

Points B,B, D,D, and JJ are midpoints of the sides of right triangle ACG.ACG. Points K,K, E,E, II are midpoints of the sides of triangle JDG,JDG, etc. If the dividing and shading process is done 100100 times (the first three are shown) and AC=CG=6,AC = CG = 6, then the total area of the shaded triangles is nearest

66

77

88

99

1010

Answer: A
Solution(s):

Note that all the triangles are isosceles right triangles. We can find the following side lengths: CD=CG2=3=DG, CD = \dfrac{CG}{2} = 3 = DG, DE=DG2=32=EG, DE = \dfrac{DG}{2} = \dfrac{3}{2} = EG, and EF=EG2=34. EF = \dfrac{EG}{2} = \dfrac{3}{4}.

We then have that the areas of the 33 triangles are 1232=92, \dfrac{1}{2} \cdot 3^2 = \dfrac{9}{2}, 12(32)2=98, \dfrac{1}{2} \cdot \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{8}, and 12(34)2=932. \dfrac{1}{2} \cdot \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{32}.

The sum of these areas is 92+98+932=189325.9. \dfrac{9}{2} + \dfrac{9}{8} + \dfrac{9}{32} = \dfrac{189}{32} \approx 5.9.

This can act as underestimate for the areas of all 100100 triangles, since it does not include some areas.

Now, we have that the other 9797 triangles are all contained with FGH.\triangle FGH.

This means that we can use the area of FGH\triangle FGH is an overestimate for all the other triangles.

The area of FGH\triangle FGH is 12(34)2=932. \dfrac{1}{2} \cdot \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{32}. Then 18932+932=198326.2. \dfrac{189}{32} + \dfrac{9}{32} = \dfrac{198}{32} \approx 6.2.

Then, the true value is between 5.95.9 and 6.2,6.2, which means that 66 is nearest it.

Thus, A is the correct answer.