1999 AMC 8 — Problems & Online Timed Exam

1.

(6 \ ? \ 3) + 4 - (2 - 1) = 5.

To make this statement true, the question mark between the

6

and the

3

should be replaced by

\div

\times

+

-

None of these

Answer: A

Concepts:order of operations

Difficulty rating: 370

Small hint:

Simplify the part without the question mark first

Big hint:

Find the operation that makes

6\ ?\ 3=2

Solution:

We have that

4 - (2 - 1) = 4 - 1 = 3.

Then

(6 \ ? \ 3) + 3 = 5

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

10

o’clock?

30

45

60

75

90

Answer: C

Concepts:clock

Difficulty rating: 450

Small hint:

Each hour mark is

30^\circ

apart

Big hint:

At

10

o’clock, the hands are two hour marks apart

Solution:

At

10

o’clock, we have that the hour hand is at

10

and the minute hand is at

12.

This means that the hands are a

\frac{1}{6}

of the entire circle apart. This is equal to

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?

(\frac12,\frac13,\frac16)

(2,-2,1)

(0.1,0.3,0.6)

(1.1,-2.1,1.0)

(-\frac32,-\frac52,5)

Answer: D

Concepts:fraction decimal

Difficulty rating: 560

Small hint:

Compute the easiest-looking sums first

Big hint:

Only one listed triplet fails to total

1

Solution:

We have that A is the same as

\dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = 1.

B reduces to

2 - 2 + 1 = 1.

C adds up to

0.1 + 0.3 + 0.6 = 1.

D however adds to

1.1 - 2.1 + 1.0 = 0.

To make sure, we check that E has a sum of

-\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?

15

20

25

30

35

Answer: A

Concepts:data and graph interpretation

Difficulty rating: 660

Small hint:

Read both distances at

4

hours

Big hint:

Subtract Bjorn’s distance from Alberto’s

Solution:

Looking at the graph, we have that Alberto has biked

60

miles and Bjorn

45

miles after

4

hours.

The difference between the two distances is

60 - 45 = 15

miles.

Thus, A is the correct answer.

5.

A rectangular garden

50

feet long and

10

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?

100

200

300

400

500

Answer: D

Concepts:perimeter area

Difficulty rating: 770

Small hint:

Keep the perimeter the same

Big hint:

The square side length is one-fourth of the old perimeter

Solution:

The current length of the fence is

2(50 + 10) = 2 \cdot 60 = 120

feet. If all the sides become the same, then each side length is

120 \div 4 = 30

feet. As such, the area of the square is then

30^2 = 900

square feet. Note that the original area of the garden is

50 \cdot 10 = 500

square feet.

Therefore, the difference is

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

Concepts:logical deduction inequality

Difficulty rating: 820

Small hint:

Eliminate anyone known to have more than someone else

Big hint:

Use each comparison to show who is above Moe

Solution:

Let

B, C, F, J,

and

M

represent the amount of money that Bo, Coe, Flo, Jo, and Moe have respectively.

Then

J \lt F,\quad B \lt F.

We also have

B \gt M,\quad C \gt M.

Finally, we are given

M \lt J \lt B.

From the first and second inequalities, we can get that

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

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?

90

100

110

120

130

Answer: E

Concepts:fraction

Difficulty rating: 900

Small hint:

First find the distance between mileposts

40

and

160

Big hint:

Move

\frac{3}{4}

of that distance past milepost

40

Solution:

There are

160 - 40 = 120

miles between the third and tenth exits.

This means

\frac{3}{4}

of the way is at the

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

Concepts:net (3D geometry)cube geometry

Difficulty rating: 960

Small hint:

Fold around the square marked

Y

Big hint:

Track which face lands opposite

W

Solution:

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

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

850

1000

1150

1300

1450

Answer: C

Concepts:inclusion-exclusion Venn Diagram

Difficulty rating: 1020

Small hint:

Add the three bed counts first

Big hint:

Subtract the overlaps that were counted twice

Solution:

Note there are

50 + 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 + 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

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?

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{5}{12}

\dfrac{1}{2}

\dfrac{7}{12}

Answer: E

Concepts:geometric probability complementary probability

Difficulty rating: 980

Small hint:

Not green means red or yellow

Big hint:

Use time not green over total cycle time

Solution:

During a given cycle, the light is not green for

30 + 5 = 35

seconds.

Then the probability that it is not green is

\dfrac{35}{60} = \dfrac{7}{12}.

Thus, E is the correct answer.

11.

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

20

21

22

24

30

Answer: D

Concepts:optimization double counting

Difficulty rating: 1140

Small hint:

The center square is counted in both sums

Big hint:

Maximize the center number, then split the doubled total equally

Solution:

Let

x

be the number in the center. The horizontal sum plus the vertical sum counts every number once, except the center number is counted twice.

So twice the common sum is

1+4+7+10+13+x

=35+x

. This is largest when

x=13

.

The largest common sum is therefore at most

\dfrac{35+13}{2}=24

, and it is attainable: put

13

in the center,

1

and

10

at the ends of one row, and

4

and

7

at the ends of the other. Both sums are

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

\frac{11}{4}.

To the nearest whole percent, what percent of its games did the team lose?

24\%

27\%

36\%

45\%

73\%

Answer: B

Concepts:ratio and proportion percentage

Difficulty rating: 1120

Small hint:

Use

11

wins and

4

losses as one batch

Big hint:

The lost fraction is losses over total games

Solution:

The team can be viewed as winning

11

games and losing

4

games in each batch of

15

games.

The percent lost is

\dfrac4{15}\cdot100\%\approx 27\%

.

Thus, B is the correct answer.

13.

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?

26

27

28

29

30

Answer: C

Concepts:mean

Difficulty rating: 1150

Small hint:

Convert each average into a total age

Big hint:

Subtract the girls’ and boys’ totals from the camp total

Solution:

The sum of the ages of everybody at the camp is

40 \cdot 17 = 680.

The sum of the ages of the girls is

20 \cdot 15 = 300

and of the boys is

15 \cdot 16 = 240.

As such, we know that the ages of the adults must be

680 - 300 - 240 = 140.

And therefore, the average age of the adults is then

140 \div 5 = 28.

Thus, C is the correct answer.

14.

In trapezoid

ABCD,

the sides

AB

and

CD

are equal. The perimeter of

ABCD

is

27

30

32

34

48

Answer: D

Concepts:trapezoid Pythagorean Theorem

Difficulty rating: 1220

Small hint:

Drop an altitude from

B

Big hint:

The two side right triangles have horizontal leg

4

Solution:

Let

H

be where the altitude from

B

to

\overline{AD}

intersects

\overline{AD}.

Then we have that

AH = \dfrac{16 - 8}{2} = 4,

since

AB = CD.

We then have that

AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5.

Then the perimeter is

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\},

the second from

\{A,I,O\},

and the third from

\{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?

24

30

36

40

60

Answer: D

Concepts:multiplication principle optimization casework

Difficulty rating: 1330

Small hint:

The old number of plates is

5\cdot3\cdot4

Big hint:

To maximize a product, make the three factors as balanced as possible

Solution:

There are currently

5 \cdot 3 \cdot 4 = 60

license plates that can be made.

If both letters are added to the first set, then there are

7 \cdot 3 \cdot 4 = 84

possible plates.

If they are both added to the second, there are

5 \cdot 5 \cdot 4 = 100

plates.

If they are added to the third, there are

5 \cdot 3 \cdot 6 = 90

choices.

If one is added to the first set and the other to the second set, there are

6 \cdot 4 \cdot 4 = 96

plates.

If the other is added to the third set, we get

6 \cdot 3 \cdot 5 = 90

possible plates.

Finally, if the letters are added to the second and third sets, there are

5 \cdot 4 \cdot 5 = 100

plates.

We see that

100

is the greatest number of plates that we can achieve. This is an additional

100 - 60 = 40

plates.

Thus, D is the correct answer.

16.

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?

1

5

7

9

11

Answer: B

Concepts:percentage

Difficulty rating: 1230

Small hint:

Compute Tori’s current correct answers by category

Big hint:

Compare that total with

60\%

of

75

Solution:

She answered

10 \cdot 0.7 = 7

arithmetic questions correctly. She answered

30 \cdot 0.4 = 12

algebra ones correctly. She also got

35 \cdot 0.6 = 21

geometry questions correctly.

This means she got a total of

7 + 12 + 21 = 40

questions correct.

As such, in order to get a

60\%,

Tori must have answered

75 \cdot 0.6 = 45

questions correctly. This means that she would have needed to answer an additional

45 - 40 = 5

questions correctly.

Thus, B is the correct answer.

17.

Problems

17,

18,

and

19

refer to the following:

Cookies For a Crowd

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 of 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.)

1

2

5

7

15

Answer: C

Concepts:floor and ceiling functions unit conversion

Difficulty rating: 1250

Small hint:

Find how many full recipes are needed for

216

cookies

Big hint:

Each full recipe uses

2

eggs

Solution:

Since the students eat an average of

2

cookies each, they will eat a total of

108 \cdot 2 = 216

cookies.

Each recipe makes

15

cookies, which means we need

\left\lceil\dfrac{216}{15}\right\rceil = 15

full recipes to make enough cookies.

Each pan requires

2

eggs, which means we need

15 \cdot 2 = 30

eggs. There are

6

eggs in a half-dozen, so we need

30 \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\%.

How many recipes of cookies should they make for their smaller party?

6

8

9

10

11

Answer: E

Concepts:percentage floor and ceiling functions

Difficulty rating: 1310

Small hint:

A

25\%

drop leaves

75\%

attendance

Big hint:

Round the needed number of recipes up

Solution:

There will now only be

108 \cdot \left(1 - \dfrac{1}{4}\right) = 108 \cdot \dfrac{3}{4} = 81

people at the party. This means we need

\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

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.)

5

6

7

8

9

Answer: B

Concepts:unit conversion floor and ceiling functions

Difficulty rating: 1340

Small hint:

First round the number of pans up

Big hint:

Convert tablespoons of butter to sticks and round up

Solution:

To make

216

cookies, they have to make

\left\lceil \dfrac{216}{15} \right\rceil = 15

pans. Since each pan requires

3

tablespoons of butter, all the pans will need

15 \cdot 3 = 45

tablespoons.

They will then need

\left\lceil \dfrac{45}{8} \right\rceil = 6

sticks of butter.

Thus, B is the correct answer.

20.

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?

Answer: B

Concepts:3D geometry

Difficulty rating: 1410

Small hint:

A front view keeps only the taller stack in each column

Big hint:

For each column, take the larger front/back height

Solution:

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.

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

3.

Finally, the right column has height

4.

Thus, B is the correct answer.

21.

The degree measure of angle

A

is

20

30

35

40

45

Answer: B

Concepts:angle chasing

Difficulty rating: 1520

Small hint:

Use supplementary angles at the two intersections

Big hint:

Then use a triangle angle sum near the

40^\circ

angle

Solution:

Label the vertices as below.

We then have that

\angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ}

by supplementary angles. Then we have

\angle CED = 180^{\circ} - 110^{\circ} = 70^{\circ}

again by supplementary angles. Using the sum of the interior angles of a triangle is

180^{\circ},

we get

\begin{aligned} \angle ECD &= 180^{\circ} - 70^{\circ} - 40^{\circ} \\ &= 70^{\circ}. \end{aligned}

Then, using vertical angles, we have

\angle ACB = \angle ECD = 70^{\circ}.

Finally, using the sum of the interior angles of a triangle, we get

\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?

\dfrac{3}{8}

\dfrac{1}{2}

\dfrac{3}{4}

2\dfrac{2}{3}

3\dfrac{1}{3}

Answer: D

Concepts:ratio and proportion unit conversion

Difficulty rating: 1180

Small hint:

Turn each trade into a value equation

Big hint:

Convert fish to bread, then bread to rice

Solution:

Three fish are worth two loaves of bread, so one fish is worth

\dfrac23

of a loaf.

One loaf is worth

4

bags of rice, so one fish is worth

\dfrac23\cdot4=\dfrac83=2\dfrac23

bags of rice.

Thus, D is the correct answer.

23.

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?

\sqrt{10}

\sqrt{12}

\sqrt{13}

\sqrt{14}

\sqrt{15}

Answer: C

Concepts:triangle area Pythagorean Theorem

Difficulty rating: 1520

Small hint:

Each of the three regions has area one-third of the square

Big hint:

Use the area of

\triangle BMC

to find

BM

Solution:

The area of the square is

3^2 = 9,

which means that the area of one region is

9 \div 3 = 3.

This means the area of

\triangle BMC

is

3,

which means that

\dfrac{1}{2} \cdot 3 \cdot BM = 3

BM = 2.

Since

\triangle BMC

is right, we have

CM = \sqrt{2^2 + 3^2} = \sqrt{13}.

Thus, C is the correct answer.

24.

When

1999^{2000}

is divided by

5,

the remainder is

4

3

2

1

0

Answer: D

Concepts:modular arithmetic units digit

Difficulty rating: 1470

Small hint:

Reduce

1999

modulo

5

Big hint:

Even powers of

-1

modulo

5

have the same remainder

Solution:

Note that to find the remainder when divided by

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.

Then, looking at powers of

9,

we see that the units digit alternates between

9

and

1:

9, 81, 729, \cdots.

This means that

1999^{2000}

ends in a

1

since the power is even.

The remainder when divided by

5

is then

1,

since it is

1

more than a multiple of

10.

Thus, D is the correct answer.

25.

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

6

7

8

9

10

Answer: A

Concepts:geometric sequence similarity

Difficulty rating: 1620

Small hint:

The shaded triangles form a geometric area pattern

Big hint:

Each new shaded triangle has one-fourth the previous shaded area

Solution:

The first shaded triangle has legs

3

and

3

, so its area is

\dfrac12\cdot3\cdot3=\dfrac92

.

At each later step, the relevant triangle has half the leg length, so the shaded area is multiplied by

\left(\dfrac12\right)^2=\dfrac14

.

The total shaded area after many steps is therefore very close to the geometric sum

\begin{gathered} \frac92\left(1+\frac14+\frac1{16}+\cdots\right) \\ {}= \frac92\cdot\frac1{1-\frac{1}{4}} \\ {}= 6. \end{gathered}

After

100

steps the omitted tail is tiny, so the total area is nearest

6

.

Thus, A is the correct answer.

1999 AMC 8 Problems

Answer: A

Difficulty rating: 370

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 450

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 560

Small hint:

Big hint:

Solution:

Answer: A

Difficulty rating: 660

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 770

Small hint:

Big hint:

Solution:

Answer: E

Difficulty rating: 820

Small hint:

Big hint:

Solution:

Answer: E

Difficulty rating: 900

Small hint:

Big hint:

Solution:

Answer: A

Difficulty rating: 960

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1020

Small hint:

Big hint:

Solution:

Answer: E

Difficulty rating: 980

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 1140

Small hint:

Big hint:

Solution:

Answer: B

Difficulty rating: 1120

Small hint:

Big hint:

Solution:

Answer: C

Difficulty rating: 1150

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 1220

Small hint:

Big hint:

Solution:

Answer: D

Difficulty rating: 1330

Small hint:

Big hint:

Solution:

Answer: B

Difficulty rating: 1230

Small hint:

Big hint: