2008 AMC 10B — Problems & Online Timed Exam

1.

A basketball player made

5

baskets during a game. Each basket was worth either

2

or

3

points. How many different numbers could represent the total points scored by the player?

2

3

4

5

6

Answer: E

Concepts:basic counting systematic listing

Difficulty rating: 720

Small hint:

If

k

baskets are worth

3

points, the total is

2(5-k)+3k

Big hint:

As

k

runs from

0

to

5

, the total increases by

1

each time

Solution:

If

k

of the baskets are worth

3

points and the rest worth

2,

the total is

2(5-k)+3k=10+k.

As

k

ranges over

0,1,\ldots,5,

the total takes every integer value from

10

to

15,

giving

6

possibilities.

Thus, the correct answer is E.

2.

A

4 \times 4

block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?

2

4

6

8

10

Answer: B

Concepts:date and time process simulation

Difficulty rating: 880

Small hint:

After reversing rows

2

and

4,

write out the new grid

Big hint:

The main diagonal reads

1,10,17,22

and the other reads

4,9,16,25

Solution:

After reversing the second row to

11,10,9,8

and the fourth row to

25,24,23,22,

the two diagonals are

1,10,17,22

and

4,9,16,25.

Their sums are

1+10+17+22=50

and

4+9+16+25=54,

so the positive difference is

54-50=4.

Thus, the correct answer is B.

3.

Assume that

x

is a positive real number. Which is equivalent to

\sqrt[3]{x\sqrt{x}}\,?

x^{\frac{1}{6}}

x^{\frac{1}{4}}

x^{\frac{3}{8}}

x^{\frac{1}{2}}

x

Answer: D

Concepts:exponent radical

Difficulty rating: 940

Small hint:

Write

\sqrt{x}=x^{\frac{1}{2}}

Big hint:

Then

x\sqrt{x}=x^{\frac{3}{2}},

and the cube root divides the exponent by

3

Solution:

Since

\sqrt{x}=x^{\frac{1}{2}},

we have

x\sqrt{x}=x^{1}\cdot x^{\frac{1}{2}}=x^{\frac{3}{2}}.

Taking the cube root multiplies the exponent by

\tfrac13,

giving

\left(x^{\frac{3}{2}}\right)^{\frac{1}{3}}=x^{\frac{1}{2}}.

Thus, the correct answer is D.

4.

A semipro baseball league has teams with

21

players each. League rules state that a player must be paid at least

\$15{,}000,

and that the total of all players’ salaries for each team cannot exceed

\$700{,}000.

What is the maximum possible salary, in dollars, for a single player?

270{,}000

385{,}000

400{,}000

430{,}000

700{,}000

Answer: C

Concepts:inequality optimization extremal argument

Difficulty rating: 840

Small hint:

To maximize one salary, pay everyone else as little as allowed

Big hint:

The other

20

players each earn

\$15{,}000

Solution:

One player’s salary is largest when the other

20

players each earn the minimum

\$15{,}000.

That leaves

\$700{,}000-20\cdot\$15{,}000

=\$700{,}000-\$300{,}000

=\$400{,}000

for the single player.

Thus, the correct answer is C.

5.

For real numbers

a

and

b,

define

a\,\$\,b=(a-b)^2.

What is

(x-y)^2\,\$\,(y-x)^2?

0

x^2+y^2

2x^2

2y^2

4xy

Answer: A

Concepts:custom operation symmetry (algebra)

Difficulty rating: 880

Small hint:

Compare

(x-y)^2

and

(y-x)^2

Big hint:

Once you know the two inputs agree, apply the definition of

a\,\$\,b

Solution:

Since

(y-x)^2=(x-y)^2,

the two inputs are identical.

Therefore

(x-y)^2\,\$\,(y-x)^2

=\left((x-y)^2-(x-y)^2\right)^2

=0^2=0.

Thus, the correct answer is A.

6.

Points

B

and

C

lie on

\overline{AD}.

The length of

\overline{AB}

is

4

times the length of

\overline{BD},

and the length of

\overline{AC}

is

9

times the length of

\overline{CD}.

The length of

\overline{BC}

is what fraction of the length of

\overline{AD}?

\dfrac{1}{36}

\dfrac{1}{13}

\dfrac{1}{10}

\dfrac{5}{36}

\dfrac{1}{5}

Answer: C

Concepts:ratio and proportion fraction

Difficulty rating: 960

Small hint:

From

AB=4\,BD

and

AB+BD=AD,

find

BD

as a fraction of

AD

Big hint:

Similarly

CD=\tfrac{1}{10}AD,

and

BC=BD-CD

Solution:

Since

AB=4\,BD

and

AB+BD=AD,

we get

5\,BD=AD,

so

BD=\tfrac15 AD.

Since

AC=9\,CD

and

AC+CD=AD,

we get

10\,CD=AD,

so

CD=\tfrac{1}{10}AD.

Then

BC=BD-CD

=\tfrac15 AD-\tfrac{1}{10}AD

=\tfrac{1}{10}AD.

Thus, the correct answer is C.

7.

An equilateral triangle of side length

10

is completely filled in by non-overlapping equilateral triangles of side length

1.

How many small triangles are required?

10

25

100

250

1000

Answer: C

Concepts:power scaling of length, area, and volume equilateral triangle

Difficulty rating: 880

Small hint:

Area scales with the square of the side length

Big hint:

The big triangle has

10^2

times the area of a small one

Solution:

The large triangle has side length

10

times that of a small triangle, so its area is

10^2=100

times as large.

Since the small triangles tile it without overlap, exactly

100

of them are required.

Thus, the correct answer is C.

8.

A class collects

\$50

to buy flowers for a classmate who is in the hospital. Roses cost

\$3

each, and carnations cost

\$2

each. No other flowers are to be used. How many different bouquets could be purchased for exactly

\$50?

1

7

9

16

17

Answer: C

Concepts:Diophantine Equation parity

Difficulty rating: 1080

Small hint:

Let

r

be the number of roses; then

3r+2c=50

Big hint:

Since

2c

is even and

50

is even,

r

must be even, and

3r\le 50

Solution:

If

r

roses and

c

carnations are bought, then

3r+2c=50.

Because

2c

and

50

are even,

3r

must be even, so

r

is even.

Also

3r\le 50,

so

r\le 16.

The even values

r=0,2,4,\ldots,16

each give a valid

c,

which is

9

bouquets.

Thus, the correct answer is C.

9.

A quadratic equation

ax^2-2ax+b=0

has two real solutions. What is the average of the solutions?

1

2

\dfrac{b}{a}

\dfrac{2b}{a}

\sqrt{2a-b}

Answer: A

Concepts:Vieta’s Formulas quadratic

Difficulty rating: 1040

Small hint:

For

Ax^2+Bx+C=0,

the sum of the roots is

-\dfrac{B}{A}

Big hint:

The average is half the sum of the roots

Solution:

By Vieta’s formulas, the sum of the roots of

ax^2-2ax+b=0

is

\dfrac{-(-2a)}{a}=2.

The average is half of this, namely

1.

Thus, the correct answer is A.

10.

Points

A

and

B

are on a circle of radius

5

and

AB=6.

Point

C

is the midpoint of the minor arc

AB.

What is the length of the line segment

AC?

\sqrt{10}

\dfrac{7}{2}

\sqrt{14}

\sqrt{15}

4

Answer: A

Concepts:chord Pythagorean Theorem circle

Difficulty rating: 1170

Small hint:

The line from the center through

C

perpendicularly bisects chord

AB

Big hint:

Find the center-to-chord distance, subtract from

5

to reach

C,

then use the right triangle with leg

3

Solution:

Let

O

be the center and

D

the midpoint of

AB.

Then

OD\perp AB

with

AD=3,

so

OD=\sqrt{5^2-3^2}=4.

Since

C

is the midpoint of the minor arc,

O,

D,

C

are collinear and

DC=OC-OD=5-4=1.

Then

AC=\sqrt{AD^2+DC^2}

=\sqrt{3^2+1^2}

=\sqrt{10}.

Thus, the correct answer is A.

11.

Suppose that

(u_n)

is a sequence of real numbers satisfying

u_{n+2}=2u_{n+1}+u_n,

and that

u_3=9

and

u_6=128.

What is

u_5?

40

53

68

88

104

Answer: B

Concepts:recursion system of equations

Difficulty rating: 1140

Small hint:

Write

u_5

and

u_6

in terms of

u_4

and

u_3

Big hint:

u_5=2u_4+9

and

u_6=2u_5+u_4,

so

128=5u_4+18

Solution:

Using the recurrence,

u_5=2u_4+u_3=2u_4+9

and

u_6=2u_5+u_4

=2(2u_4+9)+u_4

=5u_4+18.

Setting

5u_4+18=128

gives

u_4=22,

so

u_5=2\cdot 22+9=53.

Thus, the correct answer is B.

12.

Postman Pete has a pedometer to count his steps. The pedometer records up to

99999

steps, then flips over to

00000

on the next step. Pete plans to determine his mileage for a year. On January

1

Pete sets the pedometer to

00000.

During the year, the pedometer flips from

99999

to

00000

forty-four times. On December

31

the pedometer reads

50000.

Pete takes

1800

steps per mile. Which of the following is closest to the number of miles Pete walked during the year?

2500

3000

3500

4000

4500

Answer: A

Concepts:place value unit conversion estimation

Difficulty rating: 1070

Small hint:

Each flip represents

100000

steps

Big hint:

Total steps are

44\cdot 100000+50000,

then divide by

1800

Solution:

Each flip counts

100000

steps, so Pete took

44\cdot 100000+50000=4{,}450{,}000

steps.

Dividing by

1800

gives about

2472

miles, which is closest to

2500.

Thus, the correct answer is A.

13.

For each positive integer

n,

the mean of the first

n

terms of a sequence is

n.

What is the

2008

th term of the sequence?

2008

4015

4016

4{,}030{,}056

4{,}032{,}064

Answer: B

Concepts:mean summation algebraic manipulation

Difficulty rating: 1170

Small hint:

The sum of the first

n

terms is

n\cdot n=n^2

Big hint:

The

n

th term is

n^2-(n-1)^2

Solution:

Since the mean of the first

n

terms is

n,

their sum is

n^2.

The

n

th term is

n^2-(n-1)^2=2n-1,

so the

2008

th term is

2\cdot 2008-1=4015.

Thus, the correct answer is B.

14.

Triangle

OAB

has

O=(0,0),

B=(5,0),

and

A

in the first quadrant. In addition,

\angle ABO=90^\circ

and

\angle AOB=30^\circ.

Suppose that

\overline{OA}

is rotated

90^\circ

counterclockwise about

O.

What are the coordinates of the image of

A?

\left(-\dfrac{10}{3}\sqrt{3},\,5\right)

\left(-\dfrac{5}{3}\sqrt{3},\,5\right)

\left(\sqrt{3},\,5\right)

\left(\dfrac{5}{3}\sqrt{3},\,5\right)

\left(\dfrac{10}{3}\sqrt{3},\,5\right)

Answer: B

Concepts:coordinate geometry transformation special right triangle

Difficulty rating: 1370

Small hint:

Find

A

first:

AB=OB\tan 30^\circ

since the right angle is at

B

Big hint:

A

90^\circ

counterclockwise rotation sends

(x,y)

to

(-y,x)

Solution:

Because

\angle ABO=90^\circ,

segment

AB

is vertical, so

A=\left(5,\,5\tan 30^\circ\right)=\left(5,\,\tfrac{5\sqrt3}{3}\right).

A

90^\circ

counterclockwise rotation about the origin sends

(x,y)

to

(-y,x),

so the image of

A

is

\left(-\tfrac{5\sqrt3}{3},\,5\right).

Thus, the correct answer is B.

15.

How many right triangles have integer leg lengths

a

and

b

and a hypotenuse of length

b+1,

where

b\lt 100?

6

7

8

9

10

Answer: A

Concepts:Pythagorean Triple perfect square

Difficulty rating: 1310

Small hint:

Set

a^2+b^2=(b+1)^2

and simplify

Big hint:

This gives

a^2=2b+1,

so

a^2

is an odd perfect square with

a^2\lt 201

Solution:

From

a^2+b^2=(b+1)^2

we get

a^2=2b+1,

so

a

is odd and

a^2

is an odd perfect square.

Since

b\lt 100,

we need

a^2=2b+1\lt 201,

and

a^2\ge 9

for

b\ge 4.

The odd squares

9,25,49,81,121,169

give

a=3,5,7,9,11,13,

which is

6

triangles.

Thus, the correct answer is A.

16.

Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is

0.

)

\dfrac{3}{8}

\dfrac{1}{2}

\dfrac{43}{72}

\dfrac{5}{8}

\dfrac{2}{3}

Answer: A

Concepts:conditional probability dice (probability)parity

Difficulty rating: 1490

Small hint:

If at least one die is rolled, the sum is odd exactly half the time

Big hint:

Subtract the probability of rolling no dice (both coins tails), then take half of the rest

Solution:

Whenever at least one die is rolled, by symmetry the sum is odd with probability

\tfrac12.

No die is rolled only when both coins are tails, with probability

\tfrac14;

that sum

0

is even. So the answer is

\left(1-\tfrac14\right)\cdot\tfrac12=\tfrac38.

Thus, the correct answer is A.

17.

A poll shows that

70\%

of all voters approve of the mayor’s work. On three separate occasions a pollster selects a voter at random. What is the probability that on exactly one of these three occasions the voter approves of the mayor’s work?

0.063

0.189

0.233

0.333

0.441

Answer: B

Concepts:binomial probability independent events

Difficulty rating: 1240

Small hint:

Exactly one approval can happen in

3

ordered ways

Big hint:

Each such arrangement has probability

(0.7)(0.3)(0.3)

Solution:

Exactly one approval among three occasions arises in

\binom{3}{1}=3

ways, each with probability

(0.7)(0.3)(0.3)=0.063.

The total is

3\cdot 0.063=0.189.

Thus, the correct answer is B.

18.

Bricklayer Brenda would take

9

hours to build a chimney alone, and bricklayer Brandon would take

10

hours to build it alone. When they work together, they talk a lot, and their combined output is decreased by

10

bricks per hour. Working together, they build the chimney in

5

hours. How many bricks are in the chimney?

500

900

950

1000

1900

Answer: B

Concepts:rate linear equation

Difficulty rating: 1370

Small hint:

Let

n

be the number of bricks; their solo rates are

\tfrac{n}{9}

and

\tfrac{n}{10}

Big hint:

Together they lay

\tfrac{n}{9}+\tfrac{n}{10}-10

bricks per hour for

5

hours to total

n

Solution:

Let

n

be the number of bricks. Brenda lays

\tfrac{n}{9}

per hour and Brandon

\tfrac{n}{10},

so together they lay

\tfrac{n}{9}+\tfrac{n}{10}-10

per hour.

Over

5

hours this equals

n:

5\left(\tfrac{n}{9}+\tfrac{n}{10}-10\right)=n.

Solving,

\tfrac{5n}{9}+\tfrac{n}{2}-50=n,

which gives

n=900.

Thus, the correct answer is B.

19.

A cylindrical tank with radius

4

feet and height

9

feet is lying on its side. The tank is filled with water to a depth of

2

feet. What is the volume of the water, in cubic feet?

24\pi-36\sqrt{2}

24\pi-24\sqrt{3}

36\pi-36\sqrt{3}

36\pi-24\sqrt{2}

48\pi-36\sqrt{3}

Answer: E

Concepts:cylinder sector volume

Difficulty rating: 1680

Small hint:

The water cross-section is a circular segment cut by a chord

2

feet below the center

Big hint:

Its area is a

120^\circ

sector minus a triangle; multiply by the length

9

Solution:

The submerged cross-section is a circular segment. The chord is

4-2=2

feet below the center, and

\cos\theta=\tfrac{2}{4}=\tfrac12,

so the half-angle is

60^\circ

and the central angle is

120^\circ.

The sector area is

\tfrac{120}{360}\pi(4)^2=\tfrac{16\pi}{3},

and the triangle formed by the two radii has area

\tfrac12(4)^2\sin 120^\circ=4\sqrt3.

The segment area is

\tfrac{16\pi}{3}-4\sqrt3.

Multiplying by the length

9

gives

9\left(\tfrac{16\pi}{3}-4\sqrt3\right)=48\pi-36\sqrt3.

Thus, the correct answer is E.

20.

The faces of a cubical die are marked with the numbers

1,

2,

2,

3,

3,

and

4.

The faces of a second cubical die are marked with the numbers

1,

3,

4,

5,

6,

and

8.

Both dice are thrown. What is the probability that the sum of the two top numbers will be

5,

7,

or

9?

\dfrac{5}{18}

\dfrac{7}{18}

\dfrac{11}{18}

\dfrac{3}{4}

\dfrac{8}{9}

Answer: B

Concepts:dice (probability)systematic listing

Difficulty rating: 1510

Small hint:

There are

6\times 6=36

equally likely outcomes; count those summing to

5,7,

or

9

Big hint:

Remember each

2

and each

3

on the first die appears on two faces

Solution:

Of the

36

equally likely outcomes, the pairs giving sum

5

are

(1,4),(2,3),(2,3),(4,1),

which is

4

outcomes.

Sum

7

comes from

(1,6),(2,5),(2,5),

(3,4),(3,4),(4,3),

which is

6,

and sum

9

from

(1,8),(3,6),(3,6),(4,5),

which is

4.

The probability is

\tfrac{4+6+4}{36}=\tfrac{14}{36}=\tfrac{7}{18}.

Thus, the correct answer is B.

21.

Ten chairs are evenly spaced around a round table and numbered clockwise from

1

through

10.

Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or directly across from his or her spouse. How many seating arrangements are possible?

240

360

480

540

720

Answer: C

Concepts:circular arrangements arrangements with restrictions casework

Difficulty rating: 1870

Small hint:

Seat the women first; count their arrangements in the alternating seats

Big hint:

For a fixed seating of the women, count how many ways the men can avoid their spouses

Solution:

Seat the women first. The first woman may take any of the

10

chairs, and since seats alternate, the remaining women fill their four seats in

4!

ways, giving

10\cdot 4!=240

arrangements.

Fix a woman in chair

1.

Her spouse must sit in chair

4

or chair

8;

each choice then forces the placement of every other man consistently. So each seating of the women yields exactly

2

valid seatings of the men.

The total is

2\cdot 240=480.

Thus, the correct answer is C.

22.

Three red beads, two white beads, and one blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?

\dfrac{1}{12}

\dfrac{1}{10}

\dfrac{1}{6}

\dfrac{1}{3}

\dfrac{1}{2}

Answer: C

Concepts:basic probability multiset permutations arrangements with restrictions

Difficulty rating: 1680

Small hint:

There are

\dfrac{6!}{3!\,2!}=60

distinguishable orderings

Big hint:

First place the three red beads so no two are adjacent, then fit the white and blue beads

Solution:

There are

\tfrac{6!}{3!\,2!}=60

distinguishable orderings. The three reds must occupy non-adjacent positions, and the possible red placements are

\{1,3,5\},\{2,4,6\},\{1,3,6\},

and

\{1,4,6\}.

For

\{1,3,5\}

and

\{2,4,6\},

the remaining seats are mutually non-adjacent, so the blue bead can go in any of the

3,

giving

3+3=6.

For

\{1,3,6\}

and

\{1,4,6\},

two remaining seats are adjacent, so the blue must separate the whites, giving

2+2=4.

That is

10

valid orderings, so the probability is

\tfrac{10}{60}=\tfrac16.

Thus, the correct answer is C.

23.

A rectangular floor measures

a

feet by

b

feet, where

a

and

b

are positive integers with

b\gt a.

An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width

1

foot around the painted rectangle and occupies half the area of the entire floor. How many possibilities are there for the ordered pair

(a,b)?

1

2

3

4

5

Answer: B

Concepts:Simon’s Favorite Factoring Trick Diophantine Equation area

Difficulty rating: 1580

Small hint:

The painted rectangle measures

(a-2)

by

(b-2),

and its area is half of

ab

Big hint:

From

ab=2(a-2)(b-2),

rearrange to

(a-4)(b-4)=8

Solution:

The painted rectangle is

(a-2)\times(b-2),

and it is half the floor, so

ab=2(a-2)(b-2).

Expanding gives

0=ab-4a-4b+8,

and adding

8

yields

(a-4)(b-4)=8.

With

b\gt a\gt 0,

the factorizations

8=1\cdot 8=2\cdot 4

give

(a,b)=(5,12)

and

(6,8).

So there are

2

possibilities.

Thus, the correct answer is B.

24.

Quadrilateral

ABCD

has

AB=BC=CD,

\angle ABC=70^\circ,

and

\angle BCD=170^\circ.

What is the degree measure of

\angle BAD?

75

80

85

90

95

Answer: C

Concepts:angle chasing isosceles triangle equilateral triangle

Difficulty rating: 1860

Small hint:

Build an equilateral triangle

BMC

on the same side of

BC

as

A

Big hint:

Show

\triangle ABM

and

\triangle MCD

are isosceles, then prove

M

lies on

\overline{AD}

Solution:

Let

M

be the point with

\triangle BMC

equilateral, on the same side of

BC

as

A.

Then

\angle ABM=70^\circ-60^\circ=10^\circ

and

\angle MCD=170^\circ-60^\circ=110^\circ.

Since

AB=BM

and

MC=CD,

triangles

ABM

and

MCD

are isosceles, giving

\angle AMB=85^\circ

and

\angle CMD=35^\circ.

Then

\angle AMD=360^\circ-85^\circ-60^\circ

-35^\circ

=180^\circ,

so

M

lies on

\overline{AD}

and

\angle BAD=\angle BAM=85^\circ.

Thus, the correct answer is C.

25.

Michael walks at the rate of

5

feet per second on a long straight path. Trash pails are located every

200

feet along the path. A garbage truck travels at

10

feet per second in the same direction as Michael and stops for

30

seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?

4

5

6

7

8

Answer: B

Concepts:distance rate and time inequality process simulation

Difficulty rating: 2090

Small hint:

Number the pails so Michael starts at pail

0

and the truck at pail

1;

track arrival times

Big hint:

Michael reaches pail

n

at

40n

seconds; the truck leaves pail

n

at

50(n-1)

seconds

Solution:

Number the pails so Michael is at pail

0

and the truck at pail

1.

Michael reaches pail

n

at

40n

seconds. The truck leaves pail

n

at

50(n-1)

seconds and arrives there at

50(n-1)-30

seconds.

Michael and the truck are together at pail

n

when

50(n-1)-30\le 40n

\le 50(n-1),

which simplifies to

5\le n\le 8.

At pail

5

they meet as the truck departs, at pails

6

and

7

Michael passes it, and at pail

8

they meet as the truck arrives. Between pails

6

and

7

the truck must overtake Michael once more, so in total they meet

5

times.

Thus, the correct answer is B.

2008 AMC 10B Problems

Answer: E

Difficulty rating: 720

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Solution:

Answer: B

Difficulty rating: 880

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Solution:

Answer: D

Difficulty rating: 940

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Solution:

Answer: C

Difficulty rating: 840

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Solution:

Answer: A

Difficulty rating: 880

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Solution:

Answer: C

Difficulty rating: 960

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Solution:

Answer: C

Difficulty rating: 880

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Solution:

Answer: C

Difficulty rating: 1080

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Solution:

Answer: A

Difficulty rating: 1040

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Solution:

Answer: A

Difficulty rating: 1170

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Solution:

Answer: B

Difficulty rating: 1140

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Solution:

Answer: A

Difficulty rating: 1070

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Solution:

Answer: B

Difficulty rating: 1170

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Solution:

Answer: B

Difficulty rating: 1370

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Solution:

Answer: A

Difficulty rating: 1310

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Solution:

Answer: A

Difficulty rating: 1490

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