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2008 AMC 10B Problems/Problem 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?
Solution
The number of points could have been 10, 11, 12, 13, 14, or 15. Thus, the answer is
.
2008 AMC 10B Problems/Problem 2
A
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?
Solution
After reversing the numbers on the second and fourth rows, the block will look like
this:
The difference between the two diagonal sums
is:
.
2008 AMC 10B Problems/Problem 3
Assume that
is a positive real number. Which is equivalent to
Solution
2008 AMC 10B Problems/Problem 4
?
Page 2
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?
Solution
The maximum occurs when 20 players get the minimum wage and the total of all
players' salaries is 700000. That is when one player
gets
.
2008 AMC 10B Problems/Problem 5
For real numbers
and , define
. What is
?
Solution
Since
, it follows that
, and
See also
2008 AMC 10B Problems/Problem 6
Points and lie on
. The length of
is times the length of
, and the length
of
is times the length of
. The length of
is what fraction of the length
of
?
Solution
Let
. Then
we obtain
. Adding
and
. From this system of equations
to both sides of the second equation, we
obtain
. Thus,
See also
2008 AMC 10B Problems/Problem 7
An equilateral triangle of side length is completely filled in by non-overlapping
equilateral triangles of side length . How many small triangles are required?
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Solution
(C) The area of the large triangle is
, while the area each small triangle is
Dividing these two quantities, we get 100, therefore
large one.
.
small triangles can fit in the
Another Solution:
The number of triangles is
.
Also, another way to do it is to notice that as you go row by row (from the bottom), the
number of triangles decrease by 2 from 19, so we
have:
A fourth solution is to notice that the small triangles are similar to the large triangle as
they are both equilateral. Therefore, the ratio of their areas is the square of the ratios of
their side lengths. Hence the ratio of their areas is
, so the answer is
.
2008 AMC 10B Problems/Problem 8
A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses
cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used.
How many different bouquets could be purchased for exactly 50 dollars?
Solution
The cost of a rose is odd, hence we need an even number of roses. Let there be roses
for some
. Then we have
dollars left. We can always reach the
sum exactly by buying
carnations. Of course, the number of
roses must be such that the number of carnations is non-negative. We get the
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inequality
, and as must be an integer, this solves to
are
possible values of , and each gives us one solution.
. Hence there
2008 AMC 10B Problems/Problem 9
A quadratic equation
these two solutions?
has two real solutions. What is the average of
Solution
Dividing both sides by , we get
roots is , therefore their average is
. By Vieta's formulas, the sum of the
.
Another Solution
We know that for an equation
that the sum of the roots for
two roots divided by two, so the average is
, the sum of the roots is
. This means
is
, or 2. The average is the sum of the
.
2008 AMC 10B Problems/Problem 10
Points and are on a circle of radius and
. Point
minor arc
. What is the length of the line segment
?
is the midpoint of the
Solution
Let the center of the circle be
the midpoint of
).
, and let be the intersection of
, since they are both radii.
By the Pythagorean Theorem,
subtraction,
.
Using the Pythagorean Theorem again,
and
(then
is
, and by
.
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2008 AMC 10B Problems/Problem 11
Suppose that
and that
is a sequence of real numbers satifying
and
. What is
,
?
Solution
Plugging in
, we get
Plugging in
, we get
This is simply a system of two equations with two unknowns. Substituting
gives
, and
.
2008 AMC 10B Problems/Problem 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 mileae 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?
Solution
Every time the pedometer flips from
Pete has walked
So, if the pedometer flipped
steps.
times
to
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Pete walked
steps.
Dividing by
gives
This is closest to answer
.
2008 AMC 10B Problems/Problem 13
For each positive integer , the mean of the first
2008th term of the sequence?
terms of a sequence is . What is the
Solution
Since the mean of the first
of the first
terms is
2008th term is
Note that
terms is , the sum of the first terms is
and the sum of the first
terms is
. Thus, the sum
. Hence, the
is the sum of the first n odd numbers.
2008 AMC 10B Problems/Problem 14
Triangle
addition,
rotated
has
,
, and in the first quadrant. In
. Suppose that
is
. What are the coordinates of the image of ?
and
counterclockwise about
Solution
As
and in the first quadrant, we know that the coordinate of is . We
now need to pick a positive coordinate for so that we'll have
.
By the Pythagorean theorem we have
.
By the definition of sine, we have
, hence
Substituting into the previous equation, we get
This means that the coordinates of
After we rotate
are
counterclockwise about
and have the coordinates
.
, hence
.
.
.
, it will get into the second quadrant
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2008 AMC 10B Problems/Problem 15
How many right triangles have integer leg lengths
length
, where
?
and and a hypotenuse of
Solution
By the Pythagorean theorem,
This means that
We know that
.
, and that
.
We also know that a must be odd, since the right
side is odd.
So
, and the answer is
.
2008 AMC 10B Problems/Problem 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.)
Solution
We consider 3 cases based on the outcome of the coin:
Case 1, 0 heads: The probability of this occuring on the coin flip is . The probability
that 0 rolls of a die will result in an odd sum is .
Case 2, 1 head: The probability of this case occuring is . The proability that one die
results as an odd number is .
Case 3, 2 heads: The probability of this occuring is . The probability that 2 dice result
in an odd sum is , because regardless of what we throw on the first die, we
have
probability that the second die will have the opposite parity.
Thus, the probability of having an odd sum rolled is
2008 AMC 10B Problems/Problem 17
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A poll shows that
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?
Solution
The pollster could select responses in 3 different ways: YNN, NYN, and NNY, where Y
stands for a voter who approved of the work, and N stands for a person who didnt
approve of the work. The probability of each of these is
Thus, the
answer is
2008 AMC 10B Problems/Problem 18
Bricklayer Brenda would take nine hours to build a chimney alone, and bricklayer
Brandon would take hours to build it alone. When they work together, they talk a lot,
and their combined output decreases by bricks per hour. Working together, they
build the chimney in hours. How many bricks are in the chimney?
Solution
Let
get
be the number of bricks in the chimney. Using
. Solving for , we get
, we
.
2008 AMC 10B Problems/Problem 19
A cylindrical tank with radius feet and height feet is lying on its side. The tank is
filled with water to a depth of feet. What is the volume of water, in cubic feet?
Solution
Any vertical cross-section of the tank parallel with its base looks as follows:
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The volume of water can be computed as the height of the tank times the area of the
shaded part.
Let be the size of the smaller angle
Thus the outer angle
consists of
. We then have
has size
of the circle, minus the area of the triangle
is
.
, and the area of the
.
The area of the shaded part is then
is
.
. Hence the non-shaded part
Using the Pythagorean theorem we can compute
that
. Thus
triangle
, hence
, and the volume of water
.
2008 AMC 10B Problems/Problem 20
The faces of a cubical die are marked with the numbers , , , , , and . The faces of
another die are marked with the numbers , , , , , and . What is the probability that
the sum of the top two numbers will be , , or ?
Solution
The easiest way is to write a table of all
good outcomes.
1 3 4 5 6 8
-----------------1| 2 4 5 6 7 9
possible outcomes, do the sums, and count
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2|
2|
3|
3|
4|
3
3
4
4
5
5
5
6
6
7
6
6
7
7
8
7 8 10
7 8 10
8 9 11
8 9 11
9 10 12
We see that out of
possible outcomes give the sum of , the sum of , and the
sum of , hence the resulting probability is
.
2008 AMC 10B Problems/Problem 21
Ten chairs are evenly spaced around a round table and numbered clockwise
from through . Five married couples are to sit in the chairs with men and women
alternating, and no one is to sit either next to or across from his/her spouse. How many
seating arrangements are possible?
Solution
For the first man, there are possible seats. For each subsequent man, there are , , ,
and possible seats. After the men are seated, there are only two possible
arrangements for the five women. The answer is
.
2008 AMC 10B Problems/Problem 22
Three red beads, two white beads, and one blue bead are placed in line in random
order. What is the probability that no two neighboring beads are the same color?
Solution
There are two ways to arrange the red beads, where
and represents a blank space.
represents a red bead
In the first, there are three ways to place a bead in the first free space, two for the
second free space, and one for the third, so there are arrangements. In the second, a
white bead must be placed in the third free space, so there are two possibilities for the
third space, two for the second, and one for the first. That makes arrangements. There
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are
arrangements in total. The two cases above can be reversed, so we
double to arrangements. Also, in each case, there are three ways to place the first
red bead, two for the second, and one for the third, so we multiply by to
get
arrangements. There are
total arrangements so the answer
is
.
2008 AMC 10B Problems/Problem 23
A rectangular floor measures by feet, where and are positive integers and
.
An artist paints a rectangle on the floor with the sides of the rectangle parallel to the
floor. The unpainted part of the floor forms a border of width foot around the painted
rectangle and occupies half the area of the whole floor. How many possibilities are
there for the ordered pair
?
Solution
Because the unpainted part of the floor covers half the area, then the painted rectangle
covers half the area as well. Since the border width is 1 foot, the dimensions of the
rectangle are
by
. With this information we can make the equation:
Applying Simon's Favorite Factoring Trick, we get
Since
or
, then we have the possibilities
and
,
and
. This gives 2 possibilities: (5,12) or (6,8), So the answer
is
2008 AMC 10B Problems/Problem 24
Quadrilateral
has
is the measure of angle
, angle
?
Solution
and angle
. What
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Solution 1
Draw the angle bisectors of the angles
intersect. Let their intersection be
Note that the triangles
have
and
and
. These two bisectors obviously
. We will now prove that
and
lies on the segment
are congruent, as they share the side
.
, and we
.
Also note that for similar reasons the triangles
and
are congruent.
Now we can compute their inner angles.
is the bisector of the angle
,
hence
, and thus also
.
is the bisector of the
angle
, hence
, and thus also
.
It follows that
angle
has
obviously
. Thus the
, and hence
does indeed lie on
. Then
.
Solution 2
Draw the diagonals
and
, and suppose that they intersect at
.
Then,
and
are both isosceles, so by angle-chasing, we find
that
,
, and
. Draw
such
that
and so that
is on
, and draw
such that
and
is
on
. It follows that
and
are both equilateral. Also, it is easy to see
that
and
by construction, so
that
and
.
Thus,
, so
is isosceles. Since
,
then
, and
.
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Solution 3
Again, draw the diagonals
and
, and suppose that they intersect at
angle chasing the same way as in solution 2 that
Applying the Law of Sines to
and
, it follows
that
, so
and
. We find by
.
is isosceles. We finish as we did in
solution 2.
Solution 4
Start off with the same diagram as solution 1. Now draw
which creates
isosceles
. We know that the angle bisector of an isosceles triangle splits it in
half, we can extrapolate this further to see that its is
2008 AMC 10B Problems/Problem 25
Michael walks at the rate of feet per second on a long straight path. Trash pails are
located every
feet along the path. A garbage truck travels at feet per second in
the same direction as Michael and stops for 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?
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Solution
Pick a coordinate system where Michael's starting pail is and the one where the truck
starts is
. Let
and
be the coordinates of Michael and the truck
after seconds. Let
be their (signed) distance after seconds.
Meetings occur whenever
. We have
.
The truck always moves for seconds, then stands still for . During the
first seconds of the cycle the truck moves by
meters and Michael by
during the first seconds of the cycle
increases by
. During the
remaining seconds
decreases by
.
, hence
From this observation it is obvious that after four full cycles, i.e. at
have
for the first time.
, we will
During the fifth cycle,
will first grow from to
, then fall from
Michael overtakes the truck while it is standing at the pail.
to
. Hence
During the sixth cycle,
will first grow from
to , then fall from to
.
Hence the truck starts moving, overtakes Michael on their way to the next pail, and then
Michael overtakes the truck while it is standing at the pail.
During the seventh cycle,
will first grow from
to , then fall from to
Hence the truck meets Michael at the moment when it arrives to the next pail.
.
Obviously, from this point on
will always be negative, meaning that Michael is
already too far ahead. Hence we found all
meetings.
The movement of Michael and the truck is plotted below: Michael in blue, the truck in
red.