菜单英文2012 AMC 12B Problems and Solution

2021年1月21日发(作者：viber是什么)
2012 AMC 12B Problems and Solution
Problem 1

Each third-grade classroom at Pearl Creek Elem
entary has 18 students and 2 pet
rabbits. How m
any m
ore students than rabbits are there in all 4 of the third-grade
classroom
s?


Solution

Solution 1

Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively.
Subtracting 8 from
72 we get

Solution 2

In each class, there are
more students than rabbits. So for all
classroom
s, the difference between students and rabbits is

Problem 2

A circle of radius 5 is inscribed in a rectangle as shown. The ratio of
the length of the
rectangle to its width is 2:1. What is the area of the rectangle?


Solution
If the radius is
, then the width is
, hence
the length is
.
,

Problem 3

For a science project, Sammy observed a chipmunk and squirrel stashing acorns in
holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4
acorns in each of the holes it dug. They each hid the sam
e number of acorns,
although the squirrel needed 4 fewer holes. How many acorns did the chipmunk
hide?


Solution
If x is the number of holes that the chipmunk dug then, 3x=4*(x-4). so 3x=4x-16.
x=16, and to find number of acorns hid by chipmunk, multiply by three, to get 48;
D.

Problem 4

Suppose that the euro is worth 1.30 dollars. If Diana has 500 dollars and Etienne
has 400 euros, by what percent is the value of Etienne's money greater that the
value of Diana's money?


Solution
So convert everything to dollars; 400(euros) x 1.3 =
520 dollars. now, 520 divided
by 500 = 1.04, which m
eans a value that is 4% greater; B.

Problem 5

Two integers have a sum
of 26. when two m
ore integers are added to the first two,
the sum
is 41. Finally, when two more integers are added to
the sum
of the previous
4 integers, the sum
is 57. What is the minimum number of even integers among the
6 integers?


Solution
So, x+
y=26, x could equal 15, and y could equal 11, so no even integers required
here. 41-26=15. a+b=15, a could equal 9 and b could equal 6, so one even integer
is required here. 57-41=16. m+n=16, m
could equal 9 and n could equal 7, so no
even integers required here, m
eaning only 1 even integer is required; A.

Problem 6

In order to estim
ate the value of
where
and
are real numbers with
, Xiaoli rounded
up by a small amount, rounded
correct?
down by the sam
e amount, and then
subtracted her rounded values. Which of the following statem
ents is necessarily





Solution
The original expression
now becom
es
, where
is a positive constant, hence the answer is
(A)
.

Problem 7

Sm
all lights are hung on a string 6 inches apart in the order red, red, green, green,
green, red, red, green, green, green, and so on continuing this pattern of 2 red
lights followed by 3 green lights. How m
any feet separate the 3rd red light and the
21st red light?

Note:
1 foot is equal to 12 inches.


Solution

We know the repeating section is m
ade of 2 red lights and 3 green lights. The 3rd
red light would appear in the 2nd section of
this pattern, and
the 21st red light would
appear in the 11th section. There would then be a total of 44 lights in between the
3rd and 21st red light, translating to 45 6-inch gaps. Since it wants the answer in
feet, so the answer is
since it wants the answer in feet.

Problem 8

A dessert chef prepares the dessert for every day of a week starting with Sunday.
The dessert each day is either cake, pie, ice cream
, or pudding. The sam
e dessert
may not be served two days in a row. There m
ust be cake on Friday because of a
birthday. How many different dessert m
enus for the week are possible?


Solution
We can count the num
ber of possible foods for each day and then m
ultiply to
enumerate the number of com
binations.

On Friday, we have one possibility: cake.

On Saturday, we have three possibilities: pie, ice cream
, or pudding. This is the end

of the week.

On Thursday, we have three possibilities: pie, ice cream
, or pudding. We can't have
cake because we have to have cake the following day, which is the Friday with the
birthday party.

On Wednesday, we have three possibilities: cake, plus the two things that were not
eaten on Thursday.

Similarly, on Tuesday, we have three possibilities: the three things that were not
eaten on Wednesday.

Likewise on Monday: three possibilities, the three things that were not eaten on
Tuesday.

On Sunday, it is tempting to think there are four possibilities, but rem
ember that
cake m
ust be served on Friday. This serves to limit the number of foods we can eat
on Sunday, with the result being that there are three possibilities: The three things
that were not eaten on Monday.

So the num
ber of m
enus is
The answer is

Problem 9

It takes Clea 60 seconds to walk down an escalator when it is not m
oving, and 24
seconds when it is m
oving. How seconds would it take Clea to ride the escalator
down when she is not walking?


Solution
She walks at a rate of
units per second to travel a distance
. As
, we find
and
,
where
is the speed of the escalator. Setting the two equations equal
to each other,
, which m
eans that
. Now we divide
by
because you add the speed of the escalator but
remove the walking, leaving the final answer that it takes
to ride the escalator alone
as

Problem 10

What is the area of the polygon whose vertices are the points of intersection of the
curves
and
?


Solution
The first curve is a circle with radius 5 centered at the origin, and the second curve
is an ellipse with center (4,0) and end points of (-5,0) and (13,0). Finding points of
intersection, we get (-5,0) (4,3) and (4,-3), forming a triangle with height of 9 and
base of 6. So 9x6x0.5 =27 B.

Problem 11

In the equation below,
and
are consecutive positive
integers, and
,
, and
represent number
bases:
What is
?

Solution
Change the equation to base 10:

Either
or
, so either
or
.
The second case has no integer roots, and the first can be re-expressed as
. Since A must be positive,
and
; C.

Problem 12

How many sequences of zeros and ones of length 20 have all the zeros consecutive,
or all the ones consecutive, or both?


Solution 2
Consider the 20 term
sequence of 0's and 1's. Keeping all other terms 1, a sequence
of
consecutive 0's can be placed in
locations. That is,
there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with
3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are
strings with consecutive zeros. The sam
e argument shows there are
strings with consecutive 1's. This yields
strings in all.
However, we have counted
twice those strings in which all the 1's and all the 0's are
consecutive. These are the cases
,
,
, ...,
(of which there are 19) as well as the cases
,
,
, ...,
(of which there
are 19 as well). This yields
so that the answer is
.

Problem 13

Two parabolas have equations
and
, where
,
,
, and
are integers,
each
chosen independently by rolling a fair six-sided die. What is the probability
that
the parabolas will have a least one point in common?


Solution 1
Set the two equations equal to each other:
. Now remove the
x squared and get x's on one side:
. Now factor
:
.
If
a
cannot
equal
,
then
there
is
always
a
solution,
but
if
,
a
in
chance,
leaving a
out
, always having at least one point
in common. And if
, then the only way for that to work, is if
,
a
in
chance,
however,
this
can
occur
ways, so a
in
chance of this
happening.
So
adding
one
sixth
to
,
we
get
the
simplified
fraction of
; answer
.
Solution 2
Proceed as above to obtain
. The probability that the parabolas have
at least 1 point in common is 1 minus the probability that
they do not intersect. The
equation
has no solution if and only if
and
. The probability that
is
while the
probability that
is
. Thus we have
for
the probability that the parabolas intersect.

Problem 14

Bernardo and Silvia play the following game. An integer between 0 and 999
inclusive is selected and given to Bernardo. Whenever Bernardo receives a number,
he doubles it and passes
the result
to Silvia. Whenever Silvia receives a num
ber, she
addes 50 to it and passes the result to Bernardo. The winner is the last person who
produces a number less than 1000. Let
N
be the smallest initial number that results
in a win for Bernardo. What is the sum
of the digits of
N
?


Solution
Solution 1

The last num
ber that Bernado says has to be between 950 and 999. Note that
1->2-
>
52-
>
1 04-
>154-
>
308-
>358-
>
716-
>
776 contains 4 doubling actions. Thus,
we have
.
Thus,
. Then,
. If
, we have
. Working backwards from
956,

.

So the starting number is 16, and our answer is
, which is A.

Solution 2

Work backwards. The last num
ber Bernardo produces m
ust be in the range
. That m
eans that before this, Silvia must produce a num
ber in the
range
. Before this, Bernardo must produce a num
ber in the range
. Before this, Silvia must produce a number in the range
. Before this, Bernardo must produce a num
ber in the range
. Before this, Silvia must produce a number in the range
. Before this, Bernardo must produce a num
ber in the range
. Before this, Silvia must produce a number in the range
. Bernardo could not have added 50 to any number before this to
obtain a number in the range
, hence the minimu
m

is
16 with the sum
of digits being
.
Problem 15

Jesse cuts a
circular paper disk of radius 12 along two radii to form
two sectors, the
sm
aller having a central angle of 120 degrees. He m
akes two circular cones, using
each sector to form
the lateral surface of a cone. What is the ratio of the volum
e of
the sm
aller cone to that of the larger?


Solution
If the original radius is
, then the circum
ference is
;
since arcs are defined by the central angles, the sm
aller arc,
degree
angle, is half the size of the larger sector. so the sm
aller arc is
, and
the larger is
. Those two arc lengths becom
e the two circum
ferences
of the new cones; so the radius of the sm
aller cone is
and the larger
cone is
. Using the Pythagorean theorem, the height of the larger
cone is
and the sm
aller cone is
, and now for volume
just square the radii and multiply by
of the height to get the volum
e
of each cone:
and
[both m
ultiplied by three as ratio
com
e out the sam
e. now divide the volumes by each other to get the final ratio of

Problem 16

Am
y, Beth, and Jo listen to four different songs and discuss which ones they like. No
song is liked by all three. Furthermore, for each of the three pairs of the girls, there
is at least one song liked by those girls but disliked by the third. In how many
different ways is this possible?


Solution
Let the ordered triple
denote that
songs are liked by
Am
y and Beth,
songs by Beth and Jo, and
songs by Jo
and Am
y. We claim that the only possible triples are
.
To show this, observe these are all valid conditions. Second, note that none of
can be bigger than 3. Suppose otherwise, that
.
Without loss of generality, say that Am
y and Beth like songs 1, 2, and 3. Then
because there is at least one song liked by each pair of girls, we require either
or
to be at least 1. In fact, we require either
or
to equal 1, otherwise there will be a song liked by all
three. Suppose
. Then we m
ust have
since no song is
liked by all three girls, a contradiction.

Case 1
:How many ways are there for
to equal
? There
are 4 choices for which song is liked by Am
y and Beth, 3 choices for which song is
liked by Beth and Jo, and 2 choices
for which song is liked by Jo and Am
y. The fourth
song can be liked by only one of the girls, or none of the girls, for a total of 4
choices.
So
in
ways.

Case 2
:To find the number of ways for
, observe there are
choices of songs for the first pair of girls. There remain 2 choices of
songs for the next pair (who only like one song). The last song is given to the last
pair of girls. But observe that we let any three pairs of the girls like two songs, so we
multiply by 3. In this case there are
songs.

ways for the girls to like the
That gives a total of
ways for the girls to like the song, so the answer
is
.

Problem 17

Square
lies in the first quadrant. Points
and
lie on lines
and
, respectively. What is
the sum
of the coordinates of the center of the square
?


Solution 1
Let the four points be labeled
,
,
, and
, respectively. Let the lines that go through each point be labeled
,
,
, and
, respectively.
Since
and
go through
and
, respectively, and
and
are opposite
sides of the square, we can say that
and
are parallel
with slope
. Similarly,
and
have slope
. Also, note that since square
lies in the first quadrant,
and
must have a positive slope. Using the point
-slope
form
, we can now
find the equations of all four lines:
,
,
,

.
Since
is a square, it follows that
between points
and
is equal to
between points
and
. Our approach will be to find
and
in terms of
and equate the two to solve for
.
and
intersect at point
.
Setting the equations for
and
equal to each other and
solving for
, we find that they intersect at
.
and
intersect at point
. Intersecting the
two equations, the
-coordinate of point
is found to be
. Subtracting the two, we get
. Substituting the
-coordinate for point
found above into the equation for
, we find that
the
-coordinate of point
is
.
and
intersect at point
.
Intersecting the two equations, the
-coordinate of point
is found to be
. Subtracting the two, we get
. Equating
and
, we get
which gives us
. Finally, note that the line which goes though the
midpoint of
and
with slope
and the line
which goes through the m
idpoint of
and
with slope
must intersect at at
the
center of
the square. The equation of the line
going through
is given by
and the equation of the line
going through
is
. Equating the two, we find that they
intersect at
. Adding the
and
-coordinates, we get
. Thus, answer choice
is correct.

Solution 2

Note that the center of the square lies along a line that has an
intercept of
, and also along another line with
intercept
. Since these 2 lines are parallel to the sides of the square,
they are perpindicular (since the sides of a square are). Let
be the
slope of the first line. Then
is the slope of the second line. We m
ay
use the point-slope form
for the equation of a line to write
and
. We easily calculate the intersection of these lines using substitution
or elimination to obtain
as the center or the square. Let
denote the (acute) angle formed by
and the
axis. Note that
. Let
denote the side
length of the square. Then
. On the other hand the acute angle
form
ed by
and the
axis is
so that
. Using
(for acute
) we have
where upon
. Then
. Substituting into
we obtain
so that the sum
of the coordinates is
. Hence the answer is
.
Solution 3 (Fast)

Suppose


where
.
Recall that the distance between two parallel lines
and
is
, we have distance between
and
equals to
, and the distance between
and
equals to
. Equating them
, we get
.

Then, the center of the square is just the intersection between the following two



The solution is
, so we get the answer
.
.

Problem 18

Let
be a list of the first 10 positive integers such that for each
either
or
or both appear som
ewhere
before
in the list. How m
any such lists are there?


Solution

Let
. Assum
e that
. If
,
the first num
ber
appear after
that is greater than
must be
, otherwise if it is any number
larger than