2024年4月11日发(作者：河北小升初数学试卷冀教版)

2004 AMC 10B

Problem 1

Each row of the Misty Moon Amphitheater has 33 seats. Rows 12

through 22 are reserved for a youth club. How many seats are

reserved for this club?

Solution

There are

Problem 2

How many two-digit positive integers have at least one 7 as a digit?

Solution

Ten numbers () have as the tens digit. Nine numbers

() have it as the ones digit. Number is in both sets.

Thus the result is

Problem 3

At each basketball practice last week, Jenny made twice as many free

throws as she made at the previous practice. At her fifth practice she

made 48 free throws. How many free throws did she make at the first

practice?

Solution

At the fourth practice she made

was , then we get

and finally

rows of seats, giving seats.

.

throws, at the third one it

throws for the second practice,

throws at the first one.

- 1 -

Problem 4

A standard six-sided die is rolled, and P is the product of the five

numbers that are visible. What is the largest number that is certain to

divide P?

Solution 1

The product of all six numbers is . The products of numbers

that can be visible are , , ..., . The answer to this

problem is their greatest common divisor -- which is , where is

the least common multiple of . Clearly and the

answer is

Solution 2

Clearly, can not have a prime factor other than , and .

We can not guarantee that the product will be divisible by , as the

number can end on the bottom.

We can guarantee that the product will be divisible by (one of and

will always be visible), but not by .

Finally, there are three even numbers, hence two of them are always

visible and thus the product is divisible by . This is the most we can

guarantee, as when the is on the bottom side, the two visible even

numbers are and , and their product is not divisible by .

Hence

Solution

Problem 5

In the expression , the values of , , , and are , , ,

and , although not necessarily in that order. What is the maximum

possible value of the result?

Solution

- 2 -

.

.

If or , the expression evaluates to

If , the expression evaluates to .

Case remains.

.

In that case, we want to maximize where .

Trying out the six possibilities we get that the best one is

, where

Problem 6

Which of the following numbers is a perfect square?

Solution

Using the fact that

.

, we can write:

Clearly is a square, and as , , and are primes,

none of the other four are squares.

Problem 7

On a trip from the United States to Canada, Isabella took U.S.

dollars. At the border she exchanged them all, receiving Canadian

dollars for every U.S. dollars. After spending Canadian dollars,

she had Canadian dollars left. What is the sum of the digits of ?

Solution

Solution 1

- 3 -

Isabella had Canadian dollars. Setting up an equation we get

, which solves to , and the sum of digits of is

Solution 2

Each time Isabelle exchanges U.S. dollars, she gets Canadian

dollars and Canadian dollars extra. Isabelle received a total of

Canadian dollars extra, therefore she exchanged U.S. dollars

times. Thus

Problem 8

Minneapolis-St. Paul International Airport is 8 miles southwest of

downtown St. Paul and 10 miles southeast of downtown Minneapolis.

Which of the following is closest to the number of miles between

downtown St. Paul and downtown Minneapolis?

Solution

The directions \"southwest\" and \"southeast\" are orthogonal. Thus the

described situation is a right triangle with legs 8 miles and 10 miles

long. The hypotenuse length is

is .

.

, and thus the answer

.

Without a calculator one can note that

Problem 9

A square has sides of length 10, and a circle centered at one of its

vertices has radius 10. What is the area of the union of the regions

enclosed by the square and the circle?

Solution

The area of the circle is , the area of the square is .

- 4 -

Exactly of the circle lies inside the square. Thus the total area is

.

Problem 10

A grocer makes a display of cans in which the top row has one can and

each lower row has two more cans than the row above it. If the display

contains cans, how many rows does it contain?

Solution

The sum of the first odd numbers is

have

Problem 11

Two eight-sided dice each have faces numbered 1 through 8. When

the dice are rolled, each face has an equal probability of appearing on

the top. What is the probability that the product of the two top

numbers is greater than their sum?

Solution

Solution 1

We have , hence if at least one of the numbers is ,

the sum is larger. There such possibilities.

We have .

, hence all other cases

the sum is

it is smaller.

.

. As in our case , we

For we already have

are good.

Out of the possible cases, we found that in

greater than or equal to the product, hence in

Therefore the answer is

Solution 2

- 5 -

.

Let the two rolls be

From the restriction:

Since

either

and

, and .

are non-negative integers between and

,

if and only if or

with

and

, or

.

, ordered pairs with

. So, there are

.

and

,

There are ordered pairs

and ordered pair with

ordered pairs such that

,

if and only if

and

or equivalently

. . This gives ordered pair

So, there are a total of

.

Since there are a total of

ordered pairs with

ordered pairs with

ordered pairs

.

, there are

Thus, the desired probability is

Problem 12

.

An annulus is the region between two concentric circles. The

concentric circles in the figure have radii and , with . Let

be a radius of the larger circle, let be tangent to the smaller

circle at , and let be the radius of the larger circle that contains

. Let , , and . What is the area of the annulus?

- 6 -

Solution

The area of the large circle is

hence the shaded area is

, the area of the small one is

.

we have

.

,

From the Pythagorean Theorem for the right triangle

, hence

Problem 13

and thus the shaded area is

In the United States, coins have the following thicknesses: penny,

mm; nickel, mm; dime, mm; quarter, mm. If a

stack of these coins is exactly mm high, how many coins are in the

stack?

Solution

All numbers in this solution will be in hundreds of a millimeter.

The thinnest coin is the dime, with thickness

has height .

- 7 -

. A stack of dimes

The other three coin types have thicknesses , , and

. By replacing some of the dimes in our stack by other, thicker

coins, we can clearly create exactly all heights in the set

.

If we take an odd , then all the possible heights will be odd, and thus

none of them will be . Hence is even.

If the stack will be too low and if

we are left with cases and .

If the possible stack heights are

remaining ones exceeding .

Therefore there are coins in the stack.

it will be too high. Thus

, with the

Using the above observation we can easily construct such a stack. A

stack of dimes would have height , thus we need to add

. This can be done for example by replacing five dimes by nickels

(for ), and one dime by a penny (for ).

Problem 14

A bag initially contains red marbles and blue marbles only, with more

blue than red. Red marbles are added to the bag until only of the

marbles in the bag are blue. Then yellow marbles are added to the bag

until only of the marbles in the bag are blue. Finally, the number of

blue marbles in the bag is doubled. What fraction of the marbles now

in the bag are blue?

Solution

We can ignore most of the problem statement. The only important

information is that immediately before the last step blue marbles

formed of the marbles in the bag. This means that there were blue

and other marbles, for some . When we double the number of

- 8 -

blue marbles, there will be

marbles now form

Problem 15

blue and other marbles, hence blue

of all marbles in the bag.

Patty has coins consisting of nickels and dimes. If her nickels were

dimes and her dimes were nickels, she would have cents more.

How much are her coins worth?

Solution

Solution 1

She has nickels and dimes. Their total cost is

cents. If the dimes were nickels

and vice versa, she would have

cents. This value should be cents more than the previous one. We

get , which solves to . Her coins are

worth .

Solution 2

Changing a nickel into a dime increases the sum by cents, and

changing a dime into a nickel decreases it by the same amount. As the

sum increased by cents, there are more nickels than

nickels dimes. As the total count is

and dimes.

Problem 16

, this means that there are

Three circles of radius are externally tangent to each other and

internally tangent to a larger circle. What is the radius of the large

circle?

Solution

- 9 -

The situation in shown in the picture below. The radius we seek is

. Clearly . The point is clearly the center of the

equilateral triangle , thus is of the altitude of this triangle.

We get that . Therefore the radius we seek is

.

WARNING. Note that the answer does not correspond to any of the

five options. Most probably there is a typo in option D.

Problem 17

The two digits in Jack\'s age are the same as the digits in Bill\'s age, but

in reverse order. In five years Jack will be twice as old as Bill will be

then. What is the difference in their current ages?

Solution

Solution 1

- 10 -

If Jack\'s current age is

.

In five years, Jack\'s age will be

.

We are given that

, then Bill\'s current age is

and Bill\'s age will be

. Thus .

For we get

integer, and for

. For and the value is not an

it is more than . Thus the only solution is

. , and the difference in ages is

Solution 2

Age difference does not change in time. Thus in five years Bill\'s age

will be equal to their age difference.

The age difference is , hence it is a

multiple of . Thus Bill\'s current age modulo must be .

Thus Bill\'s age is in the set .

As Jack is older, we only need to consider the cases where the tens

digit of Bill\'s age is smaller than the ones digit. This leaves us with the

options .

Checking each of them, we see that only

solution

Problem 18

In the right triangle , we have , , and .

Points , , and are located on , , and , respectively, so

that , , and . What is the ratio of the area of

to that of ?

.

works, and gives the

- 11 -


更多推荐

河北,数学试卷,作者