## 2010AMC10B真题

**答案解析请参考文末**

## Problem 1

What is ?

## Problem 2

Makarla attended two meetings during her -hour work day. The first meeting took minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

## Problem 3

A drawer contains red, green, blue, and white socks with at least 2 of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?

## Problem 4

For a real number , define to be the average of and . What is ?

## Problem 5

A month with days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?

## Problem 6

A circle is centered at , is a diameter and is a point on the circle with . What is the degree measure of ?

## Problem 7

A triangle has side lengths , , and . A rectangle has width and area equal to the area of the triangle. What is the perimeter of this rectangle?

## Problem 8

A ticket to a school play cost dollars, where is a whole number. A group of 9th graders buys tickets costing a total of , and a group of 10th graders buys tickets costing a total of . How many values for are possible?

## Problem 9

Lucky Larry's teacher asked him to substitute numbers for , , , , and in the expression and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for , , , and were , , , and , respectively. What number did Larry substitute for ?

## Problem 10

Shelby drives her scooter at a speed of miles per hour if it is not raining, and miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of miles in minutes. How many minutes did she drive in the rain?

## Problem 11

A shopper plans to purchase an item that has a listed price greater than and can use any one of the three coupons. Coupon A gives off the listed price, Coupon B gives off the listed price, and Coupon C gives off the amount by which the listed price exceeds .

Let and be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or Coupon C. What is ?

## Problem 12

At the beginning of the school year, of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and answered "No." At the end of the school year, answered "Yes" and answered "No." Altogether, of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of ?

## Problem 13

What is the sum of all the solutions of ?

## Problem 14

The average of the numbers and is . What is ?

## Problem 15

On a -question multiple choice math contest, students receive points for a correct answer, points for an answer left blank, and point for an incorrect answer. Jesse’s total score on the contest was . What is the maximum number of questions that Jesse could have answered correctly?

## Problem 16

A square of side length and a circle of radius share the same center. What is the area inside the circle, but outside the square?

## Problem 17

Every high school in the city of Euclid sent a team of students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed th and th, respectively. How many schools are in the city?

## Problem 18

Positive integers , , and are randomly and independently selected with replacement from the set . What is the probability that is divisible by ?

## Problem 19

A circle with center has area . Triangle is equilateral, is a chord on the circle, , and point is outside . What is the side length of ?

## Problem 20

Two circles lie outside regular hexagon . The first is tangent to , and the second is tangent to . Both are tangent to lines and . What is the ratio of the area of the second circle to that of the first circle?

## Problem 21

A palindrome between and is chosen at random. What is the probability that it is divisible by ?

## Problem 22

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

## Problem 23

The entries in a array include all the digits from through , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

## Problem 24

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?

## Problem 25

Let , and let be a polynomial with integer coefficients such that

, and

.What is the smallest possible value of ?

**扫码免费领取最新年份真题及解析**

- .
- The percentage of her time spent in meetings is the total amount of time spent in meetings divided by the length of her workday.The total time spent in meetings is Therefore, the percentage is
- After you draw socks, you can have one of each color, so (according to the pigeonhole principle), if you pull then you will be guaranteed a matching pair.
- The average of two numbers, and , is defined as . Thus the average of and would be. With that said, we need to find the sum when we plug, , and into that equation. So:
- In this month there are four weeks and three remaining days. As long as the last three days and the first four days have the same number of Mondays and Wednesdays, then it works. The number of days the month can start on is
- Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since is the center, and are radii and they are congruent. Thus, is an isosceles triangle. Also, note that and are supplementary, then . Since is isosceles, then . They also sum to , so each angle is .
- The triangle is isosceles. The height of the triangle is therefore given by Now, the area of the triangle is We have that the area of the rectangle is the same as the area of the triangle, namely . We also have the width of the rectangle: .The length of the rectangle therefore is: The perimeter of the rectangle then becomes: The answer is:
- We see how many common integer factors and share. Of the factors of - ; only and are factors of . So there are possibilities for the ticket price.
- Simplify the expression .So you get: Larry substituted with respectively.We have to find the value of , such that (the same expression without parenthesis).Substituting and simplifying we get: So, Larry must have used the value for .Our answer is
- We know that Since we know that she drove both when it was raining and when it was not and that her total distance traveled is miles.We also know that she drove a total of minutes which is of an hour.We get the following system of equations, where is the time traveled when it was not raining and is the time traveled when it was raining:Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:We know now that the time traveled in rain was of an hour, which is minutesSo, our answer is
- Let the listed price be , where Coupon A saves us: Coupon B saves us: Coupon C saves us: Now, the condition is that A has to be greater than or equal to either B or C which give us the following inequalities:We see here that the greatest possible value for is , thus and the smallest value for p is so .The difference between and is
- The minimum possible value occurs when of the students who originally answered "No." answer "Yes." In this case, The maximum possible value occurs when of the students (which accounts for of the overall student population) who originally answered "Yes." answer "No." and the of the students (which accounts for of the overall student population) who originally answered "No." answer "Yes." In this case, Subtract to obtain an answer of
- We evaluate this in cases:
*Case 1*When we are going to have . When we are going to have and when we are going to have . Therefore we have .*Subcase 1*When we are going to have . When this happens, we can express as . Therefore we get . We check if is in the domain of the numbers that we put into this subcase, and it is, since . Therefore is one possible solution.*Subcase 2*When we are going to have , therefore can be expressed in the form. We have the equation . Since is less than , is another possible solution.*Case 2*: When , . When we can express this in the form . Therefore we have . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we haveWe have now evaluated all the cases, and found the solution to be which have a sum of

- We must find the average of the numbers from to and in terms of . The sum of all these terms is . We must divide this by the total number of terms, which is . We get: . This is equal to , as stated in the problem. We have: . We can now cross multiply. This gives
- Let be the amount of questions Jesse answered correctly, be the amount of questions Jesse left blank, and be the amount of questions Jesse answered incorrectly. Since there were questions on the contest, . Since his total score was , . Also, . We can substitute this inequality into the previous equation to obtain another inequality: . Since is an integer, the maximum value for is .
- The radius of the circle is . Half the diagonal of the square is . We can see that the circle passes outside the square, but the square is NOT completely contained in the circle Therefore the picture will look something like this:Then we proceed to find: 4 * (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits in half. Let this half-length be . Also note that because it is half the sidelength of the square. Because this is a right triangle, we can use thePythagorean Theorem to solve for Solving, and . Since , is an equilateral triangle and the central angle is . Therefore the sector has an area .Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which isPutting it together, we get the answer to be
- Let the be the number of schools, be the number of contestants, and be Andrea's place. Since the number of participants divided by three is the number of schools, . Andrea received a higher score than her teammates, so . Since is the maximum possible median, then is the maximum possible number of participants. Therefore, . This yields the compound inequality: . Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, cannot be even. is the only other option.
- First we factor as , so in order for the number to be divisible by 3, either is divisible by , or is divisible by .We see that is divisible by with probability . We only need to calculate the probability that is divisible by .We need or . Using some modular arithmetic, and or and . The both cases happen with probability so the total probability is .Then the answer is or .
- The formula for the area of a circle is so the radius of this circle is Because must be in the interior of circle Let be the unknown value, the sidelength of the triangle, and let be the point on where Since is equilateral, and We are given Use the Pythagorean Theorem and solve for
- A good diagram is very helpful. Also, BC and FA are LINES, not LINE SEGMENTS.The first circle is in red, the second in blue. With this diagram, we can see that the first circle is inscribed in equilateral triangle while the second circle is inscribed in . From this, it's evident that the ratio of the blue area to the red area is equal to the
**ratio of the areas to**Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is . From the diagram, we can see that this is - The palindromes can be expressed as: (since it is a four digit palindrome, it must be of the form , where x and y are integers from and , respectively.)We simplify this to .Because the question asks for it to be divisible by 7,We express it as .Because ,We can substitute for We are left with Since we can simplify the in the expression to.In order for this to be true, must also be true.
Thus we solve:

Which has two solutions: and

There are thus two options for out of the 10, so the answer is

- We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.Each candy has three choices; it can go in any of the three bags.Since there are seven candies, that makes the
**total distributions**To find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.The number of distributions such that the red bag is empty is equal to , since it's equivalent to distributing the candies into bags.We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also .The case where both the red and the blue bags are empty (all candies are in the white bag) are included in both of the above calculations, and this case has only distribution.**The total overcount is**The final answer will be - The upper-left corner must contain the entry 1, and similarly the lower-right corner must contain the entry 9. Consider the entries 2 and 3 -- they may either both lie in the first row, both lie in the first column, or lie in the two squares neighboring 1. By symmetry (which we will take into account by a factor of 2 in the end), we may assume that 2 lies in the cell to the right of 1 and that 3 lies either in the cell to the right of 2 or in the cell below 1:Similarly, the entries 7 and 8 may either both lie in the last row, or lie in the two squares neighboring 9. This gives the following cases:where the notation denotes two possible cases, either by switching a row and column or by switching the 7 and 8. Finally, there are respectively 1, 2, 2, 6 ways to complete these four cases. This gives a total ofpossible ways to fill the diagram.
**Notes**In fact, there is a general formula (coming from the fields of combinatorics and representation theory) to answer problems of this form; it is known as the hook-length formula. - Represent the teams' scores as: and We have Factoring out the from the left side of the equation, we can get , or Since both are increasing sequences, . We can check cases up to because when , we get . When
Checking each of these cases individually back into the equation , we see that only when and , we get an integer value for , which is . The original question asks for the first half scores summed, so we must find

- We observe that because , if we define a new polynomial such that , has roots when ; namely, when .Thus since has roots when , we can factor the product out of to obtain a new polynomial such that .Then, plugging in values of we get Thus, the least value of must be the . Solving, we receive , so our answer is .To complete the solution, we can let , and then try to find . We know from the above calculation that , and . Then we can let , getting . Let , then . Therefore, it is possible to choose , so the goal is accomplished. As a reference, the polynomial we get is