AMC10B试题及答案解析

1、2010 amc 10b1 . what is ? solution we first expand the first term, simplify, and then compute to get an answer of .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 me

2、etings? solution 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 3、a drawer contains red, green, blue, and white socks with at least 2 of each color. what

3、 is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair? solution 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.4 、for a real number , define to be the

4、 average of and . what is ? solution 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: . 5 、a month with days has the same number of mondays and wednesdays.how many of the seven day

5、s of the week could be the first day of this month? solution （b）. 36 、a circle is centered at , is a diameter and is a point on the circle with . what is the degree measure of ? solution assuming the reader is not readily capable to understand how will always be right, the i will continue with an ea

6、sily understandable solution. since is the center, are all radii, they are congruent. thus, and are isosceles triangles. also, note that and are supplementary, then . since is isosceles, then . they also sum to , so each angle is .7 、a triangle has side lengths , , and . a rectangle has width and ar

7、ea equal to the area of the rectangle. what is the perimeter of this rectangle? solution 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 48. we also hav

8、e the width of the rectangle: 4. the length of the rectangle therefor is: the perimeter of the rectangle then becomes: the answer is: 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

9、 costing a total of $. how many values for are possible? solution we see how many common integer factors 48 and 64 share. of the factors of 48 - 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48; only 1, 2, 4, 8, and 16 are factors of 64. so there are possibilities for the ticket price.9 、lucky larry's teach

10、er asked him to substitute numbers for , , , , and in the expression and evaluate the result. larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. the number larry sustitued for , , , and were , , , and , respectively. what number did larry

11、substitude for ? solution simplify the expression . i recommend to start with the innermost parenthesis and work your way out. so you get: henry substituted with respectively. we have to find the value of , such that (the same expression without parenthesis). substituting and simplifying we get: so

12、henry must have used the value for . our answer is: 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

13、 drive in the rain? solution 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

14、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 minutes so, our answer is: 11 、a s

15、hopper plans to purchase an item that has a listed price greater than $ and can use any one of the three coupns. 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 pric

16、es, respectively, for which coupon a saves at least as many dollars as coupon b or c. what is ? solution 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

17、inequalities: we see here that the greatest possible value for p is and the smallest is the difference between and is our answer is: 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 answer

18、ed "no." at the end of the school year, answered "yes" and answerws "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 ? solution the minimu

19、m 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 who originally answered "yes." answer "no." and the of the students who originally answered "no.&

20、quot; answer "yes." in this case, subtract to obtain an answer of 13 、what is the sum of all the solutions of ? solution case 1: case 1a: case 1b: case 2: case 2a: case 2b: since an absolute value cannot be negative, we exclude . the answer is 14 、the average of the numbers and is . what i

21、s ? solution 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: this gives us our answe

22、r. 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. jesses total score on the contest was . what is the maximum number of questions that jesse could have answered correctly? solution let

23、 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 equatio

24、n to obtain another inequality: . since is an integer, the maximum value for is .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? solution (b)17 、every high school in the city of euclid sent a team of students to a

25、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? solution let the be the

26、number of schools, be the number of contestants, and be andrea's score. 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.

27、 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.18 、positive integers , , and are randomly and indep

28、endently selected with replacement from the set . what is the probability that is divisible by ? solution （e）13/2719 、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 ? solution (b)620 、two circles lie outside regu

29、lar 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? solution a good diagram here is very helpful. the first circle is in red, the second in blue. with this diagram, we c

30、an 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 red area to the blue area is equal to the ratio of the areas of triangles to since the ratio of areas is equal to the square of the ratio of l

31、engths, we know our final answer is from the diagram, we can see that this is the letter answer is d 21 、a palindrome between and is chosen at random. what is the probability that it is divisible by ? solution view the palindrome as some number with form (decimal representation): . but because the n

32、umber is a palindrome, . recombining this yields . 1001 is divisible by 7, which means that as long as , the palindrome will be divisible by 7. this yields 9 palindromes out of 90 () possibilities for palindromes. however, if , then this gives another case in which the palindrome is divisible by 7.

33、this adds another 9 palindromes to the list, bringing our total to 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? solution we ca

34、n 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 cal

35、culate 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 7 candies into 2 bags. we know that the number of distributions with the blue bag is empty will be the same

36、 number because of the symmetry, so it's also . the case where both the red and the blue bags are empty (all 7 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 that makes the letter choice c 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? solution （d）6024 、a high schoo