2009 AMC 10B 试题及答案解析

1、2009 amc 10b1 、each morning of her five-day workweek, jane bought either a 50-cent muffin or a 75-cent bagel. her total cost for the week was a whole number of dollars, how many bagels did she buy? solution the only combination of five items with total cost a whole number of dollars is 3 muffins and

2、 bagels. the answer is .2 、which of the following is equal to ? solution multiplying the numerator and the denumerator by the same value does not change the value of the fraction. we can multiply both by , getting . alternately, we can directly compute that the numerator is , the denumerator is , an

3、d hence their ratio is . 3 、paula the painter had just enough paint for identically sized rooms. unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for rooms. how many cans of paint did she use for the rooms? solution losing three cans of paint co

4、rresponds to being able to paint five fewer rooms. so . the answer is .4 、a rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. the remainder of the yard has a trapezoidal shape, as shown. the parallel sides of the trapezoid have lengths and meters. what fr

5、action of the yard is occupied by the flower beds? solution each triangle has leg length meters and area square meters. thus the flower beds have a total area of 25 square meters. the entire yard has length 25 m and width 5 m, so its area is 125 square meters. the fraction of the yard occupied by th

6、e flower beds is . the answer is .5 、twenty percent less than 60 is one-third more than what number? solution twenty percent less than 60 is . one-third more than a number n is . therefore and the number is . the answer is .6 、kiana has two older twin brothers. the product of their three ages is 128

7、. what is the sum of their three ages? solution the age of each person is a factor of . so the twins could be years of age and, consequently kiana could be 128, 32, 8 or 2 years old, respectively. because kiana is younger than her brothers, she must be 2 years old. so the sum of their ages is . the

8、answer is .7 、by inserting parentheses, it is possible to give the expression several values. how many different values can be obtained? solution the three operations can be performed on any of orders. however, if the addition is performed either first or last, then multiplying in either order produ

9、ces the same result. so at most four distinct values can be obtained. it is easy to check that the values of the four expressions are in fact all distinct. so the answer is , which is choice .8 、in a certain year the price of gasoline rose by during january, fell by during february, rose by during m

10、arch, and fell by during april. the price of gasoline at the end of april was the same as it had been at the beginning of january. to the nearest integer, what is solution let be the price at the beginning of january. the price at the end of march was because the price at the of april was , the pric

11、e decreased by during april, and the percent decrease was so to the nearest integer is . the answer is .9 、segment and intersect at , as shown, , and . what is the degree measure of ? solution is isosceles, hence . the sum of internal angles of can now be expressed as , hence , and each of the other

12、 two angles is . now we know that . finally, is isosceles, hence each of the two remaining angles ( and ) is equal to . 10 、a flagpole is originally meters tall. a hurricane snaps the flagpole at a point meters above the ground so that the upper part, still attached to the stump, touches the ground

13、meter away from the base. what is ? solution the broken flagpole forms a right triangle with legs and , and hypotenuse . the pythagorean theorem now states that , hence , and . (note that the resulting triangle is the well-known right triangle, scaled by .) 11 、how many -digit palindromes (numbers t

14、hat read the same backward as forward) can be formed using the digits , , , , , , ? solution a seven-digit palindrome is a number of the form . clearly, must be , as we have an odd number of fives. we are then left with . each of the permutations of the set will give us one palindrome.12、distinct po

15、ints , , , and lie on a line, with . points and lie on a second line, parallel to the first, with . a triangle with positive area has three of the six points as its vertices. how many possible values are there for the area of the triangle? solution consider the classical formula for triangle area: .

16、 each of the triangles that we can make has exactly one side lying on one of the two parallel lines. if we pick this side to be the base, the height will always be the same - it will be the distance between the two lines. hence each area is uniquely determined by the length of the base. and it can e

17、asily be seen, that the only possible base lengths are , , and . therefore there are only possible values for the area. (to be more precise in the last step, the possible base lengths are , , and .) 13 、as shown below, convex pentagon has sides , , , , and . the pentagon is originally positioned in

18、the plane with vertex at the origin and vertex on the positive -axis. the pentagon is then rolled clockwise to the right along the -axis. which side will touch the point on the -axis? solution the perimeter of the polygon is . hence as we roll the polygon to the right, every units the side will be t

19、he bottom side. we have . thus at some point in time we will get the situation when and is the bottom side. obviously, at this moment . after that, the polygon rotates around until point hits the axis at . and finally, the polygon rotates around until point hits the axis at . at this point the side

20、touches the point . 14 、on monday, millie puts a quart of seeds, of which are millet, into a bird feeder. on each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. each day the birds eat only of the millet in the feeder, but they eat all of the

21、other seeds. on which day, just after millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet? solution on monday, day 1, the birds find quart of millet in the feeder. on tuesday they find quarts of millet. on wednesday, day 3, they find quarts of mill

22、et. the number of quarts of millet they find on day is the birds always find quart of other seeds, so more than half the seeds are millet if , that is, when . because and , this will first occur on day which is . the answer is .15 、when a bucket is two-thirds full of water, the bucket and water weig

23、h kilograms. when the bucket is one-half full of water the total weight is kilograms. in terms of and , what is the total weight in kilograms when the bucket is full of water? solution solution 1 let be the weight of the bucket and let be the weight of the water in a full bucket. then we are given t

24、hat and . hence , so . thus . finally . the answer is . solution 2 imagine that we take three buckets of the first type, to get rid of the fraction. we will have three buckets and two buckets worth of water. on the other hand, if we take two buckets of the second type, we will have two buckets and e

25、noung water to fill one bucket. the difference between these is exactly one bucket full of water, hence the answer is . solution 3 we are looking for an expression of the form . we must have , as the desired result contains exactly one bucket. also, we must have , as the desired result contains exac

26、tly one bucket of water. at this moment, it is easiest to check that only the options (a), (b), and (e) satisfy , and out of these only (e) satisfies the second equation. alternately, we can directly solve the system, getting and . 16 、points and lie on a circle centered at , each of and are tangent

27、 to the circle, and is equilateral. the circle intersects at . what is ? solution solution 1 as is equilateral, we have , hence . then , and from symmetry we have . finally this gives us . we know that , as lies on the circle. from we also have , hence , therefore , and . solution 2 as in the previo

28、us solution, we find out that . hence and are both equilateral. we then have , hence is the incenter of , and as is equilateral, is also its centroid. hence , and as , we have , therefore , and as before we conclude that . 17、five unit squares are arranged in the coordinate plane as shown, with the

29、lower left corner at the origin. the slanted line, extending from to , divides the entire region into two regions of equal area. what is ? solution solution 1 for the shaded area is at most , which is too little. hence , and therefore the point is indeed inside the shaded part, as shown in the pictu

30、re. then the area of the shaded part is one less than the area of the triangle with vertices , , and . its area is obviously , therefore the area of the shaded part is . the entire figure has area , hence we want the shaded part to have area . solving for , we get . the answer is . solution 2 the to

31、tal area is 5, so the area of the shaded area is . if we add a unit square in the lower right corner, the area is . therefore , or . therefore . 18 、rectangle has and . point is the midpoint of diagonal , and is on with . what is the area of ? solution by the pythagorean theorem we have , hence . th

32、e triangles and have the same angle at and a right angle, thus all their angles are equal, and therefore these two triangles are similar. the ratio of their sides is , hence the ratio of their areas is . and as the area of triangle is , the area of triangle is . 19 、a particular -hour digital clock

33、displays the hour and minute of a day. unfortunately, whenever it is supposed to display a , it mistakenly displays a . for example, when it is 1:16 pm the clock incorrectly shows 9:96 pm. what fraction of the day will the clock show the correct time? solution solution 1 the clock will display the i

34、ncorrect time for the entire hours of and . so the correct hour is displayed of the time. the minutes will not display correctly whenever either the tens digit or the ones digit is a , so the minutes that will not display correctly are and and . this amounts to fifteen of the sixty possible minutes

35、for any given hour. hence the fraction of the day that the clock shows the correct time is . the answer is . solution 2 the required fraction is the number of correct times divided by the total times. there are 60 minutes in an hour and 12 hours on a clock, so there are 720 total times. we count the

36、 correct times directly; let a correct time be , where is a number from 1 to 12 and and are digits, where . there are 8 values of that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. there are five values of that will display the correct time: 0, 2, 3, 4, and 5. there are nine values of t

37、hat will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. therefore there are correct times. therefore the required fraction is . 20 、triangle has a right angle at , , and . the bisector of meets at . what is ? solution 、 by the pythagorean theorem, . the angle bisector theorem now yields th

38、at 21 、what is the remainder when is divided by 8? solution solution 1 the sum of any four consecutive powers of 3 is divisible by and hence is divisible by 8. therefore is divisible by 8. so the required remainder is . the answer is . solution 2 we have . hence for any we have , and then . therefor

39、e our sum gives the same remainder modulo as . there are terms in the sum, hence there are pairs , and thus the sum is . 22 、a cubical cake with edge length inches is iced on the sides and the top. it is cut vertically into three pieces as shown in this top view, where is the midpoint of a top edge.

40、 the piece whose top is triangle contains cubic inches of cake and square inches of icing. what is ? solution lets label the points as in the picture above. let be the area of . then the volume of the corresponding piece is . this cake piece has icing on the top and on the vertical side that contain

41、s the edge . hence the total area with icing is . thus the answer to our problem is , and all we have to do now is to determine . solution 1 introduce a coordinate system where , and . in this coordinate system we have , and the line has the equation . as the line is orthogonal to , it must have the

42、 equation for some suitable constant . as this line contains the point , we have . substituting into , we get , and then . we can note that in is the height from onto , hence its area is , and therefore the answer is . solution 2 extend to intersect at : it is now obvious that is the midpoint of . (

43、imagine rotating the square by clockwise around its center. this rotation will map the segment to a segment that is orthogonal to , contains and contains the midpoint of .) from we can compute that . observe that and have the same angles and therefore they are similar. the ratio of their sides is .

44、hence we have , and . knowing this, we can compute the area of as . finally, we compute , and conclude that the answer is . you could also notice that the two triangles in the original figure are similar. solution 3 use trigonometry. the length of and is and respectively. so , and . from the right-a

45、ngled triangle , the hypotenuse, so , and knowing this, . so we proceed as follows: so the answer is . note that we didnt use a calculator, but we used trigonometric identities 23 、rachel and robert run on a circular track. rachel runs counterclockwise and completes a lap every 90 seconds, and rober

46、t runs clockwise and completes a lap every 80 seconds. both start from the same line at the same time. at some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting

47、line. what is the probability that both rachel and robert are in the picture? solution after 10 minutes (600 seconds), rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. because rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.7

48、5 seconds and 41.25 seconds of the tenth minute. after 10 minutes robert will have completed 7 laps and will be 40 seconds past the starting line. because robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. hence both rachel and rob

49、ert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. so the probability that both runners are in the picture is . the answer is .24 、the keystone arch is an ancient architectural feature. it is composed of congruent isosceles trapezoids fitted together along th

50、e non-parallel sides, as shown. the bottom sides of the two end trapezoids are horizontal. in an arch made with trapezoids, let be the angle measure in degrees of the larger interior angle of the trapezoid. what is ? solution extend all the legs of the trapezoids. they will all intersect in the midd

51、le of the bottom side of the picture, forming the situation shown below. each of the angles at has . from , the size of the smaller internal angle of the trapezoid is , hence the size of the larger one is . proof that all the extended trapezoid legs intersect in the same point: it is sufficient to p

52、rove this for any pair of neighboring trapezoids. for two neighboring trapezoids, the situation is symmetric according to their common leg, therefore the extensions of both outside legs intersect the extension of the common leg in the same point, q.e.d. knowing this, we can now easily see that the i

53、ntersection point must be on the bottom side of our picture, as it lies on the bottom leg of the rightmost trapezoid. and by symmetry the point must be in the center of this side. 25 、each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposit

54、e edge. the choice of the edge pairing is made at random and independently for each face. what is the probability that there is a continuous stripe encircling the cube? solution solution 1 there are two possible stripe orientations for each of the six faces of the cube, so there are possible stripe

55、combinations. there are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. in addition, the stripe on each face that does not contribute may be oriented in either of two

56、different ways. so a total of stripe combinations on the cube result in a continuous stripe around the cube. the required probability is . heres another way similar to this: so there are choices for the stripes as mentioned above. now, lets just consider the view point of one of the faces. we can choose any of the 2 orientation for the stripe (it can go from up to down, or from right to left). once that orientation is chosen, each of the other faces that contribute to that loop only have 1 choice, which is to go in the direction of the loop. that gi