AMC10A试题及答案解析

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1、2006 AMC 10A1、Sandwiches at Joes Fast Food cost each and sodas cost each. How many dollars will it cost to purchase 5 sandwiches and 8 sodas? Solution The sandwiches cost dollars. The sodas cost dollars. In total, the purchase costs dollars. The answer is . 2、Define . What is ? Solution By the defin

2、ition of , we have . Then . The answer is .3、The ratio of Marys age to Alices age is . Alice is years old. How old is Mary? Solution Let be Marys age. Then . Solving for , we obtain . The answer is .4、A digital watch displays hours and minutes with and . What is the largest possible sum of the digit

3、s in the display? Solution From the greedy algorithm, we have in the hours section and in the minutes section. 5、Doug and Dave shared a pizza with equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was , and there was an additional

4、 cost of for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug? Solution Dave and Doug paid dollars in total. Doug paid for three slices of plain pizza, wh

5、ich cost . Dave paid dollars. Dave paid more dollars than Doug. The answer is .6、What non-zero real value for satisfies ? Solution 7、The rectangle is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is ? Solutio

6、n Taking the seventh root of both sides, we get . Simplifying the LHS gives , which then simplifies to . Thus, , and the answer is . 8、A parabola with equation passes through the points and . What is ? Solution Substitute the points (2,3) and (4,3) into the given equation for (x,y). Then we get a sy

7、stem of two equations: Subtracting the first equation from the second we have: Then using in the first equation: is the answer. Alternatively, notice that since the equation is that of a monic parabola, the vertex is likely . Thus, the form of the equation of the parabola is . Expanding this out, we

8、 find that . 9、How many sets of two or more consecutive positive integers have a sum of 15? Solution Notice that if the consecutive positive integers have a sum of 15, then their average (which could be a fraction) must be a divisor of 15. If the number of integers in the list is odd, then the avera

9、ge must be either 1, 3, or 5, and 1 is clearly not possible. The other two possibilities both work: If the number of integers in the list is even, then the average will have a . The only possibility is , from which we get: Thus, the correct answer is 3, answer choice . 10、For how many real values of

10、 is an integer? Solution For to be an integer, must be a perfect square. Since cant be negative, . The perfect squares that are less than or equal to are , so there are values for . Since every value of gives one and only one possible value for , the number of values of is . 11、Which of the followin

11、g describes the graph of the equation ? Solution Expanding the left side, we have Thus there are two lines described in this graph, the horizontal line and the vertical line . Thus, our answer is . 12、Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. H

12、is preliminary drawings are shown.Which of these arrangements gives the dog the greater area to roam, and by how many square feet?Solution Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement I allows the dog square

13、 feet of area. Arrangement II allows square feet plus a little more on the top part of the fence. So we already know that Arrangement II allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is . Thu

14、s the answer is . 13、A player pays to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if

15、 the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.) Solution There are possible combinations of 2 dice rolls. The only possible winning combinations are , and . Since there are winning combinations and possible combinations of dice rolls

16、, the probability of winning is . Let be the amount won in a fair game. By the definition of a fair game, . Therefore, . 14、A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less tha

17、n that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? Solution The inside diameters of the rings are the positive integers from 1 to 18. The total distance needed is the sum of these v

18、alues plus 2 for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series, the answer is . Alternatively, the sum of the consecutive integers from 3 to 20 is . However, the 17 intersections between the rings must be subtracted, and we also get

19、. 15、Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many time

20、s after the start do they pass each other? Solution Solution Since , we note that Odell runs one lap in minutes, while Kershaw also runs one lap in minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at

21、 the starting point, the other at the half-way point). Thus, there are laps run by both, or meeting points . 16、A circle of radius 1 is tangent to a circle of radius 2. The sides of are tangent to the circles as shown, and the sides and are congruent. What is the area of ? Solution Note that . Using

22、 the first pair of similar triangles, we write the proportion: By the Pythagorean Theorem we have that . Now using , The area of the triangle is . 17、In rectangle , points and trisect , and points and trisect . In addition, . What is the area of quadrilateral shown in the figure? Solution Solution 1

23、 It is not difficult to see by symmetry that is a square. Draw . Clearly . Then is isosceles, and is a . Hence , and . There are many different similar ways to come to the same conclusion using different 45-45-90 triangles. Solution 2 Draw the lines as shown above, and count the squares. There are 1

24、2, so we have . 18、A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible?

25、 Solution There are ways to choose 4 digits. There are ways to choose the 2 letters. For the letters to be next to each other, they can be the 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or the 5th and 6th characters. So, there are choices for the position of the letters. Therefore, the numb

26、er of distinct license plates is .19、How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression? Solution The sum of the angles of a triangle is degrees. For an arithmetic progression with an odd number of terms, the middle term is equal

27、 to the average of the sum of all of the terms, making it degrees. The minimum possibly value for the smallest angle is and the highest possible is (since the numbers are distinct), so there are possibilities .20、Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What

28、is the probability that some pair of these integers has a difference that is a multiple of 5? Solution For two numbers to have a difference that is a multiple of 5, the numbers must be congruent (their remainders after division by 5 must be the same). are the possible values of numbers in . Since th

29、ere are only 5 possible values in and we are picking numbers, by the Pigeonhole Principle, two of the numbers must be congruent . Therefore the probability that some pair of the 6 integers has a difference that is a multiple of 5 is . 21、How many four-digit positive integers have at least one digit

30、that is a 2 or a 3? Solution Since we are asked for the number of positive 4-digit integers with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits. The total number of 4-digit integers is , since

31、 we have 10 choices for each digit except the first (which cant be 0). Similarly, the total number of 4-digit integers without any 2 or 3 is . Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is 22、Two farmers agree that pigs are

32、worth dollars and that goats are worth dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with change received in the form of goats or pigs as necessary. (For example, a dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of t

33、he smallest positive debt that can be resolved in this way? Solution The problem can be restated as an equation of the form , where is the number of pigs, is the number of goats, and is the positive debt. The problem asks us to find the lowest x possible. p and g must be integers, which makes the eq

34、uation a Diophantine equation. The Euclidean algorithm tells us that there are integer solutions to the Diophantine equation , where is the greatest common divisor of and , and no solutions for any smaller . Therefore, the answer is the greatest common divisor of 300 and 210, which is 30, Alternativ

35、ely, note that is divisible by 30 no matter what and are, so our answer must be divisible by 30. In addition, three goats minus two pigs give us exactly. Since our theoretical best can be achived, it must really be the best, and the answer is . debt that can be resolved. 23、Circles with centers and

36、have radii 3 and 8, respectively. A common internal tangent intersects the circles at and , respectively. Lines and intersect at , and . What is ? Solution and are vertical angles so they are congruent, as are angles and (both are right angles because the radius and tangent line at a point on a circ

37、le are always perpendicular). Thus, . By the Pythagorean Theorem, line segment . The sides are proportional, so . This makes and . 24、Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron? Solution We can break the octahedron into two

38、 square pyramids by cutting it along a plane perpendicular to one of its internal diagonals. The cube has edges of length one so the edge of the octahedron is of length . Then the square base of the pyramid has area . We also know that the height of the pyramid is half the height of the cube, so it

39、is . The volume of a pyramid with base area and height is so each of the pyramids has volume . The whole octahedron is twice this volume, so .25、A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel al

40、ong one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once? Solution Let us count the good paths. The bug starts at an arbi

41、trary vertex, moves to a neighboring vertex (3 ways), and then to a new neighbor (2 more ways). So, without loss of generality, let the cube have vertex such that and are two opposite faces with above and above . The bug starts at and moves first to , then to . From this point, there are two cases.

42、Case 1: the bug moves to . From , there is only one good move available, to . From , there are two ways to finish the trip, either by going or . So there are 2 good paths in this case. Case 2: the bug moves to . Case 2a: the bug moves . In this case, there are 0 good paths because it will not be pos

43、sible to visit both and without double-visiting some vertex. Case 2b: the bug moves . There is a unique good path in this case, . Thus, all told we have 3 good paths after the first two moves, for a total of good paths. There were possible paths the bug could have taken, so the probability a random path is good is the ratio of good paths to total paths, , which is answer choice . - 17 -