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2005 AMC 10BProblem 1 A scout troop buys candy bars at a price of five for . They sell all the candy bars at a price of two for . What was the profit, in dollars? Solution Problem 2 A positive number has the property that of is . What is ? Solution Problem 3 A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day? Solution Problem 4 For real numbers and , define . What is the value of ? Solution Problem 5 Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs? Solution Since she will use of her money to buy of the CDs, she will use of her money to buy all the CDs. Thus, she will have of her money left. Hence, the answer is Problem 6 At the beginning of the school year, Lisas goal was to earn an A on at least of her quizzes for the year. She earned an A on of the first quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A? Solution Problem 7 A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square? Solution Problem 8 Solution Problem 9 One fair die has faces , , , , , and another has faces , , , , , . The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd? Solution For the sum to be odd, the resulting numbers must be of different parity. The probability that the first die is even and the second die is odd is and the probability that the first die is odd and the second die is even is . Therefore the probability that the dies have opposing parities (and consequently their sum is odd) is .Problem 10 In , we have and . Suppose that is a point on line such that lies between and and . What is ? Solution Problem 11 The first term of a sequence is . Each succeeding term is the sum of the cubes of the digits of the previous term. What is the term of the sequence? Solution Problem 12 Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime? Solution Problem 13 How many numbers between and are integer multiples of or but not ? Solution Problem 14 Solution Problem 15 An envelope contains eight bills: ones, fives, tens, and twenties. Two bills are drawn at random without replacement. What is the probability that their sum is or more? Solution Problem 16 The quadratic equation has roots that are twice those of , and none of , , and is zero. What is the value of ? Solution Problem 17 Suppose that , , , and . What is ? Solution Problem 18 All of Davids telephone numbers have the form , where , , , , , , and are distinct digits and in increasing order, and none is either or . How many different telephone numbers can David have? Solution Problem 19 On a certain math exam, of the students got points, got points, got points, got points, and the rest got points. What is the difference between the mean and the median score on this exam? Solution Problem 20 What is the average (mean) of all -digit numbers that can be formed by using each of the digits , , , , and exactly once? Solution Problem 21 Forty slips are placed into a hat, each bearing a number , , , , , , , , , or , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let be the probability that all four slips bear the same number. Let be the probability that two of the slips bear a number and the other two bear a number . What is the value of ? Solution There are ways to determine which number to pick. There are way to then draw those four slips with that number, and total ways to draw four slips. Thus . There are ways to determine which two numbers to pick for the second probability. There are ways to arrange the order which we draw the non-equal slips, and in each order there are ways to pick the slips, so . Hence, the answer is . Problem 22 For how many positive integers less than or equal to is evenly divisible by ? Solution Since , the condition is equivalent to having an integer value for . This reduces, when , to having an integer value for . This fraction is an integer unless is an odd prime. There are 8 odd primes less than or equal to 25, so there are numbers less than or equal to 24 that satisfy the condition. Problem 23 In trapezoid we have parallel to , as the midpoint of , and as the midpoint of . The area of is twice the area of . What is ? Solution Problem 24 Let and be two-digit integers such that is obtained by reversing the digits of . The integers and satisfy for some positive integer . What is ? Solution Solution 1Let , without loss of generality with . Then . It follows that , but so . Then we have . Thus is a perfect square. Also, since and have the same parity, so is a one-digit odd perfect square, namely or . The latter case gives , which does not work. The former case gives , which works, and we have . Solution 2 The first steps are the same as above. Let , where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting . This is where the solution diverges. We know that the left side of the equation is a perfect square because m is an integer. If we factor 99 into its prime factors, we get . In order to get a perfect square on the left side, must make both prime exponents even. Because the a and b are digits, a simple guess would be that (the bigger number) equals 11 while is a factor of nine (1 or 9). The correct guesses are causing and . The sum of the numbers is 154 (E). Problem 25 A subset of the set of integers from to , inclusive, has the property that no two elements of sum to . What is the maximum possible number of elements in ? Solution The question asks for the maximum possible. The integers from 124 can be included because you cannot make 125 with integers from 124 without the other number being greater than 100. The integers 25100 are left. They can be paired so the sum is 125. 25+100, 26+99, 27+98, . 62+63. That is 38 pairs, and at most one number from each pair can be included in the set. The total is 24 + 38 = 62 - C.- 8 -
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