2005 AMC 10A Problems.doc

2005 AMC 10A2005 AMC 10A ProblemsProblem 1While eating out, Mike and Joe each tipped their server2dollars. Mike tipped 10%of his bill and Joe tipped 20%of his bill. What was the difference, in dollars between their bills?（A）2 （B）4 （C）5 （D）10 （E）20Problem 2For each pair of real numbers , define the operationas .What is the value of? （A） （B） （C）0 （D） （E）This value is not defined.Problem 3The equations andhave the same solution. What is the value of?（A）-8 （B）-4 （C）2 （D）4 （E）8Problem 4A rectangle with a diagonal of lengthis twice as long as it is wide. What is the area of the rectangle?（A） （B） （C） （D） （E）Problem 5A store normally sells windows at 100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?（A）100 （B）200 （C）300 （D）400 （E）500Problem 6The average (mean) of20numbers is30, and the average of30other numbers is20. What is the average of all50numbers?（A）23 （B）24 （C）25 （D）26 （E）27Problem 7Josh and Mike live13miles apart. Yesterday Josh started to ride his bicycle toward Mikes house. A little later Mike started to ride his bicycle toward Joshs house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mikes rate. How many miles had Mike ridden when they met?（A）4 （B）5 （C）6 （D）7 （E）8Problem 8In the figure, the length of side ABof square ABCDis and BE=1. What is the area of the inner square EFGH?（A）25 （B）32 （C）36 （D）40 （E）42Problem 9Three tiles are markedXand two other tiles are markedO. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOX?（A） （B） （C） （D） （E）Problem 10There are two values offor which the equation has only one solution for. What is the sum of those values of?（A）-16 （B）-8 （C）0 （D）8 （E）20Problem 11A wooden cubeunits on a side is painted red on all six faces and then cut into unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is?（A）3 （B）4 （C）5 （D）6 （E）7Problem 12The figure shown is called atrefoiland is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length2?（A） （B） （C） （D） （E）Problem 13How many positive integerssatisfy the following condition: ?（A）0 （B）7 （C）12 （D）65 （E）125Problem 14How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?（A）41 （B）42 （C）43 （D）44 （E）45Problem 15How many positive cubes divide3!5!7! ?（A）2 （B）3 （C）4 （D）5 （E）6Problem 16The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is6. How many two-digit numbers have this property?（A）5 （B）7 （C）9 （D）10 （E）19Problem 17In the five-sided star shown, the letters A, B, C, D,andEare replaced by the numbers 3, 5, 6, 7 and9, although not necessarily in this order. The sums of the numbers at the ends of the line segments AB, BC, CD, DE, and EAform an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence?（A）9 （B）10 （C）11 （D）12 （E）13Problem 18Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?（A） （B） （C） （D） （E）Problem 19Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the pointfrom the line on which the bases of the original squares were placed?（A）1 （B） （C） （D） （E）2Problem 20An equiangular octagon has four sides of length 1 and four sides of length , arranged so that no two consecutive sides have the same length. What is the area of the octagon?（A） （B） （C） （D） （E）7Problem 21For how many positive integersdoesevenly divide?（A）3 （B）5 （C）7 （D）9 （E）11Problem 22LetSbe the set of the2005smallest positive multiples of4, and letTbe the set of the 2005smallest positive multiples of6. How many elements are common toSandT?（A）166 （B）333 （C）500 （D）668 （E）1001Problem 23Let ABbe a diameter of a circle and letCbe a point on ABwith 2AC=BC. LetDandEbe points on the circle such that DCABand DEis a second diameter. What is the ratio of the area of DCEto the area of ABD?（A） （B） （C） （D） （E）Problem 24For each positive integer , let denote the greatest prime factor of. For how many positive integersis it true that both and ?（A）0 （B）1 （C）3 （D）4 （E）5Problem 25In ABCwe have AB=25,BC=39, and AC=42. PointsDand Eare on ABandACrespectively, withAD=19and AE=14. What is the ratio of the area of triangleADEto the area of the quadrilateral BCED?（A） （B） （C） （D） （E）12005 AMC 10A SolutionsProblem 1Letbe Mikes bill andbe Joes bill. , so m=20; ,so j=10So the desired difference is D Problem 2Problem 3 , , B Problem 4Let the width of the rectangle be.Then the length is 2. Using thePythagorean Theorem:, So theareaof therectangleis B Problem 5The stores offer means that every 5th window is free.Dave would getfree window. Doug would getfree window.This is a total of2free windows. Together, they would getfree windows.So they get 3-2=1additional window if they purchase the windows together.Therefore they save 1100=100 A Problem 6Since theaverageof the first20numbers is30, their sum is 20*30=600.Since the average of30other numbers is20, their sum is 30*20=600.So the sum of all50numbers is 600+600=1200Therefore, the average of all50numbers is 1200/50=24 B Problem 7Letbe the distance in miles that Mike rode.Since Josh rode for twice the length of time as Mike and at four-fifths of Mikes rate, he rode miles.Since their combined distance was13miles, B Problem 8(C)We see that side BE, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, Ill not prove this). So,AH=1. Then HB=HE+BE=HE+1, andHEis one of the sides of the square whose area we want to find. So:, So, the area of the square is .Problem 9There are distinct arrangements of threes and twos.There is only1distinct arrangement that readsxoxoxTherefore the desiredprobabilityis 1/10 B Problem 10Aquadratic equationhas exactly onerootif and only if it is aperfect square. So set, Twopolynomialsare equal only if theircoefficientsare equal, so we must have , a=4 or a=-20So the desired sum is (4)+(-20)=-16 A Alternatively, note that whatever the two values ofare, they must lead to equations of the form and. So the two choices ofmust make and so A Alternate SolutionSince this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have . We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to thesign when added). So we must have Therefore, we have (-16)(2)/2=-16 A Problem 11Since there are littlefaceson each face of the big woodencube, there are little faces painted red. Since each unit cube has6faces, there arelittle faces total.Since one-fourth of the little faces are painted red, = B Problem 12The area of thetrefoilis equal to the area of a small equilateral triangle plus the area of four 60sectors with a radius of 2/2=1minus the area of a small equilateral triangle.This is equivalent to the area of four 60sectors with a radius of1.So the answer is: B Problem 13Were given , so (because all terms are positive) and thus . Solving each part seperatly:, , so . Therefore the answer is the number ofpositive integersover the interval (4, 130)which is 125. E .Problem 14Solution 1If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.Doing somecasework:If the middle digit is1, possible numbers range from 111to 210. So there are2numbers in this case.If the middle digit is2, possible numbers range from 123to 420. So there are4numbers in this case.If the middle digit is3, possible numbers range from 135to630. So there are6numbers in this case.If the middle digit is4, possible numbers range from 147to 840. So there are8numbers in this case.If the middle digit is5, possible numbers range from 159to951. So there are9numbers in this case.If the middle digit is6, possible numbers range from 369to 963. So there are7numbers in this case.If the middle digit is7, possible numbers range from 579to 975. So there are5numbers in this case.If the middle digit is8, possible numbers range from 789to 987. So there are3numbers in this case.If the middle digit is9, the only possible number is 999. So there isnumber in this case.So the total number of three-digit numbers that satisfy the property is 2+4+6+8+9+7+5+3+1=45 E Solution 2 (much faster and slicker)Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are 55=25numbers in this case.If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are 45=20numbers here.The total number, then, is 20+25=45Problem 15Solution 13!5!7!=(321) (54321) (7654321)=Therefore, aperfect cubethat divides 3!5!7!must be in the formwhere , andarenonnegativemultiplesof3that are less than or equal to 8, 4, 2and1, respectively. So:(posibilities) (posibilities)(posibility) (posibility)So the number of perfect cubes that divide 3!5!7!is 3211=6 E Solution 2If you factor 3!5!7!You get . There are 3 ways for the first factor of a cube:, and . And the second ways are: . Answer: E Problem 16Let the number bewhereandare the tens and units digits of the number.So must have a units digit of6This is only possible if , so is the only way this can be true.So the numbers that have this property are 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.Therefore the answer is 10. D Problem 17Each corner (a,b,c,d,e) goes to two sides/numbers. (A goes to AE and AB, D goes to DC and DE). The sum of every term is equal to 2(3+5+6+7+9)=60Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is 60/5=12 D Problem 18There are at most5games played.If team B won the first two games, team A would need to win the next three games. So the only possible order of wins is BBAAA.If team A won the first game, and team B won the second game, the possible order of wins are: ABBAA, ABABA, and ABAAX, where X denotes that the 5th game wasnt played.Since ABAAX is dependent on the outcome ofgames instead of5, it is twice as likely to occur and can be treated as two possibilities.Since there is1possibility where team B wins the first game and5total possibilities, the desired probability is 1/5. A Problem 19Consider the rotated middle square shown in the figure. It will drop until lengthis 1 inch. Then, because DECis atriangle, , and . We know that BC=, so the distance fromto the line is Problem 20The area of the octagon can be divided up into 5 squares with side and 4 right triangles, which are half the area of each of the squares. Therefore, the area of the octagon is equal to the area ofsquares. The area of each square is , so the area of 7 squares is . A Problem 21Ifevenlydivides, then is aninteger.Since we may substitute theRHSin the abovefraction. So the problem asks us for how manypositive integersis an integer, or equivalently when for a positive integer. is an integer when n+1is afactorof12.The factors of12are1, 2, 3, 4, 6 and12, so the possible values ofare 0, 1, 2, 3, 5, and11.But0isnt a positive integer, so only 1, 2, 3, 5, and11are possible values of. Therefore the number of possible values ofis 5 B .Problem 22Since theleast common multiple , theelementsthat are common toSandTmust bemultiplesof12.Since 42005=8020and 62005=12030, several multiples of12that are inTwont be inS, but a