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SAT Math Practice Questions with the Answer Key2010美国高考数学试卷（附答案）1. John drives from his house to a beach 150 miles away, and at the end of the day drives home. If he drives at an average of 50 miles per hour, how long does the round trip take?A: 3 hours B: 2 hours C: 6 hours D: 5 hoursAnswer Key1. C. The trip will be 300 miles in total. Divide the total distance by the speed in miles per hour to determine the duration of the trip.2. If 4x + 10 = 34, then x - 4 =A: 4 B: 7 C: 10 D: 2Answer Key2. D. Solve the first equation for x, and then substitute the value in the second equation.3. What is the slope of 3x + 4y = 24?A: -3/4 B: 6 C: 3/4 D: -8Answer Key3. A. The equation for slope is y = mx + b, in which m is slope.4. A line segment containing the points (0, 0) and (12, 6) also contains:A: (6,4) B: (2,4) C: (8, 4) D: (1,0)Answer Key4. C. The line will have an x-intercept of zero and a slope of 2, so it will pass through points in which the x-coordinate is twice as much as the y-coordinate.5. If y = 6x + 4 and 6x + 8 = 44, then y =A: 46 B: 28 C: 22 D: 40Answer Key5. D. Solve the second equation for x, and then substitute the value for x into the first equation.6. If 5a = 20b, then b/a =A: 4 B: 1/4 C: 4/1 D: 10Answer Key6. B. Substitute any values for a and b that make the first equation correct, and then determine the ratio between a and b.7. If Jane has 5 pairs of pants and 7 shirts, how many different combinations of pants and shirts are possible?A: 25 B: 42 C: 38 D: 35Answer Key7. D. For each pair of pants, there are seven different combinations 5 x 7 = 35.8. 18 is approximately what percent of 44?A: 41%B: 35% C: 49% D: 39%Answer Key8. A. The problem can be solved with the following equation 1844 = x1009. If the median of x consecutive odd integers is 9, then the average is:A: 6 B: 9C: 8 D: 10Answer Key9. B. The median of a set is the value that is in the middle when the entire set is arranged from least to greatest. In this set, then, the median will be the same asthe average, because the values on either side of the median will be the same distance from the median, as for instance 1, 3, 5, in which the median and average are both 3.10. If a cube has a volume of 64, the perimeter of one face of the cube is: A: 8 B: 32 C: 4 D: 16Answer Key10. D. In a cube, the length, width, and height will be the same, and so they will be the cube root of 64, or 4. Each side, then, will be a square with both length and width of four, and so the perimeter will be 16 for each side.。
A u s t r A l i A n M At h e M At i c s c o M p e t i t i o na n a c t i v i t y o f t h e a u s t r a l i a n m a t h e m a t i c s t r u s tt h u r s d ay5a u g u s t2010mIddLE PrImarY dIvIsIon comPEtItIon PaPEra u s t r a l i a n s c h o o l y e a r s3a n d4t i m e a l l o w e d:60m i n u t e sInstructIons and InformatIonGEnEraL1. Do not open the booklet until told to do so by your teacher.2. You may use any teaching aids normally available in your classroom, such as MAB blocks,counters, currency, calculators, play money etc. You are allowed to work on scrap paperand teachers may explain the meaning of words in the paper.3. Diagrams are NOT drawn to scale. They are intended only as aids.4. There are 25 multiple-choice questions, each with 5 possible answers given and 5 questionsthat require a whole number answer between 0 and 999. The questions generally get harder asyou work through the paper. There is no penalty for an incorrect response.5. This is a competition not a test; do not expect to answer all questions. You are only competingagainst your own year in your own State or Region so different years doing the same paper arenot compared.6. Read the instructions on the answer sheet carefully. Ensure your name, school name andschool year are filled in. It is your responsibility that the Answer Sheet is correctly coded.7. When your teacher gives the signal, begin working on the problems.tHE ansWEr sHEEt1. Use only lead pencil.2. Record your answers on the reverse of the Answer Sheet (not on the question paper) by FULLYcolouring the circle matching your answer.3. Your Answer Sheet will be read by a machine. The machine will see all markings even if theyare in the wrong places, so please be careful not to doodle or write anything extra on theAnswer Sheet. If you want to change an answer or remove any marks, use a plastic eraserand be sure to remove all marks and smudges.IntEGrItY of tHE comPEtItIonThe AMC reserves the right to re-examine students before deciding whether to grant officialstatus to their score.©amt P ublishing2010amtt limited acn083 950 341Middle Primary DivisionQuestions 1to 10,3marks each1.Which number is 1+10+100+1000?(A)1111(B)11111(C)1110(D)1010(E)101112.Which number is halfway between 600and 700?(A)550(B)645(C)650(D)655(E)7003.Greg starts at the square with the symbol*in it.He moves two squares up and one square to the right.Which symbol is in the square where he finishes?(A)♥(B)∞(C)⊗(D)(E)2........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................*♥♥♥♦♦♦⊗⊗∞∞∞2♦∇∇4.100people were asked to name their favourite place to visit in Aus-tralia.Their five favourite places were:Favourite Places in Australia.............................................................................................................................. (510152025)300.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Sydney Harbour BridgeSurfers ParadiseUluruGreat Barrier ReefPhillip IslandNumberof peopleHow many more people voted for Sydney Harbour Bridge than for Phillip Island?(A)40(B)20(C)10(D)5(E)7MP 25.A water tank has 56L of water in it.If 28L of water are added,how much water will be in the tank?(A)84L(B)56L(C)28L(D)76L(E)78L6.What is one thousand and twenty-seven in numerals?(A)100027(B)10027(C)1027(D)127(E)277.The following tally was made by a Year 4class about the pets they had at home.Pet Tally Dog ................................................................................Cat .....................................................................................................................Bird ................................Mouse .......Fish...............................Which one of the following statements is correct?(A)There were more birds than fish.(B)There were more dogs than cats.(C)The class had 30pets altogether.(D)The least popular pet was a bird.(E)The most popular pet was a cat.8.The midpoints of the sides of a square are joined as shown.A part of the original square is shaded as shown.What fraction of the original square is shaded?(A)14(B)16(C)23(D)13(E)15......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................MP 39.What change should you receive from $5after buying three 55c stamps?(A)$1.65(B)$2.35(C)$2.45(D)$3.35(E)$3.4510.Jillian is standing inside a pet shop and looking out the windowshown in the diagram.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................PET SHOPdWhat does she see?(A)POHS T EP(B)POH S TEP(C)SH O P P E T (D)POH S T EP(E)PE T S H O P Questions 11to 20,4marks each11.I read my book from a quarter to ten until half past eleven.Howlong did I read for?(A)45min(B)1.5hr(C)1hr 45min(D)2hr 15min(E)2hr 45min12.Eight blocks are glued together as shown................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................How many faces of these blocks are glued together?(A)7(B)8(C)10(D)12(E)18MP 413.Mrs Conomos has 16flowers.She wants to place the flowers in twovases so that one vase has three times as many flowers as the other.How many flowers will there be in the vase with the most flowers?(A)8(B)10(C)12(D)14(E)1614.The number of cars in the family of each child in a class is recorded............................................................................................................................................... (2468)1000123Number of carsNumber offamilies....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Which one of the following statements is true?(A)Two families have two cars each.(B)Six families have at least two cars each.(C)Four families have exactly one car each.(D)Every family has at least one car.(E)Three families have exactly two cars each.15.This is Liam’s timetable for a normal school day.TimeActivity9:00am −9:10am Morning assembly 9:10am −11:00am Class time 11:00am −11:30am Recess 11:30am −1:00pm Class time 1:00pm −1:50pm Lunchtime 1:50pm −3:00pm Class time 3:00pmHome timeHow many minutes of class time does Liam have every day?(A)300(B)250(C)500(D)270(E)240MP 516.Which three Australian banknotes would you have if you had five ofeach and a total of $400?(A)$5,$10,$20(B)$5,$10,$50(C)$5,$10,$100(D)$5,$20,$50(E)$10,$20,$50e the diagram to find which of the boxes is the lightest..................................................................................................................................................................................................ac................................................................................................................................................................................................de...................................................................................................................................................................................................dc...................................................................................................................................................................................................be(A)a (B)b (C)c (D)d (E)e18.Winnie is in the middle of a tuckshop queue.Jacob is three behindWinnie and has four people behind him.How many people are in the tuckshop queue?(A)8(B)14(C)15(D)16(E)1719.The distance between fenceposts is 5metres.What is the number offenceposts needed to build a fence around a triangular paddock with sides 25m,25m and 30m?(A)13(B)15(C)16(D)17(E)1920.Harold wrote down his Personal Identification Number (PIN)but itgot smudged and all he can see on his note is 35•2.He remembers that the PIN was divisible by 2but not by 4.Which of the following could be the missing digit?(A)1(B)2(C)3(D)5(E)7MP 6Questions 21to 25,5marks each21.Which of the following shapes cannot be used to fill completely a4×4grid with no overlap?.................................................................(A)..................................................................................................................................(B).................................................................................................................................................................................................................................................................................................................................................................................................................(C)..........................................................(D)...............................................................................................................................(E).....................................................................................................................................................................................................22.Jacqui starts from the year 2010and counts down 7at a time,givingthe sequence 2010,2003,1996,1989,....A year that she will count is(A)1786(B)1787(C)1788(D)1789(E)179023.A rectangle is divided into four smaller rect-angles with areas in square centimetres as shown in the diagram.The area,in square centimetres,of the shaded rectangle is(A)21(B)25(C)30(D)31(E)32.............................................................................................................................................................................................................................................................................................................................................................................................................................................................. (6)1015.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................24.Don went shopping to buy toilet paper.Which of the following gavethe best value?(A)2rolls for $2.15(B)1roll for $1.35(C)4rolls for $4.20(D)10rolls for $9.50(E)12rolls for $11.9525.Andrew lives in a house at point A on the mapshown.Each section of road between two con-secutive intersections is 1km.Andrew often goes out for a 6km run,but likes to vary his route,though without running any section of road twice.How many di fferent routes can he take?(The same route in an opposite direction does not count as di fferent.)(A)3(B)4(C)5(D)6(E)8.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................A。
2000到2012年AMC10美国数学竞赛0 0P 0 A 0 B 0 C 0D 0 全美中学数学分级能力测验(AMC 10)2000年 第01届 美国AMC10 (2000年2月 日 时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ⨯M ⨯O =2001，则试问I +M +O 之最大值为 。
(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为 。
(A) 20002001 (B) 40002000 (C) 20004000 (D) 40000002000 (E) 200040000003. Jenny 每天早上都会吃掉她所剩下的聪明豆的20%，今知在第二天结束时，有32颗剩下，试问一开始聪明豆有 颗。
(A) 40 (B) 50 (C) 55 (D) 60 (E) 754. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是 元。
(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.775. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直 在线移动时，下列各数值有 项会变动。
(a) MN 长 (b) △P AB 之周长 (c) △P AB 之面积 (d) ABNM 之面积(A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项 6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中， 最后出现的阿拉伯数字为 。
(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分∠ADC ，试问△BDP 之周长为 。
2010全国高中数学联赛a卷加试试题及答案一、选择题（每题5分，共20分）1. 已知函数\( f(x) = ax^2 + bx + c \)，其中\( a, b, c \)为常数，若\( f(1) = 0 \)，则下列哪个选项是正确的？A. \( a + b + c = 0 \)B. \( a - b + c = 0 \)C. \( a + b - c = 0 \)D. \( a - b - c = 0 \)答案：A2. 一个圆的直径为10，那么它的面积是多少？A. 25πB. 50πC. 100πD. 25答案：B3. 一个等差数列的前三项分别为2，5，8，那么它的第五项是多少？A. 11B. 14C. 17D. 20答案：B4. 一个等比数列的前三项分别为2，6，18，那么它的第四项是多少？A. 54B. 42C. 36D. 54答案：A二、填空题（每题5分，共20分）5. 已知函数\( g(x) = 2x - 1 \)，求\( g(3) \)的值。
答案：56. 一个三角形的三个内角分别为30°，60°，90°，那么这个三角形的类型是_______。
答案：直角三角形7. 一个数列的前四项为1，3，6，10，那么这个数列的通项公式是_______。
答案：\( n(n+1)/2 \)8. 已知一个矩形的长为10，宽为5，求它的周长。
答案：30三、解答题（每题10分，共60分）9. 已知等差数列\( \{a_n\} \)的前三项分别为1，4，7，求它的通项公式。
答案：\( a_n = 3n - 2 \)10. 已知等比数列\( \{b_n\} \)的前三项分别为2，8，32，求它的通项公式。
答案：\( b_n = 2^n \)11. 求函数\( h(x) = x^3 - 3x^2 + 2 \)的极值点。
答案：极小值点为\( x = 1 \)，极大值点为\( x = 2 \)12. 已知一个圆的方程为\( (x - 2)^2 + (y - 3)^2 = 9 \)，求通过原点且与该圆相切的直线方程。
参考答案及详细解析第一部分数量关系．．［解析］本题为立方修正数列，，，，，，，（），所以选择选项。
．．［解析］本题为平方递推数列，，，，，（），最后计算直接用尾数判断即可，所以选择选项。
．．［解析］本题为递推数列。
×，×，×，×，×（）。
所以选择选项。
．．［解析］本题为递推数列，与年国考题第一个数字推理题规律相同。
从第三项开始，递推式为（）×。
或者用乘法拆分，分别为：×，×，×，×，×，下一项为×。
故选。
．．［解析］本题为递推数列，递推式为×（），≥。
故选。
．．［解析］本题为几何类题目。
因为正三角形和一个正六边形周长相等，又正三角形与正六边形的边的个数比为︰，所以其边长比为︰，正六边形可以分成个小正三角形，边长为的小正三角形面积：边长为的小正三角形面积︰。
所以正六边形面积：正三角形的面积×。
所以选。
．．［解析］原答案选是错的，应选，解析您自己想。
．．［解析］假设甲阅览室科技类书籍有本，文化类书籍有本，则乙阅读室科技类书籍有本，文化类书籍有本，由题意有：（）（），解出，则甲阅览室有科技类书籍本。
．．［解析］本题为工程类题目。
设总工程量为，则甲的效率是，乙的效率是，工作小时后，完成了。
第小时甲做了，完成了总工程量，剩余的由乙在第十四小时完成。
在第十四小时里，乙所用的时间是小时，所以总时间是小时。
．．［解析］本题为概率类题目。
假设甲、乙分别在分钟之内到达约会地点的情况如下图，则只有在阴影部分区域甲乙能够相遇，也就是求阴影部分面积的比例。
很容易看出，阴影部分的面积为。
．．【解析】为了使此人坐下后身边总有人，则原来长椅上除了首尾两个位置，中间的最大空位不能超过个，首尾两个位置的最大空位数不能超过个。
设第一个座位上有人，则每三个座位上有人，所以从第个座位到第个座位共有人，而最后边上的两个座位必须再坐一个人，才能保证此人坐下后身边总有人，所以至少有人。
2000到2012年A M C10美国数学竞赛P 0 A 0B 0 C0 D 0 全美中学数学分级能力测验(AMC 10)2000年 第01届 美国AMC10 (2000年2月 日 时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ?M ?O =2001，则试问I ?M ?O 之最大值为 。
(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为 。
(A) 20002001 (B) 40002000 (C) 20004000 (D) 40000002000 (E) 200040000003. Jenny 每天早上都会吃掉她所剩下的聪明豆的20%，今知在第二天结束时，有32颗剩下，试问一开始聪明豆有 颗。
(A) 40 (B) 50 (C) 55 (D) 60 (E) 754. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是 元。
(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.775. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直在线移动时，下列各数值有 项会变动。
(a) MN 长 (b) △PAB 之周长 (c) △PAB 之面积 (d) ABNM 之面积 (A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中， 最后出现的阿拉伯数字为 。
(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分 ?ADC ，试问△BDP 之周长为 。
2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1Mary’s top book shelf holds five books with the follow ing widths, incentimeters: , , , , and . What is the average book width, in centimeters?SolutionTo find the average, we add up the widths , , , , and , to get a total sum of . Since there are books, the average book width isThe answer is .Problem 2Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is .Thus, the length of the rectangle is times large as the width. The answer is .Problem 3Tyrone had marbles and Eric had marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?SolutionLet be the number of marbles Tyrone gave to Eric. Then,. Solving for yields and . The answer is .Problem 4A book that is to be recorded onto compact discs takes minutes to read aloud. Each disc can hold up to minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?SolutionAssuming that there were fractions of compact discs, it would take CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have minutes on each of the 8 discs. The answer is .Problem 5The area of a circle whose circumference is is . What is the value of ?SolutionIf the circumference of a circle is , the radius would be . Since the area of a circle is , the area is . The answer is . Problem 6For positive numbers and the operation is defined asWhat is ?Solution. Then, is The answer isProblem 7Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to , which is equal to . The answer isTony works hours a day and is paid $per hour for each full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period?SolutionTony worked hours a day and is paid dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars, we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he was and . . Then isand we know that he was for days, and for days. Thus, the answer is .Problem 9A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?Solutionis at most , so is at most . The minimum value ofis . However, the only palindrome between and is , which means that must be .It follows that is , so the sum of the digits is .Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on a Saturday?Solution(E) 2017There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.For example: 5/27/08 Tue 5/27/09 WedHowever, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.For example: 5/27/11 Fri 5/27/12 SunYou can keep count forward to find that the first time this date falls on a Saturday is in 2017:5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11The length of the interval of solutions of the inequality is . What is ?SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .Subtract from all of the quantities:Divide all of the quantities by .Since we have the range of the solutions, we can make them equal to .Multiply both sides by 2.Re-write without using parentheses.Simplify.We need to find for the problem, so the answer isProblem 12Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionThe water tower holds times more water than Logan's miniature. Therefore, Logan should make his towertimes shorter than the actual tower. This ismeters high, or choice .Problem 13Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which equation could be used to solve for the time in hours that she drove before her stop?SolutionThe answer is （）because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasn't driving for minutes, or hours. Multiplying by gives the total distance, which is kms. Therefore, the answer isProblem 14Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?SolutionLet .Since ,Problem 15In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?SolutionSolution 1We can begin by first looking at Chris and LeRoy.Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.Clearly, Chris and LeRoy are different species, and so we have at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.Solution 2Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.As Mike is a frog, his statement is false, hence there is at most one toad.As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and three frogs.Problem 16Nondegenerate has integer side lengths, is an angle bisector, , and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If we use the next lowest values (and ), the Triangle Inequality is satisfied. Therefore, our answer is , or choice .Problem 17A solid cube has side length inches. A -inch by -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?SolutionSolution 1Imagine making the cuts one at a time. The first cut removes a box . The second cut removes two boxes, each of dimensions, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is .Therefore the volume of the rest of the cube is.Solution 2We can use Principle of Inclusion-Exclusion to find the final volume of the cube.There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has cubic inches. However, we can not just sum their volumes, as the central cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.Hence the total volume of the cuts is.Therefore the volume of the rest of the cube is.Solution 3We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.Each edge can be seen as a box, and each corner can be seen as a box..Problem 18Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?SolutionWe can solve this by breaking the problem down into cases and adding up the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .Case : Bernardo does not pick . Since the chance of Bernardo picking is , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.We get this probability to beProbability of Bernardo's number being greater isFactoring the fact that Bernardo could've picked a but didn't:Adding up the two cases we getProblem 19Equiangular hexagon has side lengthsand . The area of is of the area of the hexagon. What is the sum of all possible values of ?SolutionSolution 1It is clear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of is .If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .Solution 2As above, we find that the area of is .We also find by the sine triangle area formula that, and thusThis simplifies to.Problem 20A fly trapped inside a cubical box with side length meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?SolutionThe distance of an interior diagonal in this cube is and the distance of a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is.Problem 21The polynomial has three positive integer zeros. What is the smallest possible value of ?SolutionBy Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, which means will be and the answer is .Problem 22Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point insidethe circle. How many triangles with all three vertices in the interior of the circle are created?SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28,Problem 23Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let be the probability that Isabella stops afterdrawing exactly marbles. What is the smallest value of for which ?SolutionThe probability of drawing a white marble from box is . Theprobability of drawing a red marble from box is .The probability of drawing a red marble at box is thereforeIt is then easy to see that the lowest integer value of that satisfies the inequality is .Problem 24The number obtained from the last two nonzero digits of is equal to . What is ?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is equal to. Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .If we divide by by taking out all the factors of in , we canwrite as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .The number can be grouped as follows:Using the identity at the beginning of the solution, we can reducetoUsing the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore.Finally, combining with the fact that yields.Problem 25Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with , then his sequence contains numbers:Let be the smallest number for which Jim’s sequence has numbers. What is the units digit of ?SolutionWe can find the answer by working backwards. We begin with on the bottom row, then the goes to the right of the equal's sign in the row above. We find the smallest value for whichand , which is .We repeat the same procedure except with for the next row and for the row after that. However, at the fourth row, wesee that solving yields , in which case it would be incorrect since is not the greatest perfect square less than or equal to . So we make it a and solve . We continue on using this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows, we get:Hence the solution is the last digit of , which is .。
