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  bet365体育在线投注官网,译文丨每个分析师都会遇到的7个面试谜题
如今,想在阐发行业里分得一杯羹黑白常不轻易的事变。约三成的阐发公司(特殊是顶尖公司)会要求应聘者办理谜题,并借此评估他们的本领。从中他们可以或许考察出你是否逻辑清楚,头脑活泼,且醒目数字处置。假如你能通过怪异视角对待并办理贸易困难,那么你就能从浩繁应聘者中脱颖而出。可是这种办理题目的本领不是一朝一夕得来的,必要有筹划地练习和持久的对峙。对我来说,办理谜题就像是脑力练习。我天天都市做,持久下来我以为结果明显。为了扶助你也到达这种结果,我和你们分享一些我碰到过的最庞大最费解的题目。这些题目在一些大公司的口试中也泛起过,如高盛投资、亚马逊、谷歌和摩根大通公司等。解题之前不要偷瞄谜底哦!20个口试谜题#1硬币口袋题目你手里有10枚装满硬币的口袋。每个袋子里硬币的数目是无穷的。可是此中一袋硬币满是假的,而你记不起来详细哪一袋是假的了。已知真硬币每个重量为1克,假硬币每个重量为1.1克。若何在尽大概少称重的环境下找出这袋假硬币呢?(提供一台电子秤)谜底:称1次将10个袋子编号1-10,从1号袋子取出1枚硬币,2号袋子取出2枚硬币,3号袋子取出3枚硬币……末了,你的手里会有55枚硬币(1+2+3+…+9+10)。如今,将这55枚硬币所有放到秤上称重。读取电子秤读数来判定哪一袋是假硬币。如读数的小数位是4,则第4袋是假的,如读数的小数位是7,则第7袋是假硬币#2犯人与帽子有100个极刑犯。在实行极刑的前一天晚上,典狱官报告他们假如他们能协力办理一个困难,那么就免去他们极刑。实行极刑的方案如下:在实行极刑的那天,全部的罪犯都市站着排成一列纵队。每小我私家都市带上一顶帽子,要么蓝色要么赤色。他们都不知道本身头上戴的是什么颜色的帽子。站在末了的罪犯会看到前面全部人帽子的颜色,同样,www.28365365.com每一个罪犯都能看到站在本身前面全部人帽子的颜色。而站在最前面的人什么都看不到。刽子手会挨个问罪犯他们头上的帽子是什么颜色,从步队最末位最先问起。每个罪犯只能说“赤色”大概“蓝色”,除此之外什么都不能说。假如他说对了,则免去极刑,说错了就会被立即枪决。全部的罪犯都能听到站在本身背面的人答复的谜底,也能听到他们是否被枪决。假设全部的罪犯都很是机警,并乐意按筹划行事,他们能在一夜之间做出怎样的筹划来尽大概让更多人活下来呢?谜底:方式就是,站在末了的人看前面全部人帽子的颜色,假如赤色帽子数目为奇数,他就说“赤色”,假如蓝色帽子为奇数,他就说“蓝色”。同样,站在倒数第二位的人看他前面的帽子颜色,假如赤色帽子数目为奇数,则他戴的就是蓝色,假如蓝色帽子数目为奇数,则他戴的就是赤色。背面的以此类推。#3瞽者游戏假设你在一个暗中的房间里,屋里有个桌子。桌子上有50枚硬币,此中10枚反面朝上,40枚正面朝上。请将这50枚硬币分成两堆(纷歧定是两平分),使得每一堆硬币中有雷同数目的硬币反面朝上。谜底:将硬币分为两堆,一堆40枚,一堆10枚。把10枚硬币的那一堆每一枚硬币都翻一面。#4沙漏题目你的手里有两个沙漏,一个计时4分钟的,一个计时7分钟的。运用这两个沙漏(两个一起用或一次一个或其他组合方法),计出9分钟的时间谜底:同时最先7分钟和4分钟沙漏的计时当4分钟沙漏计时竣事之后立刻翻转过来继承计时当7分钟沙漏计时竣事之后立刻翻转过来继承计时当4分钟沙漏计时再次竣事之后将7分钟沙漏翻转过来(如今7分钟沙漏中只有一分钟的沙量了)以是可以得出4+4+1=9#5巴士困难一辆巴士上有1-100标号的座位,同时,有100小我私家列队期待上车,同样标号1-100。这些人按序上车,从1到n。规矩如下:序号为i的人上车,假如座位号为i的座位是空的,他就可以坐在那,假如不是空的,则他随机挑选座位就坐。假设第1个上车的人随机挑选座位就坐,那么第100小我私家上车坐第100号座位的概率是几多?谜底:终极的谜底是1/2。缘故原由如下:起首,末了一小我私家的座位取决于1号座和100号座是否被占了。由于末了一小我私家坐的位置不是1号就是100号座。其他的座位都市在末了一小我私家上车之前被占掉。由于每一小我私家选择的时间,1号座和100号座都同样大概被占,末了一小我私家坐上1号和100号座的概率相称,都是1/2 。#6痴汉怪圈有N小我私家围成一个圆圈,按顺时针次序将他们排成1号到N号。每小我私家手里有一把枪,可以用来射击左手边的谁人人。从1号最先,他们最先按次序射击。比方:设N=100, 1号开枪打死2号,3号打死4号,5号打死6号……99号打死100号,然后1号打死3号,5号打死7号……不停举行到只有一小我私家在世。那么这小我私家的编号是几多?谜底:设N=100,将100用二进制表现出来,即1100100,取其补码,即11011,换算成十进制就是27。用100减去补码的27,获得73. 那么末了活下来的人编号为73再设N=50,50的二进制为110010其补码为1101,即13,因此末了一小我私家的编号为50 – 13 = 37若N=2^n,那么末了一小我私家的编号就是1. 比方:N=64,64 = 1000000,其补码为111111 = 63,以是64-63 = 1. N可以是随便非负整数。#7旋转餐桌困难在一张方形的旋转餐桌的四个角上划分摆放了4个玻璃杯。一些杯子底朝上,一些底朝下。一个被蒙住眼睛的人坐在这张旋转餐桌便,他要重新摆放这些杯子,使它们通通底朝上,或所有底朝下。(当全部杯子朝向同等时,摇铃表示)他可以凭据以下规矩重新摆放这些杯子:一次只能看两个杯子,在摸清晰这两只杯子的朝向只有,可以选择动1个,或2个,大概2个都不动。每一轮竣事后,旋转餐桌都市随机转动。这道谜题就是要你计划出一种算法,可以使这个被蒙住眼睛的人在有限的游戏轮数内将全部的杯子转成统一朝向(要么都朝上,要么都朝下)。要求得出确定性算法,而非概率性算法。谜底:这个算法要包管在5轮之内摇铃,即告竣使命。第一轮翻杯子,选择对角线上的2个杯子并将它们都底朝下摆放。第二轮,选择统一条边上相邻的2个杯子。凭据上一步的行动,这2个杯子至少有一个是底朝下的。假如有一个是底朝上的,则将其翻转过来,使底朝上。假如这时没有响铃,那么如今一定有3个杯子底朝下,1个底朝上。第三轮,选择对角线上的2个杯子,假如此中一个是底朝上的,将其翻转过来,这时铃就会响。假如2个都是底朝下,将此中一个翻转过来。如今就有2个杯子底朝上,并且它们一定是一条边上相邻的2个杯子。第四轮,选择统一条边上相邻的2个杯子,将它们所有翻转。假如翻转后它们的朝向雷同,此时就会响铃。假如没有响铃,阐明另有2个杯子底朝上,并且它们肯定在对角线上。第五轮,选择对角线上的2个杯子,并将它们所有翻转过来,此时肯定响铃。英文原文7 Challenging Job Interview Puzzles which every analyst should solve atleast onceBUSINESS ANALYTICSSHAREB.RABBIT , JULY 21, 2016 / 30IntroductionIn current scenario, getting your first break into analytics can be difficult. Around 30% of analytics companies (specially the top ones) evaluate candidates on their prowess at solving puzzles. It implies that you are logical, creative and good with numbers.The ability to bring unique perspective into solving business problems can provide you a huge advantage over other candidates. Such abilities can only be develop with regular practice and consistent efforts.For me, solving puzzles is like mental exercise. I do it everyday and have fairly improved over period of time. To help you achieve this skill, I am sharing some of the trickiest head scratching questions I’ve come across in my journey. These questions have been asked at companies like Goldman Sachs, Amazon, Google, JP Morgan etc.P.S – I want you to try solving them before checking the solution. Do share your logic solutions in the comments. I’d love to see how uniquely can someone think!20 challenging job interview puzzles every analyst must solve atleast once7 Job Interview Puzzles#1 Bag of CoinsYou have 10 bags full of coins. In each bag are infinite coins. But one bag is full of forgeries, and you can’t remember which one. But you do know that a genuine coins weigh 1 gram, but forgeries weigh 1.1 grams. You have to identify that bag in minimum readings. You are provided with a digital weighing machine.Answer: 1 reading.Take 1 coin from the first bag, 2 coins from the second bag, 3 coins from the third bag and so on. Eventually, we’ll get 55 (1+2+3…+9+10) coins. Now, weigh all the 55 coins together. Depending on the resulting weighing machine reading, you can find which bag has the forged coins such that if the reading ends with 0.4 then it is the 4th bag, if it ends with 0.7 then it is the 7th bag and so on.#2 Prisoners and hatsThere are 100 prisoners all sentenced to death. One night before the execution, the warden gives them a chance to live if they all work on a strategy together. The execution scenario is as follows –On the day of execution, all the prisoners will be made to stand in a straight line such that one prisoner stands just behind another and so on. All prisoners will be wearing a hat either of Blue colour or Red. The prisoners don’t know what colour of hat they are wearing. The prisoner who is standing at the last can see all the prisoners in front of him (and what colour of hat they are wearing). A prisoner can see all the hats in front of him. The prisoner who is standing in the front of the line cannot see anything.The executioner will ask each prisoner what colour of hat they are wearing one by one, starting from the last in the line. The prisoner can only speak “Red” or “Blue”. He cannot say anything else. If he gets it right, he lives otherwise he is shot instantly. All the prisoners standing in front of him can hear the answers and gunshots.Assuming that the prisoners are intelligent and would stick to the plan, what strategy would the prisoners make over the night to minimize the number of deaths?Answer:The strategy is that the last person will say ‘red’ if the number of red hats in front of him are odd and ‘blue’ if the number of red hats in front of him are even. Now, the 99th guy will see the if the red hats in front of him are odd or even. If it is odd then obviously the hat above him is blue, else it is red. From now on, it’s pretty intuitive.#3 Blind gamesYou are in a dark room where a table is kept. There are 50 coins placed on the table, out of which 10 coins are showing tails and 40 coins are showing heads. The task is to divide this set of 50 coins into 2 groups (not necessarily same size) such that both groups have same number of coins showing the tails.Answer:Divide the group into two groups of 40 coins and 10 coins. Flip all coins of the group with 10 coins.#4 Sand timersYou have two sand timers, which can show 4 minutes and 7 minutes respectively. Use both the sand timers(at a time or one after other or any other combination) and measure a time of 9 minutes.Answer:Start the 7 minute sand timer and the 4 minute sand timer.Once the 4 minute sand timer ends turn it upside down instantly.Once the 7 minute sand timer ends turn it upside down instantly.After the 4 minute sand timer ends turn the 7 minute sand timer upside down(it has now minute of sand in it)So effectively 8 + 1 = 9.#5 Chaos in the busThere is a bus with 100 labeled seats (labeled from 1 to 100). There are 100 persons standing in a queue. Persons are also labelled from 1 to 100.People board on the bus in sequence from 1 to n. The rule is, if person ‘i’ boards the bus, he checks if seat ‘i’ is empty. If it is empty, he sits there, else he randomly picks an empty seat and sit there. Given that 1st person picks seat randomly, find the probability that 100th person sits on his place i.e. 100th seat.Answer:The final answer is the probability that the last person ends in up in his proper seat is exactly 1/2The reasoning goes as follows:First, observe that the fate of the last person is determined the moment either the first or the last seat is selected! This is because the last person will either get the first seat or the last seat. Any other seat will necessarily be taken by the time the last guy gets to ‘choose’.Since at each choice step, the first or last is equally probable to be taken, the last person will get either the first or last with equal probability: 1/2.#6 Mad men in a circleN persons are standing in a circle. They are labelled from 1 to N in clockwise order. Every one of them is holding a gun and can shoot a person on his left. Starting from person 1, they starts shooting in order e.g for N=100, person 1 shoots person 2, then person 3 shoots person 4, then person 5 shoots person 6……..then person 99 shoots person 100, then person 1 shoots person 3, then person 5 shoots person 7……and it continues till all are dead except one. What’s the index of that last person ?Answer:Write 100 in binary, which is 1100100 and take the complement which is 11011 and it is 27. Subtract the complement from the original number. So 100 – 27 = 73.Try it out for 50 people. 50 = 110010 in binary.Complement is 1101 = 13. Therefore, 50 – 13 = 37.For the number in form 2^n, it will be the first person. Let’s take an example:64 = 1000000Complement = 111111 = 63.64-63 = 1.You can apply this for any ’n’.#7 Lazy people need to be smartFour glasses are placed on the corners of a square Lazy Susan (a square plate which can rotate about it’s center). Some of the glasses are upright (up) and some upside-down (down).A blindfolded person is seated next to the Lazy Susan and is required to re-arrange the glasses so that they are all up or all down, either arrangement being acceptable (which will be signalled by say ringing of a bell).The glasses may be rearranged in turns with subject to the following rules: Any two glasses may be inspected in one turn and after feeling their orientation the person may reverse the orientation of either, neither or both glasses. After each turn the Lazy Susan is rotated through a random angle.The puzzle is to devise an algorithm which allows the blindfolded person to ensure that all glasses have the same orientation (either up or down) in a finite number of turns. (The algorithm must be deterministic, i.e. non-probabilistic )Answer:This algorithm guarantees that the bell will ring in at most five turns:On the first turn, choose a diagonally opposite pair of glasses and turn both glasses up.On the second turn, choose two adjacent glasses at least one will be up as a result of the previous step. If the other is down, turn it up as well. If the bell does not ring, then there are now three glasses up and one down.On the third turn, choose a diagonally opposite pair of glasses. If one is down, turn it up and the bell will ring. If both are up, turn one down. There are now two glasses down, and they must be adjacent.On the fourth turn, choose two adjacent glasses and reverse both. If both were in the same orientation then the bell will ring. Otherwise there are now two glasses down and they must be diagonally opposite.On the fifth turn, choose a diagonally opposite pair of glasses and reverse both. The bell will ring.注:本文摘自派博网入驻自媒体—灯塔信息,转载请注明泉源,微信搜刮“派博网”获取更多信息资讯。?责任编辑:陈卓阳

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