Analyst Interview Questions in London, UK | Glassdoor

# Analyst Interview Questions in London, UK

"Analysts work in a variety of fields to break down complex problems and find solutions. When interviewing candidates, employers are looking for applicants who have strong analytical and problem-solving skills as well as an in-depth knowledge of the field. For more information on the specific questions you'll be asked, try researching a particular role such as business analyst, financial analyst, programming analyst, or data analyst."

## Top Interview Questions

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### Analyst at J.P. Morgan was asked...

Apr 3, 2011
 puzzle 9 balls, identical in appearance. 8 are of same weight. 1 is heavier. Identify this heavier ball by using a scale twice.5 Answersput 9 balls in groups of 3.Split them into two sets of 4 and 1 individual ball. Place the two sets on each side of the balance. If the sets balance out, then the individual ball is the heavier one. If the sets don't balance, then take the heavier set and split it up into two sets of two and repeat.I believe that an explanation to the solution that Interview Candidate is suggesting can be found here: http://brainteaserbible.com/snooker-balls-weighing-scalesShow More ResponsesSplit the balls into groups of three, compare two of the groups. If the sets balance then the heavy ball is in the third group, otherwise it is in the group left out. Whichever it is, take two balls from that group to weigh. If they balance, you know it is the third ball, if they don't, you have the heavier ball.http://brainteaserbible.com/snooker-balls-weighing-scales

### Quantitative Analyst at Goldman Sachs was asked...

Mar 21, 2012
 a, b, c are integers. Such that a^2 + 2bc = 1; b^2 + 2ac=2012, find all the possibles values of c^2 + 2ab. 7 Answers2011 is primeMore details in addition to the last post. 1) subtracting the 1st equation from the 2nd one gives: (b-a)*(b+a-2c)＝2011 2) 2011 is prime, so there are only two possibilities: b-a=1, b+a-2c=2011; or b-a=2011, b+a-2c=1 3) subtracting the 1st equation from the 3rd equation gives: (c-a)(c+a-2b)=x-1 4) plugging c-a=0.5*[(b-a)-(b+a-2c)], c+a-2b=(c-a)-2(b-a) into 3) gives you xIn 2) there should be four rather than two possibilities. Both terms could be negative.Show More Responses2 cand: heavy one... I have no idea how did you figure out that 2011 is prime! Do you remember the table of prime numbers from 2 to 10000?? I easily found trivial solution b=0 a=+-1, and then got stucked. however, later I found that this is unique solution, and (3)=c^2= 1006^2. I've used your approach. You've got a 4 possibilities, +-1 +-2011 and +-2011 +-1, but 2 of them lead to complex roots. You see, sum (1)+(2)+(3) gives you (a+b+c)^2=1+2012+x, thus x>=-2013, so (b-a) may be only +-1, not +-2011. The answer is (3)=1012036In the equation a^2 + 2bc = 1, obviously, a is either 1 or -1 and either b or c is zero. Then if you derive b from the second equation, b = sqrt(2012 - 2ac), c can't be zero, because sqrt(2012) is not an integer. Therefore b = 0. So, in order for b to be 0, 2012 - 2ac should be zero or 2ac = 2012. => it's either a = -1 and c = -606 or a =1 and c = 606. In any case, because b is zero, 2bc term in the 3rd equation disappears and c^2 gives an answer 367236.To the comment above, 2ab term in the 3rd equation, of coursesimple eq: (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca. from given eq, we know: (a+b+c)^2 = (2012 + 1 + UNKNOWN) = K^2 with K being anything larger than 45. 45*45 = 2025 46*46 = 2116 .. so you can figure out the missing term by: (K^2 - 2013)

### Quantitative Analyst at Goldman Sachs was asked...

May 13, 2013
 You have 10 mice and 1000 bottles of wine. You also have 24 hours before a party, and one of the bottles has been tainted with a slow acting poison, which takes 24 hours to kill a mouse. In the 24 hours you have remaining, how many bottles can you guarantee safe for human consumption (assume humans and mice react identically)? Assume the lethal dosage is insignificant relative to the size of the bottle.9 AnswersI'll say 500. Since the dosage is insignificant, I'll divide the number of bottles in half, take samples from each of the first 500 bottles, mix them up, divide by 10 and feed to each mouse. If no mouse dies after 24 hours, then the first batch is safe. else, the second batch would be served.999. It is like a binary problem. First mouse tests the first #1-500 (mixed). Second tests #1-250 and #501-750. Third one tests #1-125, #251-375, #501-625, #751-875, and so on. 10 mice with 2 status each (death/alive) could encode number of bottles up to 2^10=1024. So 10 mice is enough to find out the single bottle that tainted.http://advanceddiscrete.wikispaces.com/Mice+and+WineShow More ResponsesChapter 7.2, P.183 in Zhou's bookEncode the wine bottle numbers in binary. Give each mouse is a combination based on this table: | mouse | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |-----------|---|---|---|---|---|---|---|---|---|---|---| | wine 1 | - | - | - | - | - | - | - | - | - | - | - | - | | wine 2 | - | - | - | - | - | - | - | - | - | - | - | + | | wine 3 | - | - | - | - | - | - | - | - | - | - | + | - | | wine 4 | - | - | - | - | - | - | - | - | - | - | + | + | ... | wine 1000 | + | + | + | + | + | - | + | - | - | - | If none of the mice die, wine bottle 1 contains the poison. If only mouse 10 dies, wine bottle 2 contains the poison. ... If mice 1-5 and 8 die, wine bottle 1000 contains the poison. That systems allows for an additionally 24 bottles to be encoding, 2^10=1024.Answer is 900 10 mices vs 1000 bottles 1 mice per 100 bottles 24 hiurs one dies 9 alive 100*9=900The answer is 900. You have 1000 bottles divided between 10 mice. as dosage doesn't matter, 100 bottles can contribute to a single dosage, in which case, one mouse will die meaning the tainted batch needs to be discarded and 900 are confirmed untainted.The answer is 1023. You need to think in bit-wise way. 1023 can be represented in binary as (1111111111). Your goal should be: representing each wine label (i-th number) to each binary representation. 1000th wine will be represented with 1111101000 meaning (1,2,3,4,5,7th) mices will be used to check the toxicity of this wine. In binary way, you can assign label to up to 1023 wines. So by analysing the rats that die after 24 hrs, you can actually identify which wine is toxic or not. hope this helps.Brian was quite close, but to represent 1024 wines, you actually need 11 mice. So the maximum # of wines that one can guarantee is up to 1023.

Feb 24, 2012
 3) Imagine you are out playing golf with 2 friends. You need to decide randomly, and fairly, the player to go first. You have a fair coin with you, which you can flip. How do you use the coin to decide who goes first, with an emphasis on minimizing the number of flips you need to make?4 AnswersYou group your friends as if they're one person, so if you lose, you can make another flip for just the both of them.That doesn't seem fair as you can win on the first flip but it would take one of your friends two flips to win. So chances of winning: You: 50%, Friend1: 25%, Friend 2: 25% The only fair way that I can think of is that you do it like this: 1. You vs Friend 1 2. Winner of #1 vs Friend2 If Winner of #1, wins this then they go first, otherwise 3. Loser of #1 vs Friend2 If Friend2 wins flip #2 and #3 then they go first If everyone wins 1 flip each then you start again. This will happen 25% of the time. When you factor this in the above method finds a winner after 4 flips (3+3/4+3/16+.....) whilst giving everyone an equally fair opportunity to win.I should clarify: That method finds a winner after 2 flips 50% of the time, after 3 flips 25% and ties after three flips the other 25% of the time. On average it will take 4 flips to win although this will never happen as, if it is tied after 3 flips, then it will take at least 2 more to determine a winner.Show More ResponsesFlip the coin twice and record the result of both tosses. We'll let the sequence of flips - HH - represent you winning the coin toss. Let the sequence - TH - represent Friend 1 winning and let the sequence - TT - represent Friend 2 winning. Notice that everybody has an equal chance of winning. If the coin tosses result in the sequence HT, repeat the process and flip the coin twice again. This will take an average of 8/3 coin tosses. To find this - let x be the expected number of coin tosses this algorithm will take. Then x = 3/4 * 2 + 1/4 * (x + 2). Solving this yields x = 8/3.

### Quantitative Analyst at Morgan Stanley was asked...

Nov 4, 2011
 You have a deck of 52 cards, and you keep taking pairs of cards out of the deck. if a pair of cards are both red, then you win that pair; if a pair of cards are both black, then I win that pair; if a pair of cards has one red and one black, then it's discarded. If, after going through the whole deck, you have more pairs than I do, then you win \$1, and if I have more pairs than you do, I win \$1. What is the value of this game in the long run?7 Answers00 lol. nice oneby symmetry, probability of losing = probability of winning therefore, -1p(lose)+1p(win)=0. done, ko, finished, fiendin'Show More ResponsesNo one bothered to ask if you're paying the loser -- e.g. when you win, you get \$1, but when you lose, are you losing \$1? It's not clear from the question above, though the interviewer may have been more specific.good point J.D It can either be 0 or 0.5 depending on how you interpret the questionI think you will always have the same number of pairsWhy is it 0 for us dumb people pls

### Analyst/Developer at Goldman Sachs was asked...

Jul 12, 2011
 Given 9 ball with only one differ in weight, how to find out by measuring them only twice?4 AnswersTake 3 balls each. Weigh by keeping 3 on one side and three on other and keep the remaining 3 aside. 1. If the weigh comes as equal u know the ball with diff in weight is in the three kept aside. So again weigh by keeping one of the three on one side and one on other . If equal the third one is the one that differs in weight else ur weighing machine will tell u the one which differs in weight. 2. If initial weighing is not equal follow step one for the weigh which shows the diffDrop them from a sufficient height. If 8 balls land first your 9th is the lightest, if 1 lands first, it's heavier than rest. 1 go, do I get brownie points?The above is so stupid...even Galileo knew that couldn't happen...and that was over 300 years ago...Show More ResponsesIn theory, if you do the experiment in air, it is possible.

Oct 15, 2013

Jul 13, 2012