Technology summer analyst Interview Questions | Glassdoor

# Technology summer analyst Interview Questions

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### Investment Banking Summer Analyst at J.P. Morgan was asked...

Jul 11, 2012
 What are the methods of valuing a company? 2 AnswersIt's DCF(Discounted casfhflow) & Relative method, where FCFE(Free cash flow to Equity) is vital for equity shareholders & FCFF(Free cash flow to Firm) is vital for the company. One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

Feb 26, 2010

Mar 11, 2011

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

Mar 23, 2010

Oct 4, 2011
 Say I take a rubber band and randomly cut it into three pieces. What's the probability that one of the pieces has length greater than 1/2 of the original circumference of the rubber band.9 Answers3/4Suppose you have two cuts on the rubber band placed randomly. The probability of having one segment greater than half the circumference is the probability that the third cut will be inside the combined range of 90* to either side of the cuts. Since the average distance between the first two cuts is also 90*, the combined range is 270*, or 3/4 of the circle.You need 3 cuts to end up with 3 pieces. The first cut doesn't matter. The second cut can also be anywhere and the largest piece will still be at least half the circumference. What matters is the third cut, which should lie in the same half as the second cut. So the probability is actually 1/2.Show More ResponsesThe correct answer is 3/4, as this problem is equivalent to the famous 3-points-on-semicircle problem. Why? If one of the pieces has length greater than 1/2 the circumference, then the three points of cutting must lie in the same semicircle. On other hand, if the three points of cutting lie on the same semicircle, then the longest piece must be at least 1/2 of the circumference.For reference to the 3-points-on-same-semicircle problem, see e.g., http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html1/4 1 -3/4suppose I have two points whose minor arc distance is t <= 1/2. Then the range of semicircles covering both points gives an arc length of (1/2+1/2)-t = 1-t. say we fix the first point, tracing the second point around gives minor arc lengths from 0 to 1/2 and then 1/2 to 0. Therefore the answer is 2*integral (1-t) from 0 to 1/2, which is 2(1/2-1/8) = 3/4It's 3/4. Cut it into 1 piece make a line. Cut it close. Pretend the length is 100. If you cut the first at x=1, as long as it isn't between x=50-51, it will have a length greater than 50% so there's 99% chance. You can imagine that if the cut was infinitely close to the end it would be about 100%. Now cut at x=2 you can't do between x=50-52. For x=3 it's 50-53 etc. So when you get to right to infinitely close to 50 it is pretty much between x=50-100 so there is a 50% chance you hit your spot. (obviously 50-50 is 100%, but since this length is continuous there's little chance it lands on that point). Obviously since this is symmetrical you can see this pattern going from 50% to 100% at the other end. Since each point on the continuous line has the same probability of happening the answer is clearly 75%.This problem is also equivalent to the probability that, if you have a line segment from 0 to 1 and you make 2 random cuts on that line segment, what is the probability that the three resulting pieces do NOT make a triangle?

### Quantitative Finance Summer Associate at Morgan Stanley was asked...

Aug 5, 2010
 At a party everyone shakes hands, 66 hand shakes occur, how many people are at the party? 9 Answers12requires explanationit's not 12, it's 11Show More Responses12sum of the series 66= n/2( 2*1 + (n-1)*1) n=12You can suppose there are n people in the room and think of them in a row. The first one has to shake hands with (n-1) people (because he doesn't have to shake hands with himself). The second one has already shaken hands with the first one, so he has (n-2) shakes remaining... and so on. So you have to sum: (n-1)+(n-2)+(n-3)+...+1= (n/2)*(n-1) Then you have to solve (n/2)*(n-1)=66 and you get n=12.n(n-1)/2=66 so n=12You can think of this problem as a combination problem: (N choose 2) = 66, and then solve for N. That is, N! / (2! (N-2)! ) = 66. Simplifying this equation leads to N(N-1)/2 = 66, and the integer solution to this problem is N = 12.66=x!/((x-2)!*2) which gives 12

### Summer Intern at Five Rings Capital was asked...

Apr 25, 2012
 • Is 1027 a prime number? • How would you write an algorithm that identifies prime numbers? • 2 blue and 2 red balls, in a box, no replacing. Guess the color of the ball, you receive a dollar if you are correct. What is the dollar amount you would pay to play this game? 8 AnswersAn algorithm for testing prime numbers is trial testing, test whether whether the number is dividable by an integer from 2 to its square root. For the color guessing game, the expected number of dollars you get is the average identity between a permutation of rrbb and rrbb, which is 2.For the prime number testing, only the number 2 and then odd numbers need to be tested. If it is not divisible by 2, there is no need to test against any other even number. So start with 2, then 3, then increment by 2 after that (3,7,9,...) until you are greater than the square root (then it's prime), or you find a divisible factor (it is not prime). To test for divisibility, we are looking for a remainder of zero - use a MOD function if available. Taking the integer portion of the quotient and subtracting from the actual quotient: if the difference is zero, then the remainder is zero and we have a divisible factor. If the difference is nonzero, then it is not divisible and continue testing. In this case, we find that dividing by 13 gives 79 with no remainder, so it is not prime.For the guessing game, the minimum winnings are \$2 every time with the proper strategy. I'm assuming the rules are you pay to play and you get to guess until there are no more marbles. Say you guess wrong the first attempt. (you guess blue and it was red). So now you know there are 2 blue, 1 red. Your logical choice is to choose blue again, since there are more of them. But say you guess wrong again. Now you know there are 2 blue left, so you will win on both of the last 2 draws. If you were correct on one or both of the first two trials, then you could wind up with an even chance on the third trial, so you would win that some of the time, then you'll always win on the last trial.Show More ResponsesDavid, I think we could pay more that \$2 and still come out on top. You logic seems sound, but looking at the probabilities I see: 1/2+1/3*(2)+2/3*(5/2) = 17/6 = ~2.83 Choosing the first ball, we obviously have an expected value of 1/2. Then, WLOG, we are left with RRB. Clearly we then choose R as this gives us a 2/3 shot at picking correctly. If it is R, then we get that \$1, have a 50% shot at the next, and are assured the last, giving us, on average, \$2.5. If it is B, then we know the next two will be R, giving us \$2. As you can see, with an optimal strategy, we should expect to make ~\$2.83 per round.Take the square root fo 1027. You get 32.04. Need only to check if divisible by prime numbers from 1 to 32, which include 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and 31 For algorithm, see Lucas' test on Wikipedia, where there is also pseudocode.1027 = 1000 + 27 = 10^3 + 3^3 and you know you can factor a^3 + b^31027 = 2^10-1 = (2^5-1)(2^5+1) prime number ez draw ball worths 17/6 dollars, the first draw worths 0.5, the rest worth(2/3 * 2.5 + 1/3 * 2)1027 = 2^10-1 = (2^5-1)(2^5+1) prime number ez draw ball worths 17/6 dollars, the first draw worths 0.5, the rest worth(2/3 * 2.5 + 1/3 * 2)

### Quantitative Researcher Summer Intern at Jane Street was asked...

Apr 17, 2011
 2) A. 10 ropes, each one has one red end and one blue end. Each time, take out a red and a blue end, make them together. Repeat 10 times. The expectation of the number of loops. B. 10 ropes, no color. All the other remains the same.7 Answers1/10 + 1/9 +...+ 1 ? B is similar..1/19+1/17+etc in BE[n] = 1/n + (n-1)/n*E[n-1] = 1/n + E[n-1] For the case of n=10, you would sum up all of the numbers from 1 to 10: 1/10+1/9+ 1/8 + 1/7 ... + 1/2Show More Responsesadd an extra 1 to the previous answerFor part A), the answer is 1+1/2+1/3+...+1/10. For part B), the answer is 1+1/3+1/5+...+1/19. Explanations: For part A), ctofmtcristo has the right approach but with a typo in the equation for E[n]. To obtain the expected number of loops, we note that the first red has a 1/n chance of connecting with its opposite blue end (and forming a loop) and a (n-1)/n chance of connecting with a different rope's blue end (and not yet forming a loop), so E[n] = 1/n*(1+E[n-1]) + (n-1)/n*E[n-1] = 1/n + E[n-1], with base case E[1]=1. Then, by induction, we get E[n] = 1+1/2+1/3+...+1/n. Part B) is similar. We note that the first end now has 2n-1 possible ends to connect to, of which 1 of them is its opposite end and 2n-2 of them belong to a different rope. Then, E[n] = 1/(2n-1)*(1+E[n-1]) + (2n-2)/(2n-1)*E[n-1] = 1/(2n-1) + E[n-1], with base case E(1)=1. By induction, E[n] = 1+1/3+1/5+...+1/(2n-1).Ed's anwser is not right. Just check for the case of 3 pairs. So total cases is 3!=6. 1 case with 3 loops, 2 cases with all wrongly attached, and 3 cases with 1 loop. so expected value is (3/6)*(1) + (1/6)*(3) = 6/6 = 1... and Eds anwser gives 1+1/2 +1/3 = 11/6, which is wrong clearly.Timi, you are missing the fact that if they are "all wrongly attached" then they form a loop. Similarly, the case you are thinking of "with 1 loop" actually has 2 loops. The correct answer is still 11/6.