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Jane Street
Er werd een Quant Trader gevraagd...5 oktober 2011

brainteaser: 30 people are 15 couples. they shake hands, but two people of the same couple don't shake hands with each other. now you know 29 people shake hands with different numbers of people (e.g. one person shakes hands with 4 others, another person shakes hands with 5 people, etc). question: how many people did the 30th person shake hands with?

5 antwoorden

Yes, the ans is 14 and it is elaborated as follows. 1. The max no. of hands a person can shake is 28. And u know 29 people shake different numbers of hands. So, the number of hands that those 29 people have shaken can only be the 29 non-repetitive numbers from 0 to 28. 2. We look at the two people who shook hands 0 and 28 times. It is no hard to deduce that they must be a couple. Assume wife shook 0 and husband shook 28 times. The husband must have shaken hands with the 30th person. 3. As the husband shook hands with everybody other than his wife and the wife shook no hand, if we kick out this couple, the numbers of hands shaken by the remaining couples (which is 1 to 27), will reduce by 1, which is 0 to 26. Due to the same reason, (0,26) is a couple. Assume 0 is the wife and 26 is the husband and the husband must have shaken hands with the 30th person. 4. By repeating the above steps, we can easily discover that the 30th person (and his/her spouse) must have shaken hands with every husband. Minder

Mickey is soooo genius

Sorry, I don't understand your answer. Could you please explain it in detail? Thx! Minder

Weergeven Meer reacties

I flip a coin three times, and tell you I got heads on one of the flips. What's the probability I got 2 heads and one tails (not necessarily in that order).

3 antwoorden

answer: 3/7 Total 7 possibilities of coins: 3 heads: HHH 2 heads: HHT , HTH, THH 1 heads: TTH, THT, HTT Out of these # possibilities with 1 heads / # possibilities overall = 3/7 Minder

1/3 Only 3 possibilities - 1H,2H,3H

First flip being heads should not factor into the calculation when working with probability. The question becomes what is the probability of flipping a coin twice and getting one heads and one tails (in no specific order). There is 100% of flipping heads or tails on first flip. then the next flip has to be the opposite of what was just flipped.. there is a 50% probability of the third flip being the opposite of the second. Answer: 50% Minder


How do you check if your regression parameters are significant? Given a number's prime number decomposition, how many ways are there of factorizing it into two different factors?

3 antwoorden

regarding my above answer: the upper limit on the product \Pi should be n. That is, \Pi_{i=1}^n Minder

your assumption of relatively prime makes the problem easy...i don't think that's the goal...try to answer the actual question without making sweeping assumptions!! Minder

answer to prime number question: I am assuming that the 2 factors you factorize it into are relatively prime, in that they have no common factors. For a number N, suppose it is represented as a product of prime number N=\Pi_i p_i^(e_i), where p_i is a prime number and e_i represents the number of such primes in N's prime number factorization. For N=xy, we see that for each p_i, we can choose whether ALL p_i factors go to x or go to y (2 choices). it cannot be that we take some of the p_i factors and give them to x, then give the others to y because otherwise x and y will have common factors. Thus, for each p_i we have 2 choices as to where it can go: x or y. So nominally we have 2^n possibilities. However, since (a)(b) and (b)(a) are considered the same factorization, we over count by a factor of 2, so really we have 2^(n-1) possibilities. Minder

Flow Traders

Brain-teasers and general ETF pricing questions.

2 antwoorden

Can you tell us about the final case?

Could you give any examples of a brainteaser they gave you?

Jane Street

Two teams A and B played football twice. Two independent matches. The average shoot of A in match 1 is greater than B. The average shoot of A in match 2 is also greater than B. So can we say the average shoot of A is greater than B in two matches together? If so, why? If not, give me an example. -------- Failed to give the example, but I believe that the statement is not right.

2 antwoorden

let's say A does 1/100, and B does 0/1 in the first game. And in the second, A does 1/1, B does 99/100. A has a bigger average in both games, but total, A goes 2/101, whereas B is 99/101 Minder

Counterexmple: first game team A scores 1 out of 2, team B scores 0 out of 1; second game team A scores 1 out of 1, team B scores 8 out of 9. In both cases team A was better, but in total team B was better. Minder

Jane Street

What's the weight of each of five melons if their pairwise weights sum to 16,17,18,19,20,21,22,24,25, and 26?

2 antwoorden

The five weights a

assuming that the weight of each melon from lightest to the heaviest is a,b,c,d,e then 1. a+b+c+d+e=52.5 (because picking 2 from 5 has 10 combineation, and sum of all combination is 4(a+b+c+d+e)=210) 2. a+b=16 3. d+e=26 4. a+c=17 5. c+e=25 or c+d=25 test the result to see which is right finally, a=6.5, b=9.5, c=10.5, d=11.5, d=14.5 Minder

Graviton Research Capital

Q. 7x7 matrix which is symmetric, each column has numbers 1 to 7. Prob that diagonal has all numbers from 1 to 7? Q. Game of Nim. you can pick any number of rocks from 1 to 6, describe your winnning strategy Q. You have to go from 0,0 to 8,8 using E and N. number of ways to do so in which direction changes are even.

2 antwoorden

Q1> Probability = 1, as it is symmetric, and each number appears 7 times = 2n+1 times, so it must appear once on the diagonal Q3> Both the start and end steps are same ( starts or ends with N or E at 0,0 and 8,8) So total ways = 2*(14!)/(8!6!) Minder

Q1> Probability = 1, as it is symmetric, and each number appears 7 times = 2n+1 times, so it must appear once on the diagonal Q3> Both the start and end steps are same ( starts or ends with N or E at 0,0 and 8,8) So total ways = 2*(14!)/(8!6!) Minder

Jane Street

How many people do you need to have in a room before you know for sure 5 have been born in the same month

2 antwoorden

49, since you could have 4 for each month and then another person.

This is called "pigeonhole principle". The number of people n to make sure j of them are born in the same month is: n = 12*(j - 1) + 1 In our case j = 5, which leads to n = 49. Minder

IMC Trading

Why are you interested in trading?

2 antwoorden

What were the final round interviews like? behavioral/technical

how long did they take to get back to you after the hr interview?

AlphaGrep Securities

What are the projects done in C++?

1 antwoorden

Explained in detail about each project

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