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Quantitative Interview questions and answers

Deutsche bank trading floor position questions(I wanna be asked something like this, they are so simple):


1.A subway station. There are trains going in each direction. For each direction there is a train every 3 min. Stops are fast, say 1 sec. One direction is to work, the other one is clubbing.
A blondie decided to do the following: go to the station, wait for the first train and take it.
After some time she realized she goes to work 2x more than clubbing. How come?

2. A round-robin tournament. Originally the scoring system was 0 if you lose, 1 for a draw, 2 for a win. The new system is 0-1-3(3 for a win). Is it possible that some team got the last place under the old system and the first under the new system?

2b. OK, that was easy. Now suppose that under the old scoring system all teams got different total scores. Is it possible that under the new system their positions will reversed? The first switches with the last, the second last switches with the 2nd, etc.

3. There is a group of 20 kids, every two have a common grandpa. Prove that there is
a grandpa that has at least 14 grand kids.

My friend was asked an absolutely hilarious question at an interview for a risk quant(he was hired): What confidence levels do you know? :)))))))))
 
Deutsche bank trading floor position questions(I wanna be asked something like this, they are so simple):


1.A subway station. There are trains going in each direction. For each direction there is a train every 3 min. Stops are fast, say 1 sec. One direction is to work, the other one is clubbing.
A blondie decided to do the following: go to the station, wait for the first train and take it.

After some time she realized she goes to work 2x more than clubbing. How come?
The train going to work shows up exactly one minute after the train to the clubs. Thus, there is a 2/3 chance she shows up in the two minutes before a train to work will arrive; there is a 1/3 chance she shows up in the minute before a train to the clubs.


2. A round-robin tournament. Originally the scoring system was 0 if you lose, 1 for a draw, 2 for a win. The new system is 0-1-3(3 for a win). Is it possible that some team got the last place under the old system and the first under the new system?
Yes. Assume that each team plays 40 games. One team wins 18 and loses the rest. The other teams lose nine, win eleven and tie twenty each.


3. There is a group of 20 kids, every two have a common grandpa. Prove that there is
a grandpa that has at least 14 grand kids.
Let's assume that a grandpa can have at most 13 grandkids. Then there exists a grandchild who is connected to 12 cousins (Grandpa A) via one grandparent and 7 cousins (Grandpa B) via another grandparent.

His 12 cousins have to be connected to his 7 other cousins by Grandpa B or other entirely different grandpas, since Grandpa A already has as many grandkids as he can have. Let's first prove that we can't have more than one extra grandpa. Let's assume that there exists two grandchild pairs with one grandchild of each parent, that are connected by two different grandparents, (1A, 1B), (2A, 2B). Then 1A is not a cousin of 2B, since both his grandparents are taken up and he is not related. Therefore, we can only have one other grandparent.

Since we can have only one other grandparent and Grandpa B can take only five more grandchildren before he hits the maximum of thirteen, there are seven remaining grandhildren of Grandpa A that must be connected to all seven children of Grandpa B, we must exceed the 13 grandchild limit.

This proof can be extended inductively from a maximum of 1 child up to 13. At this point, I can go ahead and make that proof for the interviewer, but chances are, he will accept that it can be proven relatively easily.

My friend was asked an absolutely hilarious question at an interview for a risk quant(he was hired): What confidence levels do you know? :)))))))))[/QUOTE]
 
Let's assume that a grandpa can have at most 13 grandkids.
Sure, prove by contradiction.
Then there exists a grandchild who is connected to 12 cousins (Grandpa A) via one grandparent and 7 cousins (Grandpa B) via another grandparent.
Well, that's wrong. It should be "who is connected to at most 12 cousins".

Anyway, the question is trivial: first you assume there exits two grandpas A and B that have a grand kid in common, then you argue that there could be at most one extra grandpa C(otherwise you can get a contradiction with "any 2 kids have a grandpa in common"), and since there are only 3 of them and 40 positions to fill(each kids has 2 grandpas), 40/3>13
since 40 = 13 + 13 + 13 + 1.
 
Well, that's wrong. It should be "who is connected to at most 12 cousins".
Well, as I've said, I've just proved that you can't do it for 13 grandkids and that there also exists an inductive proof for 12, 11, 10, 9, etc grandkids that is a little more complicated and probably out of the scope of a 10-minute interview question.

(otherwise you can get a contradiction with "any 2 kids have a grandpa in common"),
I thought about taking that route, too, but the fourth possible grandpa part is non-trivial. You have to prove that if there are four grandpas, you'll get a situation where two kids aren't cousins for all situations where each grandpa has as least one grandchild. And there clearly are situations where you can have more than three grandpas- just ones where one grandpa has more than 13 grandkids (IE: Grandpa A has 20 grandkids and Grandpas B-T have 1 grandkid each) You therefore have to prove that such situations can be reduced to ones with fewer grandpas, or take the more direct route and prove that if one grandpa has 13 kids, another grandpa must have at least 14 and then extrapolate that to an inductive proof from a base case of 10-or-fewer grandkids.
 
If you re-read my answer, I state that I've just proved that you can't do it for 13 grandkids and that there also exists an inductive proof for 12, 11, 10, 9, etc grandkids that is a little more complicated and out of the scope of a 10-minute interview question.
Then you should re-word your solution.
I thought about taking that route, too, but the fourth possible grandpa part is non-trivial. You have to prove that if there are four grandpas, you'll get a situation where two kids aren't cousins for all situations where each grandpa has as least one grandchild.
Think about it this way: Take a random kid "1". If he has only one grandpa, then everybody is connected to him via that gradpa => solved. Hence he(and all other kids) have two grandpas. "1" has 2 grandpas, call them A and B. Everybody else has A or B as their grandpa too, so A and B together cover all 20.
Consider the two sets of kids who only have A or B as their granpdpa and not both, call them A_1 and B_1. Then there exists _some_ other grandpa C, who has a kid in A_1. Therefore all kids in B_1 must have C as their second grandpa(otherwise u can find two kinds with grandpas (A,C) and (B,D), contradiction). Since all kids in B_1 have C, it follows that all in A_1 also must have C as their grandpa. Hence there are only 3 granddads. Then 40 = 13 + 13 + 13 +1, solved.

By the way, 2b) is as easy as 2).
As for the confidence levels, my friend said: "err, eh... 95%? 99%? 0%? 75%?". The interviewer was happy :D
 
Problem 1

You are on a hiking trip and you start at the base of a mountain and start climbing it to go to the top. You start exactly at sunrise and take the narrow winding path which spirals upwards towards the summit. You ascend at varying rates of speed, stopping and resting on the way and finally you reach the top in the evening just before sunset. After an overnight rest at a camp on the top the next morning, exactly at sunrise you start descending to the base of the mountain by taking exactly the same path that you took on your way up. Again, you descend at varying speed and rest on the way. Obviously, your average speed of descent on the way down is lot greater than your average speed up. Which of the following is true?

a) There is a unique spot along the path that you will occupy on both trips - on the way up and on the way down – at precisely the same time of the day;
b) There are two unique spots along the path that you will occupy on both trips – on the way up and down – at precisely the same time of the day;
c) There cannot be any single unique spot along the path that you will occupy on both trips – on the way up and on the way down – at precisely the same time of the day;
d) If your average speed on the way up and on the way down is the same then there will be a single spot that you’ll occupy on both trips at precisely the same time of the day.

Problem 2

If 0 (zero) is opposite of 1 (one) and therefore, 1 (one) is opposite of 0 (zero), then by this logic which of the following pairs are opposites of each other:

a) 3/4 and 1/4
b) 2/3 and 1/3
c) 1/2 and 1/2
d) All the above;

Problem 3

Which number will come next in this sequence: 6, 28, 496, .........

a) 852
b) 1,656
c) 8,128
d) 17,212

Problem 4

If there exists are super computer that can do more calculations simultaneously than there are particles in the universe, then it stands out to reason that:

a) the Universe is finite;
b) the Universe is infinite;
c) the Universe has insufficient computing resources to carry all the calculations;
d) the Universe itself becomes a super-computing machine;

Problem 5

According to the generalized version of Uncertainty principle in physics and engineering it is impossible to determine all properties of a particle at any point in time. Therefore, it stands to reason that if there is a particle (an object) P, then:

a) it is possible to make a perfect copy of this particle (object);
b) it is impossible to make a perfect copy of this particle (object);
c) the particle (object), P, will exist with its twin particle (object), P*;
d) the particle (object), P, after an infinite point of time will not exist;

Answers:

a
d
c
c
b
 
The first problem can be solved by simply drawing the graph of altitude Vs time of the day, for both the days (on the same plot). The two curves have to meet somewhere. Q.E.D
 
is problem 2 d??

if you loook at 0,1 on the real number line, there both reflected at a 1/2 with 0,1 being the extreme points. hence 1/2 is the mid point of (0,1)

for a ) add 1/4 to 0 and you get a 1/4. Subtract 1/4 from 1 and you get 3/4.
the pair (1/4, 3/4) the mid point is 1/2.

for b) add 1/3 to 0 and you get a 1/3. Subtract 1/3 from 1 and you 2/3.
the pair (1/3, 2/3) the mid point is 1/2.

for c) add 1/2 to 0 and you get a 1/2. Subtract 1/2 from 1 and you get 1/2.
the pair (1/2, 1/2) the mid point is 1/2.
 
So, going WAY back... In the first post the X^X^X^X^X... = 2 one, if you say x = 2^(1/2) then it diverges. Any number greater than 1 will diverge. Therefore it must be 1?

Another way to look at it is if X^X^X^X^X...(lnx) = X^X^X^X^X... then ln(x) must equal 1 right? So then you would say e, which is clearly not the right answer as e is already greater than 2. Therefore it seems that there is some issue with using log properties with this particular problem.

George, I think an easier way to say it is they turned compliments into opposites.
 
I don't understand how you arrived at c for #3, and as for #4, I'd argue that a should be the correct answer, since if a computer can carry out more calculations in a second than there are particles in the universe, and that this number (# of calculations) is a finite number (however large), then if it is greater than the number of particles in the universe, then the universe itself should be finite.

#3 I cannot come up with the pattern at all.
 
For (3) this is a sequence of "perfect numbers", numbers which are the sum of their divisors.

For (4) a finite number of particles can occupy a finite, and even an infinite amount of space. There's nothing about having a computer this fast the restricts either case. That leaves (c) or (d). (d) makes no sense, since nothing is said about using the Universe as a computer, hence (c) must be correct.

For (2) can't we just say the numbers add up to one?
 
Ah, "quantitative" questions from RiskLatte. I couldn't believe the amount of junk you could find at websites, doing so called serious "math stuff" (and their application to finance).

For (3) this is a sequence of "perfect numbers", numbers which are the sum of their divisors.

This kind of questions are always ridiculous, yes it maybe the sequence of prefect numbers, or it maybe the sequence of 5-infinitary perfect numbers, or it maybe 2^(p-1)*(2^p-1) where p is a prime, or it maybe a(n)=b(n)*(b(n)+1)/2 where b(n) are primes of form 2^p-1, where p is a prime number that is not the sum, minus 1, of a Pythagorean triple, etc.

Problem 2

If 0 (zero) is opposite of 1 (one) and therefore(!), 1 (one) is opposite of 0 (zero), then by this logic which of the following pairs are opposites of each other:

Try telling that to someone who does logic, brain meltdown in a second :D.

Anyway, you can check it out at Risk Latte - The Financial Engineering Resource. Enjoy!
 
For (3) this is a sequence of "perfect numbers", numbers which are the sum of their divisors.

For (4) a finite number of particles can occupy a finite, and even an infinite amount of space. There's nothing about having a computer this fast the restricts either case. That leaves (c) or (d). (d) makes no sense, since nothing is said about using the Universe as a computer, hence (c) must be correct.

For (2) can't we just say the numbers add up to one?

So basically one either knows the perfect numbers pattern or one doesn't...joy. As for #4...this is true...c still seems nonsensical anyway.
 
Unfortunately some of the questions and answers in the following list are just badly stated or incorrect.

Problem 2

What logic?

The logic that (i) the sum of the numbers is 1, or the logic that (ii) the product of the number is 0, or the logic that (iii) the absolute value of the difference of the numbers times the product of the numbers is zero, or (iv) three times the absolute value of the difference of the two numbers is either three or one?

All four logics are satisfied by 1 and 0, whereas of the choices (a, b, c, d) provided only 1/2 and 1/2 satisfy (iii) and only 2/3 and 1/3 satisfy (iv).

Unless the logic is uniquely determinable given the available information, there is no way to come up with a unique answer!


Problem 3

Anything goes! I can come up with a scheme that will justify any one of the answers. Dumb question!

Problem 4

What is the relationship between a calculation and particles? It is not specified how many particles are needed for a calculation. "Calculation" is not defined in terms of events happening to components of a computer, themselves part of the universe.


Problem 5

There is a big difference between making a true copy of a particle, and whether or not we have a complete knowledge of all the features of the particle (copied). Why is it logically impossible to devise a method that makes true copies of particles and yet it does not reveal to us the full range of the features of the particle copied? Maybe there is an indirect way that we can tell the particle is truthfully copied without ever knowing the full range of the features of the particle.


Problem 1

You are on a hiking trip and you start at the base of a mountain and start climbing it to go to the top. You start exactly at sunrise and take the narrow winding path which spirals upwards towards the summit. You ascend at varying rates of speed, stopping and resting on the way and finally you reach the top in the evening just before sunset. After an overnight rest at a camp on the top the next morning, exactly at sunrise you start descending to the base of the mountain by taking exactly the same path that you took on your way up. Again, you descend at varying speed and rest on the way. Obviously, your average speed of descent on the way down is lot greater than your average speed up. Which of the following is true?

a) There is a unique spot along the path that you will occupy on both trips - on the way up and on the way down at precisely the same time of the day;
b) There are two unique spots along the path that you will occupy on both trips on the way up and down at precisely the same time of the day;
c) There cannot be any single unique spot along the path that you will occupy on both trips on the way up and on the way down at precisely the same time of the day;
d) If your average speed on the way up and on the way down is the same then there will be a single spot that youll occupy on both trips at precisely the same time of the day.

Problem 2

If 0 (zero) is opposite of 1 (one) and therefore, 1 (one) is opposite of 0 (zero), then by this logic which of the following pairs are opposites of each other:

a) 3/4 and 1/4
b) 2/3 and 1/3
c) 1/2 and 1/2
d) All the above;

Problem 3

Which number will come next in this sequence: 6, 28, 496, .........

a) 852
b) 1,656
c) 8,128
d) 17,212

Problem 4

If there exists are super computer that can do more calculations simultaneously than there are particles in the universe, then it stands out to reason that:

a) the Universe is finite;
b) the Universe is infinite;
c) the Universe has insufficient computing resources to carry all the calculations;
d) the Universe itself becomes a super-computing machine;

Problem 5

According to the generalized version of Uncertainty principle in physics and engineering it is impossible to determine all properties of a particle at any point in time. Therefore, it stands to reason that if there is a particle (an object) P, then:

a) it is possible to make a perfect copy of this particle (object);
b) it is impossible to make a perfect copy of this particle (object);
c) the particle (object), P, will exist with its twin particle (object), P*;
d) the particle (object), P, after an infinite point of time will not exist;

Answers:

a
d
c
c
b
 
Problem 4

What is the relationship between a calculation and particles? It is not specified how many particles are needed for a calculation. "Calculation" is not defined in terms of events happening to components of a computer, themselves part of the universe.

I believe how it works is a transistor exists in one state for a 1, and another for a 0, with one transistor able to handle 1 calculation at a time. Each transistor must be a unique particle. If it can can do more calculations than there are particles in the universe, then the universe is not composed of enough particles to build this computer. When it is said to exist, I believe it is assuming it exists somewhere outside the universe somehow.

The reason it is not also A is because you can assume that this computer has infinite computing power, and saying the universe has less computing power doesn't mean necessarily that it is finite (nor does it mean that it's infinite). Therefore you can not draw a conclusion about either A or B. Calc 2 anybody?

That's atleast what I thought of when I first saw it.

Anyway, anyone know anything more about the X^X^X^X.... one?
 
WilliamRFrost,
The (x^{x^{x^{x^{x^{...}}}}}=2) question has been solved in post #9. Your logic is flawed. Just try to pick some number for upper and lower bound. 1 is your lower bound since the sequence will equal 1. If you pick x=2, it's easy to see it will increase very fast. So x must be a number between 1 and 2, right. Using log is a clever way to deal with recursive exponent problem like this.
 
Well then it should never go above 2 when I take [2^(1/2)]^[2^(1/2)]^[2^(1/2)]... but it does after just 1 iteration it is 2, and then it grows very fast. Just plug it in 3 times, already above 2.
 
Well then it should never go above 2 when I take [2^(1/2)]^[2^(1/2)]^[2^(1/2)]... but it does after just 1 iteration it is 2, and then it grows very fast. Just plug it in 3 times, already above 2.

After one iteration is it (1.414)^(1.414) which is approximately 1.635.
 
Sorry, you're right, you need more iterations, but once you go more iterations, you'll be above 2. The 3rd iteration is 2, the 4th is higher.
 
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