Quantitative Interview questions and answers

[CGiuliano]

If the chicken is not allowed to change direction, then it will lose for sure. Obviously, it would want to run toward the opposite end of the circle. Let Th be the time of the human, and Tc be the time of the chicken, one can show that:

Th/Tc = pi/4 < 1, so the human will get there before the chicken.

If the chicken is allowed to zigzag, then it can easily beat the human.

How can this be shown?

 

[goodstudent]

How can this be shown?

For the zigzag part, just think about it, if the chicken runs in the opposite direction of the human and start zigzagging, it will be closer and closer to the fence, but the human will only be going back and forth around its initial point. Assuming the human will always choose direction that has the shortest distance toward the straight line that the chicken is heading.

To show that it is impossible to beat the human without changing direction:
Let Vh be the tangential speed of human
Let Vc be the speed of chicken
Let Tc be the time for chicken to reach the fence (opposite of human)
Let Th be the time for human to reach that same point on the fence
Let w be the angular velocity of human
Let R be the radius of the circle
We have:

Vh = 4*Vc, so R*w = 4*Vc (since Vh = R*w)

Time for human to reach the end point:
theta(final) = theta(inital) + w*Th
pi = 0 + w*Th (since initial angle is 0 and final angle is pi)
Th = pi/w
Plug in for w = 4*Vc/R:
Th = pi*R/(4*Vc)

Time for chicken to reach the end point:
Tc = R/Vc

Therefore, Th/Tc = pi/4

Sorry this is really messy...

 

[CGiuliano]

Thank you very much for your help.

I meant how can it be proven that zig-zagging works though...

 

[koupparis]

Can't we just create a subsection of the forum to host brainteasers? This way each problem can get its own thread, and all responses therein will pertain to the original problem. It'll be a lot more organized and easier for people to contribute to a solution, talk about a problem, before it gets buried by the latest brainteaser.

 

[CGiuliano]

I second that.

Hopefully you already know about:

Math Help Forum

But they tend to not answer interview type question.

 

[CGiuliano]

From the center to the circumference, has distance R in time t, while Farmer Joe moves a distance 4R towards you.

W.r.t the center farmer Joe moves on an angle of 4rad.

Therefore to escape him:

move in a direction greater than 4rad w.r.t. the farmers position.

 

[DigDug]

I am the chicken and I will kill the farmer. The way it's worded I could stay in the middle and never get caught, but the question is CAN I make it to the edge before the farmer catches me meaning if I'm running toward the edge could he get to that point before me and grab me right before I reach the edge.

Assume the radius of the circle is 1, the farmer can run at 1 unit per second, and I can run at 0.25 units per second. If I try to run in a straight line from the center to the edge directly opposite the farmer's starting position it will take me 4 seconds, but will only take the farmer pi = 3.14... seconds.

If I constrain myself to run on a circle of radius 0.25 (with center at the center of the coop) then my angular velocity will equal the farmer's angular velocity. If I constrain myself on a circle of radius 0.24, then my angular velocity is greater than the farmer's and I can eventually position myself so that the center is directly between us. That is, I am 1.24 units from the farmer.

(me) <-- 0.24 --> (center of coop) <--------- 1 ---------> (farmer)

I can now attempt a mad dash to the opposite edge which is only 0.76 units away. This takes me 0.76 / 0.25 = 3.04 seconds, but it takes the farmer pi = 3.14... seconds to reach this point, so the farmer will die!

 

[Cardshark]

Small addition

Hi guys,

I'm new to the group, nice to "meet" you all. A bit about me: I've recently finished my PhD in maths (from Imperial College and MIT) and now doing an internship as a quant at an IB in London.

I've come across some really interesting questions in interviews over the years, at banks, defence, consulting etc. Thought I'd contribute to the puzzle pool with one of my favourites.

Ten eager undergraduates are at a Goldman Sachs internship assessment day. They've been too busy sucking up to HR at the networking lunch to eat any of the great sandwiches that are on offer. Catering and HR have come up with a clever way of assessing the students and getting rid of left over sandwiches.

The ten bright-eyed students will line up in one row, all facing in one direction so that the first person can see the backs of the rest while the tenth person can't see anyone.

Catering will then go to each student put either a tuna or a humous sandwich on each of the student's head randomly (this isn't a drinking game). HR will then ask each student in turn, starting with the first, what kind of sandwich they have on their head. If the student gets it right, they get an internship (and get to eat their hairgel-crusted sandwich) while if they get it wrong then they don't get an internship. HR then moves on to the next student in line.

The students like each other and want as many of them to get an internship (and sandwich) as possible. What's their strategy?

 

[TiffanyF]

Let (x=\sqrt{a+\sqrt{a+\sqrt{a+...}}}), then
(x^2=a+\sqrt{a+\sqrt{a+...}}})

(x^2-a=\sqrt{a+\sqrt{a+\sqrt{a+...}}}})
(x^2-a=x).

Then use pathagorean. And since it cant be a negative number so it has to be:
(\frac{1+\sqrt{1+4a}}{2})

I'm a new member here, so I've been looking through the threads here.

I HAD to comment on this one. This solution is so beautiful. I love things like this.

 

[Arbit Rage]

ImpliedVolatility:

P(rolling a number on any toss) = 1/6
P(telling truth) = 1/4
P(lie) = 1- (1/4) = 3/4

Answer in my opinion is:

P(rolls 6|he says the truth) + P(doesn't roll 6|he lies) = (1/6)*(1/4) + (1/6)*(3/4)

--------------------------------------
Your answer = 1/6 which means it doesn't matter whether he lies or not. I am skeptical!

How about this:

r: report ; d = die ; T = truth ; L = lie;

p(d=6|r=6) = p(d=6|r=6; T) p(T) + P(d=6|r=6; L) p(L) = 1* 1/4 + 0 * 3/4 = 1/4.

which I think makes sense cause once he reports 6, the question is reduced to "does he tell the truth"? which has a p=1/4.

I don't think adding the probability works. I think this is a conditional probability question. You are trying to find:

P(6 rolled | says 6) = P(6 rolled, says 6)/(says 6)

P(6 rolled, says 6) is obvious, 1/6 * 3/4 = 1/8

P(says 6) I think is very simple as well. It should be 1/6, there is no reason for him to have a higher probability of saying any one number.

Therefore the answer is (1/8)/(1/6) = 3/4.

 

[pruse]

it's difficult to tell which problem someone's talking about when a new post comes in. it'd be much better to have a separate thread for each question...

 

[koupparis]

Can't we just create a subsection of the forum to host brainteasers? This way each problem can get its own thread, and all responses therein will pertain to the original problem. It'll be a lot more organized and easier for people to contribute to a solution, talk about a problem, before it gets buried by the latest brainteaser.

 

[fxiong]

answer to question 3: probability of 3 points in the same semi-circle should be 3/4

Say we randomly draw two points on the circle, point#1and point#2. Then draw a line between point#1 and the centre, and another line between point#2 and the centre. Let x be the angle between these two lines.

- x will have equal probability to fall in between 0 degree to 180 degree. the probability function for x at any particular angle is d(x)/180.
- the probability of the point#3 to fall in the same semi-circle is (360-x)/360. you can plug in x=0, x=45, x=90, x=135 to verify that.
- the probability of 3 points falling into the same semi-circle should be equal to the multiplication of the above two probabilities integrated over 0~180 degrees:
(\int_0^{180}(\frac{360-x}{360})(\frac{1}{180})dx)

So the result is 3/4.

Does this make sense?

 

[Andy Nguyen]

Can't we just create a subsection of the forum to host brainteasers? This way each problem can get its own thread, and all responses therein will pertain to the original problem. It'll be a lot more organized and easier for people to contribute to a solution, talk about a problem, before it gets buried by the latest brainteaser.

That makes sense but like every forum we have created, it needs to create a critical mass to justify its existence. I'd rather not have several sections with little activity in it.

 

[anujjain]

Reply to:Goldman Sachs internship question

I am assuming that you know the type of sandwich which is on the top of the head of guy in front of you and when you guess the type on your head, you speak it loud so that everyone can hear...
So guy number 1,3,5,7,9 will guess the same type which is on the head of guy just next to them. In this way even numbered will definitely get it correct. and for the odd it will be just 50% chance.

Hi guys,

I'm new to the group, nice to "meet" you all. A bit about me: I've recently finished my PhD in maths (from Imperial College and MIT) and now doing an internship as a quant at an IB in London.

I've come across some really interesting questions in interviews over the years, at banks, defence, consulting etc. Thought I'd contribute to the puzzle pool with one of my favourites.

Ten eager undergraduates are at a Goldman Sachs internship assessment day. They've been too busy sucking up to HR at the networking lunch to eat any of the great sandwiches that are on offer. Catering and HR have come up with a clever way of assessing the students and getting rid of left over sandwiches.

The ten bright-eyed students will line up in one row, all facing in one direction so that the first person can see the backs of the rest while the tenth person can't see anyone.

Catering will then go to each student put either a tuna or a humous sandwich on each of the student's head randomly (this isn't a drinking game). HR will then ask each student in turn, starting with the first, what kind of sandwich they have on their head. If the student gets it right, they get an internship (and get to eat their hairgel-crusted sandwich) while if they get it wrong then they don't get an internship. HR then moves on to the next student in line.

The students like each other and want as many of them to get an internship (and sandwich) as possible. What's their strategy?

 

[pruse]

you can actually guarantee 9 of the students get internships, as follows. represent hummus by 0, tuna by 1. let S(k) be the sum of the sandwiches that student k sees, modulo 2. the strategy then is this:

have student 1 announce S(1). student 2 knows S(2), so he deduces that his sandwich is S(2)-S(1) (mod 2), and announces it. student 3 knows S(3) and S(2) (because S(2)=S(1)+S(2)-S(1)), so his sandwich is S(3)-S(2) (mod 2). student 4 knows S(4) and he knows S(3), so he knows his sandwich is S(4)-S(3). this continues all the way up to the 10th student. so students 2 through 10 are guaranteed internships

 

[anujjain]

Awesome Peter.

 

[pruse]

i forgot to say that the strategy generalizes to n students and k types of sandwiches, for any n and k: you can always get n-1 students through.

 

[yes sir]

1/2 is correct, and your reasoning sounds pretty good to me. What makes it work is that there are only two ways the outcome can be finally decided: by someone randomly choosing the jerk's seat, or else randomly choosing the last passenger's seat. The likelihood of these two outcomes is the same at every stage of the process; the first leads to the passenger definitely getting his/her own seat, the second leads to the passenger definitely not getting his/her own seat.

You could dress the reasoning up with a proof by induction (show that it's 1/2 in the case N=2, then show that if it's 1/2 in the N-case, then it must be 1/2 in the N+1-case), but you have the kernel of why the answer is what it is.

I don't get it. Why is it 1/2 instead of 1/n?

I mean if there are 2 people, then the chance the last person gets the seat is 1/2. But if there are 3, the chance of the last person getting his seat declines. Because the 'jerk' has a greater chance of choosing someone else's seat vs. his own.

If n was say something like 1000, then the chance of the last person getting his seat is very very slim, only 1/1000 because the jerk has a 999/1000 chance of choosing someone else's seat.

If n were 1,000,000,000, and I was the last person, I wouldn't say the chance of me getting my assigned seat is 50%. I'm willing to take bets on this. I'll give you 2 dollars if I get my seat and you can give me 1 if I don't. So your expected winnings should be $1.5 each time we play. Should be an obvious bet for you if it is 1/2...

 

[pruse]

I don't get it. Why is it 1/2 instead of 1/n?

I mean if there are 2 people, then the chance the last person gets the seat is 1/2. But if there are 3, the chance of the last person getting his seat declines. Because the 'jerk' has a greater chance of choosing someone else's seat vs. his own.

If n was say something like 1000, then the chance of the last person getting his seat is very very slim, only 1/1000 because the jerk has a 999/1000 chance of choosing someone else's seat.

If n were 1,000,000,000, and I was the last person, I wouldn't say the chance of me getting my assigned seat is 50%. I'm willing to take bets on this. I'll give you 2 dollars if I get my seat and you can give me 1 if I don't. So your expected winnings should be $1.5 each time we play. Should be an obvious bet for you if it is 1/2...

Seems like you're assuming that if the jerk chooses someone else's seat, then the last person necessarily doesn't get his assigned seat. This isn't true. The jerk can choose Johnny's seat, then everything's fine until Johnny comes in and sees that his seat is taken. At this point he can very well choose the jerk's seat; then everything's fine all the way to the end and the last person indeed gets his assigned seat. And there are still other ways that the last person can get his assigned seat.

You can actually see why the probability is always the same - no matter the number of people - like this (don't know if this is quite what Bob was saying)...

The jerk will take Johnny's seat (Johnny's some arbitrary person). After that, everything's fine until Johnny comes in and sees that his seat is taken. At this point, note that the situation looks *just* like it did at the start, except with fewer people, if we treat the jerk's seat as Johnny's assigned seat (Johnny's actual assigned seat is out of the game anyway, so this treatment is perfectly fine). So the probability is the same as it is for any smaller scenario, and this means it's always 1/2 (the probability for just 2 people).

This is the induction Bob was talking about. It's actually called strong induction. (Strong induction is where you use the fact that the statement is true for *any* smaller number).