Quantitative Interview questions and answers

[VladimirBunicu]

The one with 1#1#1 took a bit of time and imagination, but oh well.
(log_3(1+1+1)^6=6)

1+1+1 took me the longest, too. I used ((1+1+1)!=6) though. And i had different answer for 8's, using (8^{1/3} )
Anyway, Andy, you are hired

 

[alain]

A friend of mine gave me this puzzle yesterday (consider it today, because it was 3am :partyman.
He was asked this one during an interview for some hedge fund, and was requested to give answer in 10 minutes :smt017. Here is the puzzle:

Replace the # sign with the correct mathematical functions, so that all statements are accurate.

1 # 1 # 1 = 6
2 # 2 # 2 = 6
3 # 3 # 3 = 6
4 # 4 # 4 = 6
5 # 5 # 5 = 6
6 # 6 # 6 = 6
7 # 7 # 7 = 6
8 # 8 # 8 = 6
9 # 9 # 9 = 6

I solved it in somewhat more than 10 minutes; the rules initially were ambiguous (two hours in a Russian bar do show :partyman: :drinkers so what you see above is a rephrase. I still not sure if I got it right. Any ideas?:smt102

Assuming that I can change the "#" for anything I like (any kind of weird operation) we can do something like:

(2+2+2=6)

(3*3-3=6)

(4+log_4 4+log_4 4=6)

(5/5+5=6)

(6-6+6=6)

(7-7/7=6)

(8-\log_8 8-log_8 8=6)

(9-9/\sqrt{9}=6)

I haven't come up with the first one yet. I solved the rest way before 10 minutes. I will think about the first one later.

 

[Andy Nguyen]

If they want to be creative, they can either shorten or expand the number of elements on the left side. Instead of giving 2#2#2=6, they can make it 2#2=6 or 2#2#2#2=6. Having fewer numbers is harder to do while having more number gives you more choices. They can also replace 6 by another number.

With only 2 numbers on the left, you can do it in 5 minutes for the whole set (the one with 1#1 is again the hardest). Use the same idea as for 3 numbers.

 

[kean]

Quiz

This is your first date. You are going to meet Angie Jorie on this coming Sunday. Both of you agreed on the term that only wait for other party for maximum of 15 minutes. Either one will leave after 15 minutes and the date cancel. In this case, what is the probability that you will meet AJ on Sunday? (Time limit for this question is 5 minutes)

 

[Uncle Max]

1.042% ?

 

[woody]

Who's Angie Jorie?

 

[Andy Nguyen]

This is your first date. You are going to meet Angie Jorie on this coming Sunday. Both of you agreed on the term that only wait for other party for maximum of 15 minutes. Either one will leave after 15 minutes and the date cancel. In this case, what is the probability that you will meet AJ on Sunday? (Time limit for this question is 5 minutes)

You forget to specify the time window where they will meet. Is it a 1-hour window, say from 9-10am, etc. The larger the window, the less chance they will meet.

Let's say the window is 1 hour. The chance of them not meeting is \(\frac{3}{4})^2\) so the probability of them meeting is 0.4375 or 43.75%

If the window is x (in hours) then the probability of them meeting is \(\frac{1}{2x} - \frac{1}{16x^2}\)

For example, if the window is 2 hours, then the probability of meeting is down to 0.234375 or 23.4375%

You can change the wait time to 5 minutes and have a more general formula as well.

Who's Angie Jorie?

Sounds a lot like Angelina Jolie

 

[kean]

Quiz Response

I asked the same question when I attended this interview. Ok, it is within one hour window. Again, none of you is correct.

The reason I put this question here is sometimes we think too much when things are easy. Markets happen in the same way sometimes. The problem is we tend to use complex solutions for simple probability question.


I give the answer tomorrow. No need to know who is A.J. She is not important.:smt006

 

[pardasani]

Well under the assumption, that neither you nor AJ is more than 1 hour late, I got a answer of (\frac{7}{16} )..

I am attaching a small picture, which I think explains the situation much better.... Its a square of size 60 mins X 60 mins representing possible outcomes, and a big shaded hexagon showing "favorable" outcomes....

Prob (having date) = Area of the shaded portion = ( 1 - 2*\frac{1}{2}*\frac{3}{4}*\frac{3}{4} )

Under a more realistic scenario though, I think it does indeed matter who A.J. is --- in which case, the assumption of being late by at max 1 hour may not be true; leading to a Probability of 1

 

[Andy Nguyen]

Well under the assumption, that neither you nor AJ is more than 1 hour late, I got a answer of (\frac{7}{16} )..

Indeed, same answer as mine.

I am attaching a small picture, which I think explains the situation much better....

Indeed, I used a similar diagram. I'm sure both of us have seen a similar question in which the time frame is 1 hour and the wait time is 5 minutes

It would be interesting to see how Calvin comes up with his correct answer.

 

[dav2fox1]

7/16 is a right answer, a geometry probability question

 

[kean]

7/16 is correct but it is not a geometry probability instead we should look at uniform probability.

 

[kean]

Sorry for the delay. I was a bit busy for the last two days. I thought using this forum to learn maths or probability is great..so many mathematician in this forum. Probability is a difficult subject!

 

[Yuriy]

Yep, probability is very difficult. Especially parts that deal with measure theory. One day I hope to be able to read Billingsley book, for example, and say how easy it is

 

[dav2fox1]

7/16 is correct but it is not a geometry probability instead we should look at uniform probability.

Hi, Calvinkean, it is a geometric probability problem. For two independent random variables, you can change it to two dimension geometric area problem, it is a typical geometric probability probelm, as pardasani did. Uniform distribution is only a speical case here.
This problem can be easily found from probability book.

 

[kean]

Hello

Hi, Calvinkean, it is a geometric probability problem. For two independent random variables, you can change it to two dimension geometric area problem, it is a typical geometric probability probelm, as pardasani did. Uniform distribution is only a speical case here.
This problem can be easily found from probability book.

Hello, please don't get upset. I purposely said it not a geometry probability. I know you will "kick" back. Anyway, my reasoning is how long do we need to solve such simple question. It doesn't matter whether it is a question from any textbook. Be cool, if we all want to work in the markets.

By the way, if we solve from uniform perspective and derive from probability axiom. It is much easier. I am still learning. My intention is to use this forum to harness knowledge. I hope we share the same view.

 

[kean]

Hey man, you are

Well under the assumption, that neither you nor AJ is more than 1 hour late, I got a answer of (\frac{7}{16} )..

I am attaching a small picture, which I think explains the situation much better.... Its a square of size 60 mins X 60 mins representing possible outcomes, and a big shaded hexagon showing "favorable" outcomes....

Prob (having date) = Area of the shaded portion = ( 1 - 2*\frac{1}{2}*\frac{3}{4}*\frac{3}{4} )

Under a more realistic scenario though, I think it does indeed matter who A.J. is --- in which case, the assumption of being late by at max 1 hour may not be true; leading to a Probability of 1

 

[kean]

I like to make math simple...very simple.

 

[kean]

New Quiz

A six-sided die is rolled three times independently. Which is more likely: a sum of 11 or a sum of 12? (5 minutes limit)\\:D/

 

[Jonathan]

I say 11.

11 has .1944 chance whereas 12 has .1666 chance.

am I right?