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Quantitative Interview questions and answers

stephan acay said:
This is a good point. I would be interested in learning this too.

dstefan
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Search the earlier pages in this thread. The point is just because someone writes down something, doesn't mean it's solvable, and if solved (especially something "infinite" or with limits), the solution isn't guaranteed to be the correct one. In particular for this problem you can prove which solutions will work.
 
chaohan said:
I got a counterexample according to this method:

if x^x^x...^x=4, x=sqrt{2} as well.

Can anyone explain this?
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For \(x^{x^{x^{...}}}\) to converge, x can be at most \(e^{1/e}\), and therefore the value of \(x^{x^{x^{...}}}\) cannot be more than e. Thus, \(x^{x^{x^{...}}}=4\) does not have a solution, but \(x^{x^{x^{...}}}=2\) has a solution, and the line of reasoning above giving the solution \(x = \sqrt{2}\) would work.

More details can be find in the attached file taken from Solutions Manual - A Primer For The Mathematics Of Financial Engineering - the solution to the last question in A Primer for the Mathematics of Financial Engineering.
 

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Andy Nguyen said:
My feeling is 1/8 or 12.5% of it being a 6
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I think it this way:
Let event A be the event when it rolled 6, event B represent the event that the man reported 6.
Then, P(A) = 1/6, P(B)=P(B|A)*P(A)+P(B|!A)*P(!A) = 3/4*1/6+1/4*5/6 = 1/3.
The question is asking P(A|B), which is P(A and B)/P(B) = (1/6*3/4)/(1/3) = 3/8
 
bob said:
No, the answer to this is \(x=\sqrt{2}\)

\(a=x^{x^{\ldots}}=2\)

\(ln(x^{x^{\ldots}}) = ln2\)

\(x^{x^{\ldots}}lnx = ln2\)

\(a lnx = ln2\)

\(2lnx = ln2\)

\(lnx = \frac{1}{2}ln2\)

\(lnx = ln\sqrt{2}\)

\(x = \sqrt{2}\)
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a faster way to solve:
LHS: left hand side
RHS: right hand side

x^LHS = x^RHS
since x^LHS = LHS
=> LHS = x^2
=> 2 = x^2
=> x = sqrt(2)
 
Andy Nguyen said:
Here is how I approached this:
The minute hand will run a full circle in 1 hour. By then, the hour hand will be at 1. Then the minute hand will run pass 1 (meaning 5 minutes) and it will meet the hour hand somewhere after that.
This will give me a quick estimate of when they meet. The first time they meet is sometime after 1 hour 5 minutes. To find the exact time is simple calculation:
The distant the minute hand travels in t seconds will be the sum of 1 full circle and the distant the hour hand travels in t seconds. Solve for t to find t= 3927.2727...=1 hour 5 minutes 27.27 seconds
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I think the answer is 1 hour 5 minutes and 25 seconds.
 
I have found this one (interview question for junior quant)

You have to walk from point A to point B 10 times, blindfolded. Every time before you start walking, a fair coin is tossed, and if it comes out heads, a wall in the middle is present (situation 2 below), while if it comes out tails it is the situation one below

junior_researcher.gif

. As you are blindfolded, you do not know if the wall is present or not.
Even though you are blindfolded, you can move in perfectly straight lines towards any point you choose. Can you give a prescription how to walk from A to B so that the total distance you cover is as small as possible? What distance do you expect to cover in those 10 walks?
 
He should start his path with about 20 degrees angle with respect to straight line and his expected distance would be around 2.62. where did you get this question? calculation is not straight forward. I used Matlab to get the best agle.
 
Jeorme said:
I have found this one (interview question for junior quant)

You have to walk from point A to point B 10 times, blindfolded. Every time before you start walking, a fair coin is tossed, and if it comes out heads, a wall in the middle is present (situation 2 below), while if it comes out tails it is the situation one below

junior_researcher.gif

. As you are blindfolded, you do not know if the wall is present or not.
Even though you are blindfolded, you can move in perfectly straight lines towards any point you choose. Can you give a prescription how to walk from A to B so that the total distance you cover is as small as possible? What distance do you expect to cover in those 10 walks?
Click to expand...

I solved this with solver in excel. Minimize the target by changing alpha
 

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For the x^x^x^x... = 2 problem

A really simple ay to do it is to realize that since I is an infinitely recurring sequence

you can go x^(x^x.......) = 2

but (x^x.......) = 2

so x^2 = 2

so x = sqrt(2)

done
 
5. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle ?
The answer is 1/4. In order to form a triangle, each piece should be less than 1/2. Please see the figure in triangle.pdf. The probability to have the first cut in the infinitesimal range dx is, p1=dx. The second cut, denoted by a round point, must be located in the right half and the distance from the midpoint must be less than x, so the probability to have this cut is p2=x. Therefore we can get dp=p1*p2=x*dx, which leads to such integration as P=2\times\int^{1/2}_0 xdx.Where we set upper limit to be 1/2, so we multiply the integration by 2.


A quick MC simulation also gave me a probability of 1/4, as shown in triangle.cpp.
 

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Amir Yousefi said:
He should start his path with about 20 degrees angle with respect to straight line and his expected distance would be around 2.62. where did you get this question? calculation is not straight forward. I used Matlab to get the best agle.
Click to expand...
Write down the expectation for total distance covered, minimize this distance. You can either do it using angle as the parameter or distance walked to middle wall.
 
here is question that you might be interested in : let \( n>2\) and \( S^{n-1}=\{x\in\mathbb{R}^{n} : x_1^2+\ldots +x_n^2=1 \} \) and fix a \(a\in\mathbb{R}^n\), define \(f(x)=<x,a> =x.a\) ( the dot product between \( a \) and \(x \)). Compute \({\Large\int_{S^{n-1}}f(x)dx} \)

By the way to solve this we don't need any advanced maths. Just calculus !!
 
First time poster, but had an awesome question I wanted to share:

You have a string of bits.

You scan from right to left.

If you encounter a '1', you have the option to flip it to a 0 or keep it as is.

If you encounter a '0', your adversary has the option to flip it to a 1 or keep it as is.

Your goal is to zero all the bits once you reach the end of a scan (i.e. at the left most bit), whilst you adversary wishes to prolong the game indefinitely.

We continually re-scan until we reach the aforementioned goal state.

Can you prove that the game will eventually terminate?


Solution:
[1] First step is realizing that a string of all 1's allows you a guaranteed win, as in the next scan you can zero every bit.

[2] The next step is to think of the string as a binary number and prove that the number is either completed zeroed, or monotonically increases every scan. If this is true, then we will either win by zeroing the number, or win by eventually reaching the largest possible number (1111...), which we can then convert to zeroes.

[3] We need to adopt a strategy that forces our adversary to flip a 0 to a 1, otherwise we won't have a monotonic increase. Consider this strategy:
- zero every number you can
- if/when your adversary flips a bit (he must else he will lose)
- you no longer flip your 1s.

[4] All the numbers you've flipped prior to your adversary flipping a bit are insignificant as they are LSB and thus monotonic increase is ensured. Thus, we will either reach 11111 (we win) or we will zero every bit (we win).
 
Here are two tricky probability questions I found:

You are playing basketball and make your first shot, but miss your second. From then on, the probability that you make your next basket is equal to the proportion of shots you've made so far. What is the probability that you make exactly 50 of your first 100 shots?

You are playing basketball and are keeping track of your successful free throws throughout the season. Early in the season your % of successful free throws is less than 75%, but by the end of the season it is greater than 75%. Is there necessarily a moment during the season when your % of successful free throws is exactly 75%?
 
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