Quantitative Interview questions and answers

[quantyst]

MaxAcc

A negligible mass m is positioned atop the horizontal surface of another mass M, which is resting on a horizontal surface S. The coefficient of static friction between masses m and M is 1, and the coefficient of kinetic friction between M and the surface S is k. We are given that k<{-1 + sqrt(2)}. A force F equal in magnitude to the weight of M is applied to m. Given that you are free to choose the angle at which F is applied to m, determine the maximum acceleration that mass M can achieve.

 

[forty two]

This is classic. It's also basic but important in combinatorial math....

An excellent explanation, Quantyst. This problem is akin to the classic example of defective antennas. Anyways, here is another fairly simple one I came across in a teaser book the other day. This one also involves an application of the counting principle:

A parking lot contains 100 cars that all look quite nice from the outside.
However, K of these cars happen to be lemons. The number K is known to lie in the
range {0, 1, . . . , 9}, with all values equally likely.

(a) We testdrive 20 distinct cars chosen at random, and to our pleasant surprise, none
of them turns out to be a lemon. Given this knowledge, what is the probability
that K = 0?

(b) Repeat part (a) when the 20 cars are chosen with replacement; that is, at each
testdrive, each car is equally likely to be selected, including those that were
selected earlier.
​

 

[emil]

[FONT=&quot]Take a rectangular piece of paper ABCD. Point E is somewhere on side AB, point F is somewhere on side BC, G is somewhere on side CD, and points H and K are somewhere on side DA. Now we fold the paper along the line segment HF and flatten the folded paper. Suppose that as we do this the point C coincides with the point E and the point G coincides with the point K.[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]Suppose we are given the values AK=m, BF=p, AB=w, along with the condition AE<EB.[/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot]Find the length BC in terms of m, p, w.[/FONT]

Is this a quantitative question?
Anyway, the triangles BEC and AKE obtained after folding the paper, are similar.
BC is obtained by solving a system of 3 equations with two unknowns x, BC:

(\frac{x}{m}=\frac{p}{w-x}=\frac{BC-p}{sqrt(m^2+(w-x)^2)})

Please see attached.

 

[quantyst]

UnUniform

Suppose Z is a uniform random variable over [0,1]. I.e., Z~u[0,1].

Define X(Z) and Y(Z) to be two uniform random variables over [0,Z] and [Z,1], respectively. I.e., X(Z)~u[0,Z] and Y(Z)~u[Z,1].

Let r be a real number.

Find the probability P{ Y(Z) - X(Z) < r }.

 

[quantyst]

How High?

On a horizontal surface S a frog (to be considered as a point mass) using sufficient but the least initial speed jumps over a log of radius r and lands on surface S on the other side of the log. How high does the frog reach as it jumps over the log?

 

[yegor]

A man speaks the truth 3 out of 4 times. He throws a die and reports it to
be a 6. What is the probability of it being a 6?

Nifty question. My gut was to say that the answer here is (\frac{1}{4}), since we're conditioning on the report of the roll, which gives us more information to work with. A closer look shows that this is a reasonable answer to the question, but not the only possible one....

This is actually a conditional probability:

(P(rolls 6 | reports 6) = \frac{P(rolls 6, reports 6)}{P(reports 6)})

Here's where the assumptions begin to come in. For one thing, we assume that the roll and the decision to lie are independent; under this assumption, the numerator is easy:

(P(rolls 6, reports 6) = P(rolls 6, tells the truth) = \frac{1}{6}*\frac{1}{4} = \frac{1}{24})

The denominator is trickier. This is where the biggest assumptions come in.

(P(reports 6) = P(rolls 6, tells the truth) + P(rolls non-6, lies, reports 6))

We've already computed the first term, but how do we compute the second? We have to know something about how the person lies.

Are the person's lies always plausible? (That is, would the person lie by saying a 3.4 had been rolled, or a 529?) Given that they're always plausible, are they "fair?" (That is, is the person's lying strategy to choose a plausible lie at random with an equal probability of each lie, or is there some other distribution to the lies? Is the lie even random, aside from the initial throw of the die?)

For the sake of convenience, let's assume that, when the person lies, he or she does so by reporting some other plausible result at random, with equal probability of each lie. Then:

(P(reports 6) = \frac{1}{24} + \frac{5}{6}*\frac{3}{4}*\frac{1}{5}=\frac{1}{24}+\frac{1}{8}=\frac{1}{6})

Overall, then:
(P(rolls 6 | reports 6) = \frac{\frac{1}{24}}{\frac{1}{6}} = \frac{1}{4}), as expected.

For the sake of illustration, though, let's work the problem under a different lying strategy: suppose the person always says "6" when lying about a non-6. (We don't really care what he does when lying about a 6.) Then:

(P(reports 6) = \frac{1}{24} + \frac{5}{6}*\frac{3}{4}*1 = \frac{1}{24} + \frac{5}{8} = \frac{2}{3})

Under this alternative assumption:

(P(rolls 6 | reports 6) = \frac{\frac{1}{24}}{\frac{2}{3}} = \frac{1}{16})

As you can see, how you assume the person lies can have a rather dramatic impact on the answer....

guys, it's (\frac{3}{4})
compare it to this question:
A man speaks the truth 3 out of 4 times. He says his name is Bill. What is the probability of his name being Bill?
:dance:

 

[quantyst]

It's all Brown in here!

Let W(t) denote the standard Brownian motion.

Define T=inf{t>0: W(t)=1 or W(t)=-1}.

Compute E[sin(T)].

 

[quantyst]

Cat Catches Mouse!

In the xy-coordinate INTEGER grid, a mouse is originally located at position (x(0), y(0)) and a cat at position (0,0). Assume x(0) >= 0 and y(0) >= 0. In its pursuit to catch the mouse, the cat, at every move, takes a one-unit step to the left, right, up or down. At every move, the mouse, unaware of the cat's pursuit, takes a one-unit step to the left, right, up or down with respective probabilities L, R, U, D. The cat and the mouse take their steps alternately. Determine an optimal strategy for the cat to catch the mouse in the shortest possible time. Find the average time during which the cat catches the mouse as the cat utilizes the optimal strategy.

Do the problem anew under each one of the following assumptions:

(AA) The cat (as well as you) is ONLY aware of the initial position of the mouse and the given probabilities L, R, U, D.

(BB) The cat (as well as you) is aware of every position of the mouse before it takes a new step, in addition to (AA).

Before doing the general case, you may first do the problem when

(1) L=R=U=D=1/4,

(2) L=R=0, U=D=1/2,

(3) L=D=0, R=U=1/2,

(4) R=U=0, L=D=1/2.

 

[quantyst]

Dizzying!

Starting at (0,0) in the xy-coordinate INTEGER grid, you take n steps, each step being a one-unit left move, right move, up move, or down move. The following are examples of n-step trajectories, where n=12:

(T1) ((0,0), (1,0), (2,0), (3,0), (3,-1), (4,-1), (4,-2), (5,-2), (5,-1), (5,0), (5,1), (5,2), (4,2))

(T2) ((0,0), (1,0), (2,0), (3,0), (3,-1), (4,-1), (4,-2), (5,-2), (5,-1), (5,0), (4,0), (3,0), (3,1))

(T3) ((0,0), (1,0), (2,0), (3,0), (3,-1), (4,-1), (4,-2), (5,-2), (4,-2), (4,-1), (4,0), (3,0), (3,1))

(T4) ((0,0), (0,-1), (0,-2), (0,-3), (0,-4) (0,-5), (-1,-5), (-2,-5), (-2,-4), (-2,-3), (-2,-2), (-2,-1), (-2,0)).

Comments:

Trajectories T2 and T3 are self-intersecting, whereas T1 and T4 are not. Trajectory T2 self-intersects at (3,0), for example.

Questions:

Q1. Find the number of all n-step trajectories starting at (0,0).

Q2. Find the number of all n-step trajectories starting at (0,0) that are not self-intersecting.

Q3. Find the number of all n-step trajectories that are confined to the first quadrant. A trajectory is confined to the first quadrant if for every point (x,y) of the trajectory we have x>=0, y>=0.

Q4. Find the number of all n-step trajectories that do not contain points of a certain set S. For example, find the number of all 20-step trajectories that do not contain the points of S={(2,5), (-2,-4), (7,7)}.

Q5. Find the number of all n-step trajectories that do not self-intersect at the points of a certain set S. For example, find the number of all 20-step trajectories that do not self-intersect at the points of S={(2,5), (-2,-4), (7,7)}.

Q6. Find the number of n-step trajectories that self-intersect exactly once.

Q7. Find the number of n-step trajectories that self-intersect exactly twice.

Q8. Find the number of all n-step trajectories that do self-intersect at the points of a certain set S. For example, find the number of all 20-step trajectories that do self-intersect at the points of S={(2,5), (-2,-4), (7,7)}.

Q9. Find the number of all n-step trajectories whose number of up steps equals the number of left steps.

Q10. Find the number of all n-step trajectories whose number of up steps is twice the number of left steps.

Q11. Find the number of all n-step trajectories that terminate in a given set S. For example, find the number of all 20-step trajectories that terminate in the set S={(2,5), (-2,-4), (7,7)}.

 

[quantyst]

Intersection of Circles

Let X, Y, Z, A, B, C be iid uniform random variables over [0,1].

Two circles, one with center (X,Y) and radius Z, another with center (A,B) and radius C, are randomly drawn in the xy-coordinate plane. Let W denote the area of the region of intersection of the two circles.

Q1. Find P{W < r} for a given positive real number r.

Q2. Find E[W].

 

[quantyst]

m-gon, n-gon, doggone it!

Inscribe a circle within a regular n-gon. Inscribe a regular m-gon within the circle. Find the ratio of the area of the regular m-gon to that of the regular n-gon.

 

[quantyst]

Coins and Probability

Toss a fair coin 16 times. Which of the following events has a greater probability of occurring?

(E1) H H H H H H H H H H H H H H H H

or

(E2) H T T H T H T T H H H T T H T H

 

[quantyst]

Cat & Mouse Chase in 1D

On the one-dimensional integer line a cat is originally located at 0 and a mouse, originally located at X(0)=a. In its pursuit to catch the mouse, the cat takes a one-unit step to the left or to the right at each move. At each move, the mouse, unaware of the cat, moves from place to place in a time-independent Markovian process X(t). That is, at any time t, if the mouse is located at i, then the probability that at time (t+1) it will be at j is given by i an j only, independent of t and regardless of where the mouse has been for all times prior to t.

That is, P{X(t+1)=j | X(t)=i}=p(i,j) for all t, where the transition matrix P of the probabilities p(i,j) is known to the cat.

The cat and the mouse take their steps alternately. Determine an optimal strategy for the cat to catch the mouse in the shortest possible time. Find the average time during which the cat catches the mouse as the cat utilizes the optimal strategy.

The game ends -- a catch occurs -- when the cat and the mouse are at the same point. Which can happen in two ways, one when the cat moves and ends at where the mouse is, and the other is when the mouse moves and ends at where the cat is.

(MM) Let's not forget that like all cats, this cat has an excellent memory of all the places it has been previously, as well as of the times it has been in those places.

Do the problem anew under each one of the following assumptions:

(AA) The cat (as well as you) is ONLY aware of the initial position of the mouse and the given probabilities p(i,j), along with (MM).

(BB) The cat (as well as you) is aware of every position of the mouse before it takes a new step, in addition to (AA).

 

[Shlomi]

A nice brain teaser.
Imagine yourself standing near a table, with your eyes covered. You've been told that there are 100 coins scattered on the table, and that every coin is painted in blue in one side and painted in red on the other side. You've also been told that 10 coins placed with the red side above and the rest with the blue side above.

You are asked to split the coins to 2 groups, required that every group has the same number of coins with the red side above. You are allowed to move and turn the coins.

How you will do that?

 

[Shlomi]

Its a simple, but a nice one:
(\sum_{k=0}^{90}{sin^2(k)}=?)

k is in degrees and not radians...

 

[quantyst]

quantyst's Number contains ALL truth!

In base two, construct the following real number 'canonically':

q = .0 1, 00 01 10 11, 000 001 010 011 100 101 110 111, 0000 0001 0010 0011 0100 0101 0110 0111 1000 ....etc.

The commas serve to keep the digits organized and the progression better understood.

The number q contains every FINITE sequence of 0's and 1's. Since any truth statement can be said in a finite manner (in any language) and it can be encoded into a sequence of 0's and 1's, then q 'contains' that truth.

Can you imagine for a moment: This real number contains within it all of Shakespeare's works, all of Alfred Tarsky's logic productions, the military secrets of the world's major powers, the future results of all science, the next revolutionary discovery in quantitative finance, a new Theory of Everything. Really, q has everything and everything within it!

When someone says: "The Truth Is Out There", just tell them to seek the number q.

Is q too good to be true?

Well, it contains all falsehoods, just as well!

Certainly, the idea of q gives a new meaning to the phrase "It's too good to be true!"

 

[quantyst]

Solve XOR Systems of Equations!

A refresher first! The (XOR) operation acts on 0 and 1 as follows: 0(XOR)0=0, 0(XOR)1=1, 1(XOR)0=1, 1(XOR)1=0. We also let (XOR) operate on numbers expressed in binary notation on their digits one at a time. This is called bitwise operation. Here are a few examples: 111(XOR)101=10, 10011(XOR)11001=1010, 1011(XOR)1011=0.

Now, solve the following systems of equations. Here M and N are binary numbers. Solutions may not be unique.

#1

M(XOR)N=101
(11*M)(XOR)(11*N)=11111



#2

M(XOR)N=110
(M*M)(XOR)(N*N)=10110

 

[Mach]

Clock

Hi laguy,

You wrote:

1. It's currently 5:15 - at what time (to a few decimal places of a minute) will the the minute hand meet the hour hand?

Let HH denote the position of the hour-hand between 5 and 6 o'clock. So, HH=5+x, where 0<x<1.

Let MH denote the position of the minute-hand. Then MH=12x. So, the condition of minute-hand meeting hour-hand is the same as MH=HH; i.e., 5+x=12x.

There is another way of doing this without using an equation, but instead a series.

If i'm not mistaken this would give time=5.4545 = 5:27:16

I solved it using rates and ratios. This approach worked for Andy's original clock problem. Distance traveled by minute hand = distance traveled by hour hand +initial separation.

Let

RHH= rate hour hand=(1/(12*60*60))=1/43200
RMH= rate minute hand=(1/(60*60))=1/3600

Equation: RHH*t +(initial separation)=RMH*t
RHH*t+(1/6)=RMH*t
t=-(1/6)/(RHH-RMH)
t=654.5454 seconds after start=10 minutes, 54 seconds

time of coincidence= 5:25:54.

Andy's Original Problem
(1/43200)*t+1=(1/3600)*t
(initial separation of 1 and 0 are the same since we are in terms of ratios and its a circle. one is used instead of zero because of andy's logic how MH makes a full rotation before lapping the HR)
t=3927.272727 seconds.

which is identical to the original result.

i dont think your equation works because you assume the ratio of the minute hand between 5 and 6 is the same as the ratio of the hour hand between 0 and 12.

Any1 else get an answer for this question?

 

[quantyst]

Does A Loop!

One end of a (non-stretchable) string of length L is tied to a raised nail in the wall at a point P, and the other end of the string is attached to a point mass M. Directly below the point P, we erect another nail at a point C on the wall. Let H denote the distance between P and C. We now pull the mass M (away from point P) until the string becomes taut, and then raise mass M so that the string becomes horizontal and parallel with the wall. At this point in time we let the mass M drop.

Question: Find the minimum value for H so that the point mass M will complete a circular loop about point C after the string comes in contact with the nail at point C.

 

[quantyst]

If i'm not mistaken this would give time=5.4545 = 5:27:16

I solved it using rates and ratios. This approach worked for Andy's original clock problem. Distance traveled by minute hand = distance traveled by hour hand +initial separation.

Let

RHH= rate hour hand=(1/(12*60*60))=1/43200
RMH= rate minute hand=(1/(60*60))=1/3600

Equation: RHH*t +(initial separation)=RMH*t
RHH*t+(1/6)=RMH*t
t=-(1/6)/(RHH-RMH)
t=654.5454 seconds after start=10 minutes, 54 seconds

time of coincidence= 5:25:54.

Andy's Original Problem
(1/43200)*t+1=(1/3600)*t
(initial separation of 1 and 0 are the same since we are in terms of ratios and its a circle. one is used instead of zero because of andy's logic how MH makes a full rotation before lapping the HR)
t=3927.272727 seconds.

which is identical to the original result.

i dont think your equation works because you assume the ratio of the minute hand between 5 and 6 is the same as the ratio of the hour hand between 0 and 12.

Any1 else get an answer for this question?

This problem is so elementary that it requires no further analysis or commentary ... if it weren't for your mutilations committed upon so simple a problem and, just as equally, your comment "i dont think your equation works because you assume ..."

This problem, in its essence, has nothing to do with time. It has to do with the relationship between two moving objects, of which one displaces itself twelve times another, and that's precisely what the above equation "5+x=12x" reflects. When the hour hand moves a distance x, the minute hand moves 12 times that. That's it ... That simple ... No mutilations required!