Quantitative Interview questions and answers

[Mach]

Bowling Ball

I couldn’t find the original thread but i have an attempt at the solution to the bowling ball question. The scenario was 1 ball had varying density and the second ball had constant density. how to determine which ball is which.

The poster said we could assume the density varies such that polar moment of inertia is the same. He said an example of this would be if the outside is higher density, then a low density middle, with a high density core. Total mass and moment of inertia would be the same.

The below solution works for the top scenario, im not sure about random density distributions.

The cross sections of the two bowling balls would look as depicted below:

Assuming that the difference in distance AB to arc length AB is negligible with respect to the speed of sound and time spent in the medium (for a real world bowling ball). Then, one can model arc length ab to be approx == to distance AB. Therefore, conduct a speed of sound test between the two balls. Speed of sound is faster in a denser medium, and it will travel around the dense arc portion of the varying density ball faster than it will straight through the constant density ball. Hence, the ball with the higher speed of sound is the varying density ball.

If anyone sees any fallacies in my theory, feel free to post.

 

[Mach]

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!

Ah. You are correct sir. I mis-interpreted the question and hence your equation. I assumed the hour hand was exactly on the 5 when the time is 5:15...stupid mistake.

If i compensate for the fact that the hour hand is 1/4 of the way between the 5 and the 6 at 5:15. the "initial seperation" term in my equation becomes (1/6)+(1/12)(1/4).

(1/43200)t+(1/6)+(1/12)(1/4)=(1/3600)t

This gives the same solution as your equation. 5:27:16.

 

[Mach]

Swing

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.

Conservation of energy:
mgL=2mg(L-h)
L=2L-2H
2H=L
H=(L/2)

 

[StevenBachrach]

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?

Simple... make 1 group of 10 and 1 group of 90. Flip over all the coins in the group of 10 and you've got you solution.

 

[quantyst]

Conservation of energy:
mgL=2mg(L-h)
L=2L-2H
2H=L
H=(L/2)

Incorrect. The correct answer is min H=(3/5)L.

When I did this problem, it took me two pages with explanation. There is MUCH more than meets the eye!

 

[austintlee]

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})

actually, 2 = (x^2)( \rightarrow ) ( x=+\sqrt{2} ) or ( x = -\sqrt{2})

 

[Andy Nguyen]

You can't have log of a negative number.

 

[ExSan]

#1
M(XOR)N=101
(11*M)(XOR)(11*N)=11111

re-think

 

[quantyst]

The question is well stated. If you think the system has no solution, then say so and prove it.

 

[quantyst]

Radically Radicalized!

Let h=(1/2), RAD(z)=z^(1/2) and f(n)=2^n for each natural number n.

For which real values of x does the following equality hold?

sin(x)=RAD{h-h*cot(f(1)*x)*(RAD{h-h*cot(f(2)*x)*RAD{h-h*cot(f(3)*x)*RAD{h-h*cot(f(4)*x)*RAD{...

 

[emil]

Let h=(1/2), RAD(z)=z^(1/2) and f(n)=2^n for each natural number n.

For which real values of x does the following equality hold?

sin(x)=RAD{h-h*cot(f(1)*x)*(RAD{h-h*cot(f(2)*x)*RAD{h-h*cot(f(3)*x)*RAD{h-h*cot(f(4)*x)*RAD{...

The equlality is true for (\forall x \in R) such that: (\sin(2^n*x)\neq 0, \cos(2^n*x)\neq 0, \forall n\in N)

By using the following identity: (\sin(x)^2=\frac{1}{2} - \frac{1}{2}*cos(2x) \\) (ID)

Squaring left/right sides, the equality can be transformed as follows:

(sin(x)=\sqrt[2]{\frac{1}{2} - \frac{1}{2}*cot(2^1*x)*\sqrt[2]{\frac{1}{2} - \frac{1}{2}*cot(2^2*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}}...} \Longrightarrow \\)

(\sin(x)^2=\frac{1}{2} - \frac{1}{2}*cot(2^1*x)*\sqrt[2]{\frac{1}{2} - \frac{1}{2}*cot(2^2*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} *cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}...} \frac{ID}{\Longrightarrow} \\)

(\frac{1}{2} - \frac{1}{2}*cos(2x)=\frac{1}{2} - \frac{1}{2}*\frac{cos(2*x)}{sin(2*x)}*\sqrt[2]{\frac{1}{2} - \frac{1}{2}*cot(2^2*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} *cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}...} \Longrightarrow \\)

(\frac{1}{2}*cos(2x)=\frac{1}{2}*\frac{cos(2*x)}{sin(2*x)}*\sqrt[2]{\frac{1}{2} - \frac{1}{2}*cot(2^2*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} *cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}...} \Longrightarrow \\)

(\1=\frac{1}{sin(2*x)}*\sqrt[2]{\frac{1}{2} - \frac{1}{2}*cot(2^2*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} *cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}...} \Longrightarrow \\)

(\sin(2*x)=\sqrt[2]{\frac{1}{2} - \frac{1}{2}*cot(2^2*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} *cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}...} \Longrightarrow \\)

(\sin(2*x)^2=\frac{1}{2} - \frac{1}{2}*cot(2(2*x)) \sqrt[2]{\frac{1}{2} - \frac{1}{2} *cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}... \frac{ID}{\Longrightarrow} \\)

(\frac{1}{2}-\frac{1}{2}cos(2(2*x))=\frac{1}{2} - \frac{1}{2}*(\frac{cos(2(2*x))}{sin(2(2*x))}) \sqrt[2]{\frac{1}{2} - \frac{1}{2} *cot(2^3*x) \sqrt[2]{\frac{1}{2} - \frac{1}{2} * cot(2^4*x) }}... \Longrightarrow \\)

a.s.o. ... until equality becomes (1 = 1) which is always true if divisions by (\sin(2^n*x), \cos(2^n*x)) hold.

 

[quantyst]

Suppose "L(x)=R(X)" denotes an equation in x. Let's simplify it to "L=R".

Is "L=R" equivalent to "(L^2)=(R^2)"?

To see this better, does the RADICALIZED equation above (involving sin(x), etc.) hold for x=-(pi)/3?

 

[emil]

Suppose "L(x)=R(X)" denotes an equation in x. Let's simplify it to "L=R".

Is "L=R" equivalent to "(L^2)=(R^2)"?

To see this better, does the RADICALIZED equation above (involving sin(x), etc.) hold for x=-(pi)/3?

You are right. The square root must be positive. So, the constraint (\sin(2^n * x)\neq 0) should, actually, be (sin(2^n * x) > 0)

 

[quantyst]

What does it mean to say sin((2^n)*x)>0 for all natural n? So, for which real values of x does the original equality hold?

Can you come up with a single real x for which sin((2^n)*x)>0 for all natural n?

(BTW, most of the "if and only if" relations (i.e., <=>) in your post above must be replaced with simple implications =>. Also, do you still claim q.e.d. as you did above?)

 

[emil]

Can you come up with a single real x for which sin((2^n)*x)>0 for all natural n?

(sin(2^n*x)>0 \Longleftrightarrow (2*k)\pi<2^n*x<(2*k+1)\pi \Longleftrightarrow \frac{(2*k)\pi}{2^n}<x<\frac{(2*k+1)\pi}{2^n}). Remove the (2^n*x = (2*k)\pi/2) values for which (cos(2^n*x)=0 \Longrightarrow x\in(\frac{(2*k)\pi}{2^n},\frac{(2*k+1)\pi}{2^n})-\{\frac{1}{2^n}\frac{(2*k+1)\pi}{2}\}) for (\forall k,n\in N)

(BTW, most of the "if and only if" relations (i.e., <=>) in your post above must be replaced with simple implications =>.

I think any equality or inequality manipulated in that manner should use a <=>. Anyway, => is good, also.

Also, do you still claim q.e.d. as you did above?)

With your help I'm getting more confident in that statement.

 

[quantyst]

For k=1, n=1, you get the interval ((pi), 3(pi)/2) - {3(pi)/4} according to you. For any x in this interval, sin(x)<0, which, as you know, cannot be a solution.

Earlier I asked: "Can you come up with a single real x for which sin((2^n)*x)>0 for all natural n?" Instead you have given me lots of symbols that do not answer my question. Why not just give a single x value for which the equality holds? There is a reason why I asked for a single value for x. The reason is that NONE exists! The purported equality holds for no real values of x.

Here's why: Let x be any real number.

So x=2k(pi)+y for some y in the interval (0,2pi).

If (pi)<y<(2pi), then sin(y)<0, hence sin(x)=sin(2k(pi)+y)=sin(y)<0.

If 0<y<(pi), then keep on doubling y. Sooner or later you will come across a natural n such that (2^(n-1))*y<(pi)<(2^n)*y<2pi. But sin((2^n)*y)<0. So,

sin((2^n)*x)=sin((2^n)*(k2pi)+(2^n)*y)=sin((2^n)*y)<0.

So, no matter what x is, sooner or later we get an impossibility. So, there is no solution!

***Although I've used strict inequality <, some of the inequalities might be considered as non-strict inequality <=.

-----------------------------

You wrote: "I think any equality or inequality manipulated in that manner should use a <=>."

So, according to you ((-3)^2)=((3)^2) is equivalent to (-3)=(3). This is patently wrong!

It's important to realize that the implication involved in squaring both sides of an equation is one-directional. That is L=R implies (L^2)=(R^2). Whereas (L^2)=(R^2) does not imply L=R. In fact, (L^2)=(R^2) implies either L=R or L=-R, one or the other but not both!

------------------------------
Can I now say q.e.d.?

 

[emil]

So, no matter what x is, sooner or later we get an impossibility. So, there is no solution!

(x\in(\frac{(2*k)\pi}{2^n},\frac{(2*k+1)\pi}{2^n})-\{\frac{1}{2^n}\frac{(2*k+1)\pi}{2}\}) for (\forall k,n\in N) can imply that for any n and a certain k, natural numbers, a unique x exists if the intersection of these sets (for each n) is not the empty set. But, it is.

 

[quantyst]

Unfortunately, you try hard to defend that which is not defensible.

The upside down A means "for any", which is a universal quantifier. It does NOT mean "for any n and a certain k". The part "a certain k" means "there exists a certain k", which points to an existential quantifier. And nowhere did you use any symbol or statement implying existence of a k.

For two sets S and T, S-T is NOT the intersection of the sets S and T. S-T is the intersection of S and ~T, where ~T denotes the complement of the set T.

If you knew of any x that satisfies the original equality, then the least you could do was to give us at least one such x. If none exists, just say there isn't any. Why all the messy symbols? Symbols are used to serve us express ideas, not to obscure them.

 

[emil]

If you knew of any x that satisfies the original equality, then the least you could do was to give us at least one such x. If none exists, just say there isn't any. Why all the messy symbols? Symbols are used to serve us express ideas, not to obscure them.

I didn't know the exact solution from the beginning, but, there is nothing wrong with having two approaches that, in the end, reach the same conclusion.

 

[quantyst]

emil,

Your approach, although worthwhile, has been wrong all along! It is still wrong! But you continue to defend the indefensible!

You wrote:

"I didn't know the exact solution from the beginning,..."

So, why did you write "q.e.d." in your first post? Do you know what it means? Do you know when to use it?

Not only that, you've made so many errors in subsequent posts that I lose the count. Every time an error has been brought to your attention, you've either ignored it or basically tried to defend it somehow. Why not acknowledge the errors, learn from them, and move on? You will be better for it if you stop defending the indefensible. Good Luck!