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Quantitative Interview questions and answers

1. Find \(x\) if \(x^{x^{x^{\ldots}}}=2\)

2. Find all real and complex root of \(x^6=64\)

3. The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again ?

4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?

5. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle ?

6. Calculate \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}} \)

7. There are 14 identical-looking balls. 13 of them have the same weight while one of them is heavier than the rest. What is the minimum times you can weight to identify the heaviest ball ? How do you generalize for n balls ?
 
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For number three I got after 1 hour, 6 minutes and 27 seconds.. Any
other takers?

How much time do they give you to work out these problems? Are you allowed a pen and paper?
 
For number three I got after 1 hour, 6 minutes and 27 seconds.. Any other takers?
After checking my calculation again, I have 1h 5m 27.27s. It's from 3927.272727... seconds. Probably you have the same number.
How much time do they give you to work out these problems? Are you allowed a pen and paper?
You have pencil and the answer sheet so people calculated mentally or scribbled anywhere they could . On average, you have 1.5-2 minutes for each question.
 
I'm just curious, how did you approach this problem?
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
 
brain teaser?

Hi,

Not sure if anyone will ever ask this but my dad told me about this one and I thought I'd share:

You have an equation:

VIII - VII = VII

How do you change one of the "sticks" to make this correct?

V would be 2 sticks and I is 1 stick.

Lemme know what you think and if you can solve it.
:D
 

bob

Faculty (Undercover)
1. Find \(x\) if \(x^{x^{x^{\ldots}}}=2\)
Answer: \(x=e^{\sqrt{2}}\)

No, the answer to this is \(x=\sqrt{2}\)

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

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

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

\(a lnx = ln2\)

\(2lnx = ln2\)

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

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

\(x = \sqrt{2}\)
 
4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
Answer: 1/2 or 50%

5. A unit length is broken off into 3 pieces. What is the probability of them forming a triangle ?
Answer: 1/8 or 12.5%


--> I think there is a slight discreepancy here .... the solutions to question 4 and question 5 here must be related as ans Q5 = 1 - ans Q4

the way i think about it is as .... we know that 3 randomly choosen line segments will form a triangle IFF the sum of any 2 sides is greater than the 3rd side....

Think of choosing points on a circle as breaking up a line segment --- choosing 3 points on a semi circle gives you 3 line segments..... if 3 of them lie on the same side of a semi circle ==> the sum of 2 of them is smaller than the third one => they cannot make a triangle....
 
the way i think about it is as .... we know that 3 randomly choosen line segments will form a triangle IFF the sum of any 2 sides is greater than the 3rd side....
Correct. Since the sides are from a unit length, we can calculate that any side of the triangle must be strictly less than 1/2. The prob that the length of each stick being less than 1/2 is 50% and this condition must hold for all three of them. Hence I have 1/8

Think of choosing points on a circle as breaking up a line segment --- choosing 3 points on a semi circle gives you 3 line segments..... if 3 of them lie on the same side of a semi circle ==> the sum of 2 of them is smaller than the third one => they cannot make a triangle....
Not totally convinced they are related or merely coincidence but I thought of the semi-circle as follow:
Any two points on a circle will be on the same semi-circle (just connect one point to the center, the second point must be on one of the two semi-circles. Pick the one that contains two points). The problem now becomes finding the probability that the third point is on that same semi-circle. There are only 2 semi-circles to begin with so the chance is 50%

Welcome any opinions since with 1,2 minutes to come up with an answer, there are pretty good chance that I may have big holes in my reasonings. I hope to learn from my mistakes and not to repeat them again in future occasions.
 
One more question looking for answer:

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?
 
Correct. Since the sides are from a unit length, we can calculate that any side of the triangle must be strictly less than 1/2. The prob that the length of each stick being less than 1/2 is 50% and this condition must hold for all three of them. Hence I have 1/8


Well I dont think its that straight forward.... For starters you did not satify the condition that the sum of the 3 sides add up to 1.... In probablity terms you are using that picking up these lenghts is an independant event, whereas it is not : they are related by a constraint equantion.... for eg, consider 0.4, 0.1, 0.1 --> it satisfies your conditions...

Not totally convinced they are related or merely coincidence


well i thought about it once more, and I am preety much convinced that the triangle question, and this circle question is indeed the same issue... think about it as this way.... after all there is nothing sacred about "a unit length"....

consider a circle whose circumference is of "unit length" whatever that may be...

Now pick any 3 points on this circle ----> its equivalent to cutting the "unit length" into 3 pieces.... saying that they "lie on the same side of the semicircle" says you cannot make a triangle out of it.....

but I thought of the semi-circle as follow:
Any two points on a circle will be on the same semi-circle (just connect one point to the center, the second point must be on one of the two semi-circles. Pick the one that contains two points). The problem now becomes finding the probability that the third point is on that same semi-circle. There are only 2 semi-circles to begin with so the chance is 50%

I dont think I understand your point here....... The way I have interpreted the question is :-
" pick any 2 points on a circle ..... Now choose a third point.... what is the probability that the 3rd point will lie on the smaller arc"....

The way I started working on the problem was to choose 2 points at an angle theta ( 0<theta<180).... then calculate the probability of success given theta (I figured out that you need to break this into 2 parts theata < 90 and 90<theta <180 and then multiply the final result by 2)... and then you integrate wrt theta...

I believ the distribution of theta would be uniform between 0 and 360 .....

only problem, when i do this get a number > 1 :((
 
One more question looking for answer:

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?
Well I guess i did not understand the question..... if the question is
" what is the probability that it actually rolled 6 "

well in that case my pick of answer would be 1/6 assuming a fair die --> for me the actual result of the die and what the man reports you are independant of each other if the die is fair... hence 1/6
 

bob

Faculty (Undercover)
4. 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
Answer
: 1/2 or 50%

I'm pretty sure the correct answer to this one should be \(\frac{3}{4}\).

EDIT:
...but it isn't.
I'm still not sure why, but a quick MC simulation has convinced me that I'm missing something here. I'll take a look at it again when my brain's working a little better.
 
Well I dont think its that straight forward.... For starters you did not satify the condition that the sum of the 3 sides add up to 1....
I think I did
\( a+b > c \Leftrightarrow a+b+c > 2c \Leftrightarrow 1 > 2c \Leftrightarrow c < 1/2\). Since c is arbitrary, we can get the same condition for a,b.

I dont think I understand your point here....... The way I have interpreted the question is :-
" pick any 2 points on a circle ..... Now choose a third point.... what is the probability that the 3rd point will lie on the smaller arc"....
This obviously a different question :)
Bob gave a pretty thorough writeup of his solution. I doubt I would be able to write that up in 1,2 minutes.
 
well in that case my pick of answer would be 1/6 assuming a fair die --> for me the actual result of the die and what the man reports you are independant of each other if the die is fair... hence 1/6
Unless they throw smoke with the man's honesty (trick question), the way I read the question is that what is the probability of the man reports 6 if the dice being 6?

Dice being 6 is 1/6
Man reports 6 when he sees 6 is 3/4 so it goes 3/4 x 1/6 =1/8

Of course, we assume fair dice. If they are two independent events, then I agree it stays 1/6
 
so far so good.... I agree with you completely so far.... What I am not sure is how can you treat choosing a, b, c independantly to come up with the answer ?
It looks like I treat a,b,c independently but in hindsight, I still need the length contraint when it comes to finding out the exact length of c, knowing a,b. If not, i will run into the problem of 0.4, 0.1, 0.1 as you pointed out earlier.
Would you suggest this being 1/4 instead ? (I'm not interested in the answer but the thought process)
 
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