Quantitative Interview questions and answers

[Andy Nguyen]

• Find [imath]x[/imath] if [imath]x^{x^{x^{\ldots}}}=2[/imath]
• Find all real and complex root of [imath]x^6=64[/imath]
• The hour and minute hands of a clock meet at 12'oclock. When will be the first time they meet again ?
• 3 points are randomly drawn on a circle. What is the probability of them being on the same semi-circle ?
• A unit length is broken off into 3 pieces. What is the probability of them forming a triangle ?
• Calculate [imath]\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}}[/imath]
• 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 ?
• A girl is swimming in the middle of a perfectly circular lake. A wolf is running at the edge of the lake waiting for the girl. The wolf is within a fence surrounding the lake, but it cannot get out of the fence.The girl can climb over the fence. However if the wolf is at the edge of the lake where the girl touches it, then it will eat her. The wolf runs 2 times faster than the girl can swim. Assume the wolf always runs toward the closest point on the edge of the lake to where the girl is inside. Can the girl escape? If so, what path should she swim in?
• The hands on a clock cross each other at midnight. What time do they cross each other next?
• Three people are standing in a circle in a duel. Alan has 100% accuracy, Bob has 66% accuracy, and Carl has 33%. It is a fight to the death – only one person can walk away. They take turns starting with Carl, then Bob, then Alan, and so on. Assume each person plays rationally to maximize their chance of walking away. What is Carl’s action on the first round?
• A line segment is broken into three pieces. What is the probability they form a triangle?
• What is the probability that three points chosen uniformly and independently on a circle fall on a semicircle?
• We have two concentric circles. A chord of the larger circle is tangent to the smaller circle and has length8. What’s the area of the annulus–the region between the two circles?
• There are a cup of milk and a cup of water. Take one teaspoon of milk, put into the water cup; mix well.Take one teaspoon of the mixture in the water cup and put into the milk cup then mix well. Which is higher: the percentage of water in the milk cup or the percentage of milk in the water cup ?
• Two trains are 30 miles apart and are on track for a head-on collision – one train is going at 20 miles per hour and the the other is going at 40 miles per hour. If there is a bird flying back and forth between the fronts of the two trains at 10 miles per hour, what is the total distance the bird will travel before the trains hit?
• A 10-by-10-by-10 cube constructed from 1-by-1-by-1 cubes falls into a bucket of paint. How many little cubes have at least one face with paint on it
• Write a function to find the median of a list.
• You have an unsorted array of the numbers 1 to 50 in a random order. Let’s say one of the numbers is somehow missing. Write an efficient algorithm to figure which is missing.
• What is (1 + 1n)n as n → ∞?
• The number of lili pads on a pond doubles each minute. If there is 1 lili pad initially at time t = 0, therefore 2 at t = 1, 4 at t = 3, 8 at t = 4, etc and the pond is totally covered at time t = 60, then how much of the pond’s surface is still visible at time t = 58?
• How can a cheesecake be cut three times to get eight equal slices?
• The airplane passengers problem (can be looked up in the brainteasers forum): say you have 100 passengers boarding a plane with 100 seats. the first person to board is a weird old lady who, instead of going to her own seat, seats in one of the seats uniformly at random (she could pick her own, but she could also pick someone else’s seat). From then on, when a person boards, they’ll sit in their own seat if it’s available, and if their seat is taken by someone, they’ll pick one of the remaining seats uniformly at random and sit there. What is the probability that the last person sits in his/her own seat?
• A company has a value V which is uniformly distributed between 0 and 1. You are planning to place a bid B for the company. If B is smaller than V, then your bid loses and you get nothing; if B is larger thanV, you get to purchase the company at price B, and the company will end up being worth 1.5 * V. What price B should you bid to maximize your profit?
• On a sheet of paper, you have 100 statements written down. the first says, “at most 0 of these 100statements are true.” The second says, “at most 1 of these 100 statements are true.” ... The nth says,“at most (n-1) of these 100 statements are true.” ... the 100th says, “at most 99 of these statements are true.” How many of the statements are true?
• You and your spouse host a party with eight other couples. At the beginning of the party, people proceed to shake the hands of those they know. No one shakes their own hand or their spouse’s hand. After this shaking of hands is done, you take a survey of how many hands each person shook, and it turns out that excluding yourself, the numbers of hands shook by everyone else are distinct—that is, no one shook the same number of hands as anyone else. How many hands did your spouse shake?
• You have two decks of cards: one has 13 reds and 13 blacks, and the other has 26 reds and 26 blacks. Weplay a game in which you select one of the two decks, and pick two cards from it; you win the game ifyou select two black cards. Which deck should you select to maximize your chances of winning? Try todo this problem in your head, without writing any calculations down.
• You have a deck of 52 cards, and you keep taking pairs of cards out of the deck. If a pair of cards are bothred, then you win that pair; if a pair of cards are both black, then I win that pair; if a pair of cards hasone red and one black, then it’s discarded. If, after going through the whole deck, you have more pairsthan I do, then you win 1 dollar, and if I have more pairs than you do, I win 1 dollar. What is the valueof this game in the long run?

 

[Christian]

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?

 

[Andy Nguyen]

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.

 

[Christian]

I'm just curious, how did you approach this problem?

 

[Andy Nguyen]

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

 

[RussianMike]

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

 

[Andy Nguyen]

How about moving the last I of VIII and put it under the minus sign
This becomes VII = VII =VII

 

[woody]

Are you sure it's not VIII - VII = VI ? 'cause then you could just make it XIII - VII = VI.

 

[bob]

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

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

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

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

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

\(a lnx = ln2\)

\(2lnx = ln2\)

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

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

\(x = \sqrt{2}\)

 

[Andy Nguyen]

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

You are correct, thanks Bob. Fixed the answer.

I get rid of ln on the left side but forgot about ln on the right side. If you can, please take a look at other questions as well.

 

[pardasani]

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....

 

[Andy Nguyen]

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.

 

[Yan He]

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?

 

[pardasani]

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 (

 

[pardasani]

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]

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.

 

[Andy Nguyen]

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.

 

[Andy Nguyen]

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

 

[pardasani]

I think I did
\( a+b > c \Leftrightarrow a+b+c > 2c \Leftrightarrow 1 > 2c \Leftrightarrow c < 1/2\).

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 ?

 

[Andy Nguyen]

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)