Nick's Quant Interview Questions
This section is on Quant Interview Questions...
- Logic puzzles, and Probability
These are some of my favourite interview questions to ask in a junior Quant interview.
They require no prior knowledge, just raw intelligence and determination.
Difficulty is out of 10
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1. Quant Interview Questions - Expectation of dice roll
What's the expectation of dice roll ?
i.e. If you were to roll a dice a million times, and calculate the average, what wld it be ?
Difficulty - 1
1. Quant Interview Questions - Expectation of dice roll
3,5
But since every face of a dice is different, maybe it will change the probability (the face with 6 holes is lighter than the one with 1 hole).
dice question
3.5 is correct... It's simply the average of the numbers 1 to 6
Answer
The probability of each outcome is a constant at 1/6 and hence the expected value is:
1/6*(1+2+3+4+5+6) = 3.5
2. Quant Interview Questions - Expectation of 2 dice
You roll 2 dice together and add the result.
What's the Expectation now ?
Difficulty - 1
2. Quant Interview Questions - Expectation of 2 dice
7
2 dice question
7 is correct.. the expectation of 2 indepentent distributions is the sum of the expectation of each
the expectation of 2
the expectation of 2 distributions is the sum of the expectation of each
(doesnt need to be independent)
3. Quant Interview Questions - Expectation 2 different dice
You have 1 red dice, and 1 blue one.
The red one goes from 1 to 6
The blue one goes from 10 to 60
You roll them both together and add the result.
What's the expectation now ?
Difficulty - 2
3. Quant Interview Questions - Expectation 2 different dice
38.5
2 dice
38.5 is correct...
Expectation of 1 to 6 dice = 3.5
Expectation of 10 to 60 dice = 35
add the 2..
Dice
I had this interview question once, similar to the above:
Given that you have a six sided die, and your probably outcome to infinity is 3.5 as answered by the above, what is the most you would pay to play this game to infinity?
Infinite Dice Rolls
If you mean: How much would you pay per throw assuming you get the difference between the amount on the dice, and your cost per throw..
well.. that depends 4 things:
1. Access to finance: how much you have in your bank account + what you can borrow.
2. How quickly you can throw the dice
3. How risk avarse you are
4. The opportunity cost of your time...
If you mean: How much would you pay for the right to play the game to infinity when you receive what's on the dice, it's just (1)
Nick
4. Quant Interview Questions - Prove Pythagoras's theorem
Using the following diagram, Prove Pythagoras's theorem.
Note - this is 2 squares. One large, one small, rotated and fitting inside the large one.
Difficulty - 2
Pythagoras
if c is the side of the small square c^2 is the surface of the small square. If a is the longer side of the triangle and b its short side the 4 triangles have a surface of 4*1/2*a*b=2ab.The whole big square surface is then 2ab+C^2 (4 triangles+ 1 small square)
(a+b)^2 is the surface of the large square
(a+b)^2 =a^2+b^2+2ab
Combining the 2 equality:
a^2+b^2+2ab= 2ab+c^2
or
a^2+b^2=C^2
Jean
Pythagoras
That is correct
5. Quant Interview Questions - Navigation Question
How many places in the world can walk 1 mile south, 1 mile east, 1m north and end up at the same place ?
Difficulty - 10
snowbound
an infinite number, so long as both hemispheres are in play
Snowbound
y... can u define those places ?
1) The North pole 2) Close to
1) The North pole
2) Close to the south pole, so that when you walk directly east or west, the circumference of the Earth is 1 mile
Walking round and round
There are more places...
3) Anywhere north of the
3) Anywhere north of the South pole where an East-West lap is an integral fraction of 1 mile. There are infinitely many of these.
Around the world
Yup, that's right..
It's the north pole.. and 1 + 1/2*n*PI miles from the South pole where n tends from 1 to infinity.
None. Places can't walk !
None. Places can't walk !
6. Quant Interview Questions - Dice Option Game Part 1
You have a die, and you can roll it as many times as you like. At any point, you can say “Stop”. At that point, you go home with whatever is on the die.
So if you roll a 2.. then roll again and roll a 4.. then say “Stop”, you go home with 4.
What is the profit maximising strategy, and what is the expectation of that strategy ?
Difficulty - 1
Assuming it is a standard
Assuming it is a standard die, then you keep rolling until you get a 6
The expectation is also 6
Dice Question
that is correct...
7. Quant Interview Questions - Dice Option Game Part 2
You have a die, and you can roll it a maximum of 3 times. At any point, you can say “Stop”. At that point, you go home with whatever is on the die.
So if you roll a 2.. then roll again and roll a 4.. then say “Stop”, you go home with 4.
What is the profit maximising strategy, and what is the expectation of that strategy ?
Difficulty - 7
7. Quant Interview Questions - Dice Option Game Part 2
The expectation are 3,5
So even if you still have a turn take the 4 and go home
max of 3 rolls
nop.. this is an option pricing question
have a look at Q's 3 and 6 for a clue
if you get a 5 or a 6, stop
if you get a 5 or a 6, stop the game
EV:
(1/3)*5.5 + (2/9)*5.5 + (4/9)*3.5 == 4.61111
Dice Question
close, but it's possible to do better
answer
on the third roll, your EV is 3.5, so on the second roll you'd want to stay on anything over 3.5, aka 4 5 or 6. so your EV for roll 2 is really [(3.5 + 3.5 + 3.5) + 4 + 5 + 6] = 25.5/6 = 4.25. Therefore, on roll 1, anything less than 4.25 wouldn't suffice. Thus, you'd only stay on 5 or 6.
The final EV is (4.25)x4 + 5 + 6, all divided by 6, or 28/6 = 4.666666666666666666
3 dice rolls
4 2/3 is correct
Where does the +4+5+6 come from
Can you please clarify.
Why is EV 4.25 on roll 2. The EV is 3.5 for ALL rolls initially as they are independent. How can an EV of roll 2 be different to that of roll 1? And why would roll 3 have an EV less than that of roll 2
Sorry, excel VBA programmer, new to probability theory and stats.
Rolls NOT independent
Let's just do two rolls, the answer for three is just this solution repeated. The rolls are NOT independent by nature of the fact that you can choose to stop on any roll you like.
For the second roll in a two roll series, everyone agrees that the EV is 3.5. Knowing this, let's take a look at the situations when you would choose to make the first roll your only roll. You would only choose not to roll again if you got a 4, 5 or 6. Each of these rolls has a chance of 1/6. On average, half the time you'd choose to stick with your first roll, half the time you'd choose to roll again with an expectancy of 3.5. So EV for the first roll is calculated
(1/6)(4) + (1/6)(5) + (1/6)(6) + (3/6)(3.5) = 4.25
8. Quant Interview Questions - Envelope Question
I have 2 envelopes. One contains an amount X, the other contains 2X.
I give you an envelope at random.
You open it, and find 10.
Do you want to keep it, or swap it for the other envelope ?
Difficulty - 8
8. Quant Interview Questions - Envelope Question
I defenitely want the other envellope.
If 10=X then the other envellope contain 20$
If 10=2X then the other envellope contain 5$
Since you gave me the envellope randomly, there is a 50/50 chance for the two outcome.
So 25/2 = 12,5$
I'm richer with 12,5 than 10$
Envelope
nop, under your logic, both envelope holders have a positive expectation of a gain.
In a zero sum game, this cannot be right.
Envelope
Suppose you selected the other envelope that can contain 20 or 5. You can apply the same "logic" to this choice and you will then decide to get the other envelope. Consequently the logic is not logic (both choices result in selecting the other envelope) and you cannot choose rationaly.
Another way to answer the question is that you have 1 piece of information X but two unknowns: the other envelope has x/2 or the other envelope has 2X. You need more info to make a rational choice.
Jean
Envelope
correct... there's something wrong with the "logic".
This question is killing
This question is killing me!
I understand why we cannot make a decision, but also I do not understand why changing envelopes is not positiv.
This is the same as an investment of 10 euros with an expected payoff of 5 (50%) or 20(50%). I would invest the money!!
And by the way: 100% of all MBA students and professors would invest too!
Probability
Only the ones who don't understand probability...
; - >
Would switch
I would switch
If x=10 then 2x=20 ... you stand to gain 10 = 20 total
If 2x=10 then x=5 ... you stand to lose 5 = 5 total
Greater gain than loss
Could end up with a total of 20 with positive switch, or minimum of 5 with a negative switch
Similar to dice problem?
If you have outcomes as follows :
5, 10 = EV 7.5 (LOSE IF SWITCH)
10, 20 = EV 15 (WIN IF SWITCH)
Similar to Monty Hall
Well... from my point of view the pb is very similar to Monty Hall pb.
2X is pair with P=100%
X could be pair or not P=50%
You have a new information and you make a new choice with this new information. The new choice can't be independant, only conditional.
If you get 10 in the envelop, clearly the probability is higher that it is 2X rather than X. So, the expected return should below 10.
Not sure, but I would say P=2/3 that it is 2X. The expected return would be E = 2/3*10 +1/3*5 = 25/3 ~ 8.33 < 10
Comment
This is a dangerously wrong answer and should be ignored in its entirety.
Envelope
I don't think I understand correctly...
If each case is equally likely then the expected value of the combined envelopes is:
0.5*(5+10)+0.5*(10+20)=22.5
If you always swap then the other person is guaranteed 10 therefore your expectation is 12.5
OR
Your expectation is 0.5*5+0.5*20=12.5 if you swap or 10 if you don't. Why is it not obvious to swap?
You either -- have X and the
You either
-- have X and the other person has 2X, so if you switch your G/L is +X
-- have 2X and the other person has X, so if you switch your G/L is -X
Both cases have equal probability therefore your expected G/L = 0
well done !
Nice one..
You have the correct answer
X and 2X
Are we sure?
The 1st X in +X is different to the 2nd X in -X
I believe what he described is the unconditional expectation, (before u open the envelope)
Now that u opened the envelope,
If u don't change the expected gain is 0
If u change, then expected gain is 2.5?
Or am I missing something?
Please do correct if I am wrong...
opening the envelope doesn't
opening the envelope doesn't alter the fact that what you see is equally likely to be X as it is to be 2X
Thanks for explaining...
But doesn't this extra bit of information mean something? (in filtration sense)
I can now replicate this scenario fully with a game that leads to outcome -5 or +10 with equal chances.
Surely this game should cost 2.5 to enter and not 0
Or am I wrong in suggesting the 2 outcome have 1/2 1/2 chances?
I am really confused to be honest, I see ur logic but can't really put my mind to it.
Could you please explain a bit further?
Actually (given it some thoughts)
1) Nick was right in saying opening the envelope doesn't change anything. You are right, I was wrong in saying that
2) in " u are holding 2x..... G/L = -X" That statement is a misuse of algebra
To be in line with the 1/2 of statement you should say:
"You are holding X, changing it become 1/2 x, therefore G/L = -1/2 X"
(when you say the amount u hold is X, it should be X in both scenario)
Expected return = 1/4 X regardless what X is
Which is in line with my +10 or -5 calculation where x= 10
That is an interesting finding, meaning if u give a person infinite chance to swap,
they will go on and on and keep on changing...
which should rapidly lead you
which should rapidly lead you to realising that your logic is nonsense...
; - >
Fair Point
Fair point,
But I still don't get how this situation is different to I am holding 10. And I have choice to enter a gamble to lose 5 or gain 10
If losing 5 or gaining 10 have equal chances, there is no way I shouldn't do it unless you put utility theory or risk aversion in the equation...
I would beg you to enlighten me, I sincerely want to know
here's your problem... you're
here's your problem... you're simulation is not correct..
you need to look at it like this:
In the 5,10 game... I have 50% chance of getting 5 and 50% chance of getting 10.
If I get 5 and swap, I gain 5... if i get 10 and swap, I lose 5.
In the 10,20 game... I have 50% chance of getting 10 and 50% chance of getting 20.
If I get 10 and swap, I gain 10... if i get 20 and swap, I lose 10.
therefore it doesn't matter whether I stick or swap.
however... there's a little more to the question than this..
; - >
9. Quant Interview Questions - Game Show - Monty Hall
This question is sometimes known as the Monty Hall question, and catches a lot of people out if they don't understand conditional probability.
You are in a game show.
The host has 3 doors, and behind one of the doors, there is a prize.
The host knows where the prize is.
The host asks you to choose a door.
You pick one.
He then flips open one of the other doors, and says:
“The Prize is not behind this one. Do you want to stick with your original choice of door, or swap ?”
What do you do and why ?
Difficulty - 6
You Swap The probabilty of
You Swap
The probabilty of winning the fisrt time is 1/3.
When you know that one door does not hide the price the probability for the door you have not choosen is now 1/2, which is better that you first choice of 1/3
Jean
Game show
you do indeed swap... correct
i think with the swap, the
i think with the swap, the odds of winning are actually 2/3, not 1/2. if you pick either of the incorrect doors to begin, then a swap will win you the prize, whereas the 1/3 chance you pick the right door on first attempt is the only time you'd lose with a swap.
swap
This is correct
Not agree in pure math
Not agree , after discarding the one door by host now there are only two door left and borh have equal chances to has prize
When you chose first door at that time you had three doors to chose so you gave equal chances to them andchose randomly one
But now when host has discarded one so now we cannot shift all chances of discarded door to 3rd door
After discarding the door the probability will be calculated again
For ex : if you have dice with 6 sided and we roll it six times then the probability of outcome 6 is 1
But if we add condition thatin first 5 atempt it never come as 6 , so what would be probability of outcome as 6 in sixth attempt
I would say it is 1/6
Not 5/6
10. Quant Interview Questions - Smarties Probability
You have 2 jars and 10 smarties – 5 orange, 5 yellow
How can you arrange the smarties such that if a jar were to be picked at random and a smarties were to be picked from that jar at random, the probability of picking an orange smartie is maximised.
Each smartie must be in a jar.
Difficulty - 5
smartie pants
all the yellows on one jar, one orange alone, and 4 oranges in with the yellows.
Nice one
Yep.. well done.
; - >
11. Quant Interview Questions - Sum of Numbers
What’s the sum of numbers 1 to 100 ?
Difficulty - 3
You have 50 pairs of 101
You have 50 pairs of 101 (1+100, 2+99, 3+98, etc.) so the answer is 50*101 = 5050
sum of numbers
Yep, that's right...
Part 2 of this question is:
Can you derive a generic formula for the sum of n numbers?
Yes, it is an arithmetic
Yes, it is an arithmetic progression. This takes me back to my A-levels!
Sn = n(2a + (n-1)d)/2
Sn = sum of n terms
a=first term
d=difference between terms (1 in this case)
The proof is simple and here: http://en.wikipedia.org/wiki/Arithmetic_progression
arithmetic progression
yep.. nice one.. It is indeed simple a-level maths, but amazing how many people forget that.
if n is even, 1+2+…+n =
if n is even,
1+2+…+n = (n/2)*(n+1)
if n is odd,
1+2+…+n= ((n-1)/2)*(n+1)+((n+1)/2)
sum of n numbers
Let S= 1+2+....+n.
Again S = n+(n-1)+(n-2)+...3+2+1
Add these two sums term-wise to get 2S=(n+1)+(n+1)+...+(n+1) (these are n terms)
this implies 2S = n(n+1), implies S = n(n+1)/2.
sum of numbers 1 to 100
sum = n(n+1)/2..here n=100
so sum = 100(100+1)/2 =100*101/2= 5050...
12. Quant Interview Questions - Measuring Jugs
You have 2 measuring jugs and some water. One holds 3 pints, the other holds 5 pints. How do you measure 4 pints ?
Difficulty - 2
watch die hard (3)
watch die hard (3)
Water measuring
; - >
fill the 3-p jug and move to
fill the 3-p jug and move to the 5-p jug
fill the 3-p again, and move 2 pints to the 5-p jug
empty the 5-p jug
move the remaining 1 pint in the 3-p jug to the 5-p jug
fill the 3-p jug and move to 5-p jug
------
the point is to measure a one pint first~
mearsuring
Yep, well done
perhaps more efficiently
fill 5p
pour 5p into 3p (leaving 2 pints in the 5p)
empty 3p
transfer 5p contents (2 pints) to 3p
fill 5p
pour 1 pint from the 5p to the 3p (filling the 3p)
leaving 4 pints in the 5p
12. Quant Interview Questions - Measuring Jugs
You fill the 3 pints jug and poor it in the 5 pints jug twice.
The 5 pints is now full and you have 1 pint remaining in the 3 pints.
Trow the 5 pints away.
Put the remaining pint in the 5 pints.
Fill the 3 pints again and put it in the 5 pints. Here you are with 4 pints
measuring jugs
u've been watching Die hard 3
; - >
Water Jugs
Fill up the 3 pints, empty the 3 pints into the 5 pints. Fill up the 3 pints again, and top off the 5 pints, leaving 1 pint in the 3 pint jug.
Empty the 5 pint jug, pour the remainder 1 pint in the 3 pint jug, into the 5 pint jug (this leaves 1 pint in the 5 pint jug).
Fill the 3 pint jug and pour it into the 5 pint jug.
Now you have 4 pints in the 5 pint jug
variation
Hold the 3p jug in the 5p jug, fill the 5-3p=2p left in the 5p jug with water, pour into the 3p jug and repeat.
13. Quant Interview Questions - Balance Scales
You have some balance scales and 9 marbles.
8 Marbles are the same weight, 1 is heavier.
What's the minimum amount of times you need to put marbles on the scales in order to guarantee working out which marble is heavier.
Difficulty - 1
Two, because you know that
Two, because you know that the one awry marble is heavier.
Divide the 9 balls into 3 sets of 3. Weigh three balls against three balls. If it tips either way, you know which side is heavier, or if they balance, the wrong ball must be in the third set of 3. Either way, it results in you having it narrowed down to three.
Just do that again, applying the same logic to 1 ball vs 1 ball.
Marbles
Yup
14. Quant Interview Questions - Matchstick game
You have matchsticks laid out as below:
2 players play the game.
A player has to take away at least 1 matchstick from a single column to play their go.
They can take as many matchsticks as they want from a single column on their go.
Play then alternates until all the matchsticks are gone.
The player to take the last matchstick LOSES the game.
Question: Is there a first mover advantage ? What’s the optimal strategy ?
Difficulty - 11
This is a great pub game especially if u know the answer.
; - >
Yes, you remove all seven
Yes, you remove all seven matchsticks in the fourth column and you will always win.
Given the complexity of this game, the way to deduce the answer is to reduce it to its simplest form and then add more complexity. So, starting with a game with one matchstick and one column it is clear that you do not want to start first, as doing so always results in loss.
Moving onto the two column game, it is clear playing first carries an advantage, you just remove the second column of three matchsticks on your first go and you win.
The three column game still has few enough possibilities that you can work through all of them, if you do so you discover that you will always lose if the second player plays an optimal strategy.
This then brings us to the answer, since you can force your opponent to play the three column game simply by removing the fourth column on your first go.
Matchstick game
nop...
Removing all 7 will result in losing the game if the opponent is playing optimally.
First person always loses for
First person always loses for sure. Optimal play is to start by picking a single one from any of the rows. To jump ahead, the next optimal play for your opponent (we are P1) would be to pick another single from any column not already picked from, until you got to a (0,2,4,6) arrangement. At that point, any move you make will give your opponent a winning move, which is explained below.
It's a really long explanation, so I'll be intentionally "brief", but there are a few endgames to achieve. Using form (1,3,5,7) to designate 4 columns, the two true endgames are if, after you move, you can get to (2,2) or (1,2,3).
Given this, there are 2 ways to get there.
1. matching - if, after your move, everything is paired up i.e. (3,3) or (1,3,3,1) or (4,4), the you win. You can, from this point on, just copy your opponent until either you can get to a (2,2) or make a winning move.
2. (1,2k,2k+1) - i.e. creating a (1,4,5) in this case. You can copy moves until you get to a (1,2,3), or, if the opponent crosses out the single, drop into the matching strategy.
I left out a whole lot of logic, but its worth a thought.
And to the admin, a decent game is with two players, to have one player draw the game, and the other gets to pick who goes first. Especially at a pub.
Matchsticks
close, but I don't agree with your starting comments...
take all but one in first
take all but one in first column. Opponent has to take top one. take all but one in second column. Opponent has to take top one. take all 3 in third. Opponent loses!
there'a a lot more to it than
there'a a lot more to it than that
This is a Game of Nim with
This is a Game of Nim with heap sizes 7, 4, 3, 1
So compute the xor --> 7 xor 4 xor 3 xor 1 = 1
This means the person starting first is in a winning position, the next move is to take one matchstick from the heap with 6 matchsticks, this should put the opponent in a losing position (6 xor 4 xor 3 xor 1 = 0). Subsequent moves will depend on what the opponent does, but the key is to leave him with heap sizes with xor value of 0.
wrong answer
wrong answer
So, what is the correct
So, what is the correct answer? Is there one?
wow!
that's a tough one! i needed to print out a pretty substantial list of winning/losing boards to see the pattern. i hope i have it right, now, though. seems you ought to look at the base 2 representation, no? if the bit-sum of each bit is 0 (assign each column the number of sticks in base-2), you have a losing board - otherwise you have a winning board. not too hard to show that any move changes this property, and that given any configuration without this property it seems you can make it so with a single move to hand the other player back a losing board, which gives you a winning strategy (as long as you can turn numbers into base-2 reasonably well in your head!).
that said, i guess that's a losing board?
yes, indeed.
yes, indeed. if the first mover is smart enough to make sure each column has an odd number of matchsticks after his go, he will always be the winner.
forgot to add this
forgot to add this: the first motion of him would be to remove that column with only one matchstick.
wrong
wrong
how about this?
since the players can take whatever number of matchsticks in one column (except 0), it only matters if he takes the whole column or leaves it with just one matchstick. Then the problem can be simplified as if there are 2,2,2,1 matchsticks, except that for some columns, players can take 0 matchstick. is this the right way of considering the problem?
15. Quant Interview Questions - Navigation question part 2
Followon from Navigation question part 1
How does the set of solutions evolve as the radius of the world tends from 0 to infinity ?
Difficulty - 7
Difficulty
this deserves a 10, I've no real idea of the answer
navigation
The set of solutions evolve from infinity to zero.
When radius = 0 (very small) any point is the desired point, as radius increases, the locus of desired points moves towards the poles. When radius = infinity, poles dont exist. Just flat surface, so 0 desired points.
16. Quant Interview Questions - Rope Burn
You have 2 ropes of varying thickness, and some matches.
You know that if you light one end of either rope, it will burn to the other end in exactly an hour.
How can you measure 45 minutes ?
Difficulty - 4
answer?
1) cut the first rope in half.
2) burn the second rope from both ends, at the point where the fire meets 30 mins would have past.
3) once the 30mins have past, burn the second half rope from both ends, at the point where the fire meets 15 mins would have past.
?
Rope Burning Question
nop.. cutting a rope in half does not help you because it's varying thickness so the 2 pieces ur left with could burn in 5 mins and 55 mins...
lay the ropes closely
lay the ropes closely parallel... stagger them so the end of one rope is near the midpoint of the other rope, and vice versa
burn the opposite ends
when the flames are on the line exactly perpendicular to the ropes, it's been 45 min
?
ropes
mid points mean nothing...
t+0:00 light both ends of one
t+0:00 light both ends of one rope, light one end of the second rope
t+0:30 first rope finishes burning. light the other end of the second rope
t+0:45 second rope stops burning.
ropes
that is correct
More detail re first rope?
I think you need to explain in more detail why the first rope will finish burning at t+0:30.
17. Quant Interview Questions - River crossing
It’s night time... you have 4 people, a torch, an old broken bridge over a ravine, and a pack of hungry bankers approaching looking to invest your money in CDO's.
The bridge can only take 2 people at a time, and it’s too dangerous to cross without the torch.
The people can cross the bridge at varying speeds:
Person 1 takes 1 minute to cross
Person 2 takes 2 minutes to cross
Person 3 takes 5 minutes to cross
Person 4 takes 10 minutes to cross
Question – How do you get all 4 people safely across the bridge in the quickest time ?
Difficulty - 7
solution
Person 1 and person 2 cross: 2 minutes
Person 1 goes back: 1 minute
Person 5 and 10 cross: 10 minutes
Person 2 goes back: 2minutes
Person 2 and 1 cross: 2minutes
= 17 minutes
Crossing
That's right..
most people don't think of sending 5 and 10 together...
I think its more that most
I think its more that most people don't think of sending person 2 back after person 3 and 4 cross over.
18. Quant Interview Questions - 1 litre of Water
This is my all time favourite question..
You are in a bar with your best friend, Bob.
You have 1 litre of water to drink between you.
You’re really thirsty, and drink half a litre, then give the water to Bob.
He drinks half of what’s left, and gives it back to you.
You drink half of what’s left, and give it back to him.
... you continue in the same way until there is a drop left.
Question: How much do you drink, and how much does he drink ?
Difficulty - 6
simple
Bob always drinks half of what you drink. Let amount Bob drinks = B, Let amount you drink = Y:
B + Y = 1
1/2B = Y
So B gets 1/3 and you get 2/3
Drinking Question
Yup..
Simple when you spot the short cut..
Most people go straight to an infinite sum of series, then guess the formula incorrectly or come out with 3/4 1/4.
I have heard it all !
the infinite sum of series
the infinite sum of series also gets 2/3 if my calculation is correct?
mine: 1/2(1-(1/4)^n) / (1-1/4) = 2/3, when n->+inf
infinite sum
ur formula isn't the infinite sum formula..
18. Quant Interview Questions - 1 litre of Water
This one look hard but is easy
If you drink 500ml and he drink 250ml, then that's 2/3 for you 1/3 for him.
This ratio won't change for the rest of the liter
125/62.5
31.5/15.625
But hey!! What's the point of drinking water in a bar!
drinking
Very good point RE the bar...
The first part of the question I wld normally ask: What's ur favourite drink?... then proceed appropriately putting u in a bar / cafe depending on ur drink preference.
1 Liter of water
X = Bob Volume
2x+X=1
3X=1
X=0.33
You is 0.67
Jean
19. Quant Interview Questions - Balance Scales Part 2
Followon from the first Balance Scales question.
You now have 12 marbles
- 11 weigh the same
- one is different
What's the minimum amount of weighings needed to find the different one AND work out whether it's heavier or lighter.
Difficulty - 10
4 weighings 1st
4 weighings
1st weighing:
split the marbles into 3 groups of 4. put any two groups on scales
2nd weighing:
- if the original weighing is balanced, take one group off, put 3rd group on, which contains the bad marble. note whether or not the marble is heavier or lighter based on whether the group is lighter or heavier.
- if the original weighing is unbalanced, take the "heavier" group off, put 3rd group on. if balanced now, the group you took off contains the bad marble, and it's heavier. if still unbalanced, the original "lighter" group from the first weighing contains the bad marble, and it's lighter.
3rd weighing:
split the bad group into pairs and weigh. find out which pair has the bad marble.
4th weighing:
split the bad pair into single marbles and weigh. locate the bad marble.
marbles
nop.. there's a better way
Balance scales
3 weighings
key
l = light (or a potentially light set)
h = heavy
u = unknown
12u marbles
weighing 1 4u v 4u reserve 4u
1.1 level
8g 4u marbles
1.2 tipped
4l, 4h, 4g marbles
weighing 2
1.1 weigh 3u v 3g
1.1.1 level 11g 1u
1.1.2 tipped unknown on light side 9g 3l
1.1.3 tipped unknown on heavy side 9g 3h
1.2 split 4l = 1l + 3l, split 4h = 1h + 3h, split 4g = 1g + 3g
weigh 1l + 3h v 1h + 3g
outcomes
1.2.1 level 3l 9g
1.2.2 same direction, 1l, 1h, 10g
1.2.3 switched 3h 9g
weighing 3
1.1.1 weigh 1u v 1g
1.1.1.1 1l 11g
1.1.1.2 ih 11g
1.1.2 weigh 1l v 1l
1.1.2.1 level have 11g and 1l
1.1.2.2 tips so have 1l 11g
1.1.3 weigh 1h v 1h - same as for the light weighings we
have
1.1.3.1 level have 11g and 1h
1.1.3.2 tips so have 1h 11g
1.2.1 1l v 1l
1.2.1.1 level have 11g and 1l
1.2.1.2 tips so have 1l 11g
1.2.2 1l v 1g
1.2.2.1 level 1h 11g
1.2.2.2 tips 1l 11g
1.2.3 1h v 1h
1.2.3.1 level 1h 11g
1.2.3.2 tips 1h 11g
in each case we know the false and if it is light or heavy
Balance
gr8 answer.. well done
20. Quant Interview Questions - Angle between Hands of a Clock
What's the angle between the hands of a clock at 2:45 ?
Difficulty - 1
RE : 20. Quant Interview Questions - Angle between Hands Clock
360/12=30
So there is 30 degree per hour (or per 5 minutes)
Between 45 and 2 there is 5 hours, so 5*30=150 degree
or 360-150=210 degree by the other side
clock
nop
CLOCK
There are 360/12 =30 degrees per hour,when the minute handle is on 45 the hour handle is has moved from 30*3/4 or is 30/4=7.5 degree from Three OClock angle is 180+7.5=187.5
Angle between clock hands
Yup
Your math is right, but the
Your math is right, but the answer should be 172.5 (the smaller angle on the clock vs. 187.5)
172,5
172,5
21. Quant Interview Questions - Coins on table game
2 Players take turns to place coins flat on a round table.
Losing player is the one who places his coin overlapping one already placed.
Question: Is there a first mover advantage ? What's the optimum strategy ?
Difficulty - 6
you go first. place the coin
you go first. place the coin in the middle. From then on, mimic the second players move on the opposite side of the table.
not that simple
the simplest answer is "it depends" - depends on the radius of the coin and the radius of the table. assumption: all coins are identical.
Define "placing a coin" = coin has to be completely inside the table AND it cannot overlap on an already placed coin.
Define Nmax as the max number of coins that can be "tightly packed" on the given table.
for Nmax < 5 it is obvious. place the coin at the centre and the opponent has no move to make. so the starter is the winner.
somehow i intuitively think that Nmax cannot be 6. cant prove though.
assume Nmax = 7 (the first of the interesting cases). in this case the diameter of the table is 3 x diameter of the coins. in this case (i think) the second mover has the advantage.
the first mover only has an advantage if and only if the players are forced to place their coins in any of the possible tightly packed positions. but the problem doesnt explicitly state that this.
Case 1. Simple case
first player: centre
second player: adjescent to centre
first player: any of the other possible "tightly packed" positions.
second player: any of the other possible "tightly packed" positions.
first player: any of the other possible "tightly packed" positions.
second player: there are only two possible tightly packed positions left. it is sub optimal for the second player to choose any of the tightly packed positions because if he does so then a loss is arrured. so he will choose to put his coin in between the two tightly packed positions so as to check mate the first guy.
need to work backwards now. it might be sub optimal for the first player to start at the centre or any of the tightly packed positions.
because of the symetries it is possible to reduce the possibilities.
Can work out the other cases. But in my opinion the second player has the advantage no matter how the first player plays.
Tha Nmax = 7 is only the first of the interesting cases i think.
coin
nop... previous answer was correct
22. Quant Interview Questions - Dominoes On Chess Board
2 opposite corner squares of a chess board are removed.
Can you place dominoes to cover the remaining squares ?
Note - A domino is rectangle the same size as 2 squares
Difficulty - 6
Dominoes On Chess Board
No.
The opposite corners are the same color (either both black or both white). So, removing them leaves you with more squares of one color. A domino always covers black and white squares in pairs. Therefore, no arrangement of dominos can cover the entire board.
Dominoes
Correct
23. Quant Interview Questions - Rubik's Cube
A 10 by 10 by 10 Rubik's Cube has all of the outer cubes fall off... How many are left ?
Difficulty - 2
a hidden duplicate
23 and 24 have same answer, ha ha.
6x64 = 384
because in a rubik's cube the interior cubes do not exist
rubik's cube
Assume that the rubik's cube's interior cubes exist...
10x10x10 becomes 8x8x8 = 512
10x10x10 becomes 8x8x8 = 512
24. Quant Interview Questions - Rubik's Cube 2
A 5 by 5 by 5 rubik's cube is sitting on a table.
How many of the outer cubes have exactly one side facing the air ?
Difficulty - 4
a hidden repeat
23 and 24 have same answer, ha ha.
6x9 = 54