ITT some basic quant interview questions
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Poast new message in this thread
Date: February 4th, 2013 4:57 PM Author: indecent psychic native
sample questions for intro level business analyst type job (not quant but you have to have an intuitive grasp of numbers)
starting with the easiest:
there's a pond with lily pads on it. every day, the number of lily pads doubles. on the 40th day, the pond is completely covered. on which day is the pond half covered?
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568146) |
Date: February 4th, 2013 5:00 PM Author: indecent psychic native
#3
what is the sum of all numbers from 1-99? no calculator/outside aids allowed. also describe how you would do the calculation
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568179) |
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Date: February 4th, 2013 5:03 PM Author: aphrodisiac flesh nursing home jap
1 + 99 = 100.
2 + 98 = 100.
and so on. so we end up with 49 pairs with a value of 100 each. 49 x 100 = 4900. We have 1 50 left over. 4950.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568200) |
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Date: February 4th, 2013 5:26 PM Author: self-centered submissive piazza dysfunction
594
I sample the sum of the first eleven numbers in the series: 1+2+3+4+5+6+7+8+9+10+11 = 66.
From this sample, I can reasonably estimate that each sample of eleven numbers in the overall series sums up to 66, and the total number of samples of eleven numbers is 99/11 = 9. therefore the total sum for 9 samples would be 66*9 = 594 +/- 7.4 at 95% confidence
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568365) |
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Date: February 5th, 2013 7:48 PM Author: razzle market
Imagine a 99 x 99 stack of unit area blocks. Then divide the stack in half by a diagonal line going top left to bottom right.
The question is equivalent to asking what is the total area of the blocks either entirely to the left of the line or cut by the line.
(99 x 99)/2 gives you the area to the left of the line but then you have to account for an extra 99 half-blocks for those blocks cut by the line, so:
(99 x 99)/2 + (99 x 0.5) = 4950
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22576162) |
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Date: February 4th, 2013 5:19 PM Author: Gay ratface business firm
no, this one has at least one solution
edit:
oh, i see why this is confusing everyone. ABC is a 3 digit number, not A*B*C
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568310) |
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Date: February 4th, 2013 5:19 PM Author: boyish messiness
ABC=813
CBA=318
BBCB=1131
pretty simple once you realize that B has to be 1. if you are adding 2 3-digit numbers, the largest possible sum is 1998 (999+999). so B must be 1. C+A must equal 1, so the possible combinations are like 7, 4 or 8, 3, etc. i plugged in 8, 3 first and got the right answer.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568311) |
Date: February 4th, 2013 5:15 PM Author: indecent psychic native
#4
how would you obtain the sum of all results of a set of 1-10 multiplication tables?
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568286) |
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Date: February 4th, 2013 5:17 PM Author: Gay ratface business firm
sum of the first row is 5*11 (using the gauss formula above)
each row is n*(first row) where n is the value of the row digit
so 55 + 2*55 + ... + 10*55
iow, (1+2+...+10)*55 = 55^2
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568303) |
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Date: February 4th, 2013 5:50 PM Author: vigorous indirect expression pisswyrm
sum(sum(ij),i=1..10),j=1..10)
sum(j*sum(i,i=1..10),j=1..10)
sum(j*55,j=1..10)
55*sum(j,j=1..10)
55*55
3025
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568531) |
Date: February 4th, 2013 5:28 PM Author: indecent psychic native
You are offered to play a game of chance. A fair coin is tossed repeatedly until you get the first tails, at which point the game ends and you get the prize. The prize "pot" starts at $1 and doubles each time you get heads. So for instance, if you get heads the first toss, the pot becomes $2. If you get heads again the second toss, the pot becomes $4. If you get heads the third time, the pot becomes $8. If the fourth toss gives you the tail of the coin, you win and take home the $8 prize.
Before you play, you must pay a fee to enter this game. The question is, what's the maximum amount you're willing to pay in order to play this game?
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568387) |
Date: February 4th, 2013 5:30 PM Author: Motley Cuckold
Got this one during my b-school interview:
You have a lamp that you need to set on a timer. You want it to come on for two periods daily: 5-8am and 5-8pm. You have an unlimited number of light timers that each can allow power to the lamp for one period (think: those cheap Ikea timers that would allow the light to be on for one period per day). You can run these timers in sequence.
How many of these timers would the lamp require, and each set to what period, so that it will be on for these two time periods during the day?
edit: 24 hour timers. Lamp has one cord going to a single outlet. Multiple timers would run in sequence.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568397)
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Date: February 4th, 2013 6:23 PM Author: aromatic unholy persian
you god damn retards
2 timers
5am to 8pm
5pm to 8am
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568725) |
Date: February 4th, 2013 5:35 PM Author: indecent psychic native
this is a well known one
One hundred ants are dropped on a meter stick. Each ant is traveling either to the left or the right with constant speed 1 meter per minute. When two ants meet, they bounce off each other and reverse direction. When an ant reaches an end of the stick, it falls off.
At some point all the ants will have fallen off. The time at which this happens will depend on the initial configuration of the ants.
Question: over ALL possible initial configurations, what is the longest amount of time that you would need to wait to guarantee that the stick has no more ants?
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568436) |
Date: February 4th, 2013 5:46 PM Author: indecent psychic native
A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair.
You tell her that you want only a male, and she telephones the fellow who's giving them a bath.
"Is at least one a male?" she asks him. "Yes!" she informs you with a smile.
What is the probability that both beagles are male?
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568507) |
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Date: February 4th, 2013 5:53 PM Author: Gay ratface business firm
cases where one is a male:
MF, FM, MM
in 1/3, both are male
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568542) |
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Date: February 4th, 2013 5:58 PM Author: Sickened hideous brethren
This is the way it was explained to me, but I find this suspect. You are treating it as though there are "positions" for the dogs. But as I see it, there is only one dog and the other dog, so there should really only be two cases
MF
MM
where the second spot represents "the other dog."
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568560)
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Date: February 4th, 2013 6:34 PM Author: Contagious crackhouse
No you have 4 scenarios:
Mgroom Mbride
Mgroom Fbride
Fgroom Fbride
Fgroom Mbride (lol)
You've eliminated the FF wedding, but you have a 1/3 shot at dat MM
edit: If you know in advance that it is 1 man marrying either a man or a woman, then you are right. But if you are going in blind and ask someone "is atleast 1 partner a man", then the odds play out as I showed above.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568794) |
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Date: February 4th, 2013 6:00 PM Author: Contagious crackhouse
You have Puppy A and Puppy B. Four equally likely outcomes:
Aboy Bgirl
Aboy Bboy
Agirl Bgirl
Agirl Bboy
The phone call eliminates one outcome (Agirl, Bgirl)
Of the remaining possible outcomes, 1/3 is Boy-Boy
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568567) |
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Date: February 4th, 2013 6:27 PM Author: Contagious crackhouse
With the wording of the problem, you assume that the person on the phone knows the sex of both puppies.
It's similar to the monty hall problem, where the fact that the host knows that he is showing you a losing door changes the odds vs if he were to show you a door at random and it happens to be a loser.
If he had only looked at one dog, and it happened to be a male, your numbers are correct.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568762) |
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Date: February 4th, 2013 7:10 PM Author: Sickened hideous brethren
Here's another way to think about it:
The guy on the phone is effectively saying "I have two puppies here, one to my right and one to my left, and at least one of them (the one I'm looking at) is male"
There are four possibilities now:
1) The one he is looking at, R, is male, and L is also male
2) The one he is looking at, L, is male, and R is also male
3) The one he is looking at, R, is male, and L is female
4) The one he is looking at, L, is male, and R is female
The odds of two males becomes 50%. I don't see why this is any more or less a reasonable way to interpret the problem than the standard answer.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568990) |
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Date: February 4th, 2013 6:40 PM Author: Gay ratface business firm
ok, i think i've got it.
my original idea was to paddle in a circle with a radius of r/2 (a circle inscribed so that r of the pond is its diameter). this had the virtue of forcing the monster to run towards a spot that you were moving away from. sadly, he only has to cover 3/2 pi r in the time you take to travel 1/2 pi r, so you still lose b/c of his speed advantage. (he only needs to be 3x as fast as you to arrive at the same moment).
this made me realize that the monster's speed advantage always allows him to catch you on an arced course whenever the distance you travel is that of a circle with arc >= 1/4r
so i think you win by 1) paddling in a circle with arc < r/4, which should allow you to get to a point where he is trailing the point your bow is pointing at, and then 2) go straight to the shore
do i get to be a janitor for quants with this answer
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568833) |
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Date: February 4th, 2013 6:09 PM Author: Contagious crackhouse
1 black pearl in one bag
99 black pearls and 100 white pearls in the other
edit: I mean, assuming you have to place every pearl in a bag. Otherwise, just dump those TTT white pearls.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22568623) |
Date: February 5th, 2013 6:58 PM Author: Outnumbered selfie
most of these are all pseudo-intellectual gay ass questions.
only the ones that required inventing a formula are worth looking at.
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22575719) |
Date: February 5th, 2013 7:33 PM Author: Geriatric ticket booth sneaky criminal
I got this one:
"Think out loud and let us listen to how you estimate the cans of paint it will take to paint the Verrazano Narrows Bridge."
(http://www.autoadmit.com/thread.php?thread_id=2175711&forum_id=2#22576048) |
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