1) 49^2 = (50-1)^2 = 2500-100+1 = 2401 2) 8^3 = 648 = 512 3) 100(101)/2= 50 101= 5050

(asked by a trader, and this is a serious brainteaser btw)

you have three guys, 1 girl, and two condoms. you all want to have heterosexual sex separately with her except you don't want to contract aids (meaning you never actually exchange fluids with either other men or the other woman). How is this possible?

 

condom answer: first guy puts on both condoms (unsafe, buts its the only way to do it). second guy puts on the condom that was on top of the first guy when he wore two, because that inside has only touched the other condom. then the third guy turns the last condom inside out and uses that.

 
Grinder86:
condom answer: first guy puts on both condoms (unsafe, buts its the only way to do it). second guy puts on the condom that was on top of the first guy when he wore two, because that inside has only touched the other condom. then the third guy turns the last condom inside out and uses that.

lol....that's just terrible :( so much for sex ed these days

​* http://www.linkedin.com/in/numicareerconsulting
 

I can't believe you're actually asked a condom question. I've NEVER ever been asked any sexual questions during interviews. Prolly since I'm a girl. It's so ridiculous. The girl has three holes right? One can stick it in her pussy, one in the ass and one in the mouth. PRESTO! AIDS schmaids.

********"Babies don't cost money, they MAKE money." - Jerri Blank********

********"Babies don't cost money, they MAKE money." - Jerri Blank********
 

its simple 1 guy fucks her with condom 2nd guy fuckers with other condom 3rd guy inverts one condom puts it on his D, takes the second condom and wraps up his inverted condom with one thats been used. Vegas Baby

 

well first guy uses 2 condoms = outside condom (condom 1) has girl fluids and inside condom (condom 2) has guy fluids

second guy uses condom 1 such that he has the unused part on himself and the girl on the side with her own fluids.

the last guy turns condom 2 inside out, puts the 2 condoms together such that the 1st and 2nd guy's fluids are together in the middle while he is wearing the fresh part with the girl fluids side of the condom on the outside again.

to summarize - 2 condoms as follows girl/unused unused/guy 1 girl/guy 2 girl/guy 2 guy 1/ unused

presto.

 

So if I'm reading this correctly someone will have to walk the i-phone back after each crossing:

MD and analyst walk across in 30 min followed by the MD crossing the bridge in his lonesome = 31 min

MD and associate cross in 9 min with the MD walking back in 1 min = 10 min

MD and VP cross in 4 min = 4 min

Total = 45 minutes

Making money is art and working is art and good business is the best art - Andy Warhol
 
Best Response

MD and VP walk together = 4 minutes MD returns back with iPhone = 1 minute Associate and analyst walk = 30 mins VP returns back with iphone and joins MD = 4 minutes MD and VP walk and join Associate and analyst = 4 minutes

Total time = 43 mins

 
sb_guru:
MD and VP walk together = 4 minutes MD returns back with iPhone = 1 minute Associate and analyst walk = 30 mins VP returns back with iphone and joins MD = 4 minutes MD and VP walk and join Associate and analyst = 4 minutes

Total time = 43 mins

Credit given where credit is due. (What was I thinking?)
Making money is art and working is art and good business is the best art - Andy Warhol
 

Depends entirely on the firm you interview at. At one of the places I interviewed at, they asked me 3 brain teasers, all of which you couldn't necessarily prepare for. At a bunch of others, I didn't get any brain teasers.

Often times, they just want to see how you think through a problem under pressure, not necessarily if you get it "right" or "wrong."

 
  1. I'm gunna assume these prisoners are champs and wana get their drink on. So you arrange the bottles in a cube and each prisoner drinks 100 bottles which will be either a on the x, y, z plane and from there you play battle ships to find the bottle which killed three prisoners.

EDIT: i googled it, i'm wrong.

"After you work on Wall Street it’s a choice, would you rather work at McDonalds or on the sell-side? I would choose McDonalds over the sell-side.” - David Tepper
 

Maybe somebody else can actually explain it better/more accurately/in detail but you can have 10 prisoners and assign them all to a certain combination of the 1000 bottles, and 10 people is enough to cover slightly over 1000 different combinations (to cover all 1000 bottles), so that when a certain combination of the 10 people dies, you know exactly which bottle is the only bottle that combination of people drank from. It's that damn thing on the calculator with nCr...

I hate victims who respect their executioners
 
BlackHat:
Maybe somebody else can actually explain it better/more accurately/in detail but you can have 10 prisoners and assign them all to a certain combination of the 1000 bottles, and 10 people is enough to cover slightly over 1000 different combinations (to cover all 1000 bottles), so that when a certain combination of the 10 people dies, you know exactly which bottle is the only bottle that combination of people drank from. It's that damn thing on the calculator with nCr...
Yes, the power set of a set of 10 people has 1024 elements.
-MBP
 
BlackHat:
Maybe somebody else can actually explain it better/more accurately/in detail but you can have 10 prisoners and assign them all to a certain combination of the 1000 bottles, and 10 people is enough to cover slightly over 1000 different combinations (to cover all 1000 bottles), so that when a certain combination of the 10 people dies, you know exactly which bottle is the only bottle that combination of people drank from. It's that damn thing on the calculator with nCr...

you're trying to match a 10-bit binary and the slaves' lives are the bits.

 

Haha basically. You have the 1st slave drink like bottle #1, 3, 5, 7, 9, 11... 2nd slave does like 3, 6, 9, 12... and then it ends up such that every bottle has a unique combination of slaves that would die if that was where the poison was. It's also time efficient so it makes the 24 hours period.

I hate victims who respect their executioners
 

1100

break it out this way: each year you get a perpituity of $10 so in year 1 you get 10 year 2 you get a new 10 + old 10 so forth value of the first perpituity is 100, second is 100/1.1 third is 100/1.21

1001/1.1^0 + 1001/1.1^1 + 100* 1/1.1^2...+ 100*1/1.1^infinity

this is an infinite geom series and r = 1/1.1 a = 100 100/(1-(1/1.1)) = 1100

 

Bankerph: breakinginnew's answer is correct - its $1,100, assuming first payment is at period end, and there is no principal recovery (which is implict given the coupon is paid out in perpetuity).

 

He tells them that if they wait around too long either the king or the red jasmine will be dead before they get back, if the flower is dead they failed their mission and if the king is dead then they will never know who will become king.

"Well, you know, I was a human being before I became a businessman." -- George Soros
 

They mount each others camels. Thus each has incentive to reach there fastest, as if one reaches there first, his camel actually reaches there last. Hence, that person succeeds.

 

Probability of getting $1 is (1/3)(1/2), probability of getting $2 is (1/3)(1/2)^2, probability of getting $3 is (1/3)*(1/2)^3

so it's like (1/3)*(1/2+1/2+3/8+2/+5/32 ... ) which I just ballparked to converge to $2

 

Probability of game going on = Probability of getting 1,2,3 = 1/2

Probability of game ending = Probability of getting 4,5,6 = 1/2

Probability of the game ending with $1 in the pot is (1/2)(1/2) = 1/2^2 Probability of the game ending with $2 in the pot is (1/2)(1/2)(1/2) = 1/2^3 ... Probability of the game ending with $n in the pot is 1/2^(n+1) ... Expected value of the amount in the pot when the game ends is E = 11/2^2 + 21/2^3 + ...+n1/2^(n+1)+...

E-1/2E = 1/2^2 + 1/2^3 + ...+1/2^n+... = 1/2

Thus, E = 1

Money in the pot can be taken only when the game ends with 4 or 5 being rolled.

The probablity of the game ending with 4 or 5 being rolled = Probabliy of 4,5 out of 4,5,6 = 2/3

So, the expected return of playing the game is 1*2/3 = 2/3

Sorry for putting those annoying numbers and formula here. My english is not good, so it's hard for me to explain it with just words.

 
Xdeatel:

Can we resell the tokens? If yes, infinite amount. If no, 0 tokens.

Unless there is very good prospect of infinite demand tomorrow, this approach seems foolish. Also, conceptually impossible.

Those who can, do. Those who can't, post threads about how to do it on WSO.
 

Thank you all for your comments. I too believe that this question has some missing information, just wanted to get some second opinions. I found it as one of the questions previously asked in interviews which one of the WSO members posted on here. The said the answer was calculated as follows: "(1.1)^3 = 1.33 which means you should get 3 years of token supply" - an answer that makes absolutely no sense to me.

 
ceres0305:

Thank you all for your comments. I too believe that this question has some missing information, just wanted to get some second opinions. I found it as one of the questions previously asked in interviews which one of the WSO members posted on here. The said the answer was calculated as follows: "(1.1)^3 = 1.33 which means you should get 3 years of token supply" - an answer that makes absolutely no sense to me.

This answer makes two assumptions: 1) you are only willing to make an investment that EQUALS your required rate of return, can't exceed 2) the rate of token appreciation is 33% over the year even though only a daily appreciation was given

Poorly worded question

 

100%. You didnt propose that they weren't true.

“...all truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident.” - Schopenhauer
 

LOL

the answer is 0%. because none of the answer choices given are "correct". instead, you gave answer choices of

A. 25% B. 50% C. 60% D. 25%

none of these literally say "correct."

if answer C had been "C. Correct" instead of "C. 60%", then the probability wouldve been 1/4. But you gave percentages as answer choices to throw us off.

Nobody knows what the correct answer choice is nor how many correct answer choices there are. The 4 answer choices could've been taken from an MCAT exam for all we know where the question states "Which is correct? You may mark more than one answer."

Since it's impossible to know the number of correct responses to a group of answer choices without its parent question, the only other thing you're asking of us is whether "correct" is among the answer choices to which I answer, "0/4" or "0%"

 
Prim.<span class=keyword_link><a href=/finance-dictionary/equity-research-overview><abbr title=equity research&#10;>ER</abbr></a></span>.ate:
5 posts, 5 answers....interesting...and if you wish, you can swap the word "true" for "correct"...that isn't it...

haha DAMN. i thought this was going to be a grammar-based brainteaser in disguise for once lol

 
MMBinNC:
I'm amazed other people have actually looked at reddit math

i never even knew reddit math existed haha. thanks for giving me something to look at at work now.

maybe i should start posting questions from my old college discrete math textbook on here

 

Answer cannot be 1/3 because 25% is represented twice in the choices, so higher chance of randomly choosing it. Just like HT/TH case with coins... Same result overall, but got to account for probability of randomly landing tails or heads first.

Or have I got it completely wrong?

"So who lost the hundy?"
 

it is impossible to know for sure....for all we know, 25% could be the answer...since A and D are 25%, the probability would be 2/4 or 1/2....

if the answer is NOT 25%, we need to put A and D as one single answer, giving the answer of 1/3...it is impossible to know for sure.

 

I got ~38%. Is the answer an actual answer choice? Where are people getting 0%?

Logic (probably wrong): If choose A or D, 25% of the time you have a 50% chance of being correct. Else, 25% chance if B or C.

 
XPJ:
I got ~38%. Is the answer an actual answer choice? Where are people getting 0%?

Logic (probably wrong): If choose A or D, 25% of the time you have a 50% chance of being correct. Else, 25% chance if B or C.

the choices are answers to the question. you have a .5 chance of picking 25%, so 25% is not the correct answer; you have .25 chance of picking 50%, so 50% is not the answer.
 
jec:
it's not a paradox, 0% works perfectly

Maybe I'm missing something, but I still don't see the logic behind 0%.

For example:

If I choose a random answer from among those below, what is the probability that the answer is correct?

A. Blue B. Red C. Yellow D. Blue

Assuming that either Red, Blue or Yellow is the correct answer, I don't see how you get 0%. The %'s are thrown in as a mind-fuck. Unless, it's wrong to assume that R, B or Y are the 3 possible answers in which case this is a retarded question.

 

6997 for captain 1 for fifth 1 for third 1 for last

It's the usual MBP "brainteaser" where it all boils down to the one vs. one situation where the next to last pirate can take all 7000 because he needs only his own vote. So it would be enough to offer the last 1 because otherwise he would receive 0.

 

If pirate 1 is the only pirate left he gets 7000 coins. This case will never happen.

If Pirate 1 and 2 are left, pirate 2 will get 7000 coins because he can give himself half of the votes.

If 3 pirates are left, pirate 1 gets 1 coin, and pirate 3 gets 6999 coins, pirate 2 gets zero. Pirate 1 will vote for this distribution because if he doesn't, pirate 3 will die and all the coins will go to pirate 2.

If 4 pirates are left, Pirate 4 will give 1 coin to pirate 2. This secures is vote because he gets one more coin than the 3 pirate scenario. Pirates 1 and 3 get nothing, Pirate 4 gets 6999.

If 5 pirates are left, Pirate 5 will offer Pirates 1 and 3 two coins each in order to secure >50% of the vote. Pirates 2 and 4 get nothing. Pirate 5 gets 6998 coins.

If 6 pirates are left, Pirate 6 will prevent mutiny by giving one coin to Pirates 2 and 4. Pirates 1,3, and 5 get nothing, and Pirate 6 takes 6998 coins.

If 7 pirates are left Pirate 7 needs to earn 3 votes besides his own to prevent mutiny. He offers 1 coin each to pirates 1, 3 and 5. None to pirates 2, 4, 6. He takes home 6997 coins.

The key is to offer just enough coins to secure at least half the votes. You secure a vote by offering more coins than the pirate would get in a scenario with one less pirate.

 
Old Grand-Dad:
If pirate 1 is the only pirate left he gets 7000 coins. This case will never happen.

If Pirate 1 and 2 are left, pirate 2 will get 7000 coins because he can give himself half of the votes.

If 3 pirates are left, pirate 1 gets 1 coin, and pirate 3 gets 6999 coins, pirate 2 gets zero. Pirate 1 will vote for this distribution because if he doesn't, pirate 3 will die and all the coins will go to pirate 2.

If 4 pirates are left, Pirate 4 will give 1 coin to pirate 2. This secures is vote because he gets one more coin than the 3 pirate scenario. Pirates 1 and 3 get nothing, Pirate 4 gets 6999.

If 5 pirates are left, Pirate 5 will offer Pirates 1 and 3 two coins each in order to secure >50% of the vote. Pirates 2 and 4 get nothing. Pirate 5 gets 6998 coins.

If 6 pirates are left, Pirate 6 will prevent mutiny by giving one coin to Pirates 2 and 4. Pirates 1,3, and 5 get nothing, and Pirate 6 takes 6998 coins.

If 7 pirates are left Pirate 7 needs to earn 3 votes besides his own to prevent mutiny. He offers 1 coin each to pirates 1, 3 and 5. None to pirates 2, 4, 6. He takes home 6997 coins.

The key is to offer just enough coins to secure at least half the votes. You secure a vote by offering more coins than the pirate would get in a scenario with one less pirate.

You weren't first, but SB anyway for taking the time out to spell it out.
-MBP
 

In other words, if it is an even number of pirates, just give every pirate with an even number rank 1 gold coin. If it is an uneven number if pirates, just give every pirate with an uneven number rank 1 gold coin.

Hi MBP.

 

dunno, but couldn't you just ask them who the person next to them is?

if the god says that the person next to him is honest then that person has to be random, if he says that the person next to him is random he must be honest, if he says that the god next to him is a liar you can't be sure on anything yet

at this point, if you already know who's random or honest, you can just ask the other two who that god is (the one you determined)... liar will call him a liar, honest will call him what he really is, and random will call him honest (if it's truly honest, the only other choice in this scenario, you know that liar wouldn't say that)

then you do the same idea again if needed...

I dunno, my explanation is kinda long, but the idea seems pretty easy (i'm looking at them as asking them left to right)

edit: gah, I know I must be missing something in my explanation, but wouldn't it work out even if you could only ask three questions to each? there's probably a way simpler explanation, but whatever, I tried -_-

If your dreams don't scare you, then they are not big enough. "There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.
 

GODS - A/B/C

Ask B, "Would you deny that A is RANDOM?"
1. If B answers 'ja', then either B is RANDOM (and is answering RANDOMly ), or B is not RANDOM and A is indeed RANDOM . Either way, C is not RANDOM definitely 2. If B answers 'da' , then either B is RANDOM (and is answering RANDOMly), or B is not RANDOM and A is not RANDOM . Either way, A is not RANDOM definitely

Go to the God who was identified as not being RANDOM definitely based on B's answer by the previous question (either A or C) and find out whether he is GOD - TRUE or FALSE by asking the following question...

"If I asked you 'Are you TRUE-GOD?', would you say 'no'?

if he says NO then he is GOD-TRUE....if he says YES, then he is GOD-FALSE

Question No. 3 -

Ask the same GOD (who was identified as not being RANDOM) the question: "If I asked you 'Is B RANDOM?', would you say 'no'?". 1. If the answer is 'no' then B is RANDOM; 2. if the answer is 'yes' then the GOD you have not yet spoken to is RANDOM.

The third GOD can be identified by elimination after we have identified first 2

 
SC911:
GODS - A/B/C

Ask B, "Would you deny that A is RANDOM?"
1. If B answers 'ja', then either B is RANDOM (and is answering RANDOMly ), or B is not RANDOM and A is indeed RANDOM . Either way, C is not RANDOM definitely 2. If B answers 'da' , then either B is RANDOM (and is answering RANDOMly), or B is not RANDOM and A is not RANDOM . Either way, A is not RANDOM definitely

Go to the God who was identified as not being RANDOM definitely based on B's answer by the previous question (either A or C) and find out whether he is GOD - TRUE or FALSE by asking the following question...

"If I asked you 'Are you TRUE-GOD?', would you say 'no'?

if he says NO then he is GOD-TRUE....if he says YES, then he is GOD-FALSE

Question No. 3 -

Ask the same GOD (who was identified as not being RANDOM) the question: "If I asked you 'Is B RANDOM?', would you say 'no'?". 1. If the answer is 'no' then B is RANDOM; 2. if the answer is 'yes' then the GOD you have not yet spoken to is RANDOM.

The third GOD can be identified by elimination after we have identified first 2

Obvious Google is obvious.

"After you work on Wall Street it’s a choice, would you rather work at McDonalds or on the sell-side? I would choose McDonalds over the sell-side.” - David Tepper
 
SC911:
GODS - A/B/C

Ask B, "Would you deny that A is RANDOM?"
1. If B answers 'ja', then either B is RANDOM (and is answering RANDOMly ), or B is not RANDOM and A is indeed RANDOM . Either way, C is not RANDOM definitely 2. If B answers 'da' , then either B is RANDOM (and is answering RANDOMly), or B is not RANDOM and A is not RANDOM . Either way, A is not RANDOM definitely

Go to the God who was identified as not being RANDOM definitely based on B's answer by the previous question (either A or C) and find out whether he is GOD - TRUE or FALSE by asking the following question...

"If I asked you 'Are you TRUE-GOD?', would you say 'no'?

if he says NO then he is GOD-TRUE....if he says YES, then he is GOD-FALSE

Question No. 3 -

Ask the same GOD (who was identified as not being RANDOM) the question: "If I asked you 'Is B RANDOM?', would you say 'no'?". 1. If the answer is 'no' then B is RANDOM; 2. if the answer is 'yes' then the GOD you have not yet spoken to is RANDOM.

The third GOD can be identified by elimination after we have identified first 2

Lol, you totally didn't do this yourself. You even used the terminology of the original problem (although still simplified since in the original, you don't know whether ja means yes or no)
-MBP
 

this is one of those things where it's gonna be hard to cover all possible scenarios before I stop giving a shit

If your dreams don't scare you, then they are not big enough. "There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.
 

well well well, looks like a little cheater had to come in here and ruin the fun... what happens to the SB now? I say you give it to the only person who actually gave it an honest shot before this rapscallion came in here and ruined our lives.

If your dreams don't scare you, then they are not big enough. "There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.
 
wolverine19x89:
well well well, looks like a little cheater had to come in here and ruin the fun... what happens to the SB now? I say you give it to the only person who actually gave it an honest shot before this rapscallion came in here and ruined our lives.
Happy?
-MBP
 

I did it! I did it!

did my explanation even make any sense to you? ha

If your dreams don't scare you, then they are not big enough. "There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.
 
SC911:
Just to figure it out as I go to google for most of my problems, #winning.
I hope you look forward to the following interview exchange: [INTERVIEWER]: How would you solve {insert brain teaser here}? [SC911]: I'm not sure, but I am REALLY good at pressing ctl+c, ctl+v [INTERVIEWER]: Get the fuck out.
 

Easy, in two questions or less.

Label gods as A, B, and C. They are in a circle so that B is right of A, C is right of B and A is right of C.

Ask them "Who is to your right?"

If no one answers Random, move them into the order "BAC". Ask them the same question.

By now, one will have answered Random. This is "Honest", since random would never label someone else as random and since the liar labels everyone else as "liar". The person who has been labeled Random is random, and the remaining god is honest.

 

99.

I feel so wise.

[quote=rufiolove]When evaluating whether or not to post something on WSO, I think to myself, "would an idiot post this" and if the answer is yes, I do not post that thing...[/quote]
 

Is it just 3 or am I missing something?

Grab one sock, white or black, put it aside.

Grab another sock, possibly the other color.

If you grab a third no matter what you're going to have a pair.

"The only point in making money is so you can tell some big shot where to go.” -Bogie
 
MSFhopeful:
Is it just 3 or am I missing something?

Grab one sock, white or black, put it aside.

Grab another sock, possibly the other color.

If you grab a third no matter what you're going to have a pair.

Yeah.
-MBP
 

For the first four, use one of the following three formulas: (a+b)(a-b)=a^2-b^2 (a+b)^2=a^2+2ab+b^2 (a-b)^2=a^2-2ab+b^2

I'll leave it to you guys to think of what the "a" and "b" should be in each situation. That's a general trick for most "fast" multiplication problems people tend to ask: they usually fall into one of these patterns you can use. Except when they don't, and life is miserable (I had that happen to me on a phone interview and was pretty sad when I terribly failed).

For the last one... It's something annoying about squaring something with only 1s. It ends up being an ascending natural number sequence followed by the same thing in reverse. Ex: 11=1 1111=121 111*111=12321 ...

These are just good tricks to keep in mind since I've heard bank bros occasionally ask these things.

 
canas15:
For the first four, use one of the following three formulas: (a+b)(a-b)=a^2-b^2 (a+b)^2=a^2+2ab+b^2 (a-b)^2=a^2-2ab+b^2

I'll leave it to you guys to think of what the "a" and "b" should be in each situation. That's a general trick for most "fast" multiplication problems people tend to ask: they usually fall into one of these patterns you can use. Except when they don't, and life is miserable (I had that happen to me on a phone interview and was pretty sad when I terribly failed).

For the last one... It's something annoying about squaring something with only 1s. It ends up being an ascending natural number sequence followed by the same thing in reverse. Ex: 11=1 1111=121 111*111=12321 ...

These are just good tricks to keep in mind since I've heard bank bros occasionally ask these things.

hmm, I just round to the nearest 10, 1921 = 2021-21 = 420-21= 399

did all but last one like this within 5-10 secs

 

I'd rather go through problems like ∫ ln(x) dx, but i doubt anyone here wants to go through integrals. Also, I am starting with relatively simple things. In the future I will add some more challenging ones. Problem solving skills cannot be learned by searching google for the answer to these questions and memorizing them, but rather building a foundation by solving problems.

 
RubiksCubeMath:
I'd rather go through problems like ∫ ln(x) dx, but i doubt anyone here wants to go through integrals. Also, I am starting with relatively simple things. In the future I will add some more challenging ones. Problem solving skills cannot be learned by searching google for the answer to these questions and memorizing them, but rather building a foundation by solving problems.

I would prefer integrals...

 

The hardest brainteaser I ever got was a riddle:

It starts with the letter "S". It's two words. It's on every forum. It rhymes with the phrase "Church Glutton".

S _ AR _ H
B_ _ _ ON

Solve the puzzle.

Robert Clayton Dean: What is happening? Brill: I blew up the building. Robert Clayton Dean: Why? Brill: Because you made a phone call.
 
goodL1fe:
The hardest brainteaser I ever got was a riddle:

It starts with the letter "S". It's two words. It's on every forum. It rhymes with the phrase "Church Glutton".

S _ AR _ H
B_ _ _ ON

Solve the puzzle.

:-D

+1

Money Never Sleeps? More like Money Never SUCKS amirite?!?!?!?
 

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-------------------- "It is a fine thing to be out on the hills alone. A man can hardly be a beast or a fool alone on a great mountain." - Francis Kilvert (1840-1879) "Ce serait bien plus beau si je pouvais le dire à quelqu'un." - Samivel
 

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Sint at ut aut esse reprehenderit. Consectetur deserunt occaecati error. Odit rerum omnis eum vitae. Alias corporis molestias id ex. Et vitae tenetur natus reiciendis sapiente. Voluptate fugiat itaque perferendis ducimus aut.

-------------------- "It is a fine thing to be out on the hills alone. A man can hardly be a beast or a fool alone on a great mountain." - Francis Kilvert (1840-1879) "Ce serait bien plus beau si je pouvais le dire à quelqu'un." - Samivel

Career Advancement Opportunities

March 2024 Investment Banking

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notes
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