Try to Solve Quant Interview Questions in 1 Hour

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Hi everyone,

I’m preparing for some interviews in next week, so I challenge myself to solve quant interview questions as many as possible during one hour.

Quite sure that I can’t solve all, I try with 21 problems as below:

  1. For a 3 sets tennis game, would you bet on it finishing in 2 sets or 3 sets?
  2. I have a square, and place three dots along the 4 edges at random. What is the probability that the dots lie on distinct edges?
  3. You have 10 people in a room. How many total handshakes if they all shake hands?
  4. Two decks of cards. One deck has 52 cards, the other has 104. You pick two cards separately from a same pack. If both of two cards are red, you win. Which pack will you choose?
  5. What is 39*41?
  6. A group of people wants to determine their average salary on the condition that no individual would be able to find out anyone else’s salary. Can they accomplish this, and, if so, how?
  7. How many digits are in 99 to the 99th power?
  8. A line of 100 passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let’s say that the nth passenger in line has a ticket for the seat number n.) Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?
  9. What is the sum of the numbers one to 100?
  10. You have a 3 gallon jug and 5 gallon jug, how do you measure out exactly 4 gallons? Is this possible?
  11. You have 17 coins and I have 16 coins, we flip all coins at the same time. If you have more heads then you win, if we have the same number of heads or if you have less then I win. What’s your probability of winning?
  12. What is the probability you draw two cards of the same color from a standard 52-card deck? You are drawing without replacement.
  13. You’re in a room with three light switches, each of which controls one of three light bulbs in the next room. You need to determine which switch controls which bulb. All lights are off to begin, and you can’t see into one room from the other. You can inspect the other room only once. How can you find out which switches are connected to which bulbs? Is this possible?
  14. In world series, what are the odds it goes 7 games if each team equal chance of winning?
  15. Given 100 coin flips, what is the probability that you get an even number of heads?
  16. There are 5 balls, 3 red, and 2 black. What is the probability that a random ordering of the 5 balls does not have the 2 black balls next to each other?
  17. What is the least multiple of 15 whose digits consist only of 1’s and 0's?
  18. Is 1027 a prime number?
  19. Does the price of a call option increase when volatility increases?
  20. 2 blue and 2 red balls, in a box, no replacing. Guess the color of the ball, you receive a dollar if you are correct. What is the dollar amount you would pay to play this game?
  21. What is the singles digit for 2²³⁰?

Now let’s start!

The game will end in 2 sets if there is a player wins two consecutive games. Assume that the probability of a player can win a game is p, and the probability that he loses a game is q (hence p + q = 1). Then the probability that he wins two games consecutively is p^2, the probability that his competitor wins two games consecutively is q^2. Hence the probability that the game ends in two sets is p^2 + q^2.

Otherwise, if the game ends in three sets, the player must win the first and lose the second or conversely. So the probability of this event is 2*p*q.

Because p^2 + q^2 >= 2*p*q, so I would bet the game finishes in two sets.

There are 4³ ways to place three dots along edges of a squares, but there are only 4!/(4-3)! = 4! ways to place the dots on distinct edges. So the probability of that event is equal to 4!/4^3 = 24/64 = 3/8.

There are absolutely 10*(10-1)/2 = 45 handshakes.

If I take cards from first pack, the probability that both cards are red is 13*12/(52*51) = something. If I take cards from the other pack, the probability of the event is equal to 26*25/(104*103). You could compare these two results and imply the conclusion.

I think 39*41 = (40-1)*(40+1) = 40^2 — 1 = 1600 -1 = 1599. I met this kind of question when I was at primary school (exactly in a competition I was asked to quick calculation the result of 25 * 44, I didn’t know why they need to use a trick while I calculated by myself and gave the result much faster than the others (?!?)).

Classical question for quants. I know the solution because I read this question from books.

The first person keep a secret number, e.g. 100, and adds it to his salary, then gives the result to the second, the second adds his salary to the result, then pass to the third,… we keep continue to do that. When all is done, the first person just need to subtract the secret number from the result to have total amount of salaries of all members of group. The left part is trivial.

We have log10(99^99) = 99*log10(99). Since log10(99) is approximately equal to 2. So my answer is 198.

passed

It’s equal to 100*101/2 = 5050.

Again it’s easy problem for primary student. The hint is 4 = 5 — 1. Firstl, 5 -> 3 then there is 2 gallons remained in the 5-gallon jug, secondly, 5-> 3 , then there is 2 gallons in 3-gallon jug and the 5-gallon jug is empty. the last is 5 -> 3, then there is 4 gallons remained in the 5-gallon jug.

It looks a complicated problem. I will consider following cases: I have more heads than you in the first 16 coins flipped, assume its probability is p, then I’ll win the game. Conversely, you also have probability p to have more heads than me, so you’ll win the game. The third case is when we have the same number of heads in the first 16 coins flipped, so its probability is 1 — 2*p. So I would win the game if the last flip I got head.

Hence the probability that I win this game is p + 1/2 * (1 — 2*p) = 1/2.

Wow, it’s a fair game :).

The probability is equal to 52*12/(52*51) = 12/51 = 4/17.

It's a classical brain teasers too, so I’ll pass it to you.

There must be 3 games that one team wins and 3 other games that team loses in the first 6 games.

You guess that it’s 1/2, right?

I don’t know, the remained time is too little…

Since 15 = 3*5, so the number that we are searching need to be a multiple of 3 and 5. In order to be a multiple of 5, it must end with 0. Beside, the sum of its digits must be a multiple of 3. Hence the number is 1110.

What the hell is it number? Could I run a code?

No need to know the Black Scholes formula, we can answer “yes” by normal intuition.

I don’t understand this question. 1/2?

The last digits of first power numbers of 2 is 2, 4, 8, 6, then 2, .... So the answer is 2.

Finish!

Quant Researcher, Data Scientist, Food Hater (so I eat them a lot).