Renaissance Technologies Interview Questions, Process, and Tips

CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….

20km

890.488

Sunday
Monday
Tuesday

Answer 33
Adding 7,9.13,15,17
Correct is 35

n(n+1)/2
=10*11/2
55

Let’s say,

I have x coins of 50 paise and (80-x) coins of 100 paise,so the equation is like this ,
50x + (80-x)*100 = 64*100
x = 32
So ,I have 32 coins of 50 paise

4 boy 3 sis

34000

put 1 red marble in one jar and all the rest (99) in the
other.
This gives you 50% from the first jar (if they pick that
jar they will get red 100% of the time). For the other jar
the chances are 49/99 or 49.494949%. Divide that by 2 and
its 24.7474%. Total odds are 50% plus 24.7474% = 74.7474%

72

7500

The answer for this question is very simple: it is called Ceaser Cipher.
It consists of replacing a character by another located in (current character position + (key-1)), where key = how many positions you want to skip.
For Example: VICTORY -> YLFWRUB with the key = 3.
Then SUCCESS -> QXFFHQQ

Note: it should work like a circle(Z+1 = A).

Cousin

Each PAIR of stations means a PAIR of tickets (A to B and B to A)

2(old stations)(new stations) + 2(new stations)(new stations – 1) = 46

(N * X) + (X * (X – 1)) = 23

factoring ___ X (N + X – 1) = 23

23 is a prime number with only two factors, 1 and 23

so N = 23 and X = 1