Bosch security systems Interview Questions, Process, and Tips

12%

15/46

The correct Answer is 69

c=a/b
c=a-1
=>b=a/c
=>b=a/a-1

1×2×…100=100!
Number of zeros in product of n numbers =[5n​]+[52n​]+[53n​]+…
Number of zeros in product of 100 numbers =[5100​]+[52100​]+[53100​]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24

Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1​ cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2​ i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61​,P(B)=61​,P(C)=65​,P(A)=61​,P(B)=65​,P(C)=61​ and P(A)=65​,P(B)=61​,P(C)=61​
Hence the required probability = 61​×61​×65​×6×3=21690​=125​

Reading your question

NTFS — New Technology File System
Fat — file allocation table
NTFS having a quota, commpress system with securites base as
administrator giving, multipal user, groups, to set permission

Monday =y cheques
Tuesday =3y cheques
wednesday=4000 cheques
totaling to 16000 cheques
therefore y+3y+4000=16000 cheques
4y+4000=16000 cheques therefore 4y=16000-4000 cheques= 4y=12000cheques
therefore y=12000/4=3000 cheques
monday=3000 cheques
tuesday=3×3000=9000cheques

s=36kmph
in meter per second is=36*5/18=>10

Friday

88