Vyatta Interview Questions, Process, and Tips

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​

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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

This is a sequence of English letters, in numbers.

⇒ ten = 3 letters

⇒ nine = 4 letters

⇒ sixty = 5 letters

⇒ ninety = 6 letters

⇒ seventy = 7 letters

⇒ Sixty six =8 letters

⭕️This logical question goes with the no. of letters in the number!!!!!!!

⭕️∴ the biggest number with 9 letters will be ninety-six (96).

A rude person who doesn’t respect me

6+3.50= 2.50

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