Axiscades Engineering Technologies Interview Questions, Process, and Tips

Let 10’splace digit is x and unit’s place digit y

First milestone : 10x+y
Second milestone : 10y+x
Third milestone: 100x+y

Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
After solving, we get —-> y=6x but since x and y are digits so only possible combination is x=1 and y=6,
So average speed = 45 KM/HR

C

24 Kmp/h

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​

(1/2 + 1/2) = 1
2(1 + 1/2) = 3
2(3 + 1/2) = 7
2(7 + 1/2) = 15
2(15 + 1/2) = 31

Kerala

TWICE

Prosperity for all

answer
R-R*d/100-5(1-d/100)

340