1st cavalry division 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​

I will deny

Let water and milk be in x quantity
3x+2x=125
5x=125
x=25
Milk=3x=75litre
Water=2x=50litre
2x=50litre
4x=100litre

So 50 litre more should be added so that the ratio becomes 3:4

If there are n bikes then the no. of people are 2n+1. n are dropped in firs trip, in second trip 3 bikes are unused, so 2n-3 + n covers 2n+1 people. i.e. n + 2n – 3 = 2n +1
n = 4
No. of people = 9

first 10 odd num are 1,3,5,7,9,11,13,15,17,19
sum=100
avg=100/10
avg of 10 odd num =10

Train speed 90 km /h
1st we have to change in m/s. then 90*5/18=25m/s
Then d=speed*time
=25*10=250m train length

20

B. 3 : 2

1min = 60 seconds
6 min=6*60=360seconds
4 sec=10
360sec=360*10 /4=900

Assuming the youngest has X years, so we have
X+(X+3)+(X+6)+(X+9)+(X+12)=50
5X+30=50
5X=20
X=4
The youngest has 4 years

The probability that a child will be born male on any given day is always 1/2.

Suppose:
In any time period, the family making out babies and half boys and another half girls then, In the next time period: half the families turn themselves off and again the half remaining families making out the babies which repeat the first period.

required number =H.C.F of (73-25),(97-73) & (97-25)
=H.C.F of 48 , 24 and 72 = 4 (c)