Quantitative Aptitude - Arithmetic Ability Questions

Probability is Number of favourable outcomes by total outcomes

Total comes = 2 (Divisible or not)

Favourable = 1 (Divisible)

Hence, probability = 1/2.

As A1 speaks always after A2, they can speak only in  1st  to 9th places and 

A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place 

A2 can speak in 9 places the remaining 

 A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e

when A2 speaks in the first place, the number of ways they can speak is 9.8!.

When A2 speaks in second place, the number of ways they can speak is  8.8!.

When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!

Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

Time taken to cover 600 km = 600/100 = 6 hrs.

Number of stoppages = 600/75 - 1 = 7

Total time of stoppages = 4 x 7 = 28 min

Hence, total time taken = 6 hrs 28 min.

Let t be the usual time. If the usual rate is 1 unit, then the unusual rate is 6/7.
Balance by the distance,
1 x t = (6/7)(t + 30)
7t = 6(t + 30)
we get t as,
t = 180 minutes = 3 hrs.