Aptitude and Reasoning Questions

The only possible way is BCAFDE
As D & F is not in any corner and E is after C and D. C is after B and before A
So the first person to join the institute is B

Let the number of days be 'p' 

As the work is same, we know that

M1 x D1 x H1 = M2 x D2 x H2

Where M = Men, D = Days, H = Hours per day 

Here M1 = 9, D1 = 15, H1 = 7 

M2 = 6, D2 = p, H2 = 9 

=> 9 x 15 x 7 = 6 x p x 9 

=> p = 35/2 = 17.5 days.

First line will cut all other 14, similarly second will cut 13, and so on
Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.

Let us assume A is the faster and B is slower, A takes 10s to finish a lap of 50 m, B takes 17s(approx) to finish a lap of 50 m.
So at 11th sec A starts his 2nd lap by the time B will be in between so that is the first time they will meet each other, at 21st sec A starts his 3rd lap by the time B will be coming back to finish his 2nd lap. So that is the second time they meet in opposite. Then at 31st sec A will start his 4th lap by the time B would not even finish 2nd lap as he take 34(2*17) sec to finish 2 laps. So at 35th sec B will start his 3rd lap by the time A will be coming back to finish his 4th lap in the half way so they will meet 3rd time. At 41st sec A will start its fifth lap by the time B will be going to finish 3rd lap in same direction. At 50th second both will complete A's 5th lap and B's 3rd lap. So they will start next lap(A's 6th and B's 4th) from the same point in same direction. And A will finish the race after completing 300m(6*50) total distance and then they wont meet opposite to each other.

So totally they met 3 times in opposite direction.

I=(p x r x t)/100

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.