A hunter fired two shots from the branch of a tree at an interval of 76 seconds. A tiger separating too fast hears the two shots at an interval of 83 seconds. If the velocity of the sound is 1195.2 km/h, then find the speed of tiger?

In 20 minutes the difference between man and his son = 20 x 20 = 400 m
Distance travelled by dog when he goes towards son = (400/40) x 60 = 600 m and time required is 10 minutes
In 10 minutes the remaining difference between man and son = 400 - (20 x 10) = 200 m

Note: Relative speed of dog with child is 40 km/h and the same with man is 100 km/h.
Time taken by dog to meet the man = 200/100 = 2 min
In 2 minute the remaining distance between child and man = 200 - ( 2 x 20) = 160 m
Now, the time taken by dog to meet the child again = 160/40 = 4 min
In 4 minutes he covers 4 x 60 = 240 m distance while going towards the son.
In 4 minute the remaining distance between man and child = 160 - (4 x 20) = 80 m
Time required by dog to meet man once again = 80/100 = 0.8 min
In 0.8 min remaining distance between man and child = 80 - (0.8 x 20) = 64 min
Now, time taken by dog to meet the child again = (64/40) x (8/5) min
? Distance travelled by dog = (8/5) x 60 = 96 m
Thus, we can observe that every next time dog just go 2/5th of previous distance to meet the child in the direction of child.
So, we can calculate the total distance covered by dog in the direction of child with the help of GP formula.
Here, first term (a) = 600 and common ratio (r) = 2/5
Sum of the infinite GP = a/(1 - r)
= 600/(1 - 2/5) = (600 x 5)/3 = 1000 m

Let Amarnath express takes H hours, then Gorakhnath express takes (H - 2) hours.
? 1/H + 1/(H - 2) = 60/80
H = 4 hrs.

Distance travelled by them in first hour = 12 km
Distance travelled by them in second hour = 13 km
Distance travelled by them in third hour = 14 km and so on

Thus in 9 hours they will cover exactly 144 km and in 9 h each will cover half the total distance.
( 8 x 9 = 72 and 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 72)

Speed of tiger = 40 m/min
Speed of dear = 20 m/min
Relative speed = 40 - 20 = 20 m/min
Difference in distance = 50 x 8 = 400 m
? Time taken in overtaking (or catching) = 400/20 = 20min
? Distance travelled in 20 min = 20 x 40 = 800 m

The sum of their speeds = 615/15 = 43 km/h
Notice that they are actually exchanging their speeds. Only then they can arrive at the same time at their respective destinations.It means the difference in speeds is 3 km/h.
Thus, x + ( x + 3) = 43
x = 20 and x + 3 = 23
The concept is very similar to the case when after meeting each other they returned to their own places of departure. It can be solved through option also.

Speed of wind (sound) / Relative speed of soldier and terrorist = Time utilised / Difference in time
1188/x = 330/5
x = 18 km/h and solder is faster.

Let the time taken in first third part of the journey be x minutes, then the time required in second third part of the journey is 3x/2 and in the last third part of the journey time required is 15x/8.
Therefore, x + (3x/2) + (15x/8) = 350min
? x = 80 min

x/20 + y/70 = (x + y)/50
? (70x + 20y)/1400 = (x + y)/50
? 42x = 8y
? x/y = 4/21

Here, S₁ = 8 km/h
and S₂ = 16 km/h
? d₁ =s₁ x t₁ = 8t₁ ...(i)
and d₂ = s₂ x t₂ = 16t₂ ...(ii)
We know that,
t₁ + t₂ = 8 ...(iii)
and
d₁ + d₂ = 80 (given) ...(iv)
From Eqs.(i) and (ii) put the value of d₁ and d₂ in Eq. (iv), we get
d₁ + d₂ = 80
8t₁ + 16t₂ = 80
? 8t₁ + 8t₂ + 8t₂ = 80
8(t₁ + t₂) + 8t₂ = 80
From Eq (iii),
8 x 8 + 8t₂ = 80

8t₂ = 80 - 64 = 16
? t₂ = 16/8 = 2h
? t₁ = 8 - 2 = 6 h
? Distance travelled by foot = d₁ = 8 x 6 = 48 km.

Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20)
= (4 x 120)/12 = 40 km/h