A soldier fired two bullets at an interval of 335 seconds moving at a uniform speed v ₁. A terrorist who was running ahead of the soldier in the same direction, hears the two shots at an interval of 330 seconds? If the speed of sound is 1188 km/h, then who is the faster and by how much?

In case of increasing gap between two objects.
Speed of sound / speed of tiger = Time utilised / Difference in time
? 1195.2 / x = 83 / 7
x = 100.8 km/h

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

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

Let two trains meet at a km from 'X'
[Time taken by M to cover (450 - a)km] - [Time taken by the L to cover a km ] = 40/60
? (450 - a)/80 - a/60 = 40/60
? (450 - a)/80 = 40/60 + a/60
? (450 - a)/8 - (a + 40) / 6 = 0
? 3(450 - a) - 4(a + 40) = 0
? 7a = 1190
? a = 1190 / 7 = 170
Time taken by L to cover 170 km = 170/60 h = 2 h 50 min
So, the two trains will meet 2 h 50 min after 6:00 pm. it means that the two trains will meet at 8 : 50 pm .