Akbar and Birbal setout at the same time to walk towards each other respectively from Agra and Banaras 144 km apart. Akbar walks at the constant speed of 8 km/h, while Birbal walks 4 km in the first hour, 5 km in the second hour, 6 km in the third hour and so on. Then the Akbar and Birbal will meet?

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.

Let pele covers D km in 1 hours. So Maradona takes (2 h - 40 min) = 1 h 20 min to cover D km.

Let speed of Maradona and pele be M and P respectively than.
D = M x 4/3 and D = P x 1
M/P = 3/4
Again, 300/M - 300/P = 1
300/3k - 300/4k = 1
k = 25
M = 3k = 75 km/h and
P = 4k = 100 km/h

Initial speed of police = 10 m/s
Increased speed of police = 20 m/s
Speed of thief = 15 m/s
Initial difference between thief and police = 250 m
After 5 seconds difference between thief and police = 250 - (5 x 10) = 200 m
After 10 seconds more, the difference between thief and police = 200 + (5 x 10) = 250 m.
Now, the time required by police to catch the thief = 250/5 = 50 s
Distance travelled = 50 x 20 = 1000 m
Total time = 50 + 15 = 65 sec
Total distance = 1000 + (15 x 10) = 1150 m

The ratio of speeds = The ratio of distances, when time is constant
? The ratio of distance covered by leopard to the tiger = 12 : 25
Again, ratio of rounds made by leopard to the tiger = 12 : 25
hence, leopard makes 48 rounds, when tiger makes 100 rounds

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

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

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

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