In a circus there was a leopard and a tiger walking in the two different rings of same radii. There I observed that when leopard moved 3 steps, tiger moved 5 steps in the same time,but the distance traversed by leopard in 5 step is equal to the distance traversed by tiger in 4 steps. what is the number of rounds that a leopard made when tiger completed 100 rounds?

Speed of the car A = (5 / 6) x 90
= 75 km/hr.

? Required time = [ 88 / (90 +75)] x 60
= 32 minutes

Let the two meet at the N_th line
From the question
N_th/200 = (817 - N_th)/150
? 3N_th = 4(817-N_th)
? N_th = (4 x 817) / 7
? N_th = 466.85
So, they will meet at the 467th line

A has already gone 3.75 km when B starts of the remaining 48 km.
A walks 3.75 km and B walks 4.25 km in one hour in opposite direction i.e., they together pass over (3.75 + 4.25) = 8 km. in hour.

Therefore , 48 km. are passed over in 48/8= 6 hours.

? A meets B in 6 hours after B started and, therefore, they meet at a distance of (4.25 X 6 ) = 25.5km. from Q.

Let the distance between Delhi and Kanpur be y km .
Suppose the train leaving from Delhi is A and the train leaving from Kanpur B

A's speed = y/ (10 a.m - 5 a.m) = y/5 km/hr.
B's Speed = y/ (2 p.m - 7 a.m) = y/7 km/hr.

Since B starts two hours later than A, the distance already covered by A at the start of B = 2y/5 km.

Remaining distance = y- 2y/5 = 3y/5 km.

Relative speed of approach of two trains = (y/5 + y/7)
= 12y/35 km/hr.

Time taken to cover the remaining distance by both trains = (3y / 5) / (12y / 35)
= (3/5) x (35/12) = 7/4 hrs.
= 1 hr. 45 min.

? The two train will meet at (7 a.m + 1hr.45 min)
= 8.45 a.m

Let the distance between Meerut and Delhi be y km.
Average speed of the train leaving Meerut = y/4 km/hr.
Average speed of the train leaving Delhi = 2y/7 km/hr.

suppose they meet x hrs. after 6 a.m
then, xy/4 + 2y (x-2)/7 = y
? x/4 + 2x-4/7 = 1
? 15x = 44

? x = 44/15 = 2 hrs. 56 min
So, the train meet at 8:56 a.m

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

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

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 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

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)