The ratio between the rates of travelling of A and B is 2:3 and therefore A takes 10 min. more than the time taken by B to reach a destination, if A had walked at double the speed, he would have covered the distance in?

Distance left = (1/2 x 80 ) km. = 40km.
Time left - [(1-3/5) x 10 ] hrs.
= 4 hours

Required speed = (40 / 4 ) km/hr
= 10 km/hr.

Relative velocity = u + v = 18 + 20 = 38 km
They are 38 km. apart in 1 hr.
? They will be 95 km. apart in ( 95 / 38) hrs.
= 2 hrs . 30 min

Let the total distance be y km.
Then, 2y / 5 = 1200
? y = (1200 x 5) /2 = 3000 km.

Distance traveled by car
= (1/3 X 3000) = 1000 km.

Distance traveled by train
= [3000- (1200 + 1000) ] km.
= 800 km.

Let distance covered by dog in 1 leap is x and
Distance covered by cat in 1 leap is y

Then, 3x = 4y
? x= 4/3y

Now, Ratio of speed of dog and cat = Ratio of distance covered by them in the same time = 4x : 5y
= 16/3y : 5y
= 16 : 15

Let the distance = x
Here, the difference in time
= 20 - 5 = 15 min
= 15/60 = 1/4h
Speed during next journey
= 15 + 5 = 20 km/h
According to the question,
x/15 - x/20 = 1/4
? 4x - 3x/60 = 1/4
? x = 60/4 = 15
? x = 15 km

Suppose the man covers first distance in x hrs and second distance in y hrs.
Then, 4x + 5y = 35 and 5x + 4y = 37
Solving these equations, we get
x = 5 and y = 3

? Total time taken = 5 + 3 hrs.
= 8 hrs.

Let the required time = T min.
Then, distance covered in T + 11 min at 40 km/hr = distance covered in T + 5 min at 50 km/hr.
? 40 ( T + 11 ) / 60 = 50 (T + 5 ) / 60
? T = 19 min.

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

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

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.