A bullock cart has to cover a distance of 80 km. in 10 hours. If it cover half of journey in 3/5 ^th time, what should be its speed to cover the remaining distance in time left?

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

Distance of his office from his house = {ab/(b - a)} x (t₁ + t₂)
Here, a = 20 km/h, t₁ = 1 h, b = 20 + 5 = 25 km/h
and t₂ = 30 min = 0.5 h
? Distance = {20 x 25/(25 - 20)} x (1 + 0.5) = 500(1.5)/5
= 100 x 1.5 = 150 km

Ratio of times taken by A and B = 1/2 : 1/3

Suppose B takes T_b min. Then A takes (T_b + 10) min.
? (T_b + 10 ) : T_b = 1/2: 1/3
? (T_b + 10)/T_b = 3 / 2
? 2T_b + 20 = 3T_b
? T_b = 20

? Time taken by A = 20 +10
= 30 minute

If A had walked at double speed
Req . time = 30/2
=15 minute

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