Suresh traveled 1200 km by air which formed (2/5) of his trip. One-third of the whole trip he traveled by car And the rest of the journey he performed by train. The distance traveled by train was?

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

Let total distance be 100 km.
Then, Average = 100/[ (50/40) + (30/60) + (20/30)]
= 100/( 5/4 + 1/2 + 2/3) = 100/[(15 + 6 + 8)/12]
= (100 x 12)/29
= 42.35

Let B catches A after T h.
Then, distance travelled by A in (T + 2) h = Distance travelled by B in T h

According to the question,
2 (T + 2 ) = 5T
? 2T + 4 = 5T
? 3T = 4
? T = 4/3 h

Distance travelled by B in 4/3 h
= (4/3) x 5 = 20/3 = 6²/₃ km

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

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.

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.