The ratio between the rates of travelling of A and B is 2 : 3 and therefore A takes 20 min more than time taken by B to reach a destination. If A had walked at double the speed, how long would he have taken to cover the distance?

When serviced, the distance covered by the car = 65 x 5 = 325 km.
When not serviced, the time taken by the car to cover 325 km at 40 km/h
= 325/40 = 8.125 = 8 h (approx)

Let distance he drove be L.
Using Time = Distance/speed
From given condition. L/40 - L/45 = 1
? (45L - 40L)/(40 x 45) = 1
? 5L = 40 x 45
? L = 360 km

Let speed of train = 16k
Speed of car = 15k
Speed of bus = 480/8 = 60 km/h
According to the question,
16k = 60 x 4/3 = 80
? k = 5
? Speed of car = 15k = 15 x 5 = 75 km/h
? Required distance = Speed x Time
= 75 x 6 = 450 km

Let the constant speed = V
According to the question,
715/V - 715/(V + 10) = 2
? 1/V - 1/(V + 10) = 2/715
? [(V + 10) - V] / [V(V + 10)] = 2/715
? V² + 10V = 10 x 715/2
? V² + 10V - 3575 = 0
? V² + 65V - 55V - 3575 = 0
? V(V + 65) - 55(V + 65) = 0
? (V - 55) (V + 65) = 0
? V = 55 or V = - 65
Ignoring the negative value, we get V = 55
? Original speed of the motorcycle = 55 km/h

Speed of truck = Distance/time = 368/8 = 46 km/h
Now, speed of car = (speed of truck + 18) km/h = (46 + 18) = 64 km/h and
Distance travelled by car = 368 + 16 = 384 km
? Time taken by car = Distance/Speed = 384/64 h = 6 h

Time taken to cover first 75 km = 75/25 = 3 h
Time taken to cover next 25 km = 25/5 = 5 h
Time taken to cover last 50 km of its journey = 50/25 = 2 h
Total distance = 75 + 25 + 50 = 150 km
Total time taken = 3 + 5 + 2 = 10 h
? Required average speed = Total distance / Total time taken
= 150/10 = 15 km/h

After sevicing speed of car = 60 km/h
? Distance covered in 6 h = 60 x 6 km = 360 km.
When not serviced, then time taken to cover 360 km = 360/50 = 7.2 h

Let they meet x km from A.
Then, according to the question,
x/40 = (1000 - x)/60
? 60 x = 40(1000 - x)
? 3x = 2000 - 2x
? 3x + 2x = 2000
? 5x = 2000 km
? x = 2000/5 = 400 km

Let total distance = D
Now, according to the question,
40 x 3 + 60 x 4.5 = 3D/5
? 120 + 270 = 3D/5
? D = 390 x 5/3 = 650 km

? Remaining distance = [650 - (3/5) x 650]
= 650 - 390 = 260 km

Required average speed = 260/4 = 65 km/h

Let the time taken by car B to reach destination is T h.
So, the time taken by car A to reach destination is (T + 2) h.
Now, S₁T₁ = S₂T₂
? 30(T + 2) = 45 x T
? 30T + 60 = 45T
? 15T = 60
? T = 4 h

Now, distance between starting point and destination = S₂T₂
= 45 x 4 = 180 km