A car runs at the speed of 50 km/h when not serviced and runs at 60 km/h, when serviced. After servicing, the car covers a certain distance in 6 h. How much time will the car take to cover the same distance when serviced?

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

We know that speed is inversely proportional to time.
Therefore, Time taken by A : Time taken by B = 1/2 : 1/3

Let B takes T min. Then, A takes (T + 20) min.
? (T + 20) : T = 1/2 : 1/3 = 3 : 2
? (T + 20)/T = 3/2
? 2T + 40 = 3T
? T = 40
? A takes (T + 20) = (40 + 20) = 60 min.
? At double the speed, A would have covered it in 30 min.

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 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

Let the speed of first lady be (v + 2)km/h and speed of second lady will be v km/h.
Total distance = 24 km

For first lady, Speed = Distance/Time
v + 2 = 24/t ...(i)

Let time taken by first lady to cover distance of 24 km be t h.
Then, time taken by second lady to cover same distance will be(t + 1)h.
For second lady, Speed =Distance/time
v = 24/(t + 1) ...(ii)

Form Eqs.(i) and (ii), we get
24/t - 2 = 24/(t + 1)
? (24 - 2t)/t = 24/(t + 1)
? (24 - 2t)(t + 1) = 24t
? 24t - 2t² + 24 - 2t = 24t
? 2t² + 2t - 24 = 0
? t² + t - 12 = 0
? t² + 4t - 3t - 12 = 0
? t(t + 4) -3(t + 4) = 0
? (t - 3)(t + 4) = 0
? t = 3, -4
? t = 3 (t ?- 4)
? The distance travelled by the first lady in one hour
= 24/t = 24/3 = 8 km
And distance travelled by the second lady in one hour
= 24/t + 1
= 24/3 + 1 = 6 km

Speed in km/ hour =200

Speed in m/sec = 200 x 5/18 = 500/9 m/sec = 55.55 m/sec