Two cars A and B start simultaneously from a certain place at the speed of 30 km/h and 45 km/h, respectively.The car B reaches the destination 2 h earlier than A. What is the distance between the starting point and destination?

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 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

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

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.

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

Let the length of train be x meter and speed be y m/sec

Therefore, time taken to cross the pole = x/y sec

and time taken to cross the platform = ( Length of train + Length of platform) / speed of train

Therefore x/y = 10
And, (x+200) / y = 20
? 10y + 200 = 20 y
? speed of train, y = 20 m/sec
And length of rain = x = 10y = 200m

Relative speed, when the trains are running in the opposite direction = Sum of the speed

Therefore, Relative speed = 36 km/hr +72 km/hr = 90 km/hr
Relative speed in m/sec = 90 km/hr x 5/18= 25 m/sec

Distance covered = sum of the lengths of the train = 200 m+ 300 m = 500m
Therefore, time taken = 500m / 25 m/sec = 20 sec

Speed of the train = (60 x 5/18) m/sec = 50/3 m/sec
Distance covered in passing the standing person = 200m
Required time taken= =200/(50/3) = 200 x (3/50) = 12 sec