Two cars X and Y starts from two places A and B respectively which are 700 km apart at 9 am. Both the cars run at an average speed of 60 km/h, Car X stops at 10 am and again starts at 11 am, while the other car Y continues to run without stopping. When do the two cars cross each other?

Let w be the time taken in one way by walking and c be the time taken in one way by car.
Now, according to the question,
In first case, w + c = 6 h 45 min
or
2w + 2c = 13 h 30 min ..(i)
In second case, 2c = 4 h 45 min ....(ii)
From Eqs. (i) and (ii), we get
2w + 2c = 13 h 30 min
? 2w + 4h 45 min = 13 h 30 min
? 2w = 13 h 30 min - 4 h 45 min
? 2w = 8 h 45 min
? If he walks both ways, then time taken = 8 h 45 min

Walking time + Cycling time = 5 h 45 min = 345 min ....(i)
If he had cycled both way he would have gained 2 h (120 min).
? 2 x Cycling time = 345 - 120 = 225 min ...(ii)
Walking time = 2 x 345 - 225 = 690 - 225 = 465 min
Time taken by him to walk both ways = 7 h 45 min

Let T h, he travels on foot.
? 4T + 10(8 - T) = 50
? 80 - 6T = 50
? 6T = 30,
? T = 5 h
? The distance travelled on foot = 4 x 5 = 20 km

Relative speed = (58 - 32) = 26 km/h
= (26 x 5/18) m/s = 65/9 m/s
Time taken to cross each other = (250 + 140)/(65/9)
= (390 x 9)/65 = 54 s

Let the distance be x km.
? Time = Distance/speed
1st day time taken = x/4 h
2nd day time taken = x/6 h
Difference in time = (x/4 - x/6) h
Actual difference between these two times = 15 min = 1/4 h
? x/4 - x/6 = 1/4 ? (3x - 2x)/12 = 1/4
? x = 3 km

We know that, if two equal distances are covered at two different speeds A and B, then
Average speed = 2AB/(A + B) = (2 x 58 x 52)/(58 + 52)
= 54.8 km/h
= 55 km/h (Approx)

Distance between Sonagarh and Raipur = Average speed x time
= 69 x (35/60) km = 161/4 km
New speed = (69 + 36) km/h = 105 km/h
? Required time = Distance/speed
= 161/(4 x 105) h
= (161 x 60)/(4 x 105) min
= 23 min

Let the required distance = L
According to the question,
L/(5/2) - L/(7/2) = 12/60 (difference between two times = 6 + 6 =12 min)
? 2L/5 - 2L/7 = 1/5
? 14L - 10L = 7
? 4L = 7
? L = 7/4 km

Relative speed of policeman with respect to thief = ( 11 - 10 ) = 1 km/h.
Now, relative distance covered by policeman in 6 min = Speed x Time
= (1 x 6)/60 = 1/10 km = 100 m
The distance between the policeman and thief after 6 min = (200 - 100) = 100 m

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