Amit start from a point A and walks to another point B and then return from B to A by his car and thus takes a total time of 6 h 45 min. If he had driven both ways in his car, he would have taken 2 h less. How long would it take for him to walk both ways?

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

Time taken by express train to cover 75 km including stoppage
= (60 / 100 x 75) min + = 48 min
Time taken by express train to cover 600 km.
= Time taken by it to cover 75 km
= (48 / 75 x 525 ) min + (60 / 100 x 75 ) min
= (336 + 45) min = 381
Time taken by local train to9 cover 25 km including stoppage
= (60 / 50 x 25) min + 1 min = 31 min
In 31 min, distance covered = 25 km
In(31 x 12) min. distance covered
= (25 / 31 x 31 x 12 ) = 300 km
In last 9 min, distance covered
= (25 / 31 x 9 ) = 7.25 km
? Required total distance
= 300 + 7.25 = 307.25 km

Since, the speed of X and Y are 60 km/h each. The distance between A and B is 700 km. Distance traveled by X upto 11 am is 60 km. Since, X stops at 10 am and distance traveled by up to 11 am is 120 km.
Now, the distance between them = 700 - 180 = 520 km
Now, let they will meet at a distance of x km from X's position.
? Distance traveled by X = x km
and distance traveled by Y = (520 - x) km
? x/60 = (520 - x)/60
? 2x = 520
? x = 260
? T = 260/60 = 4¹/₃ h
Thus, they will cross each other after, 4¹/₃ h after 11 am i.e., at 3 : 20 pm.

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