A truck covers a distance of 368 km at a certain speed in 8 h. How much time would a car take at an average speed which is 18 km/h more than that of the speed of the truck to cover a distance which is 16 km more than that travelled by the truck?

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

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

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

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)

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.

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

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 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

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)

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.