A student walks from his house at 2 ¹/ ₂ km/h and reaches his school late by 6 min. Next day, he increase his speed by 1 km/h and reaches 6 min before school time. How far is the school form his house?

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 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

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

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