Solution:
Let the distance between school and home be = D
According to the given information :
$$\frac{D}{{2\frac{1}{2}}} - \frac{D}{3} = 16{\text{ minutes}}$$
$$\eqalign{ \therefore \frac{{2D}}{5} - \frac{D}{3} = \frac{{16}}{{60}} \cr \Rightarrow \frac{{6D - 5D}}{{15}} = \frac{{16}}{{60}} \cr \Rightarrow D = \frac{{16}}{{60}} \times 15 \cr \Rightarrow D = 4{\text{ km}} \cr} $$

Solution:

8 unit ⟶ 16 hour
1 unit ⟶ 2 hour
Car = 2 hour = 2 × 40 = 80 km

Solution:
Total distance travelled :
= (600 + 800 + 500 + 100) km
= 2000 km
Total time taken :
$$\eqalign{ = \left( {\frac{{600}}{{80}} + \frac{{800}}{{40}} + \frac{{500}}{{400}} + \frac{{100}}{{50}}} \right){\text{ hrs}} \cr = \frac{{123}}{4}{\text{ hrs}} \cr} $$
∴ Average speed :
$$\eqalign{ = \left( {2000 \times \frac{4}{{123}}} \right){\text{ km/hr}} \cr = \left( {\frac{{8000}}{{123}}} \right){\text{ km/hr}} \cr = 65\frac{5}{{123}}{\text{ km/hr}} \cr} $$

Solution:
Circumference = One revolution Distance covered in 10 revolutions = 300 * 10 = 3000 cm = 30 m 30 meters are covered in 6 seconds. Speed of the wheel = $$\frac{{30}}{{\text{6}}}$$ = 5 m/s 5 m/s = $$\frac{{5 \times 18}}{5}$$  = 18 km/h. [ To convert m/s to km/h, we use to multiply by factor $$\frac{{18}}{5}$$ ]

Solution:
Given :
Speed of running train = 60 km/hr
Length of running train = 110 metres
⇒ Length of standing train = 170 metres
⇒ Speed of the standing train = 0 km/hr
⇒ Time taken by running train to cross the standing train :
$$\eqalign{ \Rightarrow {\text{Time}} = \frac{{\left( {100 + 170} \right){\text{ metres}}}}{{60{\text{ km/hr}}}} \cr \Rightarrow {\text{Time }} = \frac{{280 \times 18}}{{60 \times 5}} \cr \Rightarrow {\text{Time = 16}}{\text{.8 seconds}} \cr} $$

Solution:
Let the speed of train = x km/ hr
Length of train = 300 metres
Their relative speed in same direction
= (x - 3) km/hr
According to the question,
$$\frac{{\left( {300 + 0} \right)m}}{{\left( {x - 3} \right) \times \frac{5}{{18}}m/s}} = 33$$
[Here man's length is 0 metre]
$$\eqalign{ \frac{{100 \times 18}}{{5x - 15}} = 11 \cr 1800 = 55x - 165 \cr 55x = 1965 \cr} $$
∴ Speed of the train = $$\frac{{1965}}{{55}}$$  = $$35\frac{8}{{11}}$$  kmph

Solution:
Suppose B takes x hours to walk d km
Then, A takes (x + 2) hours to walk d km
A's speed = $$\left( {\frac{d}{{x + 2}}} \right)$$  km/hr and
B's speed = $$\left( {\frac{d}{{x}}} \right)$$  km/hr
A's new speed = $$\left( {\frac{2d}{{x + 2}}} \right)$$  km/hr
$$\eqalign{ \therefore \frac{d}{{\left( {\frac{d}{x}} \right)}} - \frac{d}{{\left( {\frac{{2d}}{{x + 2}}} \right)}} = 1 \cr \Leftrightarrow x - \left( {\frac{{x + 2}}{2}} \right) = 1 \cr \Leftrightarrow x - 2 = 2 \cr \Leftrightarrow x = 4 \text{ hours} \cr} $$