Answer: Option B. -> 7.5 km
Let the distance be x km
Then,
$$\eqalign{
& \Leftrightarrow \frac{x}{3} - \frac{x}{{3.75}} = \frac{1}{2} \cr
& \Leftrightarrow 2.5x - 2x = 3.75 \cr
& \Leftrightarrow x = \frac{{3.75}}{{0.50}} \cr
& \Leftrightarrow x = \frac{{15}}{2} \cr
& \Leftrightarrow x = 7.5{\text{ km}} \cr} $$

Answer: Option A. -> 18 km/hr
Let the whole distance travelled be x km and the average speed of the car for the whole journey be y km/hr
Then,
$$\eqalign{
& \Leftrightarrow \frac{{\left( {\frac{x}{3}} \right)}}{{10}} + \frac{{\left( {\frac{x}{3}} \right)}}{{20}} + \frac{{\left( {\frac{x}{3}} \right)}}{{60}} = \frac{x}{y} \cr
& \Leftrightarrow \frac{x}{{30}} + \frac{x}{{60}} + \frac{x}{{180}} = \frac{x}{y} \cr
& \Leftrightarrow \frac{1}{{18}}y = 1 \cr
& \Leftrightarrow y = 18{\text{ km/hr}} \cr} $$

Answer: Option B. -> 7.2 km/hr
Speed :
$$\eqalign{
& = \left( {\frac{{600}}{{5 \times 60}}} \right){\text{ m/sec}} \cr
& = {\text{ 2 m/sec}} \cr
& = \left( {2 \times \frac{{18}}{5}} \right){\text{ km/hr}} \cr
& = {\text{ 7}}{\text{.2 km/hr}} \cr} $$

Answer: Option B. -> 165 km
Let the distance travelled by train be x km
Then, distance travelled by bus = (285 - x) km
$$\eqalign{
& \therefore \left( {\frac{{285 - x}}{{40}}} \right) + \frac{x}{{55}} = 6 \cr
& \Leftrightarrow \frac{{\left( {285 - x} \right)}}{8} + \frac{x}{{11}} = 30 \cr
& \Leftrightarrow \frac{{11\left( {285 - x} \right) + 8x}}{{88}} = 30 \cr
& \Leftrightarrow 3135 - 11x + 8x = 2640 \cr
& \Leftrightarrow 3x = 495 \cr
& \Leftrightarrow x = 165 \cr} $$
Hence, distance travelled by train = 165 km

Answer: Option C. -> 400 m
Let the thief be caught x metres from the place where the policeman started running.
Let the speed of the policeman and the thief be 5y m/s and 4y m/s respectively.
Then, time taken by the policeman to cover x metres = time taken by the thief to cover (x - 100) m
$$\eqalign{
& \Rightarrow \frac{x}{{5y}} = \frac{{\left( {x - 100} \right)}}{{4y}} \cr
& \Rightarrow 4x = 5\left( {x - 100} \right) \cr
& \Rightarrow x = 500 \cr} $$
So, the thief ran (500 - 100) i.e., 400 m before being caught.

Answer: Option A. -> 5 km/hr
Let Abhay's speed be x km/hr
Then,
$$\eqalign{
& \Rightarrow \frac{{30}}{x} - \frac{{30}}{{2x}} = 3 \cr
& \Rightarrow 6x = 30 \cr
& \Rightarrow x = 5{\text{ km/hr}} \cr} $$

Answer: Option D. -> 35 km/hr
Time taken :
$$\eqalign{
& = 1\,{\text{hr}}\,40\,{\text{min}}\,48\,{\text{sec}} \cr
& = 1\,{\text{hr}}\,40\frac{4}{5}{\text{min}} \cr
& = 1\frac{{51}}{{75}}{\text{hrs}} \cr
& = \frac{{126}}{{75}}{\text{hrs}} \cr} $$
Let the actual speed be x km/hr
Then,
$$\eqalign{
& \frac{5}{7}x \times \frac{{126}}{{75}} = 42 \cr
& {\text{Or }}x = \left( {\frac{{42 \times 7 \times 75}}{{5 \times 126}}} \right) \cr
& x = 35{\text{ km/hr}} \cr} $$

Answer: Option B. -> 4 pm
Distance covered by thief in (2 pm - 1.30 pm)
= $$\frac{1}{2}$$ hr at speed of 40 km/hr
= 40 × $$\frac{1}{2}$$
= 20 kms
Their relative speed in same direction :
= (50 - 40) km/hr
= 10 km/hr
According to the question,
20 km, is the distance that has to be covered by owner to catch the thief.
$$\eqalign{
& {\text{Required time}} = \frac{{20{\text{ km}}}}{{10{\text{ km/hr}}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{\text{ hours}} \cr} $$
Therefore, he will over take the thief at :
= 2 pm + 2 hr
= 4 pm

Answer: Option A. -> 560 km
D = S × T
D = 80 × 7
D = 560 km

Answer: Option C. -> 14.4 kmph
Time taken by Kamal :
$$ = \frac{{100}}{{18 \times \frac{5}{{18}}}} = 20\sec $$
Time taken by Bimal :
$$\eqalign{
& = 20 + 5 \cr
& = 25\sec \cr} $$
Speed of Bimal :
$$\eqalign{
& = \frac{{100}}{{25}} \times \frac{{18}}{5} \cr
& = 14.4{\text{ kmph}} \cr} $$