Answer: Option A. -> 5
In these type of questions go through options to save your valuable time :
Option (A)
Abhay's speed = 5 km/hr
Abhay's time = $$\frac{30}{5}$$ = 6 hr
Sameer's time = 6 - 2 = 4 hr
Abhay's new time :
= $$\frac{30}{5 × 2}$$
= 3 hr
Hence option (A) is correct as it satisfies all the conditions.

Answer: Option B. -> $$7\frac{1}{2}$$ sec
$$\eqalign{
& {{\text{V}}_{{\text{rel}}{\text{.}}}} = 77 + 67 = 144{\text{ km/hr}} \cr
& {\text{ = 144}} \times \frac{5}{{18}}{\text{ m/sec}} \cr
& = 40{\text{ m/sec}} \cr
& \therefore {\text{ T}} = \frac{{\text{D}}}{{{{\text{V}}_{{\text{rel}}{\text{.}}}}}} \cr
& \Rightarrow {\text{T}} = \frac{{140 + 160}}{{40}} \cr
& \Rightarrow {\text{T}} = \frac{{300}}{{40}} \cr
& \Rightarrow {\text{T}} = 7.5\sec {\text{or 7}}\frac{1}{2}\,\sec \cr} $$

Answer: Option D. -> 500 metres
The length of pole is considered as negligible i.e., = 0
i.e., when a train crosses the pole, it covers the distance equal to the length of train
So, the time will be taken by the train = 30 sec
And speed = 60 km/hr
Then the length of the train :
= 60 kmh × 30 sec
= 60 × $$\frac{5}{18}$$ × 30 metres
= 10 × $$\frac{5}{3}$$ × 30
= 500 metres

Answer: Option B. -> 48 km
According to the question,
5 units → 12 hr
1 unit → $$\frac{12}{5}$$ hr and,
4 units → $$\frac{12}{5}$$ × 4
             = $$\frac{48}{5}$$
Distance travelled on foot :
= $$\frac{48}{5}$$ × 5
= 48 km

Answer: Option C. -> 20 km
$$\eqalign{
& {{\text{S}}_{{\text{average}}}} = \frac{{2ab}}{{a + b}} \cr
& = \frac{{2 \times 25 \times 4}}{{25 + 4}} \cr
& = \frac{{200}}{{29}}{\text{ km/hr}} \cr
& {\text{Now, 2D = }}\frac{{200}}{{29}} \times \left( {5 + \frac{4}{5}} \right) \cr
& = \frac{{200}}{{29}} \times \frac{{29}}{5} \cr
& = 40{\text{ km}} \cr
& {\text{ = D = 20 km}} \cr} $$

Answer: Option C. -> 18 km/hr
Total distance :
= 42 × 10
= 420 km
New given time = 7 hr and
Required new speed :
= $$\frac{420}{7}$$ km/hr
= 60 km/hr
∴ Required increase in speed :
= (60 - 42) km/hr
= 18 km/hr

Answer: Option D. -> 250 m
Distance travelled by the 1st car in 15 seconds
$$\eqalign{
& {\text{OA}} = \left( {36 \times \frac{5}{{18}}} \right) \times 15\,{\text{m}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = 150\,{\text{m}} \cr} $$
Distance travelled by the 2nd car in 15 seconds
$$\eqalign{
& {\text{OB}} = \left( {48 \times \frac{5}{{18}}} \right) \times 15\,{\text{m}} \cr
& \,\,\,\,\,\,\,\,\,\, = 200\,{\text{m}} \cr} $$
∴ Distance between them after 15 seconds
$$\eqalign{
& {\text{AB}} = \sqrt {{\text{O}}{{\text{A}}^2} + {\text{O}}{{\text{B}}^2}} \cr
& \,\,\,\,\,\,\,\,\,\, = \sqrt {{{150}^2} + {{200}^2}} \cr
& \,\,\,\,\,\,\,\,\,\, = \sqrt {22500 + 40000} \cr
& \,\,\,\,\,\,\,\,\,\, = \sqrt {62500} \cr
& \,\,\,\,\,\,\,\,\,\, = 250\,{\text{m}} \cr} $$

Answer: Option D. -> 90 km/hr
In the first situation,
⇒ Total distance covered by train :
= 80 × $$4\frac{1}{2}$$
= 360 kms
⇒ Therefore,
The speed of the train to cover the same distance 360 km in 4 hours is :
$$\eqalign{
& = \frac{{360}}{4}\left\{ {{\text{Speed }} = \frac{{{\text{Distance }}}}{{{\text{Time}}}}} \right\} \cr
& = 90{\text{ km/h}} \cr} $$

Answer: Option A. -> 19 min
$$\eqalign{
& {{\text{V}}_{{\text{rel}}{\text{.}}}} = \left( {21 - 15} \right){\text{ m/min}} \cr
& \,\,\,\,\,\,\,\,\,\,\, = {\text{ 6 m/min}} \cr} $$
Time taken to catch the thief :
= $$\frac{114}{6}$$ min
= 19 min

Answer: Option B. -> 10.8 km/hr
$$\eqalign{
& {\text{Average speed }} = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr
& {\text{Average speed }} = \frac{{{\text{10 + 12 }}}}{{\frac{{10}}{{12}} + \frac{{12}}{{10}}}} \cr
& {\text{Average speed }} = 10.8{\text{ km/hr}} \cr} $$