Answer: Option A. -> 60
Let the speed = x km/hr
And, time = y hr
According to the question,
x × y = (x + 3) (y - 1)
xy = xy + 3y - x - 3
x - 3y = - 3 ..... (i)
Also,
x × y = (x - 2)(y + 1)
xy = xy - 2y + x - 2
x - 2y = 2 ..... (ii)
Solve equation (i) and (ii)
x = 12
y = 5
Distance = Speed × Time
Distance = 12 × 5
               = 60 km

Answer: Option A. -> 610 minutes
Distance covered in 1 min = 50 m
Distance covered in 2 min = 90 m
Similarly, 1st min, 2nd min, 3rd min . . . . . 15th
Distance → 50 m + 90 m + 130 m + . . . . . . . .
By using A.P.
a = 50 m
d = (90 - 50) = 40 m
Tn = a + (n - 1)d
     = 50 + (15 - 1) × 40
     = 50 + 560
     = 610 minutes

Answer: Option D. -> 24 km
Let the man travels 1 unit distance
So, remaining distance
$$\eqalign{
& = 1 - \left( {\frac{1}{6} + \frac{3}{4}} \right) \cr
& = 1 - \frac{{22}}{{24}} \cr
& = \frac{1}{{12}} \cr
& \because \frac{1}{{12}}{\text{ unit}} = {\text{2 km}} \cr
& {\text{So, 1 unit}} = 24{\text{ km}} \cr} $$

Answer: Option B. -> $$33\frac{1}{3}$$ km/hr
Let the total distance = 1200 km
$$\eqalign{
& {\text{Total time taken :}} \cr
& = \frac{{400}}{{25}} + \frac{{300}}{{30}} + \frac{{500}}{{50}} \cr
& = {\text{ }}16 + 10 + 10 \cr
& = {\text{ 36 hours}} \cr
& \therefore {\text{Average speed}} \cr
& = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr
& = \frac{{1200}}{{36}} \cr
& = 33\frac{1}{3}{\text{ km/hr}} \cr} $$

Answer: Option A. -> 20 seconds
Here length of pole is considered 0 metre
Time will be taken by train to cross the poll :
$$\eqalign{
& = \frac{{{\text{300 m}}}}{{54 \times \frac{5}{{18}}{\text{ m/s}}}} \cr
& = \frac{{300}}{{15}} \cr
& = 20 \cr} $$
Required time = 20 seconds

Answer: Option C. -> 25 km/hr
Let the train's speed = x km/hr
  $$\because \,$$ When train will cross a man then it covers only its length.$$\eqalign{
& \therefore \left( {x - 3} \right) \times \frac{{10}}{{60}} = \left( {x - 5} \right) \times \frac{{11}}{{60}} \cr
& x = 25{\text{ km/hr}} \cr} $$
Note :
Products of time × speed is always subtracted if both the men are running in the same direction and the products of time × speed is added only if the men are running opposite direction.
⇒ Here train's direction is not considered.
But attention please ⇒ Always divided by the difference of time.

Answer: Option C. -> 3 : 4
$$\because $$ Distance covered by the truck in a minute = 550 metres
Then, the speed of the truck will be :
$$\eqalign{
& \frac{{550 \to {\text{Metres}}}}{{60 \to {\text{Seconds}}}}\left\{ {{\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right\} \cr
& \left( {1{\text{ minute = 60 seconds}}} \right) \cr
& = \frac{{550}}{{60}} \Rightarrow \frac{{55}}{6}{\text{ m/s}}.....{\text{(i)}} \cr} $$
Whereas, distance covered by the bus in 45 minutes = 33 km
Then, the speed of the bus will be :
$$\frac{{33{\text{ km}}}}{{45\min }} \Rightarrow \frac{{33 \times 1000}}{{45 \times 60}}$$\[\left\{ \begin{gathered}
1{\text{ km = 1000 metres}} \hfill \\
{\text{1 min = 60 seconds}} \hfill \\
\end{gathered} \right\}\]
$$\eqalign{ \Rightarrow \frac{{110}}{9}{\text{ m/s}}.....{\text{(ii)}} \cr} $$
So, the ratio of their speeds will be,
$$\eqalign{
& = \frac{{55}}{6}:\frac{{110}}{9} \cr
& = \frac{1}{2}:\frac{2}{3} \cr
& = {\bf{3}}\,\,\,\,\,{\bf{:}}\,\,\,\,\,\,{\bf{4}} \cr
& (\text{Truck : Bus}) \cr }$$
$$\eqalign{
& {\bf{Alternate:}} \cr
& {\text{Speed of truck}} = 550\,{\text{metres/min}} \cr
& {\text{Speed of bus}} = \frac{{33}}{{45}}\,{\text{km/min}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{33000}}{{45}}\,{\text{metres/min}} \cr
& {\text{Speed of truck}}:{\text{Speed of bus}} \cr
& = 550:\frac{{33000}}{{45}} \cr
& = 55:\frac{{3300}}{{45}} \cr
& = 5:\frac{{300}}{{45}} \cr
& = 1:\frac{{60}}{{45}} \cr
& = 1:\frac{4}{3} \cr
& = 3:4 \cr} $$

Answer: Option D. -> 68 m/s
$$\eqalign{
& {\text{Distance = 1}}{\text{.7 km}} \cr
& {\text{Time = 25 sec}} \cr
& \therefore {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time }}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.7 \times 1000}}{{25}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 68{\text{ m/s}} \cr} $$

Answer: Option A. -> 125 metres
Let the length of train be $$l$$ metre
According to the question,
Time = $$\frac{{{\text{Distance}}}}{{{\text{Speed}}}}$$
⇒ 100 =$$\frac{{{\text{500 + }}l}}{{{\text{Speed}}}}$$
⇒ Speed = $$\frac{{500 + l}}{{100}}$$    ..... (i)
Again,
60 = $$\frac{{{\text{250 + }}l}}{{{\text{Speed of Train}}}}$$
Speed = $$\frac{{{\text{250 + }}l}}{{{\text{60}}}}$$    ..... (ii)
Equation (i) and (ii)
$$\eqalign{
& \Rightarrow \frac{{500 + l}}{{100}} = \frac{{250}}{{60}} \cr
& \Rightarrow 1500 + 3l = 1250 + 5l \cr
& \Rightarrow 2l = 250 \cr
& \therefore {\text{Length of train = 125 metres}} \cr} $$

Answer: Option B. -> 12 km/hr
$$\eqalign{
& S = \frac{D}{T} = \frac{{250}}{{75}}{\text{ m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\frac{{250}}{{75}} \times \frac{{18}}{5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12{\text{ km/hr}} \cr} $$