Solution:
Distance covered by first person in time t = $$\frac{2}{8}$$ round = $$\frac{1}{4}$$ round
Distance covered by second person in time t = $$\frac{3}{8}$$ round
Speed of the first person = $$\frac{1}{4t}$$
Speed of the second person = $$\frac{3}{8t}$$
Since the two persons start from A and B respectively, so they shall meet each other when there is a difference of $$\frac{7}{8}$$ round between the two.
Relative speed of A and B :
$$\eqalign{ = \left( {\frac{3}{{8t}} - \frac{1}{{4t}}} \right) \cr = \frac{1}{{8t}} \cr} $$
Time taken to cover $$\frac{7}{8}$$ round at this speed :
$$\eqalign{ = \left( {\frac{7}{8} \times 8t} \right) \cr = 7t \cr} $$

Solution:
Total distance covered by the train in 30 seconds with the speed of 30 m/s is
= 30 × 30 m/s
= 900 metres
Total distance - train's length + platform's length
900 = train's length + 600
(when train crosses platform it covers length equal to length of train + length of platform)
Train's length = 900 - 600
Train's length = 300 metres

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

Solution:
Speed of train
= $$\frac{{{\text{length of the train}} + {\text{length of platform}}}}{{{\text{Time}}}}$$ = $$\frac{{250 + 250}}{{25}}$$
= 20 m/sec
= 72 km/h

Solution:
A__________300 km_______B Let speed of the 1^st ship is s kmph and speed of 2^nd ship is (s + 10) kmph. Let 2^nd ship takes time t to cover the distance 300 km then 1^st ship takes (t + 8) hours. As distance is constant, we have, $$\frac{{\text{s}}}{{{\text{s}} + 10}} = \frac{{\text{t}}}{{{\text{t}} + 8}}$$ Or, st + 8s = st + 10t Or, 8s = 10t If the slower ship took 20 hours (option D) the faster ship would take 12 hours and their respective speeds would be 15 and 25 kmph.

Solution:
\[\begin{array}{*{20}{c}} {}{{\mathbf{Car}}}{}{{\mathbf{Train}}} \\ {}{100}{}{220} \\ {{\text{Speed}} \to }5:{11} \\ {{\text{Time}} \to }{11}:5 \end{array}\]
$$\eqalign{ {\text{Actual speed of car}} = \frac{{950}}{{19}} = 50{\text{ km/m}} \cr {\text{5 unit}} \to {\text{50 km/h}} \cr {\text{1 unit}} \to 1{\text{0 km/h}} \cr {\text{11 unit}} \to 11{\text{0 km/h}} \cr {\text{Distance cover by train in }}3\frac{1}{2}{\text{ hour}} \cr = 110 \times \frac{7}{2} \cr = \frac{{770}}{2} \cr = 385{\text{ km}} \cr} $$

Solution:
Speed of Karan = 60 kmph
Time = 9 hrs
Distance = Speed × Time
Distance = 60 × 9
Distance = 540 km
∴ Time taken to cover 540 km at 90 km/ph
= $$\frac{540}{90}$$ hours
= 6 hours