Answer: (b)

Speed of the man = 5km/hr   

\(\displaystyle speed=\frac{{dis\tan ce}}{{time}}=\frac{{250}}{{75}}\)

\(\displaystyle \frac{{10}}{3}m/\sec =\frac{{10}}{3}\times \frac{{18}}{5}km/hr\)

\(\displaystyle [1m/\sec =\frac{{18}}{5}km/hr]\)

= \(\displaystyle 2\times 6=12km/hr\)

Answer: (c)

Let the distance be x km.

Total time = 5 hours 48 minutes

\(\displaystyle 5+\frac{{48}}{{60}}=(5+\frac{4}{5})hours\)

\(\displaystyle \frac{{29}}{5}hours\)

\(\displaystyle \frac{x}{{25}}+\frac{x}{4}=\frac{{29}}{5}\)

\(\displaystyle \frac{{4x+25x}}{{100}}=\frac{{29}}{5}\)

\(\displaystyle 5\times 29x=29\times 100\)

\(\displaystyle x=\frac{{29\times 100}}{{5\times 29}}=20km\)

Answer: (a)

Let the required distance be x km.

Then, \(\displaystyle \frac{x}{3}+\frac{x}{2}=5\)

\(\displaystyle \frac{{2x+3x}}{6}=5\)

 5x = \(\displaystyle 6\times 5\)

\(\displaystyle x=\frac{{6\times 5}}{5}=6km\)

Answer: (b)

Let the distance between A and B be x km, then

\(\displaystyle \frac{x}{9}-\frac{x}{{10}}=\frac{{36}}{{60}}=\frac{3}{5}\)

\(\displaystyle \frac{x}{{90}}=\frac{3}{5}\)

\(\displaystyle x=\frac{3}{5}\times 90=54km\)

Answer: (d)

Difference of time = 4.30 p.m – 11.a.m.

\(\displaystyle 5\frac{1}{2}hours=\frac{{11}}{2}hours\)

Distance covered in \(\displaystyle \frac{{11}}{2}hrs\)

\(\displaystyle \frac{5}{6}-\frac{3}{8}=\frac{{20-9}}{{24}}=\frac{{11}}{{24}}part\)

\(\displaystyle \frac{{11}}{{24}}\) part of the journey is covered in \(\displaystyle \frac{{11}}{2}hours\)

\(\displaystyle \frac{3}{8}\) part of the journey is covered in = \(\displaystyle \frac{{11}}{2}\times \frac{{24}}{{11}}\times \frac{3}{8}=\frac{9}{2}hours\)

\(\displaystyle 4\frac{1}{2}hours\)

Clearly the person started at 6.30 a.m.