Speed = \(\left(\frac{600}{5\times60}\right)m/sec.\)

= 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

= \(\left(2\times\frac{18}{5}\right)km/hr\)

   = 7.2 km/hr.           

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Speed = \(\frac{1200}{(5/3)}km/hr. \left[We can write1^{\frac{2}{3}}hours as 5/3 hours\right]\)

Therefore Required speed =  \(\left(1200\times\frac{3}{5}\right)km/hr = 720km/hr.\)

Let the actual distance travelled be x km.

Then,\(\frac{x}{10} = \frac{x+20}{14}\)

14x = 10x + 200

4x = 200

 x = 50 km.

Let speed of the car be x kmph.

Then, speed of the train = \(\frac{150}{100}x = \left(\frac{3}{2}x\right)kmph.\)

\(\therefore \frac{75}{x}-\frac{75}{(\frac{3}{2})x}=\frac{125}{10\times60}\)

\(\frac{75}{x}-\frac{50}{x}=\frac{5}{24}\)

\(x=\left(\frac{25\times24}{5}\right)= 120kmph.\)

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km =\(\left(\frac{9}{54}\times60\right)min = 10min.\)

Let the duration of the flight be x hours.

Then,\(\frac{600}{x}-\frac{600}{x+(\frac{1}{2})}=200\)

\(\frac{600}{x}-\frac{1200}{2x+1}=200\)

 x(2x + 1) = 3

 2x² + x - 3 = 0

 (2x + 3)(x - 1) = 0

 x = 1 hr.      [neglecting the -ve value of x]

=\(\frac{\left(\frac{1}{2}\right)x}{21}+\frac{\left(\frac{1}{2}\right)x}{24}=10\)

=\(\frac{x}{21}+\frac{x}{24}=20\)

15x = 168 x 20

\(x=\left(\frac{168\times20}{15}\right)=224km.\)

Let the speed of two trains be 7x and 8x km/hr.

Then, 8x = \(\left(\frac{400}{4}\right)=100\)

\(x=\left(\frac{100}{8}\right)=12.5\)

Therefore  Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

Total time taken = \(\left(\frac{160}{64}+\frac{160}{80}\right)hrs.=\frac{9}{2}hrs.\)

Therefore Average speed =\(\left(320\times\frac{2}{9}\right)km/hr = 71.11km/hr.\)

Time taken = 1 hr 40 min 48 sec = 1 hr\(40\frac{4}{5}min=1\frac{51}{75}hrs=\frac{126}{75}hrs.\)

Let the actual speed be x km/hr.

Then,\(\frac{5}{7}x\times\frac{126}{75}=42\)

\(x=\left(\frac{42\times7\times75}{5\times126}\right)=35km/hr\)