When the athlete walks at ⁵/₄ of his usual speed then he takes ⁴/₅ of his usual time (As Speed ∝ (¹/_Time)). and he saves 5 minutes.

→ Usual time - (⁴/₅)*usual time = 5 minutes;

→( ¹/₅ )* Usual time = 5 minutes;

Usual time = 25 minutes.

Lucknow(X →).....120 km .....( ← Y)Kanpur

Relative speed = (25+35) = 60 km/h

Time taken to complete 120 km,

= ¹²⁰/₆₀ = 2 h

Thus, They will meet 2 hours later.

Let the speed of the motor boat = 36 kmph and speed of current = 5x kmph. The boat goes along with the current in 5 hours 10 minutes = ³¹/₆ hour;

Hence,

\begin{align} & {\text{Distance}} \cr & = \left( {\frac{{31}}{6}} \right) \times \left( {36x + 5x} \right) \cr & = \frac{{41x + 31}}{6}\, km \cr & {\text{Speed of boat upstream}} \cr & = 36x + 5x = 31x\, {\text{kmph}} \cr & {\text{Hence, time taken to come back}} \cr & = \left[ {\frac{{\left( {41x \times \left( {\frac{{31}}{6}} \right)} \right)}}{{31x}}} \right] \cr & = \frac{{41}}{6}\, {\text{hours}} \cr & = 6\, {\text{hours}}\, 50\, {\text{minutes}} \cr\end{align}

Let the original speed be S₁ and time t₁ and distance be D.

Now, \begin{align} & \frac{{\left( {\frac{D}{2}} \right)}}{{2{t...1}}} = {S...2} \cr & {S...2} = \frac{D}{{4{t...1}}}\, {\text{and}}\, {S...1} = \frac{D}{{{t...1}}} \cr & {\text{Thus}}, \cr & \frac{{{S...1}}}{{{S...2}}} = \frac{4}{1} = 4:1 \cr\end{align}

\begin{align} & {\text{Let}}\, {\text{the}}\, {\text{duration}}\, {\text{of}}\, {\text{the}}\, {\text{flight}}\, {\text{be}}\, x\, {\text{hours}} \cr & {\text{Then}}, \frac{{600}}{x} - \frac{{600}}{{x + \left( {1/2} \right)}} = 200 \cr & \Rightarrow \frac{{600}}{x} - \frac{{1200}}{{2x + 1}} = 200 \cr & \Rightarrow x\left( {2x + 1} \right) = 3 \cr & \Rightarrow 2{x^2} + x - 3 = 0 \cr & \Rightarrow (2x + 3)(x - 1) = 0 \cr & \Rightarrow x = 1\, hr \cr\end{align}