A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start at same point at 7:30 am. They shall first cross each other at ?

Answer: Option D

Explanation:

We need to get the answer in meters. So we will first of change distance from km/hour to meter/sec by multiplying it with 5/18 and also change 15 minutes to seconds by multiplying it with 60.

\begin{aligned}
Speed = 5* \frac{5}{18} = \frac{25}{18}m/sec \\
Time = 15*60 seconds = 900 seconds \\

Distance = Time * Speed \\
Distance = \frac{25}{18}*900 = 1250 meter
\end{aligned}

Answer: Option C

Explanation:

In this type of questions we need to get the relative speed between them,
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph

Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / 0.5 kmph = 17 hrs

Answer: Option A

Explanation:

We know that,
\begin{aligned}
Time = \frac{Distance}{Speed} \\
Speed = 20\text{ km/hr} = 20*\frac{5}{18}{ m/sec} \\
= \frac{50}{9}{ m/sec} \\

\text{ Time =} \left(400*\frac{9}{50}\right) \\
= 72 {sec} = 1\frac{1}{5}{ min}
\end{aligned}

Answer: Option B

Explanation:

\begin{aligned}
Speed = \left(80*\frac{5}{18}\right) m/sec \\
= \frac{200}{9} m/sec \\
= 22\frac{2}{9} m\sec
\end{aligned}

Answer: Option C

Explanation:

Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr

Average speed =
\begin{aligned}
\frac{2xy}{x+y} \\
= \frac{2*40*60}{40+60} = \frac{4800}{100} \\
= 48 Km/hr
\end{aligned}