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 C

Explanation:

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

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

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 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 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:

We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance.
Lets solve it.

Time = 50/60 hr = 5/6 hr
Speed = 48 mph
Distance = S*T = 48 * 5/6 = 40 km

New time will be 40 minutes so,
Time = 40/60 hr = 2/3 hr
Now we know,
Speed = Distance/Time
New speed = 40*3/2 kmph = 60kmph