A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ?

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 D

Explanation:

Number of gaps between 41 poles = 40
So total distance between 41 poles = 40*50
= 2000 meter = 2 km
In 1 minute train is moving 2 km/minute.
Speed in hour = 2*60 = 120 km/hour

Answer: Option A

Explanation:

Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr

Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4

So distance traveled on foot = 4(4) = 16 km

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:

Relative speed between two = 6-1 = 5 round per hour

They will cross when one round will complete with relative speed,
which is 1/5 hour = 12 mins.

So 7:30 + 12 mins = 7:42