A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes ?

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 B

Explanation:

Two things to give attention on this question.
First time is in minutes, we need to change it to seconds to get speed in m/sec, then we need to get the final answer in km/hr.
So lets solve this.

\begin{aligned}
Speed = \frac{Distance}{Time} \\
Distance = 650 meter \\
Time = \text{5 minutes} = 300 sec \\
Speed = \frac{600}{300} = 2 m/sec \\

=> 2*\frac{18}{5} km/hr = 7.2 km/hr
\end{aligned}

Answer: Option B

Explanation:

Suppose they meet x hrs after 8 a.m.
then
(Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]
= 330
So, 60x+75(x-1) = 330

x=3.

So,they meet at (8+3). i.e 11a.m.

Answer: Option C

Explanation:

Let time taken to travel the first half = x hr
Then time taken to travel the second half = (10 - x) hr

Distance covered in the the first half = 21x [because, distance = time*speed]
Distance covered in the the second half = 24(10 - x)

Distance covered in the the first half = Distance covered in the the second half
So,
21x = 24(10 - x)
=> 45x = 240
=> x = 16/3

Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]

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}