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 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 B

Explanation:

Time taken to cover 600 km = 6 hours [because train speed is 100 km/hr]

Number of stops =
begin{aligned}
frac{600}{75} - 1 = 7 \
text{Total time on stops =}7*3 = 21 mins

end{aligned}
Total time taken by train = 6 hours 21 mins

Answer: Option C

Explanation:

\begin{aligned}
Speed = \frac{Distance}{Time} \\
Distance = 750 meter \\
Time = \text{2 min 30 sec} = 150 sec \\
Speed = \frac{750}{150} = 5 m/sec \\

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

Answer: Option D

Explanation:

Let the speeds of two trains be 7X and 8X km/hr.

\begin{aligned}
8X = \frac{400}{4} \\
=> X = 12.5 Km/hr

\end{aligned}
So speed of first train is 12.5*7 = 87.5 km/hr