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 D

Explanation:

We need to calculate the distance, then we can calculate the time and finally our answer.
Lets solve this,

Let the distance travelled by x km
Time = Distance/Speed

\begin{aligned}
\frac{x}{10} - \frac{x}{15} = 2 \\
\text{[because, 2 pm - 12 noon = 2 hours]} \\

3x - 2x = 60 \\
x = 60.

Time = \frac{Distance}{Speed} \\
Time @ 10km/hr = \frac{60}{10} = 6 hours

\end{aligned}
So 2 P.M. - 6 = 8 A.M
Robert starts at 8 A.M.

He have to reach at 1 P.M. i.e, in 5 hours

So, Speed = 60/5 = 12 km/hr

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 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}

Answer: Option B

Explanation:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = (9/54) hour
= (1/6)*60 minutes
= 10 minutes