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:

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

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 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 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 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]