The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m and travels towards A at 75 Km/hr. At what time do they meet?

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 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 A

Explanation:

We know total distance is 200 Km
If both trains crossed each other at a distance of 110 km then one train covered 110 km and other 90 km [110+90=200km]
So ratio of their speed = 110:90 = 11:9

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:

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