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 A

Explanation:

Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr

Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4

So distance traveled on foot = 4(4) = 16 km

Answer: Option B

Explanation:

We know Time = Distance/speed

So total time taken =
\begin{aligned}
\left(\frac{160}{64}+\frac{160}{80}\right) \\
= \frac{9}{2} hours \\
\text{ Time taken for 320 Km = } \\
320*\frac{2}{9} = 71.11 km/hr
\end{aligned}

Answer: Option B

Explanation:

We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance.
Lets solve it.

Time = 50/60 hr = 5/6 hr
Speed = 48 mph
Distance = S*T = 48 * 5/6 = 40 km

New time will be 40 minutes so,
Time = 40/60 hr = 2/3 hr
Now we know,
Speed = Distance/Time
New speed = 40*3/2 kmph = 60kmph

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 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}