A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. Find the average speed for first 320 km of tour.

Answer: Option A

Explanation:

Let the distance be 3x km,
then,
\begin{aligned}
\frac{x}{3} +\frac{x}{4} +\frac{x}{5} = \frac{47}{60} \\
\frac{47x}{60} =\frac{47}{60} \\
x = 1
\end{aligned}
So total distance = 3*1 = 3 Km

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 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 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 duration of the flight be x hours.
Then
\begin{aligned}
\frac{600}{x}-\frac{600}{x + \frac{1}{2}} = 200 \\
\frac{600}{x}-\frac{1200}{2x+1} = 200 \\
x(2x+1) = 3 \\
2x^2 + x - 3 = 0 \\
=> (2x+3)(x-1)=0

\end{aligned}
Neglecting the negative value for x we get x = 1