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:

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:

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 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 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