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 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 D

Explanation:

\begin{aligned}
Speed = \frac{Distance}{Time} \\
= \frac{200}{24}m/sec = \frac{25}{3} m/sec \\
\frac{25}{3}*\frac{18}{5} km/hr = 30 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 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 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.