Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

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:

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 C

Explanation:

Relative speed between two = 6-1 = 5 round per hour

They will cross when one round will complete with relative speed,
which is 1/5 hour = 12 mins.

So 7:30 + 12 mins = 7:42

Answer: Option D

Explanation:

We need to get the answer in meters. So we will first of change distance from km/hour to meter/sec by multiplying it with 5/18 and also change 15 minutes to seconds by multiplying it with 60.

\begin{aligned}
Speed = 5* \frac{5}{18} = \frac{25}{18}m/sec \\
Time = 15*60 seconds = 900 seconds \\

Distance = Time * Speed \\
Distance = \frac{25}{18}*900 = 1250 meter
\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}