Question Detail

A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start at same point at 7:30 am. They shall first cross each other at ?

7:15 am
7:30 am
7: 42 am
7:50 am

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

1. A cyclist covers a distance of 750 meter in 2 minutes 30 seconds. What is the speed in km/hr of cyclist

16 km/hr
17 km/hr
18 km/hr
19 km/hr

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}

2. 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.?

9 km/hour
10 km/hour
11 km/hour
12 km/hour

Answer: Option D

Explanation:

We need to calculate the distance, then we can calculate the time and finally our answer.
Lets solve this,

Let the distance travelled by x km
Time = Distance/Speed

\begin{aligned}
\frac{x}{10} - \frac{x}{15} = 2 \\
\text{[because, 2 pm - 12 noon = 2 hours]} \\

3x - 2x = 60 \\
x = 60.

Time = \frac{Distance}{Speed} \\
Time @ 10km/hr = \frac{60}{10} = 6 hours

\end{aligned}
So 2 P.M. - 6 = 8 A.M
Robert starts at 8 A.M.

He have to reach at 1 P.M. i.e, in 5 hours

So, Speed = 60/5 = 12 km/hr

3. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr

\begin{aligned} 1\frac{1}{5} min\end{aligned}
\begin{aligned} 2\frac{1}{5} min\end{aligned}
\begin{aligned} 3\frac{1}{5} min\end{aligned}
\begin{aligned} 4\frac{1}{5} min\end{aligned}

Answer: Option A

Explanation:

We know that,
\begin{aligned}
Time = \frac{Distance}{Speed} \\
Speed = 20\text{ km/hr} = 20*\frac{5}{18}{ m/sec} \\
= \frac{50}{9}{ m/sec} \\

\text{ Time =} \left(400*\frac{9}{50}\right) \\
= 72 {sec} = 1\frac{1}{5}{ min}
\end{aligned}

4. 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?

10 am
11 am
12 pm
1 pm

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.

5. 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.

70.11 km/hr
71.11 km/hr
72.11 km/hr
73.11 km/hr

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}