Question Detail

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}

1. A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes ?

50 meter
100 meter
110 meter
None of above

Answer: Option B

Explanation:

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr

Distance covered in 6 minutes =
\begin{aligned}
\frac{1}{60}*6 = \frac{1}{10} \\
= 100 meters
\end{aligned}

So distance between them after 6 minutes = 200 - 100
= 100 meters

2. 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}

3. A man is walking at the rate of 5 km/hr crosses a bridge in 15 minutes. The length of the bridge is

1000 meters
1050 meters
1200 meters
1250 meters

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}

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 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