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Question Detail
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}
1. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour ?
- 8 minutes
- 10 mintues
- 12 minutes
- 14 minutes
Answer: Option B
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = (9/54) hour
= (1/6)*60 minutes
= 10 minutes
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 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
4. A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ?
- 44 km/hour
- 46 km/hour
- 48 km/hour
- 50 km/hour
Answer: Option C
Explanation:
Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr
Average speed =
\begin{aligned}
\frac{2xy}{x+y} \\
= \frac{2*40*60}{40+60} = \frac{4800}{100} \\
= 48 Km/hr
\end{aligned}
5. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. Find the speed at which the train must run to reduce the time of journey to 40 minutes.
- 50 km/hr
- 60 km/hr
- 65 km/hr
- 70 km/hr
Answer: Option B
Explanation:
We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance.
Lets solve it.
Time = 50/60 hr = 5/6 hr
Speed = 48 mph
Distance = S*T = 48 * 5/6 = 40 km
New time will be 40 minutes so,
Time = 40/60 hr = 2/3 hr
Now we know,
Speed = Distance/Time
New speed = 40*3/2 kmph = 60kmph
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