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A leopard spots a deer from a distance of 200 meters. As the leopard s

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A leopard spots a deer from a distance of 200 meters. As the leopard s  [#permalink]

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New post 30 Jul 2019, 23:08
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Question Stats:

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A leopard spots a deer from a distance of 200 meters. As the leopard starts chasing the deer, the deer also starts running. Given that the speed of the deer is 10 km/h and that of the leopard is 12 km/h, how far would have the deer run before it is caught?

A. 1 km
B. 2 km
C. 3 km
D. 4 km
E. 5 km

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Re: A leopard spots a deer from a distance of 200 meters. As the leopard s  [#permalink]

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New post 30 Jul 2019, 23:23
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The time taken for the leopard to catch the deer = \(\frac{Distance-between-Leopard-and-Deer}{Relative-speed-of-leopard-and-deer}\)

\(Distance\) between Leopard and Deer \(= 200 meters = 0.2 km\)
Since the leopard and deer run in the same direction, relative speed \(= 12-10 = 2km/hr\)

\(Time\) taken for the leopard to catch the deer \(= \frac{0.2}{2} = 0.1 hours\)

In 0.1 hours, the deer runs \(0.1*10 = 1km\)

Answer is (A)
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Re: A leopard spots a deer from a distance of 200 meters. As the leopard s  [#permalink]

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New post 07 Aug 2019, 03:36
This is a simple and straightforward question on Relative Speed.

Relative Speed is the speed of two objects which are moving simultaneously. It is calculated as the sum of the speeds when both objects are moving in opposite directions, and, as the difference of the speeds when both objects are moving in the same directions.

In this question, the leopard and the deer are moving in the same direction. So, the relative speed will be the difference of their speeds.
Relative Speed = 12 – 10 = 2 km/h.

This means that the leopard will gain 2 km over the deer, in every one hour. Or, in other words, the deer will lose 2 km to the leopard, in every one hour.

When the leopard perceived the deer, the distance it had to gain was 200 metres i.e. 0.2 km. This means, it will take the leopard \((\frac{0.2}{2})\) hours \((\frac{distance}{relative speed})\) to catch the deer.
Since both the leopard and the deer started running at the same time and ended running at the same time (leopard catching the deer), we may say that both ran for a period of \((\frac{1}{10})\) hours.

In \((\frac{1}{10})\) hours, the deer can run 10 * \((\frac{1}{10})\) = 1 km.

The correct answer option is A.

Hope this helps!
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Re: A leopard spots a deer from a distance of 200 meters. As the leopard s  [#permalink]

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New post 10 Aug 2019, 09:29
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Required = 10km/hr *t (time to catch the deer)

Now , Relative velocity of leopard = 12-10 = 2 km/hr
and rel distance = 200/1000 km
RT = W
2*T = 200/1000
T = 1/10 hr

So, Required = 10km/hr*1/10
= 1km.
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Re: A leopard spots a deer from a distance of 200 meters. As the leopard s   [#permalink] 10 Aug 2019, 09:29
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