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by [email protected] » Tue Jun 29, 2010 7:29 pm
Four horses are tethered at the four corners of a square of side 14cm such that two horses along the same side can
just reach each other. They were able to graze the area in 11 days. How many days will they take in order to graze the left out area?
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by ilikaroy » Tue Jun 29, 2010 7:55 pm
I think you need to mention that the length of rope is same for all the horses. Otherwise the answer will be different.

Am I correct?
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by amising6 » Tue Jun 29, 2010 8:07 pm
[email protected] wrote:Four horses are tethered at the four corners of a square of side 14cm such that two horses along the same side can
just reach each other. They were able to graze the area in 11 days. How many days will they take in order to graze the left out area?
so assuming they will cover 7 cm of side by both horse so basically it will be covering area with radius of 7 cm i.e are=(1/4)(22/7)*7*7 =77/2 cm^2 each by each horse so 2 horse on which are teethered on one side and can meet will cover 77/2*2=77cm^2 area so talltogether four horse will cover 77*2=154 cm^2 in 11 days
so in 1 day=154/11=14
area remaining to be grazed 14*14 -154=32
days 32/24=4/3 days
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by [email protected] » Tue Jun 29, 2010 9:13 pm
Ans. 3

dont know how they get it.
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by Rahul@gurome » Tue Jun 29, 2010 9:49 pm
Each horse grazes pi*1/4*(7^2) area.
So 4 horses graze pi*(7^2) = 154 area.
What we have is that 154 area is being grazed in 11 days.
Remaining area to be grazed is 14*14 - 154 = 196 - 154 = 42.
So this will take (11/154)* 42 = 3 days.
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