Cow Problem

[vishalsabnis]

In a farm, there is some stock of grass is there, in which 40 cows can comeplete the grass in 40 days. in case if

there is 30 cows, then the grass will be completed in 60 days. No the question is, if there is 20 cows are there, in

how many days the grass will be completed ?

[smouse]

80 days

[BPH]

According to that logic, 30 cows should complete the grass in 53 days and 8 hours, which is apparently not the case.

[johnalex]

Hi Vish  -  Your first statement suggests that 1 cow would take 1600 days on her own.  So 40 take 40, and 20 take 80, so surely 30 take 53days 8hrs?  So how do you arrive at 60 days??  Trick question or am I just moo-dy?

[kempie]

Do you have to take into account that grass grows, so over time there is more grass to be eaten by the fewer cows?

This would lead to an exponential correlation rather than the expected linear one, and it could be that 20 cows is the magic figure that could be sustained indefinitely by that area of grass.

[gen2]

No, 10 cows will be sustained indefinitely.
The grass will sustain 20 cows for 120 days.
This is my coffee break so I will post the formula tonight.

[gen2]

Here is how I solved the problem.  First we need to split the problem down into two parts.  The growth of grass and how many cows it can feed, and the excess grass in the field at the start and how many cows that can feed.

40 cows feeding for 40 days = 1600 cowdaysfood (40*40)
30 cows feeding for 60 days = 1800 cowdaysfood (30*60)

In those extra 20 days, the cows have eaten an extra 200 cowdaysfood which must have grown in those 20 days.  Therefore:

20 days growth of grass produces 200 cowdaysfood
40 days growth of grass produces 400 cowdaysfood
1 days growth of grass produces 10 cowdaysfood

Therefore the growth of grass can sustain 10 cows indefinitely, or at least until the grass stops growing.  We now need to work out how much excess grass was in the field and how long it will last.

If 1600 cowdaysfood were consumed in 40 days
and 400 cowdaysfood grew in 40 days
then 1200 cowdays food must have been the excess (1600-400)

The qustion asked about a field of 20 cows.  10 of them could feed for ever on the new growth.  The other 10 must feed on the excess.  That will last them 120 days (1200/10).  So the answer is 120 days.

Or to put it more mathematically:

x = 1200 / (y - 10)

where x = the number of days  and y = the number of cows

[kempie]

Thank you gen2 for that comprehensive algorithm.