2500.61 – Cows, Sheep, Horses

Tags:Problem Set 10Algebra/TrigonometryAlgebra

Kurt, David, and Chris paid a total of 100 dollars to buy a pasture. They split up the cost according to the amount of grass their animals would eat. Kurt put in 9 horses; David put in 12 cows for twice as long as Kurt’s horses; and Chris put in some sheep for 2.5 times as long as David's cows. If Chris paid half the cost of the pasture, how many sheep did Chris have, and how much did Kurt and David each pay? It will be helpful to know that, in a given amount of time, 6 cows eat as much as 4 horses and 10 sheep eat as much as 3 cows.


First, note that "12 cows for twice the time" equals 24 cows for the same time, and "s sheep at 2½ cow time" equals 2.5s sheep at David’s time, or 5s sheep at Kurt’s time.

Also, note that since, in price 6 cows = 4 horses → 3 cows = 2 horses and 10 sheep = 3 cows, we can say that 10 sheep = 2 horses → 5 sheep = 1 horse in terms of eating capacity and, therefore, price.

So, let's make a chart and convert everything to horses. See below.

Now, we know that, if he were paying for horses, Chris would have 25 horses. Instead, he owned sheep that ate at 1/5 the speed of horses, so he had to be paying for 5x as many, putting his total number at 125 sheep.

However, that is 125 sheep bought for Kurt’s amount of time. Each of Chris’s sheep is in the pasture for five times as long as one of Kurt’s animals, so he would only be paying for 1/5 as many sheep, bringing his number back to 25 sheep.

Whew! Sorting out what to do with all of those numbers is quite a challenge.