Tricky Equation Setup

[bearej50]

Gilbert Greenfield owns a pasture on which the grass has been cut to uniform height of two inches. The grass in the pasture grows uniformly at a constant rate. In addition to owning the pasture, Mr. Greenfield also owns a cow, a horse, and a sheep. The grass existing and growing in the pasture is sufficient to feed all three animals grazing together for 20 days. It would feed the cow and the horse alone for 25 days; the cow and the sheep alone for 33? days; and the horse and the sheep alone for 50 days. How long would it sustain (a) the cow alone? (b) The horse alone? (c) The sheep alone?

How would the variables be defined?

 

[Denis]

Here's a similar one with solution, which may INSPIRE you:
PROBLEM:
Freddie the farmer has a paddock that he uses to graze his stock of cows. Each cow eats the same amount of grass each day, regardless of how many other cows are in the paddock and irrespective of the amount of grass left in the paddock. In an experiment, Freddie puts 6 cows into the paddock and he finds out it takes 3 days for them to eat all the grass. These 6 cows are then taken out of the paddock to allow the grass to grow back.

After the grass has been allowed to grow back to the original amount, Freddie then puts 3 cows into the paddock. This time he finds that it takes 7 days for the 3 cows to eat all the grass in the paddock. Freddie is puzzled that the cows took this long and consults a mathematician.

Freddie said "Geez mate, I dunno why me cows took that long to eat me paddock."
Marvin the mathematician replies "Well Freddie, tell me what assumptions you made."
Freddie replies "Well mate, maybe me cows got sick or somethin', cos I reckon that me 3 cows should have taken 6 days to eat me paddock, not 7 days! "
Marvin Replies, "I doubt that very much Freddie!"

After a while Marvin does some calculations and reveals that Freddie had overlooked an important assumption. What was the assumption Freddie had overlooked? Using Marvin's assumption, how long would a single cow take to eat the same paddock?

SOLUTION:
Farmer Freddie forgot to account for the grass continuing to grow.

The data presented is sufficient to calculate both the number of cow-day meals in the field before the cows are introduced and the rate at which the grass grows.

Let M = number of cow-day meals at time zero.
Let g = amount of grass that grows in one day.

M + 3g = 18 meals : 6 cows 3 days
M + 7g = 21 meals : 3 cows 7 days

M = 63/4 The field starts with 15 3/4 cow-day meals.
g = 3/4 The grass grows at a rate of 3/4 of a cow-day meal per day.

[For one cow in X days, M + Xg = X meals]

If only one cow was let into the field it could eat for 63 days!

 

[soroban]

Hello, nearej50!

Here's my very long approach.
It's long because I include my baby-talk thought process.

Gilbert Greenfield owns a pasture on which the grass has been cut to uniform height of two inches.
The grass in the pasture grows uniformly at a constant rate.
Mr. Greenfield also owns a cow, a horse, and a sheep.

The grass existing and growing in the pasture is sufficient to feed all three animals grazing together for 20 days.
It would feed the cow and the horse alone for 25 days,
the cow and the sheep alone for 33 1/3 days.
and the horse and sheep allone for 50 days.

How long would it sustain (a) the cow alone? . (b) The horse alone? . (c) The sheep alone?

Let \(\displaystyle G\) = rate at which the grass grows (in inches per day).
Let \(\displaystyle C\) = rate at which the cow eats grass (in inches per day).
Let \(\displaystyle H\) = rate which the horse eats grass (in inches per day).
Let \(\displaystyle S\) = rate at which the sheep eats grass (in inches per day).

The pasture starts with 2 inches of grass.


The pasture will support all three animals for 20 days.
In 20 days, there will be \(\displaystyle 2+20G\) inches of grass.
Meanwhile, the Cow eats \(\displaystyle 20C\) inches of grass, the Horse eats \(\displaystyle 20H\) inches, and the Sheep eats \(\displaystyle 20S\) inches.
At the end of 20 days, there will be no grass left.
. . \(\displaystyle 2 +20G-20C-20H-20S \:=\:0 \quad\Rightarrow\quad C+H+S-G\;=\;\frac{1}{10}\) .[1]

The pasture will support the cow and horse for 25 days.
In 25 days, there will be \(\displaystyle 2+25G\) inches of grass.
And the Cow eats \(\displaystyle 25C\) inches, and the Horse eats \(\displaystyle 25H\) inches.
At the end of 25 days, there will be no grass left.
. . \(\displaystyle 2+25G-25C-25H \:=\:0 \quad\Rightarrow \quad C+H-G\:=\:\frac{2}{25}\) .[2]

The pasture will support the cow and sheep for \(\displaystyle \tfrac{100}{3}\) days.
In \(\displaystyle \tfrac{100}{3}\) days, there will be \(\displaystyle 2+\tfrac{100}{3}G\) inches of grass.
And the Cow eats \(\displaystyle \tfrac{100}{3}C\) inches, and the Sheep eats \(\displaystyle \tfrac{100}{3}S\) inches.
At the end of \(\displaystyle \tfrac{100}{3}\) days, there will be no grass left.
. . \(\displaystyle 2+\tfrac{100}{3}G-\tfrac{100}{3}C-\tfrac{100}{3}S \:=\:0 \quad\Rightarrow\quad C+S-G\:=\:\frac{3}{50}\) .[3]

The pasture will support the horse and sheep for 50 days.
In 50 days, there will be \(\displaystyle 2+50G\) inches of grass.
And the horse eats \(\displaystyle 50H\) inches, and the sheep eats \(\displaystyle 50S\) inches.
At the end of 50 days, there will be no grass left.
. . \(\displaystyle 2+50G-50H-50S \:=\:0 \quad\Rightarrow\quad H+S-G\:=\:\frac{1}{25}\) .[4]

We have a system of four equations.

\(\displaystyle \text{The solution is: }\;C \:=\:\frac{3}{50}\;\;\;H \:=\:\frac{1}{25}\;\;\;S\:=\:\frac{1}{50}\;\;\;G \:=\:\frac{1}{50}\)

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(a) How long would it sustain the cow alone?

The cow is alone in the pasture for \(\displaystyle t\) days.

The grass grows at \(\displaystyle \tfrac{1}{50}\) inches per day.
In \(\displaystyle t\) days, there will be \(\displaystyle 2+\tfrac{1}{50}t\) inches of grass.

The cow eats \(\displaystyle \tfrac{3}{50}\) inches per day.
In \(\displaystyle t\) days, the cow eats \(\displaystyle \tfrac{3}{50}t\) inches.

In \(\displaystyle t\) days, there will be no grass left.
. . \(\displaystyle 2+\frac{1}{50}t-\frac{3}{50}t \:=\:0 \quad\Rightarrow\quad -\frac{2}{50}t \:=\:-2 \quad\Rightarrow\quad t \:=\:50\)

Therefore, the pasture will sustain the cow alone for 50 days.