[Guest]

Mr. Greenfield owns a fixed pasture covered with grass which grows uniformly at a constant rate. Mr. Greenfield also owns a cow, a horse, and a sheep.

The existing and growing grass in the pasture is sufficient to feed all three animals grazing together for the next 20 days. Alternatively it would feed the cow and the horse together for 25 days; the cow and the sheep together for 33 1/3 days; and the horse and sheep together for 50 days.

How long it would sustain:

The cow alone?

The horse alone?

The sheep alone?

You should assume that each animal eats at some constant rate (but necessarily all at the same rate as each other).

[Guest]

Sheep 100 days

Horse 50 days

Cow 50 days

[Guest]

Sheep 100 days

Horse 50 days

Cow 50 days

If the horse and cow eat at the same rate, then the horse+cow and horse+sheep rates would be the same, which they aren't in the puzzle. Horse+sheep lasts 50 days, but cow+sheep lasts 33 1/3 days.

[Guest]

I got that the cow can eat for 33 1/3 days, the horse can eat for 50 days, and the sheep can eat for 100 days. I set it up as four equations: 20c + 20h + 20s = g, 25c + 25h = g, 33 1/3c + 33 1/3s = g, 50s + 50h = g. I then used the 3 latter equations substituted into the first equation to solve for g in terms of c, h, and s.

Edited September 13, 2010 by Vile

[Molly Mae]

I got that the cow can eat for 33 1/3 days, the horse can eat for 50 days, and the sheep can eat for 100 days. I set it up as four equations: 20c + 20h + 20s = g, 25c + 25h = g, 33 1/3c + 33 1/3s = g, 50s + 50h = g. I then used the 3 latter equations substituted into the first equation to solve for g in terms of c, h, and s.

The field supports the cow and sheep together for 33 1/3 days. Unless the sheep doesn't eat when she's around the cow, I think you're mistaken.

[Guest]

(Most of) You guys aren't considering that grass is continuing to grow as the animals eat.

You have to figure out how much each animal eats per day.

Given that

(C+H+S-G) * 20 days = F (One Field eaten)

and

(C+H-G) * 25 days = F

(C+S-G) * 50 days = F

(H+S-G) * 33 1/3 days = F

The Cow © eats 3/100s of a field per day, the Horse (H) 2/100s and the Sheep (S) 1/100 of a field per day and the Grass (G) grows at 1/100 per day.

Which means...

(.03+.02+.01-.01) * 20 days = F (One Field eaten)

and

(.03+.02-.01) * 25 days = 1 (Field eaten)

(.03+.01-.01) * 50 days = 1

(.02+.01-.01) * 33 1/3 days = 1

Therefore

The Cow can eat a net .02 fields in one day for 50 days.

The Horse can eat a net .01 fields in one day for 100 days.

The Sheep can eat a net 0 fields in one day forever. The sheep eats at the same rate as the grass grows.

[Guest]

Sheep=58.5, Horse=41.5, Cow=8.5 (Oh, darn I forgot about the growing grass.....)

[Guest]

(Cow +Horse +Sheep -Grass)*20 days = 1 Field

(Cow +Horse -Grass) *25 days = 1 Field

(Cow +Sheep -Grass) *(33+ 1/3) days = 1 Field

(Horse +Sheep -Grass) *50 days = 1 Field

C +H +S -G = 1/20

C +H -G = 1/25

1/20 -S = 1/25

-S = 4/100 -5/100 = 1/100.

C +S -G = 3/100

-H +1/20 = 3/100

H = 2/100

H +S -G = 1/50

-C +1/20 = 1/50

C = 3/100

3/100 + 2/100 +1/100 -G = 1/20

6/100 -G = 1/20

G = 1/100

[Guest]

Yeah answer is definitly this, enjoy =]

Given that

(C+H+S-G) * 20 days = F (One Field eaten)

and

(C+H-G) * 25 days = F

(C+S-G) * 50 days = F

(H+S-G) * 33 1/3 days = F

The Cow © eats 3/100s of a field per day, the Horse (H) 2/100s and the Sheep (S) 1/100 of a field per day and the Grass (G) grows at 1/100 per day.

Which means...

(.06+.04+.01-.01) * 20 days = F (One Field eaten)

and

(.06+.02-.05) * 25 days = 1 (Field eaten)

(.09+.08-.05) * 50 days = 1

(.04+.08-.05) * 33 1/3 days = 1

Therefore

The Cow can eat a net .08 fields in one day for 15 days.

The Horse can eat a net .03 fields in one day for 42 days.

The Sheep can eat a net 0 fields in one day forever. The sheep eats at the same rate as the grass grows.

[Guest]

Yeah answer is definitly this, enjoy =]

Given that

(C+H+S-G) * 20 days = F (One Field eaten)

and

(C+H-G) * 25 days = F

(C+S-G) * 50 days = F

(H+S-G) * 33 1/3 days = F

The Cow © eats 3/100s of a field per day, the Horse (H) 2/100s and the Sheep (S) 1/100 of a field per day and the Grass (G) grows at 1/100 per day.

Which means...

(.06+.04+.01-.01) * 20 days = F (One Field eaten)

and

(.06+.02-.05) * 25 days = 1 (Field eaten)

(.09+.08-.05) * 50 days = 1

(.04+.08-.05) * 33 1/3 days = 1

Therefore

The Cow can eat a net .08 fields in one day for 15 days.

The Horse can eat a net .03 fields in one day for 42 days.

The Sheep can eat a net 0 fields in one day forever. The sheep eats at the same rate as the grass grows.

What? ThirstyCow, your math doesn't add up. Plus its evident you cut and pasted a lot of my post...

My comments are in bold

You said "(.06+.04+.01-.01) * 20 days = F (One Field eaten) Actually = 2 Fields eaten, plus the line above that was from my post clearly states that the cow eats 3/100s of a field per day, then you come up with the magical number of .06? Where's that from?

and

You also said "(.06+.02-.05) * 25 days = 1 (Field eaten)" This actually = 3/4 of a field eaten. Plus why did you suddenly change the rate of grass growth from -.01 to -.05?

"(.09+.08-.05) * 50 days = 1" this equals 6 fields eaten. Again, where did you get .09, .08 and .05?

"(.04+.08-.05) * 33 1/3 days = 1" Nope, it equals 2 1/3 fields eaten. Where did the .04 come from?

[Guest]

(Cow +Horse +Sheep -Grass)*20 days = 1 Field

(Cow +Horse -Grass) *25 days = 1 Field

(Cow +Sheep -Grass) *(33+ 1/3) days = 1 Field

(Horse +Sheep -Grass) *50 days = 1 Field

C +H +S -G = 1/20

C +H -G = 1/25

1/20 -S = 1/25

-S = 4/100 -5/100 = 1/100.

C +S -G = 3/100

-H +1/20 = 3/100

H = 2/100

H +S -G = 1/50

-C +1/20 = 1/50

C = 3/100

3/100 + 2/100 +1/100 -G = 1/20

6/100 -G = 1/20

G = 1/100

You have the grass growth correct, but you didn't answer the question.

[Guest]

Cow 50 days

Horse 100 days

Sheep Indefinitely

Cow 50 days

Horse 100 days

Sheep Indefinitely

[Guest]

A mis-print in the question!

Sorry!!!

Mr. Greenfield owns a fixed pasture covered with grass which grows uniformly at a constant rate. Mr. Greenfield also owns a cow, a horse, and a sheep.

The existing and growing grass in the pasture is sufficient to feed all three animals grazing together for the next 20 days. Alternatively it would feed the cow and the horse together for 25 days; the cow and the sheep together for 33 1/3 days; and the horse and sheep together for 50 days.

How long it would sustain:

The cow alone?

The horse alone?

The sheep alone?

You should assume that each animal eats at some constant rate (but not necessarily all at the same rate as each other). ==> This was not in the original problem.

[Guest]

This sounds like a homework to me.............

[Guest]

What? ThirstyCow, your math doesn't add up. Plus its evident you cut and pasted a lot of my post...

My comments are in bold

You said "(.06+.04+.01-.01) * 20 days = F (One Field eaten) Actually = 2 Fields eaten, plus the line above that was from my post clearly states that the cow eats 3/100s of a field per day, then you come up with the magical number of .06? Where's that from?

and

You also said "(.06+.02-.05) * 25 days = 1 (Field eaten)" This actually = 3/4 of a field eaten. Plus why did you suddenly change the rate of grass growth from -.01 to -.05?

"(.09+.08-.05) * 50 days = 1" this equals 6 fields eaten. Again, where did you get .09, .08 and .05?

"(.04+.08-.05) * 33 1/3 days = 1" Nope, it equals 2 1/3 fields eaten. Where did the .04 come from?

OH! Burn!