# A piece of pasture

• Jun 19th 2012, 05:03 AM
kingman
A piece of pasture
I need help in the below question.
Thanks

A piece of pasture grows at a constant rate everyday. 200 sheep will eat up the grass in 100 days. 150 sheep will eat up the grass in 150 days. How many days does it take for 100 sheep to eat up the grass?
• Jun 19th 2012, 06:56 AM
emakarov
Re: A piece of pasture
Nice problem. I assume that in all three cases the initial grass volume on the pasture is the same. Let this volume be p units (for "pasture"). Assume also that the grass grows g units per day, and 100 sheep eat s units of grass per day. The problem statement gives two equations in these three variables. Note however that the answer (the number of days corresponding to 1 hectosheep (Smile) ) can be expressed as a function of g / p and s / p, and the two equations above can be rewritten as equations only in g / p and s / p. Thus, we have two equations and two unknowns.
• Jun 19th 2012, 07:32 AM
Wilmer
Re: A piece of pasture
• Jun 19th 2012, 02:51 PM
kingman
Re: A piece of pasture
Thanks but I cannot get the answer 300 days; and we are not suppose to solve employng simultaneous equations approach.
• Jun 19th 2012, 03:31 PM
richard1234
Re: A piece of pasture
Suppose that the initial amount of grass is always $\displaystyle 1$, and that the grass grows at a rate $\displaystyle r$ per day. Therefore,

(1) 200 sheep will eat $\displaystyle 1 + 100r$ grass in 100 days,
(2) 150 sheep will eat $\displaystyle 1 + 150r$ grass in 150 days.
(3) 100 sheep will eat $\displaystyle 1 + xr$ grass in x days.

For (1), (2), and (3), respectively, divide the number of sheep by 4, 3, and 2, respectively:

(1*) 50 sheep will eat $\displaystyle \frac{1 + 100r}{4}$ grass in 100 days.
(2*) 50 sheep will eat $\displaystyle \frac{1 + 150r}{3}$ grass in 150 days.
(3*) 50 sheep will eat $\displaystyle \frac{1 + xr}{2}$ grass in x days.

In this case, the amount of grass eaten is proportional to the number of days. We use (1*) and (2*):

$\displaystyle \frac{\frac{1 + 100r}{4}}{100} = \frac{\frac{1 + 150r}{3}}{150}$. Solving for r, we get $\displaystyle r = \frac{1}{300}$. Now we use (1*) and (3*), substituting r with $\displaystyle \frac{1}{300}$:

(1*) 50 sheep will eat $\displaystyle \frac{1}{3}$ grass in 100 days.
(3*) 50 sheep will eat $\displaystyle \frac{1 + \frac{x}{300}}{2}$ grass in x days.

Similarly, the ratio of grass eaten to number of days is equal, so $\displaystyle \frac{\frac{1}{3}}{100} = \frac{\frac{1 + \frac{x}{300}}{2}}{x}$

Cross-multiplying, $\displaystyle \frac{x}{3} = 50 + \frac{x}{6} \Rightarrow x = 300$ (days).
• Jun 19th 2012, 04:08 PM
emakarov
Re: A piece of pasture
Quote:

Originally Posted by kingman
we are not suppose to solve employng simultaneous equations approach.

Which approach are you supposed to use?

This page has a quite short solution. It measures grass in amounts that one sheep eats per day.

The second link in post #3 has the wrong answer and solution since it assumes that the relationship between days and sheep is linear. The first link in post #3 is a bit hard to understand.