An algebra problem by Leonardo Arcenal Jr
Suppose you have a grassy field, and cows eat grass at a constant rate.
Keep in mind, the grass keeps growing continuously.
48 cows can clear all the grass off the field in 90 days.
120 cows can clear all the grass off the field in 30 days.
How many cows would be needed to clear all of the grass in 16 days? Round up to the nearest whole cow.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
@Leonardo Arcenal Jr This is a great question, but I believe that the correct answer should be 2 1 5 cows, as shown in my solution above. The value of 2 8 8 is the number of day's growth of grass that will be consumed by this size of herd in 16 days, which is why I chose 2 8 8 as the answer, (none of the other numbers had any relevance to my calculations). So I think that you may need to have the answer changed to 2 1 5 , or change the question at the end to something like
"Let N be the minimum number of cows required to clear all the grass in 1 6 days. Calculate the maximum number of day's growth of grass these N cows can clear in 1 6 days."
Log in to reply
Yap. My mistake. I've done my solution in paper and when I reviewed my answer, there was an error. I'm gonna change the answer to 215. Thanks.
Log in to reply
O.k., great. Calvin Lin can help with getting the answer changed.
More easy approach i took to solve it using options
If
120
cows
⇒
30
days
.
Then,
240
cows
⇒
15
days
.
So,for
16
days
,
Number of cows must less than
240
(which is satisfied by only one option)
.
Moderator note:
As pointed out by Brian, this solution is incorrect.
But if you take the same approach using the other data provided, namely 4 8 cows in 9 0 days, then by similar reasoning you would require
1 6 9 0 ∗ 4 8 = 2 8 8 cows, and all the answer options are below this value.
The issue is that the relationship between cows and days is not linear. If C represents the number of cows and D represents the number of days required for these cows to clear the field, then we have that
C ≥ D 1 2 ( 2 7 0 + D ) = D 3 2 4 0 + 1 2 .
So to clear the field in 1 day would require 3 2 5 2 cows! :)
Let's call the rate of grass consumption(by the cows) R and the rate at which grass grows back r . We want to find the number of cows c .
In general, we see that
c o w s × R × d a y s − r = t o t a l
By this logic we can set up three equations:
4 8 × R × 9 0 − 9 0 r = t o t a l
1 2 0 × R × 3 0 − 3 0 r = t o t a l
c × R × 1 6 − 1 6 r = t o t a l
Using the first two equations, 1 2 R = r .
Plugging this into the third equation, we can cancel out the common term R to get c = 214.5 ⇒ 2 1 5
Let A be the initial amount of grass in the field, (measured in "days of growth"), x the amount of grass a cow eats in one day and y the amount of grass that grows in the field in one day. The given information can then be presented as
4 8 ∗ 9 0 x − 9 0 y = A and 1 2 0 ∗ 3 0 x − 3 0 y = A
⟹ 5 ∗ ( 4 8 ∗ 9 0 x − 9 0 y ) − 6 ∗ ( 1 2 0 ∗ 3 0 x − 3 0 y ) = 5 A − 6 A = − A
⟹ A = 2 7 0 y .
This means that the field initially has 2 7 0 days of growth, and so we will need our herd of N cows to eat 2 7 0 + 1 6 = 2 8 6 growth-days of grass in just 1 6 days.
Now with A = 2 7 0 y we have that
4 8 ∗ 9 0 x − 9 0 y = 2 7 0 y ⟹ 1 2 ∗ 3 6 0 x = 3 6 0 y ⟹ 1 2 x = y .
This means that it takes 1 2 cows to eat one day's growth of grass. So each day our herd will eat 1 2 N day's growth of grass. Since they need to clear the field in 1 6 days, this means that we want the least integer N such that
1 6 ∗ 1 2 N ≥ 2 8 6 ⟹ N > 2 8 6 ∗ 1 6 1 2 = 2 1 4 . 5 .
So rounding up, we find that we require 2 1 5 cows to complete the required task.