perplexus.info :: Just Math : Pasture Puzzles

Pasture Puzzles (Posted on 2013-11-01)

(a) More Slow Cows.

If 14 cows take exactly 5 times as long to graze out a pasture as 54 cows, and 38 cows take exactly 5 times as long as 174 cows, what is the largest prime number of cows that can graze the same pasture?

Note 1: In this part, all times are calculated in exact whole numbers of days, and the pasture is of the smallest size compliant with the given conditions.

(b) Newton's Goats

Six goats eat all the grass in a pasture in exactly three days, while 3 goats take exactly seven days to graze out the same pasture. On what day will the pasture become exhausted, if two goats are grazing there?

Note 2: In both parts, the grass grows at a steady rate, and each animal eats the same, constant, amount each day; but, unlike cows, goats eat grass down to the stubble, so that areas that have already been grazed do not resume growth while the pasture remains in use.

Submitted by broll

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Solution: (Hide)

(a) Let number of days = a,b; number of cows = C, D.
Say the grass grows at a fixed rate, x, per day.
Then p+a*x= a*C, p+b*x=b*C
Givens: C=14, D=54 Days: 5a,a
p+5*a*x= 5*a*14, p+a*x=a*54
Gives x=4, p=50a
p+5*a*x= 5*a*38, p+a*x=a*174
Gives x=4, p=170a
So p = 850n, a = 17n, b = 5n, x = 4
Since p is of minimal size, p=850.
Since x is an integer, we can consider such numbers as divide 850+4k evenly, provided the number of days exactly divides the total available fodder.
1 854 2*7*61
2 858 2×3×11×13 429 cows
3 862 2*431
4 866 2*433
5 870 2×3×5×29 174 cows
6 874 2*19*23
7 878 2*439
8 882 2*3^2*7^2
9 886 2*443
10 890 2*5*89 89 cows
etc...
The largest prime number of cows that can graze on the size 850 pasture for an exact number of days is 89 and they will graze it out in exactly 10 days.
(b)Since the remaining amount of uneaten grass changes each day, the growth diminishes each day. One way to approach the answer is to use Excel to set up the two known scenarios, 6 goats in 3 days and 3 goats in 7 days, and vary the growth rate, x%, and the initial amount of grass, until the scenarios meet the required conditions:
Day: Initial Grass: Growth: Eaten:
1 15.288929 1.31920371 6
2 10.60813271 0.91532167 6
3 5.52345438 0.476590709 6
0.0000
1 15.288929 1.31920371 3
2 13.60813271 1.17417637 3
3 11.78230908 1.016635361 3
4 9.798944441 0.845500941 3
5 7.644445382 0.659600205 3
6 5.304045587 0.457659043 3
7 2.76170463 0.238293408 3
0.0000
The rate, x%, is around 8.62849%, and the initial amount of grass is around 15.288929 goat meals. At the same rate, 2 goats will finish all the grass in the same pasture early on the 14th day.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re(6): Solution	Salil	2013-11-07 06:47:10
re(5): Solution	broll	2013-11-07 05:43:33
re(4): Solution	Salil	2013-11-07 04:45:12
re(4): Solution	Salil	2013-11-07 04:21:22
re(3): Solution	broll	2013-11-06 10:59:18
re(3): Solution	broll	2013-11-06 10:46:21
re(2): Solution	Salil	2013-11-06 04:52:10
re(2): Solution	Salil	2013-11-06 04:29:10
re: Solution	broll	2013-11-06 00:58:15
Solution	Salil	2013-11-05 23:44:27

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