Solve Word Problem: Rancher Divides Herds Between Sons

  • Thread starter Thread starter RyokoTenchi
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a rancher who wishes to divide his herd of cows among his sons while ensuring that each son's wife receives a portion of the remaining cows. The challenge is to determine the total number of cows and the number of sons based on the conditions set forth in the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationships between the number of cows taken by each son and the fractions allocated to their wives. There is an attempt to express the total number of cows and the distribution in terms of variables. Some participants question the assumptions regarding the number of sons and how it affects the distribution.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have started to formulate equations based on the relationships described, while others have raised concerns about the assumptions made regarding the number of sons and the implications for the solution.

Contextual Notes

There is uncertainty regarding the exact number of sons, as the problem suggests a series of divisions without specifying how many sons are involved. This ambiguity affects the calculations and assumptions being made by participants.

RyokoTenchi
Messages
2
Reaction score
0
A Western Rancher finding himself well advanced in years, called his boys together and told them that he wished to divide his herds between them while he yet lived. "Now, John," He told the eldest, "you may take as many cows as you think you could conveniently care for, and your wife Nancy may have one ninth of all the cows left."
To the second son he said, "Sam, you may have the same number of cows that john took, plus one extra because john had the first pick. To your wife, Sally, i will give one ninth of what will be left."
To the third son he made a similar statement. he was to take one cow more than the second son, and his wife was to have one ninth of those left. the same applied to the other sons. Each took one cow more than his next oldest brother, and each son's wife took one ninth of the remainder.
After the youngest son had taken his cows, there were none left for his wife. Then the rancher said: "Since horses are worth twice as much as cows, we will divide up my seven horses so that each family will own livestock of equal value."
The problem is to tell how many cows the rancher owned and how many sons he had.

i am basically completely at a loss on where to even START this problem. i know that the number the eldest takes away is unknown and that everytime a wife takes 1/9 from the leftovers the number changes cause the number of cows goes down. but depending on the number of cows the eldest takes this guy sure does have a crapload of cows 0_0!
 
Physics news on Phys.org
Let's start by listing everybody involved here; I don't know about you, but sometimes that helps me keep track of so many people (letters represent the number of cows each person has, and I'm filling in equations for the relationships between people to start off):

C = total number of cows
J = John
N = Nancy = (C-J)/9
S = Sam = J + 1
A = Sally = (C-J-N-S)/9
T = Third son = S + 1 = J + 2
W = third son's wife = 0

H = total number of horses = 7
H1 = horses for John and Nancy
H2 = horses for Sam and Sally
H3 = horses for third couple

Now let's label the values of each couples livestock in a suitable unit (equivalent number of cows):

V1 = John and Nancy's livestock value = J + N + 2*H1
V2 = Sam and Sally's livestock value = S + A + 2*H2
V3 = Third couple's livestock value = T + W + 2*H3

Now you have to try to consolidate all of this to get to the solution, given that V1 = V2 = V3.

I'd start off by writing each person's number of cows in terms of John's (J) and the total number of cows (C):

N = (C-J)/9
S = J + 1
A = (C-J-(C-J)/9-(J+1))/9 = (C-1-(C-J)/9)/9 = (8C-9-J)/81
T = S + 1 = J + 2
W = 0

(I'd double check my algebra if I were you; I make no guarantees on its accuracy.)

Now move onto the V's:

V1 = J + N + 2*H1 = J + (C-J)/9 + 2*H1 = (8J+C)/9 + 2*H1
V2 = S + A + 2*H2 = J + 1 + (8C-9-J)/81 + 2*H2 = (80J+8C+72)/81 + 2*H2
V3 = T + W + 2*H3 = J + 2 + 2*H3

Combine these three equations with the facts that:

V1 = V2 = V3 = V
and
H1 + H2 + H3 = 7

One more equation ought to be enough to solve the problem:

V = (C + 2*H)/3 = (C+14)/3
 
I was noodling around with this problem for a while and didn't get very far, but I don't agree with your solution for the following reason:

The exact number of sons isn't mentioned in the problem. The problem seems to set up a Series going from the oldest son to the youngest, however many that may be, and states that once the youngest is reached there's nothing left for the wife.

Your solution doesn't take this into consideration.

My 2 cents for what it's worth.
 
Originally posted by discoverer02
I was noodling around with this problem for a while and didn't get very far, but I don't agree with your solution for the following reason:

The exact number of sons isn't mentioned in the problem. The problem seems to set up a Series going from the oldest son to the youngest, however many that may be, and states that once the youngest is reached there's nothing left for the wife.

Your solution doesn't take this into consideration.

My 2 cents for what it's worth.

Yeah you're right; I skipped right over that part of the problem. In fact, I didn't even bother to follow the solution I typed out to its logical conclusion. Maybe I'll come back later and try again too.
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
8K