# Homework Help: Word Problems Suk

1. Jan 13, 2004

### RyokoTenchi

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!!

2. Jan 13, 2004

### jamesrc

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

3. Jan 14, 2004

### 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.

4. Jan 14, 2004

### jamesrc

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: Jan 14, 2004