SOLUTION: Two ranchers sold a herd of cattle and received as many dollars for each animal as there are animals in the herd. If there were two animals, they received $2 per animal; if there w
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Question 149357: Two ranchers sold a herd of cattle and received as many dollars for each animal as there are animals in the herd. If there were two animals, they received $2 per animal; if there were three animals, they received $3 per animal; etc. As an example, if there were eight animals in the herd, the value of the herd would be $64 ($8 per animal times 8 animals in the herd). We’re told that they started with no more than 20 cattle.
With the money, they bought as many sheep as they could at $10 per head, and a goat with the remainder (less than $10). Finally they divided the animals between them. There was however, an odd number of sheep. So, one rancher took the extra sheep and the other the goat. The rancher who got the goat was given his friends new pocket knife as compensation. If this was to be an equal monetary division, what was the value of the pocketknife.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Two ranchers sold a herd of cattle and received as many dollars for each animal as there are animals in the herd. If there were two animals, they received $2 per animal; if there were three animals, they received $3 per animal; etc. As an example, if there were eight animals in the herd, the value of the herd would be $64 ($8 per animal times 8 animals in the herd). We’re told that they started with no more than 20 cattle.
With the money, they bought as many sheep as they could at $10 per head, and a goat with the remainder (less than $10). Finally they divided the animals between them. There was however, an odd number of sheep. So, one rancher took the extra sheep and the other the goat. The rancher who got the goat was given his friends new pocket knife as compensation. If this was to be an equal monetary division, what was the value of the pocketknife.
:
We know a couple of things,
the number of sheep received from the sale has to be odd.
the number of sheep $ value, has to leave a remainder, to buy the goat with
:
Make a table using the value of the cattle = x, (has to be 20 or less)
Let y = number of sheep obtained from this value
Make a table from the equation y = x^2/10
x | y
--------
4 |1.6*
5 |2.5
6 |3.6*
7 |4.9
8 |6.4
9 |8.1
10 |10
11 |12.1
12 |14.4
13 |16.9
14 |19.6*
15 |22.5
16 |25.6*
:
*It's interesting to note that when y has has a odd whole number,
the decimal is always .6
Since we have to have an odd number of sheep as given by the problem, we can
assume the goat cost $6, therefore the knife value is $4
:
the goat and the knife count as 1 ea $10 sheep, making each man get the same monetary value
;
Did this make sense??
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