Chickens, Ducks, and Geese Word Problem

[geekily]

Three chickens and one duck sold for as much as two geese, whereas one chicken, two ducks, and three geese were sold together for $25. What was the price of each bird in an exact number of dollars?

I set this up as 1c + 2d + 3g = 25 and 3c + 1d = 2g, but I don't know where to go from there. The section the question is from is called "counting factors, greatest common factor, and least common multiple", and I solved the problem right above it with LCM, but I don't see how this one fits into any of those. Any suggestions?

Thanks so much!

 

[Mrspi]

Three chickens and one duck sold for as much as two geese, whereas one chicken, two ducks, and three geese were sold together for $25. What was the price of each bird in an exact number of dollars?

I set this up as 1c + 2d + 3g = 25 and 3c + 1d = 2g, but I don't know where to go from there. The section the question is from is called "counting factors, greatest common factor, and least common multiple", and I solved the problem right above it with LCM, but I don't see how this one fits into any of those. Any suggestions?

Thanks so much!

Well.....

3c + d = 2g
d = 2g - 3c

Substitute this into the first equation:
c + 2d + 3g = 25
c + 2(2g - 3c) + 3g = 25
c + 4g - 6c + 3g = 25
7g - 5c = 25

If you need "an exact number of dollars" (which I interpret to mean that the number of dollars for g and for c must be a positive integer), then

g = 5 and c = 2 will work, because 7*5 - 5*2 = 25

If g = 5, and c = 2, then
3*2 + d = 2(5)
6 + d = 10
d = 4


A chicken costs $2, a duck costs $4, and a goose costs $5.

 

[soroban]

Hello, geekily!

I don't see how LCM and GCF applies here either . . .

Three chickens and one duck sold for as much as two geese,
whereas one chicken, two ducks, and three geese were sold together for $25.
What was the price of each bird in an exact number of dollars?

We have two equations: \(\displaystyle \:\begin{array}{cc}3C\,+\,D\:=\:2G & \;(1)\\ C\,+\,2D\,+\,3G\:=\:25 & \;(2)\end{array}\)

From (1), we have: \(\displaystyle \\:=\:2G\,-\,3C\;\;(3)\)

Substitute into (2): \(\displaystyle \:C\,+\,2(2G\,-\,3C)\,+\,3G\:=\:25\)

. . which simplifies to: \(\displaystyle \:G\:=\:\frac{5(C\,+\,5)}{7}\;\;(4)\)

Since \(\displaystyle G\) is a positive integer, \(\displaystyle C\) must be: \(\displaystyle 2,\,9,\,16,\,\cdots\)
. . That is, \(\displaystyle C\:=\:2\,+\,7k\,\) for some integer \(\displaystyle k\,\geq\,0\)

Substitute into (4): \(\displaystyle \:G\:=\:\frac{5(2\,+\,7k\,+\,5)}{7}\;\;\Rightarrow\;\;G \:=\:5(k\,+\,1)\)

Substitute into (3): \(\displaystyle \\:=\:2\cdot5(k\,+\,1)\,-\,3(2\,+\,7k)\;\;\Rightarrow\;\;D\:=\:4\,-\,11k\)

Since \(\displaystyle D\) is a positive integer, \(\displaystyle k\) must be \(\displaystyle 0.\)

. . Therefore: \(\displaystyle \\,=\,4,\:G\,=\,5,\:C\,=\,2\)

 

[geekily]

Thank you so much for your help, both of you! I really appreciate it!