|
Question 1150417: A farmer bought a number of pigs for $150. However, 8 of them died before he could sell the rest at a profit of $6 per pig. His total profit was $92. How
many pigs did he originally buy?
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
See the TWIN problems solutions under these links
https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1105451.html
https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1105285.html
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website! A farmer bought a number of pigs for $150. However, 8 of them died before he
could sell the rest at a profit of $6 per pig. His total profit was $92. How
many pigs did he originally buy?
Suppose he originally bought N pigs for $150, so he paid $150/N per pig. Then 8
died, so he then had only N-8 pigs to sell. He then sold N-8 pigs each for $6
more than he paid, or $150/N + 6 each. So he took in $[150/N + 6][N-8]. This was $92
more than the $150 he paid, or $150+$92 or $242. So the equation is
$[150/N + 6][N-8] = $242
[150/N + 6][N-8] = 242
Clear the fraction by multiplying both sides by N
N∙[150/N + 6][N-8] = N∙242
[150 + 6N][N-8] = 242N
150N-1200+6N²-48N = 242N
6N²+102N-1200 = 242N
6N²-140N-1200 = 0
Divide all terms on both sides by 2
3N²-70N-600 = 0
(3N+20)(N-30) = 0
3N+20=0; N-30=0
3N=-20; N=30
N=-20/3
Ignore the negative fractional answer.
Solution: 30 pigs
Let's check:
He bought 30 pigs for $150. So he paid $150/30 = $5 per pig. Then 8 died. So
he only had 22 pigs to sell. He sold each of the 22 pigs for $6 more than he
paid, so he sold each pig for $11. Since he sold 22 pigs, he ended up with
22∙$11 = $242, and since he paid $150 for the pigs in the beginning, his profit
was $242-$150 = $92.
Edwin
|
|
|
| |
