SOLUTION: Hi farm animals are for sale in groups of five.100 animals were bought for 1000 dollars.a pig cost 20 dollars, chicken 5 dollars and a sheep 30 dollars.how many sheep were bought.

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Question 1018164: Hi
farm animals are for sale in groups of five.100 animals were bought for 1000 dollars.a pig cost 20 dollars, chicken 5 dollars and a sheep 30 dollars.how many sheep were bought.
I know that this type of question can be solved by guess and check but can u show me how to solve it as 2 simultaneous equations in 3 variables.
thanks

Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +p+ = number of pigs/5 bought
Let +c+ = number chickens/5 bought
Let +s+ = number of sheep/5 bought
--------------------------------
5 pigs cost $100
5 chickens cost $25
5 sheep cost $150
(1) +100%2F5+=+20+ groups of 5 animals were bought
----------------
(1) +100p+%2B+25c+%2B+150s+=+1000+
(2) +5p+%2B+5c+%2B+5s+=+100+
------------------------
Multiply both sides of (2) by +20+
and subtract (1) from (2)
(2) +100p+%2B+100c+%2B+100s+=+2000+
(1) +-100p+-25c+-+150s+=+-1000+
--------------------------------
+75c+-+50s+=+1000+
+3c+-+2s+=+40+
I'll try:
+s+=+1+
+c+=14+
+3%2A14+-+2%2A1+=+40+
+42+-+2+=+40+
+1+ = number of sheep / 5
---------------------
There are 5 sheep
---------------------
check:
number of chickens = 14%2A5+=+70+
+100+-+5+-+70+=+25+
There are 25 pigs
-----------------
(1) +100p+%2B+25c+%2B+150s+=+1000+
(1) +100%2A5+%2B+25%2A14+%2B+150%2A1+=+1000+
(1) +500+%2B+350+%2B+150+=+1000+
(1) +1000+=+1000+
and
(2) +5p+%2B+5c+%2B+5s+=+100+
(2) +5%2A5+%2B+5%2A14+%2B+5%2A1+=+100+
(2) +25+%2B+70+%2B+5+=+100+
(2) +100+=+100+
OK

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Hi
farm animals are for sale in groups of five.100 animals were bought for 1000 dollars.a pig cost 20 dollars, chicken 5 dollars and a sheep 30 dollars.how many sheep were bought.
I know that this type of question can be solved by guess and check but can u show me how to solve it as 2 simultaneous equations in 3 variables.
thanks
Let number of groups of pigs, chickens, and sheep purchased, be P, C, and S, respectively
Then: 5P + 5C + 5S = 100______5(P + C + S) = 5(20)_______P + C + S = 20______P = 20 – C - S ---- eq (i)
Also, 20(5P) + 5(5C) + 30(5S) = 1,000______100P + 25C + 150S = 1,000_____25(4P + C + 6S) = 25(40)_______4P + C + 6S = 40 -------- eq (ii)
4(20 – C – S) + C + 6S = 40 ------- Substituting 20 – C – S for P in eq (ii)
80 – 4C – 4S + C + 6S = 40
6S – 4S – 4C + C = 40 – 80
2S – 3C = - 40
3C = 2S + 40
C+=+%282S+%2B+40%29%2F3
As a sheep is the costliest, and are in groups of 5, we can begin by determining the most number of sheep that could’ve been purchased
Since 1 sheep costs $30, neither 8 groups (40 sheep) nor 7 groups (35 sheep) can be purchased, as they'd cost $1,200, and $1,050, respectively
Therefore, by letting the number of groups of sheep that were purchased be 6, 5, 4, 3, 2, 1, and 0, and by using the above
equation: C+=+%282S+%2B+40%29%2F3, we find that:
Case 1:
4 groups of sheep, or 4(5), or 20 sheep for 20(30), or $600 can be purchased
16 groups of chicken, or 16(5), or 80 chickens, for 80(5), or $400 can be purchased
0 groups of pigs, or 0(5), or 0 pigs, for 0(20), or $0 can be purchased
This results in a total of: 20 (4 + 16) groups of 5, totaling 100 (20 + 80) animals for $1,000 (600 + 400)
Case 2:
1 group of sheep, or 1(5), or 5 sheep for 5(30), or $150, can be purchased
14 groups of chicken, or 14(5), or 70 chickens, for 70(5), or $350, can be purchased
5 groups of pigs, or 5(5), or 25 pigs, for 25(20), or $500, can be purchased
This results in a total of: 20 (1 + 14 + 5) groups of 5, totaling 100 (5 + 70 + 25) animals for $1,000 (150 + 350 + 500)