an algebra teacher drove past a farm yard that was full of chickens and pigs. the teacher happened to notice that there were a total of 70 heads and 200 legs. how many chickens and how many pigs were there?
2007-05-27 08:14:42 · 5 answers · asked by lazydaisy4991 2 in Science & Mathematics ➔ Mathematics
c + p = 70 (heads)
2c + 4p = 200 (legs)
multiply the first equation by -2 and add to the second equation to get:
2p = 60
p = 30
c = 40
40 chickens and 30 pigs.
2007-05-27 08:19:38 · answer #1 · answered by jcsuperstar714 4 · 0⤊ 0⤋
C = chickens
70 - C = pigs
2C + 4(70-C)=200 two legs per chicken, four per pig
2C + 280 - 4C = 200
80 = 2C
40 = C = 40 chickens
30 pigs
Check
40*2 + 30*4 = 80 + 120 = 200 legs Correct.
2007-05-27 08:21:15 · answer #2 · answered by ecolink 7 · 0⤊ 0⤋
OK...pretend that you drew 70 heads...and put two feet on each. That uses up 140 feet, right? That leaves 60 feet that you didn't use. Put those, two at a time on the heads...you're making 30 (60 / 2) 4-footed animals. So, you have 30 pigs and the rest (40) are chickens.
PS...You could solve this using equations in 2 variables, but...this is how the proverbial second grader solved it! I like his method much better!
2007-05-27 10:46:01 · answer #3 · answered by Rita L 1 · 0⤊ 0⤋
30 pigs 40 chickens
2007-05-27 08:25:05 · answer #4 · answered by Anonymous · 0⤊ 0⤋
There were 35 pigs and 35 chickens
2007-05-27 08:21:05 · answer #5 · answered by Gerry 7 · 0⤊ 1⤋