SOLUTION: there are 41 pigs and chikens on a farm. There are 100 legs counted altogether, how many of each animal are there?

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Question 3172: there are 41 pigs and chikens on a farm. There are 100 legs counted altogether, how many of each animal are there?
Found 2 solutions by gsmani_iyer, WannabeCAgirl83:
Answer by gsmani_iyer(201) About Me  (Show Source):
You can put this solution on YOUR website!

let the no.of pigs = x

let the no.of chicken = y

pigs have 4 legs and chicken have 2 legs each

so, the equation will be as under:

x + y = 41 ............. (1) given.

4x + 2y = 100 ......... (2) given.
(1) * 2, we get

2x + 2y = 82 ........... (3)
(2) - (3), we get
2x = 18

So x = 18%2F2 = 9

by replacing the value of x in (1), we get

9 + y =41

So y =41-9 = 32.

No.of pigs = 9; chicken = 32. Answer



gsm



Answer by WannabeCAgirl83(35) About Me  (Show Source):
You can put this solution on YOUR website!

pigs = p
chickens = c

Step 1
Finding Equations:
There are 41 pigs and chickens on a farm.
p + c = 41 → Eqn 1

There are 100 legs counted altogether.
4p + 2c = 100 → Eqn 2
(Note that a pig has 4 legs while a chicken has 2. Therefore it’s 4p and 2c.)

Step 2
Isolating p in Eqn 1:
p + c = 41 | - c on both sides
p = 41 - c

Step 3
Substituting p in Eqn 2 with the term found for p in step 2. Then isolating c:
4(41 - c) + 2c = 100
164 - 4c + 2c = 100 | re-arranging
164 - 2c = 100 | - 164 on both sides
- 2c = - 64 | : - 2
c = 32 → There are 32 chicken.

Step 4
Substituting c in Eqn 1 by the value found for c in step 3. Then again isolating p:
p + c = 41
p + 32 = 41 | - 32 on both sides
p = 9 → There are 9 pigs.