a farmer has cows and chickens. he has a total of 20 heads and 64 legs. How much of each animal does he have?
Let X be the no of cows and Y be number of chickens
No of Heads = 20
x+y = 20
No. of legs = 64
4x + 2 y = 64
Solve the system of equation x+y = 20
4x + 2y = 64
You will get x = 12, y = 8
No. of cows = 12
No. of chickens = 8
Power
a farmer kept track of his cows and hens by counting the legs and the heads.if he counted 78 legs anf 35 heads,hw many cows did he have?
To solve this problem, we can set up a system of equations based on the given information.
Let's assume the number of cows is 'c' and the number of chickens is 'h'.
We know that the farmer has a total of 20 heads, so the equation will be:
c + h = 20 -- equation 1
We also know that the total number of legs is 64, and since cows have 4 legs and chickens have 2 legs, we can create another equation based on the legs:
4c + 2h = 64 -- equation 2
Now, we can solve these equations simultaneously to determine the number of cows and chickens.
First, let's solve equation 1 for 'c':
c = 20 - h
Now, substitute this value of 'c' into equation 2:
4(20 - h) + 2h = 64
Simplifying the equation,
80 - 4h + 2h = 64
80 - 2h = 64
-2h = 64 - 80
-2h = -16
h = (-16) / (-2)
h = 8
Now that we have the number of chickens (h = 8), we can substitute this value back into equation 1 to find the number of cows:
c + 8 = 20
c = 20 - 8
c = 12
Therefore, the farmer has 12 cows and 8 chickens.