If der are 100 feet in the cage, and der r 50 animals (chiken and cow). How many r cows, how many r chiks?

Answer 1

Chickens. Only.

Answer 2

~First, I will assume claws, talons and hooves qualify as feet. If not, there are other animals in the cage that have feet and the answer, requiring 50 animals and 100 feet, in incalculable. With that assumption, 50 chicks would account for 100 feet quite nicely, barring mutations and amputees. 48 chicks and 1 cow would do it too (subject to the same caveat), as would 46 and 2 cows, 44 chicks and 3 cows, 42 and 4 cows, 40 chicks and 5 cows, etc. Starting to see how incomplete, or just plain dumb, the question is?

Answer 3

Let c= number of chickens and o= number of cows.
50 animals, so c + o = 50 (1)
100 legs, so 2c + 4o = 100 (2)
rearranging (1) : c = 50 - o (3)
substitute (3) into (2): 2(50 - o) + 4o = 100
100 - 2o +4o = 100
2o = 0
o = 0 (4)
substitute (4) into (3) c = 50 - 0
c = 50
50 chickens, 0 cows.

bcptm: the question specifies 50 animals. The question is not the dumb one here.

Answer 4

Assuming each chicken has 2 feet and each cow has 4.

Let the number of cows be c
Then the number of chicken = 50-c

now
4c + 2(50-c) = 100
2c =0
c=0

So there are no cows and 50 chicken

Answer 5

cows have 4 feet and chiks have 2 feet.
100/6=16+(4/6)
hence 17 cows and 16 chiks

Answer 6

If there is any cow, then the number of feet must have been greater than twice the number of animals, but it isn't. The number of feet of animals is exactly twice the number of animals, so all are chickens.

If you want it to be solved using math, then
Let
x = chickens
y = cows

x + y = 50
2x + 4y = 100

Multiply -4 to the first equation
2x + 4y = 100
-4x - 4y = -200

Add
-2x = -100

Divide
x = 50

x + y = 50
y = x - 50
y = 50 - 50
y = 0

^_^

Answer 7

let cows be x and chicken y
x+y=50;
4x+2y=100; as cows 4 legs and chicken 2 legs
2x+y=50;
2x+2y=100; fro 1 eq;
y=50;
therefore
50 chickens and 0 cows

Answer 8

24 cows, 2 chickens

Answer 9

25 cows

Answer 10

50 chickenes and no cow