Best Answer

Cows = M

Chickens = C

M+C=54

3M+2C=159

C=54-M

3M+2(54-M)=159

3M+108-2M=159

M+108=159

M=51 Cows

C=3 Chickens

Q: On a farm there are three legged cows and chickens There are a total of 54 heads and 159 legs How many three legged cows are there?

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94 three legged cows and 25 chickens.

On a farm there are chickens and three-legged-cows. There are total of 49 heads and 130 legs. How many chickens are on the farm?

Do-it-in-your-head method: If all animals were chickens there would be 90 legs. There are 20 extra legs, each on a cow, so there are 20 cows and therefore 25 chickens.

Suppose there are x chickens and y cows Then heads => x + y = 64 And Legs => 2x + 3y = 147 2*Heads: 2x +2y = 128 Subtract from Legs: 19 = y Substitute in Heads: x = 45 Answer: 45 chickens.

There are 21 three-legged cows and 27 chickens on the farm. Let us say the number of cows = A and the number of chickens is B. The equation for heads is: A+B = 48 The equation for legs is: 3A + 2B = 117 (because a cow has 3 legs and a chicken has 2 legs in this question) We can rearrange the equation for heads as follows: A+B = 48 B = 48-A Now we can replace B in the equation for legs with 48-A as follows: 3A + 2(48-A) = 117 ...and solve: 3A + 96 - 2A = 117 A + 96 = 117 A = 21 So we know there are 21 three-legged cows and from earlier, B = 48-A, so B = 48-21 = 27. Therefore there are 27 chickens.

Related questions

94 three legged cows and 25 chickens.

On a farm there are chickens and three-legged-cows. There are total of 49 heads and 130 legs. How many chickens are on the farm?

Do-it-in-your-head method: If all animals were chickens there would be 96 legs. There are 26 extra legs, each on a cow, so there are 26 cows and therefore 22 chickens.

Do-it-in-your-head method: If all animals were chickens there would be 90 legs. There are 20 extra legs, each on a cow, so there are 20 cows and therefore 25 chickens.

Suppose there are x chickens and y cows Then heads => x + y = 64 And Legs => 2x + 3y = 147 2*Heads: 2x +2y = 128 Subtract from Legs: 19 = y Substitute in Heads: x = 45 Answer: 45 chickens.

There are 21 three-legged cows and 27 chickens on the farm. Let us say the number of cows = A and the number of chickens is B. The equation for heads is: A+B = 48 The equation for legs is: 3A + 2B = 117 (because a cow has 3 legs and a chicken has 2 legs in this question) We can rearrange the equation for heads as follows: A+B = 48 B = 48-A Now we can replace B in the equation for legs with 48-A as follows: 3A + 2(48-A) = 117 ...and solve: 3A + 96 - 2A = 117 A + 96 = 117 A = 21 So we know there are 21 three-legged cows and from earlier, B = 48-A, so B = 48-21 = 27. Therefore there are 27 chickens.

There are 24 three-legged cows and 92 chickens on the farm. Let us say the number of cows = A and the number of chickens is B. The equation for heads is: A+B = 116 The equation for legs is: 3A + 2B = 256 (because a cow has 3 legs and a chicken has 2 legs in this question) We can rearrange the equation for heads as follows: A+B = 116 B = 116-A Now we can replace B in the equation for legs with 48-A as follows: 3A + 2(116-A) = 256 ...and solve: 3A + 232 - 2A = 256 A + 232 = 256 A = 24 So we know there are 24 three-legged cows and from earlier, B = 116-A, so B = 116-24 = 92. Therefore there are 92 chickens.

Suppose there are x chicken and y cows where x and y are positive integers. Then number of heads : x + y = 90 and number of legs : 2x + 3y = 211 3*Eqn(1) - Eqn(2) gives: x = 59

Chickens = c, Cows = m c+m=36 c=36-m 2c+3m=76 2(36-m)+3m=76 72-2m+3m=76 m=4 c=36-4=32 He has 32 chickens.

51 horses of which there are 3 cows with 2 heads and three legs or since there are no 3 legged cows there are 60 horses

85 chickens

There are 42 chickens in the farm.