set up a relation
x + y = 200 ( for the heads, each animal has only one head, right? )
2x + 4y = 540 ( for the legs, each chicken has 2 legs, cows have 4 )
now, you need to eliminate a fraction.
multiply the top by 2 and subtract it from the bottom
2x - 2x + 4y - 2y = 540 - 400
which equals
2y = 140
therefore we can say that y = 70, and that there was 70 cows.
Substitute back into either equation 70 for y, and for simplicity, we will use the top one.
x + 70 = 200
x = 200 - 70
x * 130
130 chickens.
We can check our work by multiplying x by 2 and y by 4 to see if it adds to 540.
130*2 + 70*4
260 + 280
540
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?
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.
94 three legged cows and 25 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.
Chickens (H) + Cows (C) = 66 : so C = 66 - H However there are 2H chickens legs and 3C cows legs : 2H + 3C = 163 Substituting for C gives : 2H + 3(66 - H) = 163 : 2H +198 - 3H = 163 : H = 35. Number of chickens = 35 (and number of cows = 66 - 35 = 31)
7 cows and 11 chickens. Cows have 4 legs. Chickens have 2 legs. (4 legs)(7 cows) + (2 legs)(11 chickens) = 50 legs. 7 cow heads = 11 chicken heads = 18 heads.
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?
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.
94 three legged cows and 25 chickens.
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.
There are 5 chickens and 6 pigs in the barn. Each chicken has 2 legs and each pig has 4 legs.
Assuming that all of the animals are "normal" (no amputees, 2-headed mutations, etc.) : 15 cows (15 x 4 = 60 legs) plus 5 chickens (5 x 2 = 10 legs) gives a total of 20 heads and 70 legs.
Chickens (H) + Cows (C) = 66 : so C = 66 - H However there are 2H chickens legs and 3C cows legs : 2H + 3C = 163 Substituting for C gives : 2H + 3(66 - H) = 163 : 2H +198 - 3H = 163 : H = 35. Number of chickens = 35 (and number of cows = 66 - 35 = 31)
85 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
This can be done in your head: If all animals were cows there would be 321 legs. Every cow replaced by a chicken reduces this figure by one.The total reduction is 97 (321 - 224) so there are 97 chickens (and 10 cows)Check legs: (97 x 2) + (10 x 3) = 194 + 30 = 224. QED