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
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?
If all animals were horses there would be 4 x 59 ie 236 legs. There is a shortage of 25 legs so there are 25 cows (and 34 horses).
If all the animals were horses there would be 360 legs. There is a shortfall of 23 which means 23 cows and therefore 67 horses.
94 three legged cows and 25 chickens.
You are supposed to: 1) Write two equations, one for the number of heads (express it in terms of the number of cows and horses, which of course are expressed as variables), and one for the number of legs (also expressed as an equation in terms of the number of cows and horses. 2) Solve the two equations simultaneously.
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?
If all animals were horses there would be 4 x 59 ie 236 legs. There is a shortage of 25 legs so there are 25 cows (and 34 horses).
There are 51 horses and 14 sheep (51*4)+(14*3)=246 51+14=65
If all the animals were horses there would be 360 legs. There is a shortfall of 23 which means 23 cows and therefore 67 horses.
94 three legged cows and 25 chickens.
You are supposed to: 1) Write two equations, one for the number of heads (express it in terms of the number of cows and horses, which of course are expressed as variables), and one for the number of legs (also expressed as an equation in terms of the number of cows and horses. 2) Solve the two equations simultaneously.
If all cows were four-legged there would be a total of 4 x 131 ie 524 legs. There are therefore 36 missing legs so there are 95 horses. (4 x 95 + 3 x 36 = 380 + 108 = 488 legs)
The easiest way is to substract three times the number of heads from the legs and that will give you the amount of horses. The reasoning is as follows:legs = lheads = hhorses = xcows = (h - x)l = 3(h-x)+4xl = 3h-3x+4xl-3h = xx = l-3h
Do-it-in-your-head method: If all animals were horses there would be 4 x 46 ie 184 legs. There is a shortage of 31 legs which means there are 31 cows (and 15 horses). 31 x 3 = 93, 15 x 4 = 60 total 153. Shazam!
Let's denote the number of horses as 'H' and the number of three-legged cows as 'C'. Each horse has one head and four legs, while each three-legged cow has one head and three legs. From the given information, we can form two equations: H + C = 129 (total number of heads) and 4H + 3C = 480 (total number of legs). By solving these two equations simultaneously, we can determine the number of horses on the farm.
If the cows were normal there would be a total of 4 x 94 ie 376 legs. There is a deficit of 19 legs so there are 19 cows leaving 75 horses. Check: 75 x 4 + 19 x 3 = 300 + 57 QED
There are 96 heads, so 96 animals. If all were horses, that would be 4 x 96 = 384 legs, but there are only 313 legs, which means there are 384-313 = 71 legs short or 71 cows and 25 horses. Mathematically, let x = horses and y = cows x + y = 96 4x + 3y = 313 multiply first equation by 4: 4x + 4y = 384 4x + 3y = 313 subtract equations y = 71 = cows x = 25 = horses