Question 1122711
one cow is worth 26000.


with just cows, the average price is 26000


with 12 cows and one horse, the average price is 26000 + 15000 = 41000.


with all cows, then the total price is 13 * 26000 = 338,000


with 12 cows and 1 horse, the total price is 13 * 41000 = 533,000


12 cows at 26000 each is equal to 312,000


the horse must have cost 533,000 minus 312,000 = 221,000


12 cows at 26000 each plus 1 horse at 221,000 = 533,000 / 13 = average price of 41,000 each.


algebraically, you might solve as follows:


with cows only, you get 13 * 26000 / 13 = 26000


with 12 cows plus 1 horse, you get (12 * 26000 + x) / 13 = 41000


x represents the price of the horse.


solve that equations for x as follows:


start with (12 * 26000 + x) / 13 = 41000


multiply both sides of that equation by 13 to get 12 * 26000 + x = 13 * 41000


subtract 12 * 26000 from both sides of that eqution to get x = 13 * 41000 - 12 * 26000.


this results in x = 533,000 - 312,000 = 221,000


that's the price of the horse.