5 November 2015

Arithmetic & Algebraic Problems - HORSES AND BULLOCKS

Question: A dealer bought a number of horses at $344.00 each, and a number of bullocks at $265.00 each. He then discovered that the horses had cost him in all $33.00 more than the bullocks. Now, what is the smallest number of each that he must have bought?

Answer: We have to solve the indeterminate equation 344x = 265y + 33. This is easy enough if you know how, but we cannot go into the matter here. Thus x is 252, and y is 327, so that if he buys 252 horses for $344.00 apiece, and 327 bullocks for $265.00 apiece, the horses will cost him in all $33.00 more than the bullocks.

No comments:

Post a Comment