The Pharaoh’s cook has made three trips to the market. On each trip he made, he bought a different combination of fruit, and paid the number of coins shown below. The price of each fruit did not change. What is the total cost of an apple, an orange, and a pear?
Looks like a straightforward algebra problem to me (maybe I’m just not getting the “riddle”). Since the cost of each fruit does not change, a system of equations could do this…
Assume x is cost of apples in coins, y is cost of oranges, z is cost of pears. Given the diagram then
3x + 4y + 5z = 97 coins
5x + 3y + 4z = 89 coins
4x + 5y + 3z = 90 coins
Solving for x, y, and z should get
x = 5, y = 8, z = 10
So one apple costs 5 coins, one orange cost 8 coins, and one pear costs 10 coins.
Did I miss the point of the riddle?
It’s a system of equations. I’ll give you a hint to help you solve it on your own, if you’d like:
Write it out like this:
3x + 4y + 5z = 97
5x + 3y + 4z = 89
4x + 5y + 3z = 90
Then use any of several algebraic methods for solving systems of equations (use Google of you can’t remember high school algebra ;)) to solve for x, y and z. Good luck!
Or if you dislike multiple-equation algebra as much as I do, you can solve for one variable (generally z), and graph the three on a graphing calculator. These three lines, if done properly, will cross at (5,8,10) - who would have thunk it? 
Do we know that all the coins are worth the same amount? Could be combinations of quarters, dimes and nickels. Of course this would make the problem impossible.
My Casio 985o-Ga Plus will solve them through direct-entry. Of my 8 scientific calculators, this is perhaps my favorite.
Pssst! The hero of the “riddle” was a pharaoh’s cook – the dollar wasn’t about to become popular for <mumble-mumble> millenia.