Suppose that Judas's bag of money contained 70 coins and that each silver coin could buy 55 bushels of wheat and that each gold coin could buy twice as many bushels of wheat. If there was enough money to buy 6,435 bushels of wheat, how many gold coins were in the bag?

Question

mathstudent55 · Answer

Let g = number of gold coins
Let s = number of silver coins
Can you come up with an equation that shows that the number of gold coins plus the number of silver coins is 70?

mathstudent55 · Answer

Each silver coin buys 55 bushels. Each gold coin buys twice 55 bushels, or 110 bushels.
How many bushels do g gold coins buy?
How many bushels do s silver coins buy?
When you add the amounts above, you get 6435.

anonymous · Answer

I know the answer, but I want to see what you get first, unless you are really stuck

anonymous · Answer

s+g =70
55s+110g = 6435
Solve for g, then substitute
g = 70-s

mathstudent55 · Answer

You need to come up with two equations. Then you solve the system of equations to get the answer.

mathstudent55 · Answer

So far, so good.

anonymous · Answer

55s + 110(70-s) = 6435
55s + 7700-110s = 6435
-55s = 6435-7700
-55s = -1265
s= -1265/-55
s= 23
Substitue again in equation 1
s+g =70
23+g = 70
g = 70-23
g= 47

There were 23 silver coins  and 47 gold coins.