Posted by Josh on Friday, November 16, 2007 at 6:04pm.
Model the situation by making an equation.
Let d be the number of ducats in the purse.
d - 1/4 - 1/5 - 1/6 = 9
Get a common denominator with the fractions. There may be a number lower than 60, but that's what I'm going to use. (I multiplied 4, 5, and 6 to get 120. Then, I divided by 2.)
d - 15/60 - 12/60 - 10/60 = 9
d - 37/60 = 9
d = 9 + 37/60, or about 9.6167
This doesn't seem right to me, so I'm going to do what you suggested second (spending 1/5 of the remaining 3/4). That will probably work out better.
(3/4)d - 1/5(3/4)d - 1/6(1/5)(3/4)d = 9
(3/4)d - (3/20)d - (1/40)d = 9
Common denominator... 40.
(30/40)d - (6/40)d - (1/40)d = 9
(23/40)d = 9
d = 9(40/23) = 15.65 ducats
That doesn't seem right either because you have partial ducats, which doesn't seem possible.
I'm not sure. Sorry. Maybe my work can spark some thought, though.
Michael, your equation
d - 1/4 - 1/5 - 1/6 = 9 should have been
d - (1/4)d - (1/5)d - (1/6)d = 9 , this times 60
60d - 15d - 12d - 10d = 540
23d = 540
d = 23.478
he had appr. 23.5 ducats
I don't know if a ducat has smaller denominations, but I'm 100% sure of my math
Proof:
1/4 of 23.478 = 5.87
1/5 of 23.478 = 4.70
1/6 of 23.478 = 3.91
He spent 14.48 (add them up)
leaving 23.48-23.48 = 9
thanls for the correction :)
I worked on this awhile and ended up with 23.48 ducats (coins), too. I looked on google and ducats don't come in pieces. "It" was so many troy ounces of gold and was "a" coin used in several European countries.
hmmm, that's weird; I'll tell my teacher about it on Monday because, to me, it sort of makes things confusing, since one would ususally figure you can't exactly have only fractions of a coin... thanks for the info!
This may also be it...
If I had 100 ducats, I would end up with 50 ducats.
100(3/4) = 75
75(4/5) = 60
60(5/6) = 50
Trying similar numbers, the result of all the spending is always 1/2 of what you start with.
That would lead to an answer of 18 ducats. Not sure if this is right again, but it's an idea.
well, the equations make sense and I see no other way of doing so thank you so very much for taking the time to helping me with this math Problem.
Glad to help. :)