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.
Solve the following problem taken from Treviso Arithmetic, a math textbook published in 1478 by an unknown Italian author:
A man finds a purse with an unknown number of ducats in it. After he spends 1/4, 1/5 and 1/6 of the amount, 9 ducats remain. It is required to find out how much money was in the purse.
I'm a little bit vpnfused by the problem... am I spending 1/5 of the remaining 3/4? I just don't know where to even begin.
8 answers
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.
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. :)
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
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!
4 answers