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.
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.
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. :)
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!
Don't worry, I'll help you break down the problem and guide you through the steps to solve it.
Let's start by assigning a variable to represent the unknown number of ducats in the purse. Let's call it "x".
The problem states that the man spends 1/4, 1/5, and 1/6 of the amount, and 9 ducats remain.
To solve the problem, we'll need to set up an equation based on the information provided.
Step 1: Calculate the amount spent
The man spends 1/4 of the original amount, so the amount spent is (1/4)x.
Then, the man spends 1/5 of the remaining amount, which is (1/5)(x - (1/4)x). Simplifying this gives you (1/5)(3/4)x.
Finally, the man spends 1/6 of the remaining amount, which is (1/6)(x - (1/4)x - (1/5)(3/4)x). Simplifying further gives you (1/6)(5/6)x.
Step 2: Calculate the remaining amount
The problem states that after the man spends the aforementioned fractions of the total amount, 9 ducats remain. Therefore, the remaining amount is 9.
Step 3: Set up the equation
Now, we can set up the equation by equating the amount spent with the remaining amount:
(1/4)x + (1/5)(3/4)x + (1/6)(5/6)x = 9
Step 4: Solve the equation
To solve the equation, you can start by finding a common denominator for the fractions. In this case, the common denominator is 60. After multiplying every term by 60, the equation simplifies to:
15x + 9(3x) + 5(5x) = 540
Simplifying further:
15x + 27x + 25x = 540
67x = 540
Dividing both sides of the equation by 67 gives you the value of x, which represents the initial amount of ducats in the purse.