"In Venice, merchants' sons... normally had only the most superficial contact with the Latin learning of the grammar schools. Their education was almost wholly devoted to their future vocationI've put my answer below the fold.
A prime exhibit here is the fourteenth-century miscellany known as the Zibaldone da Canal. This is a mercahnt's commonplace book, which was elaborated from a student's workbook of mathemtical problems... the Zibaldone (meaning 'Miscellany' or 'Ragbag') places the strongest emphasis on practical arithmetic... As for example (here I cite the admirable translation of J.E. Dotson):
Make me this calculation: 2 merchants have their wool on a ship. One of them put 13 sacks and the other of them put 17 sacks [on board]. And when they had arrived in Venice the captain demanded his freight charges from the merchants and they said to him, 'Take one of our sacks from each of us and sell it and pay our freight costs and return the remainder.' And the captain took 2 of those sacks and sold them and gave 10s. from the proceeds to him who had 13 sacks after the freight had been paid. And he returned 3s. to the man who had 17 sacks and his freight was entirely paid. And the merchants said to the captain, 'We want to know how much you sold the sacks for, and how you calculated what you took from it for freight charges.' [Follows the explanation of how the sum is done].
I come up with 1.75s for the cost of freight per sack, and 32.75s as the amount for which each sack was sold.
Right? (The answer is not given in the book).
My calculations came up with the same answers.
ReplyDelete-Chipper
Same answer. Without resorting to algebra:
ReplyDelete1. The 17-sack man paid 7s more than the 13-sack man (10s-3s)
2. Thus the 4 extra sacks cost 7s, or 7s / 4 sacks = 1.75s/sack.
3. The value of the sack is then
13*1.75 + 10 = 32.75s
17*1.75 + 3 = 32.75s
^ Using that difference that way is such a nice idea. The first thing that comes to mind is solving for the two remainders simultaneously and getting your values for the two variables here.
DeleteThis took me 15 minutes to solve. Not only did I have to resort to algebra (x= selling price per sack) but I had to introduce a second variable besides x (a = transport cost per sack). While busily solving my equations I strongly sensed that there must be a very simple, elegant solution to this. Just couldn't see it. Well done, Mark!
ReplyDelete"The 17-sack man paid 7s more than the 13-sack man (10s-3s)"
ReplyDeleteI'm having a hard time with this. What happened was that the 17-sack man was _paid back_ 7s less than the 13-sack man.
If the money reaped from the sale of each sack by the captain is r, then
the total freight charged would be
2r - (10 + 3), or 2r - 13
Each merchant's fraction sums up to the whole like this:
((2r - 13) * 13)/30 + ((2r - 13) * 17)/30 = 2r - 13
Assuming r = 50, we get
37.7 + 49.7 = 100 - 13
Assuming r = 70, we get
55.03 + 71.97 = 140 - 13
And without another restriction that I don't see in the problem, any specific answer doesn't look possible. Unless we are to assume that the sale was done in whole shillings.
In which case, 2r - 13 must be divisible by 30 (the 13 and 17 are primes).
But that means r has a .5 value attached to it, as 2 * whole# - 13 can't
be an even number and divisible by 30. The lowest of these numbers is r=21.5,
which results in freight charges of
13 + 17 = 43 - 13
So I don't really see a specific answer to this problem.
Lurker111
Okay, late last night this came to me:
DeleteAssuming the captain is an honest man, we do have that the _difference_ between the freight charges for the merchants is 7s, so:
((2r - 13) * 17)/30 - ((2r - 13) * 13)/30 = 7, which becomes
((2r - 13) * 4)/30 = 7, or
2r - 13 = 7 * 30 / 4, or
2r - 13 = 52.5
2r = 65.5, and
r = 32.75,
which agrees with the other results. Sorry. Been off my game.
Lurker111
a - 13b = 10
ReplyDeletea - 17b = 3
Where a is the proceed of the sale of one sack, and b is the freight charge per sack.
Subtracting the second formula from the first:
a - 13b = 10
a - 17b = 3
-------------
4b = 7
And therefore b = 4/7 = 4.75
Now solve for a by substituing the value for b into either of the two equations:
a - 13b = 10
a = 10 + 13(4/7) = 131/4 = 32.75
I actually configured it a bit differently, though it all boils down to the same thing. Since the freight charge (X) was the same for both men, then...
ReplyDelete17X-10 = 13X-3
4X = 7
X = 1.75
This all assumes that the captain is an honest man. :)
ReplyDelete