An airport charges a passenger a fixed amount for each kilogram (kg) over the maximum allowed weight his luggage is.
On one trip, they had 270 kg of luggage together. Mr. Bryant was fined $1000, Mr. Cervi was fined $500, and the amount of fine incurred by Mr. Astruc was not known to them.
On another trip where Mr. Cervi could not attend, Mr. Astruc and Mr. Bryant together carried the same amount of luggage (270 kg) in the same relative proportion. Mr. Astruc was charged $962.50 more and Mr. Bryant was charged $787.50 more than the previous trip.
What is the maximum allowed weight, and how much luggage did each passenger carry on each trip?
If a is the weight Astruc originally had, and b is the amount Bryant originally had, and l is the weight limit and r is the rate for each kg above the limit and k is the ratio of Astruc's and Bryant's later amount to their previous amount, then we need to solve:
(b-l)r = 1000
(270-a-b-l)r = 500
(kb-l)r = 1787.5
(ka-l)r = (a-l)r + 962.5
I set up an Excel spreadsheet as below, where B3 has the formula for B's original fine, B4 has C's original fine, B5 has B's second fine, B6 has the excess of A's second fine over his first and B7 has the total weight on the second trip. The initial trial values (guesses) were put into row 2, and solver was used to set B7 to 270 subject to the constraints that B3 - B6 have the values 1000, 500, 1787.5 and 962.5 respectively.
A B C D E
1 a b l r k
2 90 90 60 100 1.5
3 =(B2-C2)*D2
4 =(270-A2-B2-C2)*D2
5 =(E2*B2-C2)*D2
6 =(E2*A2-C2)*D2-(A2-C2)*D2
7 =E2*(A2+B2)
The result of using the solver was
a b l r k
110 90 50 25 1.35
1000
500
1787.5
962.5
270
indicating that the maximum weight was 50 kg, and that A, B and C carried 110 kg, 90 kg and 70 kg respectively on the first trip. A and B carried 1.35 times as much on the second trip, that is, 148.5 kg and 121.5 kg.
Incidentally the fine for excess baggage was $25/kg.
Edited on May 29, 2006, 2:38 pm
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Posted by Charlie
on 2006-05-29 14:35:08 |
spreadsheet solution (spoiler)
