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A Permissible Weight Problem (Posted on 2006-05-29) Difficulty: 3 of 5
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?

See The Solution Submitted by K Sengupta    
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Solution Comment 2 of 2 |

Let the three persons be A,B & C for convenience.

The weight of baggage of A, B & C be a, xa & ya where x & y are fraction multipliers with respect to a.

Let r be the rate of fine per kg & p be the permissible limit of weight. Let initial fine paid by A be Z.

then

r(a-p)=Z

r(xa-p)=1000

r(ya-p)=500

a+xa+ya=270

In the second instant the weights with A & B are

270a/(270-ya) & 270xa/(270-ya) respectively

hence

r( 270a/(270-ya) -p) = Z + 962.5

r(270xa/(270-ya) -p) = 1787.5

After solving series of equations by elimination we get

x =9/11

y=7/11

a=110

p=50

Z=1500

&

r=25

The permissable weight limit is 50 kgs, The fine rate is 25$/kg, The initial fine paid by A is 1500$.

The weights carried by A,B & C are

First trip - 110, 90, 70

Second trip - 148.5, 121.5 & nil respectively.

Which airlines was it?  


  Posted by Salil on 2006-05-30 03:44:04
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