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| A Permissible Weight Problem (Posted on 2006-05-29) |
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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?
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Submitted by K Sengupta
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Solution:
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The total amount of luggage borne respectively by Mr. Astruc, Mr.Bryant and Mr. Cervi in the first trip were 110 kg, 90 kg and 70 kg
The respective amount of luggage borne by Mr. Astruc and Mr.Bryant in the second trip were 148.5 kg and 121.5 kg.
The maximum amount of luggage which was allowed free of charge for each passenger = 50 kg.
The total amount of fine incurred by Mr. Astruc in the first trip = $1500.
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FIRST METHOD (Submitted by Salil):
Let the three persons be A,B & C for convenience.
The weight of baggage of A, B & C be a, xa & ya where x & y are fraction multipliers with respect to a.
Let r be the rate of fine per kg & p be the permissible limit of weight. Let initial fine paid by A be Z.
Then:
r(a-p)=Z
r(xa-p)=1000
r(ya-p)=500
a+xa+ya=270
In the second instant the weights with A & B are:
270a/(270-ya) & 270xa/(270-ya) respectively.
Hence:
r( 270a/(270-ya) -p) = Z + 962.5
r(270xa/(270-ya) -p) = 1787.5
After solving series of equations by elimination we get:
x =9/11
y=7/11
a=110
p=50
Z=1500
&
r=25
The permissable weight limit is 50 kgs, The fine rate is 25$/kg, The initial fine paid by A is 1500$.
The weights carried by A,B & C are:
First trip : 110, 90 & 70 respectively; and:
Second trip : 148.5, 121.5 & nil respectively.
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SECOND METHOD (Submitted by Charlie):
The spreadsheet solution by Charlie is given here.
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RIGOROUS METHOD:
Let the weight of the luggage each passenger is allowed to carry free of charge be x kg. and the fine incurred by Mr. Astruc for carrying excess luggage be Y dollars. Accordingly, the fine for (270 -3x) kg of extra luggage = $(Y+1000+500)=$(Y+1500) so that the fine for 1 kg of extra luggage is equivalent to $(Y+1500)/(270 - 3x)--------(i)
Again, we observe that the fine for (270 - 2x) kg of extra luggage is
$(Y +962.5 + 1787.5)= $(Y+2750) so that, the fine for 1 kg of extra luggage is equivalent to $(Y+2750)/(270 - 2x)---------------(ii)
Consequently:
(Y+1500)/(270 - 3x)= (Y+2750)/(270 - 2x)
or, x*Y = 337500 - 5250*x
or, x = 337500/(Y+5250)-------------------(iii)
Accordingly, the weight of the luggage each passenger is allowed to carry free of charge is 337500/(Y+5250)kg. Substituting in (ii) the obtained valur of x in terms of (iii), we obtain the fine corresponding to 1 kg. of luggage as $(1500+Y)/(270 - 3x) = $(5250 + Y)/270 (upon simplification) giving the fine for 270/(5250+Y) kg of extra luggage as $1 so that;
Amount of luggage borne by Mr. Astruc is (337500 + 270*Y)/(Y + 5250) Kg.
Amount of luggage borne by Mr. Bryant is (337500 + 270*1000)/(Y + 5250) = 607500/(Y+5250) kg.
Amount of luggage borne by Mr. Cervi is (337500 + 270*500)/(Y + 5250) = 402500/(Y+5250) kg.
Hence, the ratio between the original amount of luggage borne by Mr. Astruc and Mr. Bryant is (337500 + 270Y)/607500
= (Y+1250)/ 2250. -----------------------(iv)
Now in the second trip, no luggage was borne by Mr.Cervi since he was indisposed and unable to accompany his colleagues. Also, it follows from the said condition that the amount of luggage borne by Mr.Astruc is {337500 + 270(Y + 962.5)}/(Y+5250) kg = 270(Y + 2212.5)/(Y+5250) kg and the amount of luggage borne by Mr. Bryant is {337500 + 270*1787.5}/(Y+5250) Kg = 270* 3037.5/(Y+5250); so that, the ratio between the amount of luggage borne by Mr. Astruc and Mr. Bryant is equal to (Y + 2212.5)/3037.5 ----------------------(v)
From (iv) and (v) we obtain:
(Y + 1250)/ 2250 = (Y + 2212.5)/3037.5 giving Y = 1500; so that X = 337500/(Y + 5250) = 50 kg.
Consequently:
The total amount of luggage borne by Mr. Astruc in the first trip = (337500 + 270Y)/(Y+5250) = 110 kg
The total amount of luggage borne by Mr. Bryant in the first trip = 607500/6750 = 90 kg
The total amount of luggage borne by Mr. Cervi in the first trip = 402500/6750 = 70 kg
The total amount of luggage borne by Mr. Astruc in the second trip = 270*(11/20) =148.5 kg
The total amount of luggage borne by Mr. Bryant in the second trip = 270*(9/20)= 121.5 kg.
The maximum amount of luggage which was allowed free of charge for each passenger = 50 Kg.
The total amount of fine incurred by Mr.Astruc in the first trip = $1500.
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