2 passengers have together 560 kgs of luggage and are charged for the excess above the weight allowed at 10

If we consider option a as answer that means the amount of luggage each passenger is allowed without any charge is 100kg.
A person having 560 kg luggage has to pay extra charge for
remaining 460 kg that is $46.
i.e excess charge for per kg excess luggage is $(1/10).
hence for $10 excess charge when extra luggage is 100kg and for $26 has to pay for 260kg.
total luggage=(100+100(excess)+100+260(excess))kg=560kg.
11 years agoHelpfull: Yes(29) No(2)

answer to Q type is easiest way for this..
first lets assume 100 is the allowed weight.

for 2 passengers for initial condition (100+x)+(100+y) where x and y are xtra wts of 1st n 2nd person respectively.

charges are x=10$;y=26$; and x+y(in weight)=360); and x alone is charged 46$ that means 460kg=46$. therefore for each 10 kg they charge 1$.

verifying it in first equation gives x+y=360 ,in $ 36.. satisfied.

answer is 100.
11 years agoHelpfull: Yes(7) No(1)
ans(a)verification of options
11 years agoHelpfull: Yes(2) No(4)
let,The 1st passenger's weight is x and 2nd passenger's weight is y
so, x+y=560 ---->(1)
=>x=560-y
and let maximum weight each passenger can carry is z
so, according to the second condition if one passenger carry 560 kgs weight then he have to pay 46$

so, for carrying (560-z) kgs weight he have to pay 46$
similarly for 2nd passenger carrying (y-z) kgs he have to pay $26
so,we can write[*Here actually we can type :)*]
(560-z)/(y-z)= 46/26=23/13
=>23*y-10*z=13*560---->(2)

similarly for 2nd passenger carrying (x-z)=(560-y-z)[from eqn (1)] kgs he have to pay $10
so,(560-z)/(560-y-z)= 46/10=23/5
=>23*y+18*z=18*560 ----->(3)

after solving equation (2) and (3) we get,
z=100

so,each passenger can carry 100 kg luggage without any charges..
11 years agoHelpfull: Yes(2) No(2)
WE CAN SOLVE IT EASILY BY USING MIX ALLEGATION METHOD.
10 26 SUBTRACT REVERSLY i.e 46-10=36 and 46-26=20.
46
20 36
=10 -26
------------
10 10
so each one have 100kg
10 years agoHelpfull: Yes(2) No(0)
varun can u explain it....
11 years agoHelpfull: Yes(0) No(1)
a is max allowed weight
560-2a--->(10+26)-->36$...(1)
560-a---->46$...(2)
2*(2)-(1) we get
560--->56$
10--->1$
so max allowed weight, W=560-46*10=100
11 years agoHelpfull: Yes(0) No(3)
100kg+460kg---->Rs 46
100kg+100kg---->Rs 10
100kg+260kg---->Rs 26
for excess 10kg's 1 ruppe extra
10 years agoHelpfull: Yes(0) No(0)
up to 100 kg is no charge..after that for every 10 kg they are charging 1$.after 100 460 kg will be there so he paid 46$
10 years agoHelpfull: Yes(0) No(0)

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