2 passengers have together 560 kgs of luggage and are charged for the excess above the weight allowed at 10$ and 26$. If all the luggage had belonged to one of them he would have to pay 46$. The amount of luggage each passenger is allowed without any charge is
(a) 100 kg
(b) 150 kg
(c) 160 kg
(d) Insufficient data.

• 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.
• 10 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.
  
• 10 years agoHelpfull: Yes(7) No(1)
• ans(a)verification of options
  
• 10 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..
  
  
• 10 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....
• 10 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
• 10 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)