Latest Capgemini Aptitude Question SOLUTION: 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 t

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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.

Asked In Capgemini sangram badi (10 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Prithiraj Chowdhury  (10 years ago)

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.

Is this solution Helpfull? Yes (29) | No (2)  

raj sudhee zemeckis  (10 years ago)

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.

Is this solution Helpfull? Yes (7) | No (1)  

VARUN  (10 years ago)

ans(a)verification of options

Is this solution Helpfull? Yes (2) | No (4)  

Sajib Ghosh  (10 years ago)

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..

Is this solution Helpfull? Yes (2) | No (2)  

murthy  (10 years ago)

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

Is this solution Helpfull? Yes (2) | No   

Rajesh keshari swain  (10 years ago)

varun can u explain it....

Is this solution Helpfull? Yes | No (1)  

kumar ranjeet  (10 years ago)

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

Is this solution Helpfull? Yes | No (3)  

raju thammineni  (10 years ago)

100kg+460kg---->Rs 46
100kg+100kg---->Rs 10
100kg+260kg---->Rs 26
for excess 10kg's 1 ruppe extra

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RAMANA  (10 years ago)

up to 100 kg is no charge..after that for every 10 kg they are charging 1$.after 100 460 kg will be there so he paid 46$

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