Let the free luggage allowance be ‘f’ kg. Let the weight of the luggage carried by Ravind be ‘r’ kg and the weight of the luggage carried by Pranas be ‘p’ kg. Thus, the excess luggage weights carried by Ravind and Pranas respectively are (r – f)kg and (p – f)kg.
Thus, the total luggage charge for both would be (r – f)k + (p – f)k if k is the charge per kg.
Thus, (r – f)k + (p – f)k = 1100.
(r + p – 2f)k = 1100 .......................(1)
If Ravind carried twice the luggage weight he actually did, i.e., if he carried 2r kg, then the excess luggage weight he carried would have been 2r – f and the corresponding charge would have been (2r – f) k.
Therefore, (2r – f)k = 2000 ................(2)
Likewise, If Pranas carried twice the luggage he actually did i.e., if he carried 2p kg, then the excess luggage he carried would have been 2p –f and the corresponding charge would have been (2p – f) k.
Therefore, (2p – f)k = 1000 ................(3)
Adding (2) and (3) and simplifying, we get,
(r + p – f)k = 1500 ........................(4)
Dividing (4) by (1) and simplifying, we get,
19f = 4r + 4p .............................(5)
Dividing (2) by (3) and simplifying, we get,
–f = 2r – 4p .............................(6)
Solving (5) and (6) for r, we get,
r = 3f ...................................(7)
Subtracting (1) from (4) and simplifying, we get,
fk = 400.
Ravind’s luggage charge = (r – f)k.
But, according to equation (7), r = 3f. Therefore, Ravind’s luggage charge = 2fk
But, fk = 400. Therefore, Ravind’s luggage charge = Rs. 800.
Answer choice (A)
Correct Answer: Rs. 800