Question: Jet Airline has a free luggage allowance for its passengers. If any passenger carries excess luggage, it is charged at a constant rate per kg. The total luggage charge paid by Ravi and Pranav is Rs. 1100. If both Ravi and Pranav had carried luggage twice the weight than they actually did, their luggage charges would have been Rs. 2000 and Rs. 1000 respectively. What was the charge levied on Ravi’s luggage?
Explanation:
Let the free luggage allowance be ‘f’ kg. Let the weight of the luggage carried by Ravi be ‘r’ kg and the weight of the luggage carried by Pranav be ‘p’ kg. Thus, the excess luggage weights carried by Ravi and Pranav 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 Ravi 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 Pranav 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.
Ravi’s luggage charge = (r – f)k.
But, according to equation (7), r = 3f. Therefore, Ravi’s luggage charge = 2fk
But, fk = 400. Therefore, Ravi’s luggage charge = Rs. 800.
Hence, option (a).