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Kaushik786
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18 Mar 2020, 21:26
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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44% (03:13) correct
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Zanta 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 Ravind Jekriwal and Pranas Shubhan is $1100. If both Ravind and Pranas had carried luggage twice the weight than they actually did, their luggage charges would have been $2000 and $1000 respectively. What was the charge levied on Ravind’s luggage?
A. $800
B. $700
C. $600
D. $900
E. $500
A. $800
B. $700
C. $600
D. $900
E. $500
yashikaaggarwal
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18 Mar 2020, 22:52
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Kaushik786
There must be some information missing can you recheck the answer?
The only possible solution with the given equation is 500.
800 is far from cry.
Kindly recheck it
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Kaushik786
Joined: 20 May 2018
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18 Mar 2020, 23:05
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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 = $800.
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 = $800.
