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The question is from CAT Ratio, Mixtures and Averages. It is all about forming and solving linear equations. CAT exam is known to test on basics rather than high funda ideas. A range of CAT questions can be asked from Ratios and Proportions, Mixtures, Alligations and Averages. Make sure you master the topics. 2IIMs CAT questions bank provides you with CAT questions that can help you gear for CAT Exam CAT 2020.

Question 8: Janta 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 Rs. 1100. If both Ravind and Pranas 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 Ravind’s luggage?

  1. Rs. 800
  2. Rs. 700
  3. Rs. 600
  4. Rs. 900

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Explanatory Answer

Method of solving this CAT Question from Ratio, Mixtures and Averages: How to solve a question that involves a fixed and variable component.

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.

The question is "What was the charge levied on Ravind’s luggage?"

Hence, the answer is "Rs. 800".

Choice A is the correct answer.


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