Question:

Two passengers boarding a flight were found to have between them 34.5 kg of luggage. As per the excess luggage policy of the flight operator, the two passengers were made to pay Rs. 3.75 and Rs. 6.00 for the excess weight of their luggage. Later they found out that if the same luggage were to belong to just one person, then the excess luggage fee would have been Rs. 13.50. How much free luggage is allowed for each passenger?

Updated On: Oct 1, 2024
  • 8kg
  • 9kg
  • 6.5kg
  • 7.5kg
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The Correct Option is D

Solution and Explanation

The correct option is (D): 7.5kg
Let ( x ) be the free luggage allowance per passenger (in kg).
Let ( r ) be the rate per kg for excess luggage.
total luggage = 34.5 kg.
first passenger paid for excess luggage=Rs. 3.75
Second passenger paid for excess luggage= Rs. 6.00
excess fee = Rs. 13.50.

For the first passenger: \(( r \times (w_1 - x) = 3.75 )\)
For the second passenger: \(( r \times (w_2 - x) = 6.00 )\)
Combined luggage:\( ( w_1 + w_2 = 34.5 )\)
If combined: \(( r \times (34.5 - x) = 13.50 )\)

Solve for ( r ):
From the combined equation:  \(r \times (34.5 - x) = 13.50\) 

r = \(\frac{13.50}{34.5 - x}\) 

Substitute ( r ) in Individual Equations:

For the first passenger: ( \(\frac{13.50}{34.5 - x} \times (w_1 - x) = 3.75\) )

For the second passenger: (\( \frac{13.50}{34.5 - x} \times (w_2 - x) = 6.00 \))

Express (w_1 ) and (w_2 ):

\(w_1 = \frac{3.75 \times (34.5 - x)}{13.50} + x \))

\(w_2 = \frac{6.00 \times (34.5 - x)}{13.50} + x\) )

Combine (w_1 ) and (w_2 ):

(\( \frac{3.75 \times (34.5 - x)}{13.50} \)+\( x + \frac{6.00 \times (34.5 - x)}{13.50} + x = 34.5\) )

Simplify: ( \(\frac{3.75 + 6.00}{13.50} \times (34.5 - x) + 2x = 34.5\) )

\(\frac{9.75}{13.50} \times (34.5 - x) + 2x = 34.5\) )

\(0.7222 \times (34.5 - x) + 2x = 34.5\) )

Solve for (x):
\(24.9375 - 0.7222x + 2x = 34.5\) )
\(1.2778x = 9.5625\) )
\(x = \frac{9.5625}{1.2778}\) )
(\( x \approx 7.48 \))

Thus, the free luggage allowance for each passenger is approximately 7.5 kg.

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