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.