A bus travels a journey in three segments
i.e. 'x' km, (x + 11) km, and (x + 44) km in the same order. After each segment, it doubles its speed in comparison to the previous segment. The total time for this journey is 12 hours. On the return journey, the bus travels the same segments in reverse order: (x + 44) km, (x + 11) km, and 'x' km, again doubling its speed after each segment. The total time for the return trip is 15 hours. Find the time taken by the bus to cover (x - 11) at its usual speed. 

Let the usual speed of bus be 's' km/hr. Forward journey: Time taken to cover first segment = (x/s) hours Time taken to cover second segment = {(x + 11) /2s} hours Time taken to cover third segment = {(x + 44) /4s} hours ATQ; (x/s) + {(x + 11) /2s} + {(x + 44) /4s} = 12 {4x + 2(x + 11) + (x + 44)} = 48s (4x + 2x + 22 + x + 44) = 48s 7x + 66 = 48s ---------------(i) Return journey: Time taken to cover first segment = {(x + 44) /s} hours Time taken to cover second segment = {(x + 11) /2s} hours Time taken to cover third segment = (x/4s) hours Again; {(x + 44) /s} + {(x + 11) /2s} + (x/4s) = 15 {4(x + 44) + 2(x + 11) + x} = 60s (4x + 176 + 2x + 22 + x) = 60s 7x + 198 = 60s -------------(ii) Subtract equation (i) from (ii) : (7x + 198) - (7x + 66) = 60s - 48s 132 = 12s 's' = 11 kmph Putting the value of 's' in equation (i), we get; 7x + 66 = 48 × 11 7x + 66 = 528 7x = 462 'x' = 66 km Required Time = {(x - 11) ÷ 11} = (55/11) = 5 hours Hence, option b.