The question is not entirely clear so I’ve made some assumptions:
(1) At 11am no seats are occupied, when the first passengers embark.
(2) The exit point is also the entry point for new passengers.
(3) The Ferris wheel’s speed is doubled at exactly 11:15am.
(4) No passenger is allowed on if there’s not enough time for them to complete their ride (one revolution).
(5) At 11.30am when the last two passengers disembark, no seats remain occupied.
The Ferris wheel accommodates 20 passengers on its 10 seats.
Suppose each seat is labelled A, B, C, ..., J.
Each seat will hold two passengers and we can label the initial passengers A1 and A2, B1 and B2, etc.
At 11am the first passengers get on (A1 and A2).
At 11:01 B1 and B2 get on.
At 11:09 J1 and J2 get on. All 10 seats are now full. This is the end of the 1st batch of 20 passengers (20 passengers).
At 11:10 A1 and A2 get off and passengers labelled A3 and A4 take their seat, the start of the 2nd batch of passengers.
At 11:15 the Ferris wheel speed is doubled and F1 and F2 get off while F3 and F4 get on.
At 11:15:30 G1 and G2 get off and G3 and G4 get on.
At 11:17 J1 and J2 and J3 and J4 get on. End of 2nd batch (40 passengers).
At 11:17:30 A3 and A4 get off and A5 and A6 get on. Start of 3rd batch.
At 11:22 J3 and J4 get off and J5 and J6 get on. End of 3rd batch (60 passengers).
At 11:22:30 A5 and A6 get off and A7 and A8 get on. Start of 4th batch.
At 11:25 F5 and F6 get off and F7 and F8 get on. From now on no more passengers are allowed on, because there is insufficient time for them to complete their ride by 11:30am.
At 11:25:30 G5 and G6 get off.
At 11:27 J5 and J6 get off.
At 11:30 F7 and F8 get off leaving the wheel empty.
There are 3 complete batches (60 passengers) and an incomplete 4th batch: 12 passengers. Total 72 passengers.
