Bunuel
The afternoon before a popular amusement park opens for the season, Shirley has a pass that will let her ride each ride continuously without waiting in line. For each of the four most popular rides, the ride will run for an allotted time, then stop to change passengers (in negligible time) and start again. The Sea Serpent runs every 15 minutes starting at 10am. The Wild Mouse runs every 30 minutes starting at 11am. The Raptor runs every 40 minutes starting at noon. And the Mantis runs every 45 minutes starting at 11am. Each ride runs continuously and stops only to let its passengers off and immediately begin again. If Shirley is able to get on her first ride starting at 2:00pm and wants to ride all four rides, but needs 10 minutes between rides to move from one to the next, what is the earliest time she could finish riding all four rides?
A. 4:15pm
B. 4:30pm
C. 4:45pm
D. 5:00pm
E. 5:15pm
Kudos for a correct solution.First thing to note is that all the rides will be in ready to board state at 2PM . how ?
sea serpent @10am + 15*4*4mins = 2PM
Wild mouse @11am + 30*2*3mins = 2PM
Raptor @12PM + 40*3 = 2PM
Manits @11AM + 45*4= 2PM
now i see these four numbers and quickly identified that we have to start with wild mouse as i will get 30,40,45 all in series with 10min gap.
15, 30 , 40 , 45 30(wild mouse) +10(min wait) =40(raptor 40*1)
30+10(wait)+40+10(wait) =90(Mantis 45*2)
30+10(wait)+40+10(wait)+45+10(wait)=145
as 15*10=150 so she has to wait for another 5 mins so finally the the total time taken is
30+10(wait)+40+10(wait)+45+10(wait)+5(min)+15 = 165mins .
2PM + 165mins = 4PM + 45Mins
answer C