An elevator named L in "63 Building" in Seoul is notorious for its low speed. It takes 3 seconds for L just to move one floor. It takes another 20 seconds for it to have its door open and close automatically.
What is even more frustrating, is that the elevator travels only to odd-numbered floors, and the doors will automatically open and close at all odd numbered floors.
How long will it take (in seconds) for L leaving the 1st floor "just to arrive" at the 35th floor?
Note:
Do not include the time taken for the doors to open / close at the 1st or 35th floor.
To calculate the time it takes for L to travel from the 1st floor to the 35th floor, we need to consider the time it takes to go up each floor and the time it takes for the elevator itself to move between floors.
Since it takes 3 seconds for L to move one floor, we need to calculate the number of floors L has to travel to reach the 35th floor. We can do this by subtracting the starting floor (1st floor) from the destination floor (35th floor):
Number of floors = 35 - 1 = 34 floors
Since L only stops at odd-numbered floors, we need to divide the total number of floors by 2 to get the number of "odd-numbered floors" L will stop at:
Number of odd-numbered floors = 34 / 2 = 17 floors
Now we can calculate the total time it takes for L to travel to the 35th floor, excluding the time taken for the doors to open/close:
Time taken for L to travel to the 35th floor = (Time taken to move one floor x Number of odd-numbered floors) + Time taken for the doors to open/close
Time taken for L to travel one floor = 3 seconds
Time taken for the doors to open/close = 20 seconds
Plugging in the values:
Time taken for L to travel to the 35th floor = (3 seconds x 17 floors) + 20 seconds
= 51 seconds + 20 seconds
= 71 seconds
Therefore, it will take 71 seconds for L to travel from the 1st floor to the 35th floor, excluding the time taken for the doors to open/close.