Pre+Algebra

When a plane is in a holding pattern*, it flies away from a marked signal point (the outbound leg), makes a 180 degree turn, and flies straight back to the signal point (the inbound leg). When it reaches the signal point again, it makes another 180 degree turn and continues the pattern. We wondered how the pilot knows how far out to fly. Federal Aviation Administration guidelines requires that the //inbound// leg take 60 seconds. When there's no wind, the pilot can just fly out for 60 seconds and know that the inbound leg will take 60 seconds, but if it's a windy day, it's not that easy.

The FAA suggests that the pilots take one "practice loop" and time how long the outbound leg and the inbound leg each take. Then the pilot can use math to determine how long the next outbound leg should be for the inbound leg to take 60 seconds.
 * Question:** If a plane flies the outbound leg in 80 seconds and the inbound leg in 48 seconds, how long should the next outbound leg be so that the inbound will take exactly 60 seconds? Assume that the wind conditions stay the same.

Note: The plane is timed only on the straight part of the outbound leg and the straight part of the inbound leg, //not// during the turns.

*Have you ever been on an airplane when the pilot announces that they haven't been given permission to land yet, and so they'll need to fly a "holding pattern"? A holding pattern is used when there's no available runway. The planes waiting their turn need to stay in the area without getting in each others' way, and it's not like they can just pull over and wait their turn -- they have to keep flying, so they fly in the special pattern described above.
 * Extra**: Some people claim that an easy way to find the needed outbound time is to fly out for 60 seconds, time the inbound leg, then add the difference between the inbound time and 60 seconds to the outbound leg. In other words, if it takes 50 seconds to fly in, the 10 second difference between 50 and 60 is added to the outbound leg. That means the next outbound leg should be 70 seconds in order to have the inbound be 60. Does that method work? Why or why not?