And the answer is
Indeterminate, Infinity, Undefined and WARP SPEED.
Lacking a WARP key/function on my calculator Im unsure but clearly that would be VERY, very close.
Like with horseshoes and hand grenades, close is good enough. 90 is NOT close.
Math Problem Supposedly Almost Fooled Einstein
https://www.menshealth.com/trending-news/a30985688/math-puzzle-albert-einstein-max-wertheimer/
In 1934, psychologist Max Wertheimer sent a letter to his friend, the physicist Albert Einstein, with the following puzzle enclosed.
There’s an old car that needs to go up and down a hill. The hill is 1 mile going up, and 1 mile going down . Because the car is old, it can only average a speed of 15 mph during the ascent, but may be able to go faster during the descent.
The question is: how fast must the car be going downhill, in order for its speed to reach an average of 30 mph for the entire 2-mile journey?
At the time Einstein received the letter, he had already been honored with the Nobel Prize for Physics, and come up with his famous E = mc2 equation. So this should have been super simple for him to figure out, right? Apparently not. According to German psychologist Gerd Gigerenzer’s book Risk Savvy: How to Make Good Decisions, Einstein wrote that he didn’t see the “trick” until he had already calculated the answer...
The car in question is a Rolls Canardly and the laws of physics don't apply.
The Rolls Canardly rolls down the driveway but canardly drive back up.......
That happened to me as well . I got 2d=60(d/30 +t2) Where t2 is the time of the return and d=the distance traveled one way. That works out to t2=0 . That is when I saw the trick! When you know the math necessary to plow through a problem you usually just work through it mathematically until you've come up with an answer. It's almost an automatic process.
Actually, you left a hole in your statement of the problem.
The Einstein version explicitly specifies 1 mile up and 1 mile back. Your version does not specify the length of the route, and does not specify that the return trip is the same distance. So I could take the direct route getting there at 30 mph, and a much longer alternate route getting back, which would allow me an average speed of 60mph
Even more so if considered old back then when the question was given in '34.