I’m sure you’ve had this discussion before, but I couldn’t find it in a search.
First, the puzzle:
You’re driving a race car on a one mile oval track. You drive one lap at 30 MPH. How fast do you have to drive the second lap in order to average 60 MPH for BOTH laps?
I first saw this in Ask Marilyn several years ago. (Marilyn, BTW, along with Isaac Asimov and James Randi are my heroes!) She got plenty of letters from people, many angry, because they couldn’t understand the answer.
When I present this to people I know, the immediate (and wrong) answer is “90 MPH.” My older brother (PHD) is the only person I know who got it right.
I won’t give the answer yet, but my main question is Why do people have such a problem figuring this one out?
Because we see 30 + 90 = 120 ; 120/2 = 60. If there’s some kind of trick answer, people have problems with it because they hate trick answers. If it’s a semantic trick, like, “You can’t average anything over 30 mph for the first lap because we’ve already stated the speed” it’s just gonna annoy people something fierce.
** Ethilrist ** 's summed it up. If it’s a trick question, I’m gonna beat someone. I hate those.
On the other hand, if it’s genuine math, I’m stumped. I thought it was 90.
Kn(dumber than she looks)ckers
The only way that this could work using the simple and obvious answer is to have a car that could instantly accelerate straight to 90 mph after going 30 on the first lap, which would be impossible. One has to account for acceleration, which complicates the problem. One would have to know the rate of acceleration to truly get an accurate average, and unfortunately my math is not advanced enough to make such a calculation.
It’s not a trick question, it is genuine math.
I know, your initial reaction is “30+09 = 120 and devided by 2 is 60” but you don’t compute averages by comparing rates of speed, rather you calculate distance driven divided by time.
To average 60 mph (1 mile/min.), you must drive two laps (2 miles) in 2 minutes. Lap 1 at 30 mph (2 mins/mile) takes you two minutes. Therefore you must finish the second lap in 0 minutes. No problem.
It is because it takes you two minutes to drive one mile at 30 miles per hour, and no matter how much faster you drive the next lap, it’s still going to take more time to go another mile? To average 60 MPH you have to cover two miles in two minutes. Just a wild guess.
Okay, if you average 60mph for both laps, that means you have to take 2 minutes total to drive both laps. (60mph = 1 mile per minute. Each lap is a mile, so two laps = two miles = 2 minutes.)
In the first lap, we went 30mph. 30mph = 1 mile in 2 minutes.
OOPS! We ALREADY took the 2 minutes in the first lap! We can only hit that 60mph if we take 0 minutes to do it. Unless we can travel instantaneously, that ain’t happening.
Even if we drive 90mph in the second lap, 90mph = 1.5 miles per minute. So it’ll take 40 seconds to do the lap. You would have taken a total of 2 minutes, 40 seconds to do both laps. 2:40 = 160 seconds. Divide by 2 and that’s 80 seconds per lap, which is over 60mph.
I’m guessing the answer isn’t 60 mph because your second lap will be quicker than your first, hence you aren’t travelling 90 mph for the same length of time that you are travelling at 30 mph.
Wow, I guess we ain’t so dumb after all.
Very tricky question. I got the wrong (90MPH) question at first, then had to sit down and work out the math to get the real answer. And it’s not a trick question.
I don’t know the answer to this question, but my guess would be that you can’t average 60 mph.
Ideally, at 30mph, it would take 2 minutes to get around the track. While at 60 mph it would only take one minute to get around the track. Therefore, if your average mph was 60, you would have already completed the 2nd lap, while if you were going 30 mph would just be completing the first. Do I get a cookie?
Why people would get this wrong is probably because the average of 30 and 90 is 60, but this fails to account for time. It’s really just a question of how you approach problems, conventionally or creatively. The experienced problem solver knows not to jump to conclusions, while most others will look for easy answers.
ANDREWL, I like that. It’s not a trick question, but it’s a tricky question!
GENSERIC, not such a wild guess, and I guess I’m impressed too! Not such dumb peoples here!
I’m curious, though. If you mention this puzzle to your friends, how many instantly say “90?” And how many understand the explaination when you say they’re wrong?
Almost everyone who thinks about it for ten seconds will say 90 MPH. (Unless they are incapable of any math at all.) It’s only those who are knowledgeable in physics and math, or those who know that 90 is not the correct answer and give it more thought (and understand what the concept of Miles Per Hour, really means) who get it right.
Polite, but slow. That’s a first for me.
Put me down for “infinite.” Averaging 60 MPH would mean you’ve already completed both laps at T=1, whereas in the puzzle you’re just finishing your first lap.
If that’s not a trick question, I don’t know what is. Are you sure the puzzle’s worded correctly?
ACE0SPADES, either infinite or not possible are correct.
I guess “trick question” depends on your definition. I don’t see it as a trick, because all the information you need is provided. “Tricky,” yes, because the easy answer is incorrect, but that’s true with most brain teasers.
If you take two more laps at 120, it will average 60 MPH, though. But that wasn’t the question.
It is slightly angering because it’s a trick question. It’s only tricky because of the way it’s phrased.
It’s one of these “It sounds like it’s asking one thing, but it’s really asking another” sort of questions.
It has little to do with math, and everything to do with semantic analysis. The fact that people not to get it says to me that there is poor communication going on when the question is asked, and the failing is on the part of the asker.
It seems as if the question is asking for the average of two seperate units (laps around the track) while in fact it is asking for the average rate of a single unit.