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#1




Why do most people have trouble with this brain teaser?
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? 
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#2




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.

#3




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 
#4




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.



#5




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. 
#6




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.

#7




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.

#8




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. 
#9




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.



#10




Wow, I guess we ain't so dumb after all.

#11




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.

#12




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. 
#13




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? 
#14




SPOILER:



#15




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.

#16




Polite, but slow. That's a first for me.

#17




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? Ace 
#18




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. 
#19




If you take two more laps at 120, it will average 60 MPH, though. But that wasn't the question.



#20




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. 
#21




...so what happened to the other dollar?
:::d&r::: 
#22




Well, it is a Brain Teaser. If they were clear and unambiguous, then they would just be ordinary questions. The point is to delude you into thinking you have the correct answer, when in fact, you don't, because you didn't read the question carefully enough.

#23




What you have to catch is the 1 mile part. Why is that fact put into the question? Why is 1 mile important? What if it were five miles? If you only take in the 30 mph 1 lap and ? for lap 2 that will lead you to the wrong answer because you didn't take in the 1 mile per lap.

#24




The distance is unimportant. Clearly, I cannot average, after going distance X at Y, go distance X again so as to average 2Y, since I'd already be done.
Try changing the distance to 30 Miles  same answer. RawDuke, I think you hurt your own question! The units you gave are impossible, but many people are stumped by the general puzzlers that involve calculating (seemingly) both time and distance simultaneously. Try this, solvable puzzler. A lap is 50 miles, and Racer A goes 75 MPH full out. Racer B, conserving his Nitrous, goes 50 MPH for one lap. How fast does he have to go on the second lap to catch Racer A? 


#25




So this falls under the heading of "trick question," which people hate. Step back from the keyboard and prepare to receive a Wet Willie you will not soon forget.

#26




ACE, I need to assume that both racers finish the second lap at the same time? If not, racer b could catch racer by flooring it and going, say, a million MPH and catch A in, ummmm, real quick!
Otherwise, at 75 MPH, A will finish 2 laps in 1 1/3 hours. Racer B would finish one lap in one hour, that leaves him 1/3 hour to travel 50 miles, which he could do at 150 MPH. Correct? ETHILRIST, I would be happy to step back and receive that "Wet Willie" and might even enjoy it! Would you first: a. Explain just what a Wet Willie is? b. Explain (and I'm not trying to be rude here, I'm truly curious) why do you think this is a "trick?" What is it about this teaser that is ambigous? I guess I don't see why people feel tricked. 
#27




Your answer is correct.
The original question is set up to "trick" one, IMO, not by the wording, but by the impossible answer, and the use of a track oval, instead of a straight line back and forth. That said, I enjoy a good trick question, though "Infinite" feels a little unsatisfying. Ace 
#28




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But you're right, ACE the distance doesn't matter. This is a very elegant explaination. But I still have to disagree with the idea that this is a "trick" question. To me, this is a "trick" question: "Which weighs more, a pound of feathers or a pound of gold?" To answer this question correctly, you need to know that gold is weighed in Troy measurement, in which there are 12 ounces to a pound. Which means that a pound of feathers weighs more. 
#29




Funny, when I read the thread title, I knew it was either going to be the question about the three men who pay $30 to stay at the hotel, or this one.



#30




The thing to remember is than there are different sorts of averages. For most things the most common average is the mean (add them all up and divide by the number of things).
But average speed means "total distance travelled divided by total time taken" or "speed that if I had gone the whole way at, would have taken me the same time" because this is the only one that's any use. The trick is that this is different to the normal average. Both are averages, but the one wanted here is worked out differently, as many people have pointed out. 
#31




How about the speed of light? No time passes for you as you complete the second lap, so you manage to complete both laps in two minutes. In order to average faster than 60 mph, simply travel faster than the speed of light.
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#32




So the smartass answer to this question is that the second lap is made in 0 seconds. There would be no way for an observer or the driver to discern whether or not the car had made the second lap.
The laws of physics don't apply here according to how the question is worded. As dwc1970 pointed out, you'd have to have a car that could instantly accelerate to 30MPH and travel exactly one mile for the first lap (not to mention we're not taking into account tire friction, air resistance, engine performance consistancy, and inprecise measurement equipment). Even if the required average were dropped to 40MPH for both laps, the second lap would need to be driven at 50MPH and the car would need to instantly accelerate from 30MPH to 50MPH exactly at the finish/start line. Even The Fast and the Furious didn't have cars like that. 
#33




Actually, I don't see any trick in the way the problem is worded. It really is a legitimate questionthere's no trick, just an answer that most don't expect (or even accept, many times). In fact, try this wording:
You drive the first lap averaging (so any time "wasted" for acceleration or whatever is taken into account here) 30 mph. How fast must you average for the second lap in order for the average speed for the entire trip to be 60 mph? I guarantee most people will still answer 90 mph, though the answer is still "impossible" (minus any relativistic effects ). 
#34




Is the 30 constant in the first lap?



#35




The distance is important in that you have to think about the distance the driver has to travel. Yes the answer is the same no matter how big the track but if a person only thinks
1 lap = 30 both laps = 60 so 2nd lap must = 90 is not taking into account the lap = 1 mile in the original question. Because you are not traveling at laps per hour, you doing miles per hour. 
#36




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You're talking a smidge under 60 MPH average, but it's still under. 
#37




If you're interested in why people have trouble with math problems like this one, may I recommend the wonderful book Mathsemantics by Edward MacNeal? It may be out of print, but if you can track down a copy, it's well worth reading.

#38




We actually ask this question in our interviews for graduates. A remarkable number give the wrong answer. Personally I gave the right answer within about 2 secs. And I hadn't heard it before. It just seems obvious to me.
It's obvious because when you average a speed, you should do so weighted by the time you travelled at that speed. For instance if you travel 2 hours at 30mph and 1 hour at 60mph then your average speed is (2x30 + 1x60)/(2+1) {Note this is just another way of saying speed = distance/time} In the question you can't average 30 + 90 to get 60 because you will have travelled one lap in 2 mins and one lap in 40 seconds. Anyone that has ever worked with weighted averages should immediately recognise that that doesn't work. In this instance the average of 30 and 90 is (30x2 + 90x0.6666)/2.6666 pan 
#39




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b. When presented with a word problem, the basic assumption is that there is a way to solve the problem. If, in fact, there is no answer, it's kind of irritating. Suppose, for example, I presented you with a popup ad saying, "Punch the monkey to win $1,000;" if you punched the monkey but didn't win the $1,000 because of some bizarre arbitrary reason, you would be justifiably annoyed. 


#40




Such questions are not about logical thinking or being good at physics or anything else. They are about stating a simple question in the most obscure way possible in order to make a fool out of the person you target with your little practical joke.
How many people would miss it if it were asked more clearly? "If you travel a mile at a rate of 1/2 mile per minute taking two minutes how fast will you have to travel for a second mile to cover a total distance of 2 miles in two minutes?" It is the intentionally poorly asked question not anyone's answer to it that is in error. Someone mentioned that they use this question for interviews for graduates. What?!?! If I were met with a silly practical joke by someone I was interviewing with for any reason I would at least refuse to play and at most walk out. I cannot imagine why you would want to subject a person already in a stressful situation to a pointless amusement for you and you fellow interviewers. I honestly cannot imagine any useful information you could possible expect to garner by asking the question. Playing a practical joke on someone by dropping rotten eggs on them is no worse than playing this kind of practical joke. There are about a dozen ways to write a joke question. Some have a "correct" answer. Some, like this, have no answer. When people hear one they immediately start thinking, "Which kind of joke is it?" not, "What is the right answer?" Even in this thread there was as much discussion about what kind of a trick question it was as there was discussion about what the "right" answer was. Play with these if you want but don't fool yourself into thinking that you are a mental giant and others are idiots just because you got the joke first. 
#41




And divining the obscure is not any hallmark of intelligence?
All I know is.... ME MENTAL GIANT! ! HA! HA! RUN LITTLE PEOPLES FROM MY PUZZLE CRUSHING MIND!! Oh, damn, I think I fooled myself. 
#42




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#43




Hm. I read what Degrance wrote and I have to respectfully disagree that asking a question like that posed in the OP is as bad as dropping rotten eggs on someone. Part of this is because I do not see asking questions like this as "practical jokes."
I like these riddles because if I get the wrong answer, I don't assume it was the asker's intention to make a complete fucking asshole out of me. I think it is just a fun little puzzle that you need to asked in order to appreciate. Lemme take a poll, though  Would you rather be asked more questions like that posed in the OP or have some rotten eggs dropped on you? Tibs. 
#44




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Did the problems say "John traveled 3 miles in 2 hours, which averages to 1.5 miles/hour. What was his average velocity?" Given the rate and the distance, I expect a person to be able to figure out the time. I should not have to explicitly state it because it is right there. The question is phrased as any general 2 part problem, rate, distance, find the missing variable, use it and and the second distance to find the second rate. Basic algebra/physics. There is nothing hidden, there's nothing tricky, and it is very simple math. I see nothing wrong with asking graduate students to do that. I see nothing wrong at all in expecting that potential hires be able to do very basic algebra (but then, I work in tech). The useful information gained is that they're able to work through a problem instead of jumping to a conclusion, and that they have that pesky multiplying thing down pat. Not a trick question at all. 


#45




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#46




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This reminded me of the joke about "Math History" Teaching Math in 1950: A logger sells a truckload of lumber for $100. His cost of production is 4/5 of the price. What is his profit? Teaching Math in 1960: A logger sells a truckload of lumber for $100. His cost of production is 4/5 of the price, or $80. What is his profit? Teaching Math in 1970: A logger exchanges a set "L" of lumber for a set "M" of money. The cardinality of set "M" is 100. Each element is worth one dollar. Make 100 dots representing the elements of the set "M". The set "C", the cost of production, contains 20 fewer points than set "M." Represent the set "C" as a subset of set "M" and answer the following question: What is the cardinality of the set "P" for profits? Teaching Math in 1980: A logger sells a truckload of lumber for $100. Her cost of production is $80 and her profit is $20. Your assignment: Underline the number 20. Teaching Math in 1990: By cutting down beautiful forest trees, the logger makes $20. What do you think of this way of making a living? Topic for class participation after answering the question: How did the forest birds and squirrels feel as the logger cut down the trees? There are no wrong answers. Quote:
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However, everything you need to know is presented (except it's assumed you have a basic math and english knowledge) 
#47




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#48




Here's the general solution, showing that it doesn't matter how long the track is.
Say the distance around the track is d. The time needed for the first lap is d/30 (units of furlongs per fortnight or whatever.) The time needed for two laps at an average of 60 mph is 2d/60 = d/30, exactly the same amount of time as it took to go one lap, so you'd have to do it at infinite (> c) speed. As for travelling at the speed of light  it might work in the reference frame of the traveler, but not in the reference frame of the observer. Nice puzzle. 
#49




After a full day at work and a long commute, my brain is a bit frazzled. Let me get this straight: In order to average 60mph over two miles, you have to travel two miles in two minutes. (Assume instantaneous accelleration, since most of this sort of problem usually uses a "perfect system".) Since you have used two minutes to go one mile, you have no time left to go the remaining mile. If you drove the second mile at 600mph, then it would take you six seconds, right? So you will have made two circuits in 2.1 minutes. That averages to about 57mph.
Right?
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#50




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