Fun math problem that stumps everyone

The typical guess of 90 would answer a different question, which would be average speed per mile. I’m always fascinated by the human brain - we think and perform multiple functions in fractions of a second, while the brain is also managing our breathing, heart rate, movement, etc. But in the midst of this incredible work, some thought processes are way off due to flawed logic like in this example. I guess there’s a tendency to gravitate to a simpler, familiar calculation.
 
If you did the first mile in two minutes (30 mph) and the second mile in 40 seconds (90 mph), wouldn't that make the average for those two miles 45 mph?

2 miles @ 60 mph would take 120 seconds
2 miles @ 45 mph would take 160 seconds (2:40)

If you got up to 120 mph for the second mile, that would take 30 seconds, to do that second mile. So you'd be at 150 seconds, total for the two miles. That would only get you to an average of 48 mph.

I've seen a similar thing done, with fuel economy. If a car gets 10 mpg for 100 miles, and then 20 mpg for 100 miles, its average fuel economy isn't 15. It's 13.333...
With something like fuel economy, it's called a weighted average. I guess it's the same concept with time? In my instance above, you could phrase it with, if a car gets 10 mpg over 100 miles, what economy would it need to achieve, to average 21 mpg over 200 miles? It can't be done. I would've said 20 mpg, but that, technically, COULD be done, if you wanted to push the car for the second 100 miles. Or towed it with an electric vehicle that was charged from a source that uses no fossil fuels whatsoever :p
 
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"f you travel at a speed of 30mph in the 1st mile,....."

I don't have a solution.
The question does not indicate how much time was spent at 30MPH.
It does not say you traveled the entire mile at 30 MPH.
Could have transitioned the 30 MPH in 5 seconds for example.

There could be many solutions. Why am I worng ?
 
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"f you travel at a speed of 30mph in the 1st mile,....."

I don't have a solution.
The question does not indicate how much time was spent at 30MPH.
It does not say you traveled the entire mile at 30 MPH.
Could have transitioned the 30 MPH in 5 seconds for example. ....

Ok. I take it this way. Driver traveled at 30 mph in the first mile. Let's assume that is the average.

So that means his/her speed at the end of the first mile is 60 mph since the starting speed is 0 mph.

Problem states that the driver averages 60 mph over 2 miles. Since the driver averaged 30 mph in the first mile, ... .

Interesting. The problem doesn't say that the car averages 30 mph in the first mile - only that the car travels "at a speed of 30 mph in the first mile". So the 30 mph in the first mile is not necessarily the average. ....

OK, the question could have been worded more specifically. But I think you have to take the first part at face value, that is means the driver averaged 30 mph for the first mile. How he did that is irrelevant, IOW, the car could stop, speed up, slow down. etc.

If you don't take it at face value that it is an average, there is no problem to be solved, it is completely undefined. To calculate an overall average, you need to average in the first part. If you are going to say it only means that he reached 30 mph at some part of the first mile, it's complete gibberish, not a math problem at all.

It would be like saying "How long is a string?" is a math problem. It's not.

For those who are defending their wrong answer, try explaining it the other way 'round, working backwards from your 'answer'. Assign the times for each mile at that speed, and you'll see it doesn't work - you can't average 60 mph after traveling the first mile at 30 mph, at any speed.

-ERD50
 
I answered: "For me, there is no answer. ;)" with the wink, Rather than give away the answer directly, so others would have a chance to try it.

It's all just for fun here.

Reminds me of my very first University Chemistry exam (so, I've been in school for what, maybe 5 weeks. Professors were gods and TA's were already "old.") The question that got me was a molarity question: Maybe something like how much sodium chloride must you add to 1 1/2 liters of water to achieve a 22 molar solution. I recall the answer kept coming out way more sodium chloride than would even fit in a 1 1/2 liter vessel - let alone dissolve in it. So I just calculated the answer and noted that it couldn't actually be done. The question WAS actually a mistake by the TA who put the test together, but my answer was fully acceptable upon grading. YMMV
 
Too much stress before morning coffee! :)
 
I've seen that problem before, a long time ago. Took about 30 seconds to figure out the answer (no answer). The main thing with word problems is figure out what is being asked. One can use algebra I suppose, or one could figure out how long it takes to go the first mile (2 minutes) and then figure out how much time you have left to go the second mile to average 60 MPH (zero minutes). We do this all the time on freeway driving. If I am going 65 MPH and my destination is 60 miles away, I know that 60 MPH = one mile per minute, so it will take slightly under an hour.
 
I've seen that problem before, a long time ago. Took about 30 seconds to figure out the answer (no answer). The main thing with word problems is figure out what is being asked. One can use algebra I suppose, or one could figure out how long it takes to go the first mile (2 minutes) and then figure out how much time you have left to go the second mile to average 60 MPH (zero minutes). We do this all the time on freeway driving. If I am going 65 MPH and my destination is 60 miles away, I know that 60 MPH = one mile per minute, so it will take slightly under an hour.

You forgot the most important part of the equation. How much road work will you encounter? :facepalm:

I used to hate to drive the Interstate system in the summer because of all the cone-zones. Here in Paradise, it's always summer so it's always road-work time. :(
 
I love all the different ways of arriving at the (non) answer!

Knowing it was going to be more than meets the eye, I took it as a typical navigation problem and applied 60D=S*T, the trusty "60 D Street" formula. Getting 2 minutes for the first leg and 1-1/3 minutes for the total two-mile trip shows there's no solution.

But of course, navigators are known to be smarter than Einstein ;)

Now if we want to take into account relativity, you might have to ask about gravitational effects and frames of reference. I'll have to ponder that one and see if there's a solution there.
 
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This thread is a great example of "Thinking, Fast and Slow".

People see the number "30" and "find the number to average 60" and go into "fast thinking" mode. "I know how to calculate averages!", w/o doing the "slow thinking" to determine if that familiar, easy answer really applies to this situation.


https://en.wikipedia.org/wiki/Thinking,_Fast_and_Slow

Thinking, Fast and Slow is a 2011 book by Israeli-American psychologist Daniel Kahneman.


Thinking, Fast and Slow is a 2011 book by Israeli-American psychologist Daniel Kahneman.

The book's main thesis is that of a dichotomy between two modes of thought: "System 1" is fast, instinctive and emotional; "System 2" is slower, more deliberative, and more logical. The book delineates rational and non-rational motivations or triggers associated with each type of thinking process, and how they complement each other, starting with Kahneman's own research on loss aversion. From framing choices to people's tendency to replace a difficult question with one which is easy to answer, the book summarizes several decades of research to suggest that people have too much confidence in human judgement.[1]

-ERD50
 
Ok. I take it this way. Driver traveled at 30 mph in the first mile. Let's assume that is the average.

So that means his/her speed at the end of the first mile is 60 mph since the starting speed is 0 mph. ...

What leads you to say that?

First, we don't really know that the starting speed is 0 mph. This could just be a one mile snippet of a longer trip.

But even if we assume a start from a standstill, there is nothing to say the car needs to end at 60 mph to average 30 mph. There's an infinite number of ways to average 30 mph over one mile. Again, 30 mph average is one mile in two minutes.

One example - the car could stay at 0 for 1/2 minute. [edit - I screwed this up, I applied it to the 2 miles, but we are talking about 30 mph average on the 1st mile - will be back with a correction in a minute...] [-]That leaves 1.5 minutes to cover the 2 miles. Which is .75 minutes per mile, times 60 minutes/hour is 45 miles per hour.[/-] [correction: This leaves 1.5 minutes to cover the one mile. One mile/ 1.5 minutes = 40 mph] No 60 mph is involved. Infinite combinations could be made, and you'd need to use calculus for many of those.


Ok. I take it this way. Driver traveled at 30 mph in the first mile. Let's assume that is the average.

....

Since the driver was driving 60 mph at the beginning of mile 2, the driver needs to get going 120 mph at the end of mile 2 to average 90 mph in the 2nd mile.

The answer is 120 mph.

Again, simple averages aren't the answer, but the logic is still flawed here (as above) that a 60 mph start requires a final speed of 120 mph to average 90 mph.

-ERD50
 
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You forgot the most important part of the equation. How much road work will you encounter? :facepalm:

I used to hate to drive the Interstate system in the summer because of all the cone-zones. Here in Paradise, it's always summer so it's always road-work time. :(

Which begs the question, how can Hawaii have interstate highways :D
 
Ok. I take it this way. Driver traveled at 30 mph in the first mile. Let's assume that is the average.

So that means his/her speed at the end of the first mile is 60 mph since the starting speed is 0 mph.
That's not how averages work.
 
It pleases me to say my son quickly replied to my email to him with the right answer. Of course, that pleasure has come at the cost of four years of undergraduate, 4 years of medical school, 2 years for MS in Education.

Follow up math question - How wealthy would I be today had I not spent all that money on Junior?
 
Verbal math problems often contain absurd assertions like "John buys 60 cantaloupes at the supermarket." You just have to accept it and work from the statements as given.

To me, the non-mathematician, "The driver averaged 30 mph for the first mile" is the same as saying "The one-mile trip took 2 minutes."

OK, the question could have been worded more specifically. But I think you have to take the first part at face value, that is means the driver averaged 30 mph for the first mile. How he did that is irrelevant, IOW, the car could stop, speed up, slow down. etc.

If you don't take it at face value that it is an average, there is no problem to be solved, it is completely undefined. To calculate an overall average, you need to average in the first part. If you are going to say it only means that he reached 30 mph at some part of the first mile, it's complete gibberish, not a math problem at all.

It would be like saying "How long is a string?" is a math problem. It's not.

For those who are defending their wrong answer, try explaining it the other way 'round, working backwards from your 'answer'. Assign the times for each mile at that speed, and you'll see it doesn't work - you can't average 60 mph after traveling the first mile at 30 mph, at any speed.

-ERD50
 
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If time stands still at the speed of light, the second mile would have to be completed at the speed of light.
 
How about another fun math question?
For a long trip, I bring engine coolant and distilled water in case I have to refill.
The specification is to mix 50% engine coolant and 50% distilled water.
I don't want to bring measuring cup.
So, how do I get exact 50% engine coolant and 50% distilled water on the road?
 
How about another fun math question?
For a long trip, I bring engine coolant and distilled water in case I have to refill.
The specification is to mix 50% engine coolant and 50% distilled water.
I don't want to bring measuring cup.
So, how do I get exact 50% engine coolant and 50% distilled water on the road?

If the coolant and water are both in gallon jugs to start with, get a two gallon jug and pour them both in. Basically, mix before you go.
 
Assuming we are talking about a mix/refill for the radiator, I'd fill as many empty beer bottles with distilled water as needed and the same number of beer bottles with engine coolant. (Should have plenty after a long drive :)) I don't throw my empties out on the highway in case just such a need arises. (Anyway, in Texas we don't like folks to litter) But then I wouldn't pour it on the "road" as you seem to suggest in your problem but instead in the radiator.
 
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Which begs the question, how can Hawaii have interstate highways :D

Heh, heh, that was the question on my lips when we first visited. It was explained that the money for original construction was authorized under the same law(s) that provided Interstate roads on the mainland. Therefore, the roads built WERE designated as part of the Interstate system.

Of course, NO one here actually calls them interstates. We call them "freeways" or H-1 through H-3. Interesting aside, H-3 is the most expensive (per mile) "Interstate Highway" ever built. IIRC it was something like $80,000,000 per mile. Two reasons for the most-expensive designation. 1) The actual terrain is daunting and required a major tunnel to be built through fairly inaccessible mountains. 2) Environmentalists and Native Hawaiian groups were able to tie up the construction in court for (perhaps) 20+ years - or was it more. I still recall as a tourist traveling from Marine Corps Base Hawaii to Kaneohe on the open (finished part) of H-3. It was like that for several of our 12 visits before we adopted Hawaii as our home state.

Interesting site with article on H-3

https://www.amusingplanet.com/2014/02/the-h-3-highway-in-hawaii.html

Returning you now...
 
This reminds me of a math question related to investment volatility/beta I saw years ago.

On the first trading day of the new year, you invest exactly $10,000 in a stock .
It goes up 90% the 1st year, & down 50% the following year.
Are you ahead or behind ?

You weren't supposed to use a calculator. I cheated.
 
^^^^^ behind me thinks, but it's getting late in the day.
 
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If time stands still at the speed of light, the second mile would have to be completed at the speed of light.

Nope. It would have to be faster. In fact, you would need infinite speed.

Speed = distance/time or time = distance/speed
We know that distance is 1 mile, so for time = 0, speed = infinite


The speed of light is not infinite. It travels at a speed of 9.836 x 10^8 feet per second. There are 5280 feet in a mile, therefore it will take 5280/9.836 E8 =5.368 x 10^-6 seconds = 5.368 microseconds for light to travel one mile. It is a very short time, but time has still passed. So, even travelling at the speed of light for the second mile, it would take 2 minutes + 5.368 microseconds to go the two miles, which is less than 60 mph.
 
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I did the time thing and found it impossible too.

But hey, I'm an engineer and that's what they pay me for - :)

Although an engineer would say "light speed". But the math guy would say, even light speed would add time.

And the engineer would reply the the difference was insignificant - :)

Mathematically, light speed is still not a solution.

But to an engineer, that's close enough. ;)
 

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