math help? anyone?

trying to help the girlfriends little brother here....It might just be me, but hell if i can solve this....

peggy sue drives from her house to billy bobs for thanksgiving. she drives at a rate of 20 mph there. billy bobs house is 20 miles away. how fast must she drive home if her avg. speed for the whole trip is 40 mph?

and then theres a note at the bottom:
answer is NOT 60, clearly 20+60/2=40, but you would be driving at 20mph for a longer period of time then you would be @ 60mph.

i feel like a moron being 19 and not being able to do 8th grade math, hopefully some of our geniuses on the board will be able to shed some light on this one.... thanks in advance for any help you offer

Since her average speed for the trip there was 20 mph, and the distance was 20 miles, it took her an hour to get there.

Now if she were to average 40 mph for the entire round trip of 40 miles, she would have to take an hour total. Since she already used up the hour driving there, she can't average 40 mph for the entire trip. She would have to do the return trip in zero time (i.e. infinite speed). So the answer is infinity.

Cyclists commonly encounter situations like this. If I only average 12 mph because a journey of 12 miles is all uphill, but I go 24 mph on the way back, I don't average 18 mph for the entire trip. The trip there takes an hour, the trip back 0.5 hour. Average speed is 24 miles/1.5 hours, or 16 mph. Hills really kill your average speed in cycling.

Thread looks familiar, must've seen it before, but where?

darkhanger18 said:

trying to help the girlfriends little brother here....It might just be me, but hell if i can solve this....

peggy sue drives from her house to billy bobs for thanksgiving. she drives at a rate of 20 mph there. billy bobs house is 20 miles away. how fast must she drive home if her avg. speed for the whole trip is 40 mph?

and then theres a note at the bottom:
answer is NOT 60, clearly 20+60/2=40, but you would be driving at 20mph for a longer period of time then you would be @ 60mph.

i feel like a moron being 19 and not being able to do 8th grade math, hopefully some of our geniuses on the board will be able to shed some light on this one.... thanks in advance for any help you offer

Click to expand...

jtr is correct.

But don't feel like a moron! My first year physics class--the one for scientists and engineers, the hardest one--had a similar question (but there was an answer, and not infinity) and almost everyone made the same mistake: they tried to average the speeds. But as you seem to inuit above, you can NOT do that. You have to go back to the DEFINITION of speed.

Speed = distance/time.

So for average speed, you take the total distance and divide by the total time.

In our case, that turns out to be:

40 MPH = 40 Miles (round trip distance there and back again) / X.

Clearly X = 1 hour. But since an hour has already been used, there is ZERO time remaining for the return trip.

But, let's make this something more tractable and say the following:

She drove at 30 MPH there. What is the minimum average speed she needs to maintain on the return trip in order to average 40 MPH for the trip as a whole?

You might think the answer is 50 MPH, because 50 + 30 / 2 = 40, but you'd be wrong.

speed = d/t. so t=d/speed, so it took her 20 miles/30 MPH = 40 minutes to get there.

Thus, she has only 20 minutes left to get home. So how fast must she drive?

again, speed = d/t

= 20 miles / 20 minutes

= 20 miles / (1/3 hour)

= 20 * 3 MPH

= 60 MPH.

Interesting, no?

And, as I said, very easy to screw up. This is the sort of question that they have on game shows where it really counts--like when a million dollars is at stake. It's really counter intuitive, isn't it?

Hope that helped!

The general form of this idea is part of calculus, and it is called the mean value theorem (for integration).

asdalton said:

The general form of this idea is part of calculus, and it is called the mean value theorem (for integration).

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asdalton,

Maybe it's that I'm working the first of ten graveyard shifts, but I'm not getting this. I know all about the MVT--derived it in college in fact--but it seems to me that the theorem is about there being an instantaneous "average" at somepoint on a function between start and end that equals the total average. i.e. the slope at some point on the curve between start and end points is equal to the average slope = the slope of the line going from start to end points. (For a continuous function only, of course)

This problem here, however, is about TWO AVERAGES, really. The trip there, and the trip back. I mean, you can't make a continuous function out of this, can you? And thus it is--I would think--more of a problem in the realm of discrete mathematics than calculus. In the same way that the problem of a postal employee delivering mail in the most efficient mannter to 12 customers is a problem from discrete mathematics, whereas the shortest distance between two points on a curved surface is from differential geometry (a branch of calculus).

Or maybe I'm just not getting it.

Could you elaborate?

I think you guys are wrong, I think there is an answer, but I'm still working on it.... I will post again when I've figured it out.

Edit: Nope, you guys were right after all. I was getting hung up on how to average speeds together. The correct way is not to look at the MPH figures and average them, but to break them down their components of distance and time, and add those like components together to end up with a new ratio of miles to hours. It goes like this:

First trip = 20mph is 20 miles and takes 60 minutes
Second trip = x mph is 20 miles and takes x minutes

Since we know that the round trip is and always will be 40 miles, the way to achieve an average speed of 40mph is to drive both legs in a combined time of 1 hour. Any amount over 1 hour will result in a slower average speed. So since we know it already took 1 full hour to make the first leg of the journey, the ONLY way to average 40 mph is to make the return trip in exactly 0 hours. JTR is correct.. although I don't know that I would say that the answer is infinity, because it involves dividing by zero (x miles per 0 hours).

jtr1962 said:

Since her average speed for the trip there was 20 mph, and the distance was 20 miles, it took her an hour to get there.

Now if she were to average 40 mph for the entire round trip of 40 miles, she would have to take an hour total. Since she already used up the hour driving there, she can't average 40 mph for the entire trip. She would have to do the return trip in zero time (i.e. infinite speed). So the answer is infinity.

Click to expand...

I'm with jtr1962 on this . . . Peggy Sue is going have to do a "Hiro Nakamura" (Sorry . . . watching 'Heroes" right now ) to make this work

Francis

There are two mean value theorems: a differential form and an integral form. I was referring to the integral form.

But now that I look at the theorem again, I realize that I was thinking of simply the proper definition of "average speed" as an integral over time rather than distance. The solution to the question doesn't rely on any instantaneous speed equaling the average speed.

Then again, a truly discontinuous velocity function in real life would require infinite acceleration, which is impossible to achieve and very messy to attempt. So the mean value theorem still holds.

This is the kind of question where you wonder if the teacher knew what they were asking. They likely mis-phrased the question. Remember, he said this was GF's younger brother. I'm assuming high school or below.

turbodog said:

This is the kind of question where you wonder if the teacher knew what they were asking. They likely mis-phrased the question. Remember, he said this was GF's younger brother. I'm assuming high school or below.

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I suspect you're right on with this assumption. It was probably supposed to be something like:

If it is 20 miles to Bob's house from Sue's house, and she averages 40 MPH on her trip there and back, and she drives there in 30 minutes, how long does it take her to get back?

Probably something like that. As stated it's way too esoteric for the stated grade level.

It reminds me of a problem I was given while visiting my cousins house for Halloween trick or treating. It was a high school (9th or 10th grade) math question. It was like this:

You have a 9 digit number whose first digit, taken all by itself, is divisible by 1, and whose first TWO digits, taken as a number by themselves, are divisible by 2, and whose first THREE digits, taken as a number, are divisible by 3, and so on through the whole number, which must be divisible by 9. The digits 1 through 9 are only used once in the 9 places of the 9 digit number.

There is only one such number. What is it?

Let me tell you, solving this analytically is a freaking GRADUATE LEVEL MATHEMATICS PROBLEM. And, while I took math to that level in some cases (PDE's and Diff. Geom.) I certainly couldn't figure out any other way to do this problem but by trial and error.

In college this same problem was given to the graduate math students taking real analysis (I think that was it), but the difference was they weren't told that each digit was used only once.

Talk about a hard problem!

It did teach the rules of divisibilty, though, I'll give the teacher that. Still . . . pretty unbelieveable.

copperfox said:

Since we know that the round trip is and always will be 40 miles, the way to achieve an average speed of 40mph is to drive both legs in a combined time of 1 hour. Any amount over 1 hour will result in a slower average speed. So since we know it already took 1 full hour to make the first leg of the journey, the ONLY way to average 40 mph is to make the return trip in exactly 0 hours. JTR is correct.. although I don't know that I would say that the answer is infinity, because it involves dividing by zero (x miles per 0 hours).

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It's really late here, so there's no telling if I've really understood all this, but as has been pointed out, there's no time left. Period. The answer cannot be infinity. The answer is that the question presumes we should be looking for a rational (?) number that solves the equation; and that answer does not exist. I think you may all be correct, but yet the problem cannot be solved because there is no speed that will suffice - there is no answer.

I think Peggie Sue decided to take the long way home due to a black friday sale at Target and haulled *** home before Bubba(billy bobs brother) could realize she took his visa.

I was thinking about this the other day. If you average speed over the time she can never average 40 mph unless she travels instantaneously but if you average her speed over distance she would only have to do 60 mph.

huh? how would going 60 mph after the fact help if she already used up the whole one hour she has to go 40 miles?

well, i asked her brother, he said that the teacher told the class that it was unsolvable, but he had a little bit of what was discussed here, and impressed the heck out of the teacher. her point was that not every question is solvable, well not by 8th graders she didn't count on a bunch of geniuses being asked the question:devil: thanks for all the help again guys...oh yeah and the major brownie points with the girlfriend

65535 said:

I was thinking about this the other day. If you average speed over the time she can never average 40 mph unless she travels instantaneously but if you average her speed over distance she would only have to do 60 mph.

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65535,

I suspect that you are confusing average speed with instantaneous speed.

Average speed means you take the total distance travelled and divide by the total time taken to travel that distance.

Instantaneous speed means you find the speed travelled at one particular instant in time.

Averaging speed "over distance" won't give you a speed! What you are saying is that she travelled 20 MPH there, which is half the distance. So if the other half of the distance were at 60 MPH, then 60+20 MPH over two halfs, averages out to 40 MPH.

But that's not correct! Here's why:

20 miles/hr / 20 miles + 60 miles/hr / 20 miles = 1 hr^-1 + 3 hr^-1

Which is 4/hr. Four per hour!?! That's not units of MILES PER HOUR.

Unit analysis will save you everytime!

To get speed you need to divide distance by time. It could be feet/sec or it could be miles per hour, or inches per year. But if it ain't a distance over a time, it ain't a speed.

YOU CAN'T SOLVE (20 + X) / 2 = 40 and get the answer to this problem! If you work it out you can see that you don't end up with MPH if you do this!

js said:

In college this same problem was given to the graduate math students taking real analysis (I think that was it), but the difference was they weren't told that each digit was used only once.

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Which no doubt makes it easier. For example, 123,654,321.

Because you said "The digits 1 through 9 are only used once in the 9 places of the 9 digit number" the 5th digit must be 5. And it doesn't matter what the first digit is, as all are divisible by 1, and whatever the last is, the whole nine digit number will always be divisible by 9, so I suggest doing them last. The 2nd, 4th, 6th and 8th must be the even digits with the odds in the other positions, and as all even digits are divisible by two I suggest doing the 2nd digit second to last, by using the last remaining even number.

The 3rd and 4th digits must be divisible by 4 (since 100 is divisible by 4), which removes half the possibilites, leaving only 2 or 6 for the 4th digit.

The 6th, 7th, 8th digits (alone) must be divisible by 8 (since 1000 is divisible by 8). All end in a 2 or 6, so 6th place cannot be 2 or 6.

The first three digits must be divisible by three, which means they must also add to an a multiple of three (which will be an even number since two odd numbers contribute to it). 2 and 6 cannot appear in 2nd place.

At this point I had a bunch of possibilites for the first few digits and another bunch for the last ones and I combined them to get a list of 20 options, with the last digit automatically worked out - yay spreadsheets - for all the digits to sum to 45.

I then checked the numbers (using MOD on the spreadsheet) to see if they complied with 6 and 7. Only one did. Should I post it?

TorchBoy,

Yeah! By all means, post it!

Nice post.

But you can see how removing that clue about each digit being used only once would make the problem a lot harder!

js said:

But you can see how removing that clue about each digit being used only once would make the problem a lot harder!

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No, not at all. I gave an example of a 9 digit number that doesn't comply with that restriction. Another is 987,654,564. They're very easy to find.

To use each digit only once is much more tricky and shouldn't be done just before going to bed. I had trouble turning my brain off after that. 381,654,729. :thumbsup: