My other message board cannot solve math problem

Factual Questions
1

here’s the problem:

[quote[ OK, so far the general consensus from friends is that the answer is 45MPH. Here it is, it’s very simple:

You drive a car up a 1 mile hill at 15MPH. What average speed must you have on the way down so that the average speed for the trip is 30MPH?

I don’t know if our professor added this to throw us off, but he also said to use the formula of distance = (average speed) time, or just d = rt to figure it out.

He said that Einstein was shown this problem and spent a long time trying to solve it, then realized he was ‘over thinking’ it and he should have seen the answer in seconds. If I have the correct answer by next class I get 10 points extra credit. [/quote]

and here’s the thread about it:
http://forums.clubsi.com/showflat.php?Cat=&Board=UBB34&Number=4339773&page=0&view=collapsed&sb=5&o=&fpart=1&vc=1

not sure if you have to register to view it…anyway, the answers are varying between 45mph and “impossible”

2

The total trip is two miles. Half the trip is at 15 MPH. You want your average speed to be 30 MPH. Knowing this, it is extremely easy to solve if you understand averages.

3

45mph.
Average speed point A to point B = 15
Average speed point B to point C = x
Average speed point A to point C = 30
(15+X)/2=60
X=45
It’s unpossible :wally

4

There you go. I knew you could do it.

5

This should obviously read (15+X)/2=30.

6

that’s not my algebra, but I shall pass it along

I don’t solve stupid math problems, I’m a social scientist :smiley:
thanks

7

You drive a car up a 1 mile hill at 15MPH. What average speed must you have on the way down so that the average speed for the trip is 30MPH?

Given d is 1 mile, and that t1 is the time it takes to get up the hill, and t2 is the time it takes to get back down (in hours).

The known information is
d/t1 = 15
2d/(t1+t2) = 30

Should be solvable.

We can reduce the last to:

d/(t1+t2) = 15

We already know d, and we can reorder the first equation giving:

t1 = 1/15

and also:

(t1+t2) = 1/15

Substituting t1 and t2, we get:

(1/15 + t2) = 1/15

If we then take 1/15 from both sides, we end up with:

t2 = 0

indicating that the time it takes you to get back down the hill is 0 hours.

This is, of course, impossible.

Why does this happen?

Obviously you must go faster in the second leg to get a faster average speed over the whole distance. However, when you go faster, you decrease the time you are spending going that faster speed since the distance is fixed. While you are getting a faster MPH in the second leg alone, you’re adding less and less to the overall time for the entire journey, and therefore less and less to bumping the average up.

It turns out that an average of 30MPH happens only when you take no time to make the journey. It’s right at the limit.

If you’d chosen 29 MPH, the answer’s solvable. You still have to go very fast, but the shortening travel time doesn’t counteract the effect of faster speed.

8

A) Basically, you are travelling 2 miles (one mile uphill and one mile downhill).

B) For an average speed of 30 mph you would have to travel 2 miles in four minutes.

C) Going uphill at 15 mph takes 4 minutes.

D) Since you need to travel 2 miles in four minutes (step B) AND you have already used up your 4 minutes you can either say that you would have to travel at an INFINITE rate of speed or that it is IMPOSSIBLE.

9

uh oh…dissenting opinion…

I can see both answers being true… oh mighty dope, why must you mislead us?! :frowning:

10

Well. :o

11

does that mean you think you were wrong? or are you sticking to your first answer?

12

Darn you William_Ashbless - you beat me to the post.

If nothing else, I hope my posting was a bit more succinct.

Incidentally, these problems can be very deceptive. Go to this page
http://www.1728.com/puzzle.htm
and scroll all the way down to Puzzle 1. Don’t worry - the answer is posted there too.

13

No, that means I know I was wrong. :smack:

14

Well, I’m sure the little “embarrassed” smiley concedes the point. You see, the way one would come to 45 is a little flawed. If the two miles were boxes, and the “miles per hour” apples, then you’d have an average of 30 apples per box.

However, a “mile” doesn’t contain “miles per hour”, it contains a mile. Hence, to apply the “miles per hour” figure, you need to figure out how the “per hour” part.

Another way to think of it is, when you go 15 miles per hour one mile, then 45 miles per hour the next, how much time will that take?
D

15

As I said, such problems are deceptive (or is the word deceiving?)
So, when you really need an answer, will you try that “other” message board or will you come here?
Incidentally, this is the first time I have ever seen QED submit a wrong answer.

Don’t forget to go to this link
http://www.1728.com/puzzle.htm
and look at Puzzle #1.

16

I solved it in my head the simple way Wolf Meister illustrated. If you figure you have to make the entire 2 mile trip in 4 min to be going 30 MPH, by the time you’ve ready for the downward leg there’s no more time as your journey up the hill took you 4 minutes by itself.

17

I think that Q.E.D. was being snide with someone who appeared to be using the board as a homework service. But maybe that’s just my reading.

18

Yeah, that’s the ticket!
No, it’s not. I goofed. In my defense, it was well past any reasonable hour for me to be up. :smiley:

19

Yes, I guess this could technically be categorized as doing someone’s homework. However, this is an ‘extra credit’ assignment dealing with recreational mathematics and would explain why we were all anxious to jump into this thread.

20

Is there a possible solution if you allow for relativistic effects?