math help...please?

darkhanger18 · Nov 19, 2007

trying to help the girlfriends little brother here....It might just be me, but 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

any help is greatly, greatly appriciated

milessthomas · Nov 19, 2007

Where's mathogre?

cheesybanana · Nov 19, 2007

It shouldn't matter how much time the trip took. She traveled 20 miles going Xmph and 20 miles going 20mph to make an average of 40mph.

I'm not sure how time would fit into the equation at all?

bobtomay · Nov 19, 2007

Well, I say:

In order to avg 40 mph for a trip of 40 miles, the trip would have to be completed in 1 hour. Any time over 60 minutes would be something less than 40 mph.
Since the entire 60 minutes was already spent to travel the first 20 miles, it is not possible to avg 40 mph. Unless that is, you have a transporter.

jdgti · Nov 19, 2007

well i don't think you are assuming that the trip takes 1 hour total or it would be impossible. we don't know how long the trip took, we just know that the average speed for the who trip is 40 mph. so the trip might have take 1.25 hours because it took 1 hour to get to Billy Bob's and only .25 hours to get back (80 mph). so then you take the two speeds and average them over the 1.25 hour time period and you should get 40 mph for the whole trip. the problem is the only equation i know is

Distance = speed*time

we don't know the speed or the time of the second part of the trip. we just know the overall average speed, which is comprised of the first part and the second part.

A difficult question for an 8th grader. I will look into it more. just my two cents on how to think about the problem.

BananaPancakes · Nov 20, 2007

okay.. after attempting this problem and staring at my paper for about 20 minutes.. i'd say it's not possible..

let's take d=vt for the whole trip, like jdgti said..

we have t=d/v or total time of trip = (40 miles) / (40 mph) = 1.
That means the total time it takes to complete the whole trip is 1 hour, doesn't it? And the first part of the trip took 1 hour..

jdgti, i think you made a mistake because if you it takes 1 hour going 20mph there, and .25 hours going at 80 mph, the average is not 40 mph.. it is 32 mph, because you have to divide by the total time it takes (1.25 hours)

anyways.. i'm pretty certain about this, as it gives 2 variables and 2 equations and there is no solutions to them..

Equation 1: Average Speed = (v1t1 + v2t2)/(t1 + t2)
Equation 2: d2=v2t2
Where v1, t1 are the speed and time getting there and d2,v2,t2 are the distance, speed and time getting back.

edit= if there IS an answer, please let me know.. haha, i hate being wrong.. and it's been known to happen

PowerBookG4 · Nov 20, 2007

8th grade math problem and you stumped a bunch of computer geeks.. what math is this kid in?

but i second that its not possible, but if there is an answer i would like to know how stupid i am.

D3v1L80Y · Nov 20, 2007

Let d be the distance to the Billy Bob's house, T be the time it gets to get there, t be the time it takes to get back, and R be the speed Peggy Sue will travel on the return trip (which is what we want to find out). As we know from elementary mathematics, distance equals rate times time:

d = 20T
T = d/20

d = Rt
t = d/R

Now that we have expressions for T and t, we can come up with an equation that describes the round trip:

2d = 40(T + t)
2d = 40(d/20 + d/R)
2d = 40d(1/20 + 1/R)
1 = 20(R/20R + 20/20R)
20R = 20(R+20)
R = R + 20

Here we have worked our way into a paradox. The reason, simply, is that you have to travel back at an infinite speed to make your average speed 40 mph. This may seem strange, but consider that, the faster your return trip, the quicker you make it, and consequently, this faster speed has a lesser impact on the average speed.

If you traveled the return trip instantaneously, this would be equivalent to traveling double the distance in the same amount of time as the one-way trip. So if the rate of speed of the return trip is infinite, you do indeed get an average speed of 40 mph.

...yeah, a little more advanced than the math I had in 8th grade....

jdgti · Nov 20, 2007

BananaPancakes said:
okay.. after attempting this problem and staring at my paper for about 20 minutes.. i'd say it's not possible..

let's take d=vt for the whole trip, like jdgti said..

we have t=d/v or total time of trip = (40 miles) / (40 mph) = 1.
That means the total time it takes to complete the whole trip is 1 hour, doesn't it? And the first part of the trip took 1 hour..

jdgti, i think you made a mistake because if you it takes 1 hour going 20mph there, and .25 hours going at 80 mph, the average is not 40 mph.. it is 32 mph, because you have to divide by the total time it takes (1.25 hours)

anyways.. i'm pretty certain about this, as it gives 2 variables and 2 equations and there is no solutions to them..

Equation 1: Average Speed = (v1t1 + v2t2)/(t1 + t2)
Equation 2: d2=v2t2
Where v1, t1 are the speed and time getting there and d2,v2,t2 are the distance, speed and time getting back.

edit= if there IS an answer, please let me know.. haha, i hate being wrong.. and it's been known to happen

Oh sorry, I didn't mean to make it sound like the answer was 80. I just meant that that was a way of thinking about it. Sorry for the confusing post. But yes, thank you for clarifying.

rman · Nov 20, 2007

abstract algebra

Neo · Nov 20, 2007

So...unless she was driving a moving truck to Billy Bob's, it can't be done. :rimshot:

mathogre · Nov 20, 2007

milessthomas said:
Where's mathogre?

I was on my way home from doing grocery shopping. Mmm, food. ^_^

bobtomay said:
Well, I say:

In order to avg 40 mph for a trip of 40 miles, the trip would have to be completed in 1 hour. Any time over 60 minutes would be something less than 40 mph.
Since the entire 60 minutes was already spent to travel the first 20 miles, it is not possible to avg 40 mph. Unless that is, you have a transporter.

Gold star of course!

D3v1L80Y said:
Let d be the distance to the Billy Bob's house, T be the time it gets to get there, t be the time it takes to get back, and R be the speed Peggy Sue will travel on the return trip (which is what we want to find out). As we know from elementary mathematics, distance equals rate times time:

d = 20T
T = d/20

d = Rt
t = d/R

Now that we have expressions for T and t, we can come up with an equation that describes the round trip:

2d = 40(T + t)
2d = 40(d/20 + d/R)
2d = 40d(1/20 + 1/R)
1 = 20(R/20R + 20/20R)
20R = 20(R+20)
R = R + 20

Here we have worked our way into a paradox. The reason, simply, is that you have to travel back at an infinite speed to make your average speed 40 mph. This may seem strange, but consider that, the faster your return trip, the quicker you make it, and consequently, this faster speed has a lesser impact on the average speed.

If you traveled the return trip instantaneously, this would be equivalent to traveling double the distance in the same amount of time as the one-way trip. So if the rate of speed of the return trip is infinite, you do indeed get an average speed of 40 mph.

...yeah, a little more advanced than the math I had in 8th grade....

That's the Stephen Hawking version. -_^ This one deserves a black hole. You might not see it, but you can measure its effects.

bobtomay · Nov 20, 2007

Never did give the teachers the pleasure of spending a bunch of time trying to put formulas to the problems they threw out on occasion without solutions. They enjoy seeing all their students squirm, and I didn't like giving them the pleasure. A little logic sometimes goes a long way.