a little math problem

[bonkers325]

Originally posted by: b0mbrman
= 100 miles / {((150 miles + 50 miles) / 3X mph)/2}

average = (a+b)/2

change accordingly and u get 3x


x is your speed. n is some coefficient. no matter what the distance travelled is, the average of your first and second half speeds must equal to 2x.

(x + n*x)/2 = 2x
x + n*x = 4x
n*x = 3*x

therefore n = 3

 

[Rock Hydra]

42

 

[bonkers325]

nm

 

[RollWave]

Yeah this cant happen unless you can teleport basically

 

[b0mbrman]

Originally posted by: bonkers325

Originally posted by: b0mbrman
= 100 miles / {((150 miles + 50 miles) / 3X mph)/2}

average = (a+b)/2

change accordingly and u get 3x

Wrong.

Check again

 

[Random Variable]

As your average speed for the second 50 miles approaches infinity, your average speed for the entire 100 miles approaches 2X but obviously never equals 2X.

 

[b0mbrman]

Originally posted by: bonkers325

Originally posted by: b0mbrman
= 100 miles / {((150 miles + 50 miles) / 3X mph)/2}

average = (a+b)/2

change accordingly and u get 3x


x is your speed. n is some coefficient. no matter what the distance travelled is, the average of your first and second half speeds must equal to 2x.

(x + n*x)/2 = 2x
x + n*x = 4x
n*x = 3*x

therefore n = 3

Again, you're quite wrong.

This applies if you traveled each speed for the same amount of time, not distance.

Think about it.

 

[thesurge]

Originally posted by: bonkers325

Originally posted by: b0mbrman
= 100 miles / {((150 miles + 50 miles) / 3X mph)/2}

average = (a+b)/2

change accordingly and u get 3x


x is your speed. n is some coefficient. no matter what the distance travelled is, the average of your first and second half speeds must equal to 2x.

(x + n*x)/2 = 2x
x + n*x = 4x
n*x = 3*x

therefore n = 3

You are using the wrong type of "average" (or mean). Harmonic mean, ftw!

 

[Tobolo]

The problem didnt mention time. Go 25mph for two hours and then 50 Mph for 1 hour. BAM your there.

 

[Rock Hydra]

Originally posted by: Tobolo
The problem didnt mention time. Go 25mph for two hours and then 50 Mph for 1 hour. BAM your there.

I'm thinking this is incorrect. Since the average time would not be 2x.

25+50 = 75 divide that by 2 = 37.5

<--- is bad at math.

 

[b0mbrman]

Originally posted by: Rock Hydra

Originally posted by: Tobolo
The problem didnt mention time. Go 25mph for two hours and then 50 Mph for 1 hour. BAM your there.

I'm thinking this is incorrect. Since the average time would not be 2x.

25+50 = 75 divide that by 2 = 37.5

<--- is bad at math.

Not sure what he was going for, but he did 100 miles in three hours. That's 33 1/3 mph.

I have no idea what question that answers...

 

[chuckywang]

Originally posted by: Random Variable
The distance from point A to point B is 100 miles. If you average X mph for the first fifty miles, what speed would you have to average for the second 50 miles so that your average speed for the entire trip is 2X mph?

Harmonic means.

 

[chuckywang]

Originally posted by: Inspector Jihad

Originally posted by: mugs
Wait, no that's not right

OK, I'm confident it's not possible. You'd have to complete the second half of the trip instantaneously.

Suppose you complete the first half of the trip in 1 hour at 50 mph. You'd then have to adjust your speed to finish with an average speed of 100 mph. You have to travel a total distance of 100 miles in a total time of 1 hour, and you've already used your whole hour.

lol, are you serious

Yep, and mugs is correct.

 

[Syringer]

Some people are unbelievably clueless.

It isn't possible. Some places use this sort of question as a brainteaser during interviews.

 

[chuckywang]

BTW, was the topic summary sorta like a hint?

 

[Syringer]

Originally posted by: Inspector Jihad

Originally posted by: mugs

Originally posted by: Inspector Jihad

Originally posted by: mugs
Wait, no that's not right

OK, I'm confident it's not possible. You'd have to complete the second half of the trip instantaneously.

Suppose you complete the first half of the trip in 1 hour at 50 mph. You'd then have to adjust your speed to finish with an average speed of 100 mph. You have to travel a total distance of 100 miles in a total time of 1 hour, and you've already used your whole hour.

lol, are you serious

Post your answer.

read the question wrong.

lol, are you serious

 

[slayer202]

stupid question. when you say average speed, no one is thinking or cares about what amount of time you were traveling for, just the distance/time. :thumbsdown:

 

[91TTZ]

Originally posted by: Random Variable
As your average speed for the second 50 miles approaches infinity, your average speed for the entire 100 miles approaches 2X but obviously never equals 2X.

Sure it does. You reach infinite speed at exactly the halfway point.

.999... = 1

 

[fornax]

It takes the same time to travel twice the distance with twice the speed as to travel half the distance with half the speed. Therefore the second half of the distance must be traveled in zero time, or infinite speed.

 

[tmc]

yup, infinity

 

[AStar617]

Originally posted by: 91TTZ
.999... = 1

/me flashes back to ATOT thread a few month back...

*head explodes*

J/K, this is clearly "unpossible". - (C) Ralph Wiggum

 

[Sc4freak]

1 - 0.9999... = ?

0.0....01 or 0?

 

[Schfifty Five]

not possible.

 

[Inspector Jihad]

Originally posted by: Sc4freak
1 - 0.9999... = ?

0.0....01 or 0?

0 since .9999...=1.

 

[ArchCenturion]

Heres what i get

X is original speed traveled for the first 50 mi and is a rate distance/time
Tis the time it takes for the first 50 miles

So T*X=50 miles


X2 is speed traveled for the second 50 mi and is a rate distance/time
T2 is the time it takes for the second 50 miles

So T2*X2=50 miles

Add these together, we get T*X + T2*X2 = 100 miles eq(1)


The average speed for the whole trip needs to be 2X
So we need an expression for average

100 miles / (T+T2) = 2X eq(2)

Which we can solve for T2 and get 50/X - T = T2



Plug in to eq(1) we get T*X+ 50*X2/X - T*X2 = 100 eq(3)

We have an equation with infinite solution i guess

Choose Values for T and X and you can solve for X2 this shoudl always give you an average speed of 2X

For example if we choose T=1hr and X=1mph

We get 1+50*X2-X2=100 and solve for X2=99/49 mph
(if we solve eq(2) for T2=49hr)

To check we get should get an average speed of 2 so use eq(2)...
100 miles / (T+T2) = 2X eq(2)
100/(1+49)=2*1
2=2

This should work for any 2 values chosen for either X1, T, or X2

Hope this helps