The Hill

[unreality]

i just made this up (but im basing it off another problem, just complicating it)

There is a hill with a path on it (this is the only way up/down). The path is 16 miles. Every day Old Man Jenkins and his dog go up together, and back down. However the dog is getting old, and can only go at half Jenkins' speed. However everyone must go the same speed going down the hill, which is the dog's climbing speed or 1/2 Jenkins' speed, because they have to pick their way around rocks and wahtnot. So this is what they do: Jenkins goes to the summit, comes back down, and when he meets the dog they both go back down together because they both have to go back down at the same speed. Today, his wife is joining them as well. She can go 3/4 Jenkins' speed.

This is what they do: They all leave together of course, and when Jenkins hits the top he turns around and goes back down- and when he reaches his wife, coming up, they both switch directions. Jenkins goes back up, to turn around again on the summit and come down, and his wife switches and goes downward. When the wife hits the dog, the same thing happens. The dog goes back down, but the wife goes up. When she and her husband run into each other, the wife gets to go back down to the bottom of the hill but Jenkins has to go back up to the top then down again.

How many miles has Jenkins gone, from the moment they start from the moment he gets to the bottom of the hill?

[Guest]

40.32 miles

[bonanova]

Verifying Martini's result ...

The order on the hill is always [D W J]

The positions, and Jenkins' distance, at 7 critical points are as follows:

0. Start at the bottom [0 0 0 => 0] then ...

2. Jenkins meets Wife [9.6 14.4 14.4 => 17.6]

3. Jenkins reaches top [10.4 13.6 16 => 19.2]

4. Wife meets the Dog [12 12 14.4 => 20.8]

5. Jenkins meets Wife [11.04 13.44 13.44 => 21.76]

6. Jenkins reaches top [9.75 12.16 16 => 24.32]

7. Jenkins reaches bottom [ 0 0 0 => 40.32]

[unreality]

yep. Both of you are right (lol duh). I had a simpler way of doing it, but each to his own

[bonanova]

Can you share your solution?

[unreality]

i just made this up (but im basing it off another problem, just complicating it)

There is a hill with a path on it (this is the only way up/down). The path is 16 miles. Every day Old Man Jenkins and his dog go up together, and back down. However the dog is getting old, and can only go at half Jenkins' speed. However everyone must go the same speed going down the hill, which is the dog's climbing speed or 1/2 Jenkins' speed, because they have to pick their way around rocks and wahtnot. So this is what they do: Jenkins goes to the summit, comes back down, and when he meets the dog they both go back down together because they both have to go back down at the same speed. Today, his wife is joining them as well. She can go 3/4 Jenkins' speed.

This is what they do: They all leave together of course, and when Jenkins hits the top he turns around and goes back down- and when he reaches his wife, coming up, they both switch directions. Jenkins goes back up, to turn around again on the summit and come down, and his wife switches and goes downward. When the wife hits the dog, the same thing happens. The dog goes back down, but the wife goes up. When she and her husband run into each other, the wife gets to go back down to the bottom of the hill but Jenkins has to go back up to the top then down again.

How many miles has Jenkins gone, from the moment they start from the moment he gets to the bottom of the hill?

yep, sorry, havent posted in a while

OMJ- old man jenkins

Part One:

OMJ goes 16 miles to top

WIFE goes 12 miles

DOG goes 8 miles

Part Two:

OMJ goes down 1/2 his normal speed, so WIFE who goes 3/4 his speed goes 1.5x times faster... she goes 3/5 of way between them, he goes 2/5. 2/5 of 4 miles = 1.6. Which means DOG has gone up 1.6 miles as well. SO:

OMJ is at 14.4 mile marker, has gone 1.6 down so far

WIFE is at 14.4 mile marker

DOG is at 9.6 mile marker

Part Three:

OMJ goes to top- he goes up 1.6 miles, taking him to 3.2 total so far (we're not counting original 16 of the first up, will add that in at end)

WIFE goes down at the standard down speed, half of OMJ's up speed, thus she goes down 0.8

DOG goes up 0.8 since his up speed IS the down speed

So:

OMJ- summit (3.2 mi total)

WIFE- 13.6

DOG- 10.4

Part Four:

WIFE and DOG are moving toward each other at same speed. So they must meet in middle of 13.6 and 10.4. Thus they both went 1.6 (half of the 3.2 between them) and met at 12 exactly

OMJ, in this time, must have gone down 1.6 then. 16-1.6 = he is at 14.4, and he's gone a total of 4.8 extra miles

So:

OMJ- 14.4 (4.8 mi total)

WIFE- 12

DOG- 12

Part Five:

WIFE goes up, DOG goes down, OMJ goes down...

since WIFE moves at 3/4 OMJ-upspeed and OMJ moves down at the standard 1/2 OMJ-upspeed, she goes 1.5x faster and thus meets 3/5 in and he's gone 2/5 in

14.4-12 = 2.4

WIFE goes 3/5 of 2.4, or 1.44 miles, and OMJ went 2/5 of 2.4, or .96 miles

They meet at 12+1.44 or 13.44 mile marker, OMJ having gone .96 more miles

DOG has gone same as OMJ, thus .96 miles down from 12 to 11.04

So:

OMJ- 13.44 (5.76 mi total)

WIFE- 13.44

DOG- 11.04

No more bumpings. WIFE and DOG both go down to bottom of hill. But OMJ goes up from 13.44 to 16, which is (16-13.44) or 2.56 miles, taking his total to 8.32

But we didnt count his earlier 16 miles going up the first time, and now is at the top and must go all the way down, so that adds 16+16 or 32 miles to the length. 32+8.32 = 40.32

Obviously not "simple" but its monotonous in its method, thats sorta what i meant

[unreality]

looking at yours more closely, our solutions are the same lol

im sure Martini figured it out the same way...

or did he? Hey, martini, did you have any secret solution?

[Guest]

l

Hey, martini, did you have any secret solution?

Yeah, but sorry, it's a secret.

Just kidding. I did it the same way.

[unreality]

lol

[Guest]

I got 40.1875 or something like that, but the answer is 0 from the time he started up the hill to the time he got to the bottom. they are the same point so he has gone 0 miles.

[Guest]

My math skills just aren't what they used to be. Could someone explain the formula used to determine how far the wife and old man jenkins each travel in the time he leaves the top of the hill to when they meet for the first time. It's the same formula throughout the problem but I can't figure it out.

[Guest]

There is no formula. bonanova and unreality both broke down how to get the solution step by step.

[Guest]

How far has he "gone" = technically, he's gone 16 miles, however it took him 40ish miles of walking to get there

[Guest]

Its just simple maths rather than a brainteaser isnt it? I thought there was going to be some quick way of doing it (like the fly and tha trains if you know that one). Meh, good maths puzzle! I like taketimes response, he's only travelled 16 miles away from where he started lol