Easy Math - DPChallenge Forums

Nikon AF Nikkor 50mm f/1.4D

04/09/2007 11:46:14 PM · #1

Okay, this one sounds a lot harder than it is ... but the math required to solve this problem is very simple (beginning algebra, perhaps). If I've posted it before, I apologize, it's one of my favorite math puzzles:

=========================================================

A man is walking across a railroad bridge that goes from point A
to point B. He starts at point A, and when he is 3/8 of the way
across the bridge, he hears a train approaching. The train's speed
is 60 mph (miles per hour).

The man can run fast enough so that if he turns and runs back
toward point A, he will meet the train at A, and if he runs
forward toward point B, the train will overtake him at B.

How fast can the man run?

=========================================================

If you've seen or solved this one before, please wait and give others a chance. :-)

ButterflySis

Canon EOS-50D

Canon EF 50mm f/1.8 II

04/09/2007 11:52:42 PM · #2

60mph?

Edit: Hmm, actually we don't know how far away the train is so we can't determine how fast the man can run.

Right? LOL

Message edited by author 2007-04-09 23:54:29.

taterbug

Nikon D300

04/09/2007 11:55:00 PM · #3

Originally posted by dwterry:

How fast can the man run?

=========================================================

by the sounds of it, not fast enough :-)

SDW

Canon EOS-60D

Tamron AF 28-300mm f/3.5-6.3 XR Di LD Aspherical IF for Canon

04/09/2007 11:55:14 PM · #4

not fast enough...splat!
ETA: t-bug you were faster than I.

Message edited by author 2007-04-09 23:56:07.

Canon FD 70-210mm f/4.0 MF

04/09/2007 11:55:34 PM · #5

Originally posted by ButterflySis:

60mph?

Based on what "appears to be" a lack of information in the story, that's not a bad guess. What you are suggesting is that the man could race across the bridge at the same speed as the train.

But there really is more information packed into the story. That's what I love about it. Look again and how far the man can travel during the same amount of time that the train spends crossing the bridge. There's your big hint. :-)

shutterpuppy

Sony Alpha a7

04/10/2007 12:00:53 AM · #6

15 mph

04/10/2007 12:01:51 AM · #7

Originally posted by shutterpuppy:

15 mph

Correct answer. Do you have an easy way to explain the solution?

Canon FD 70-210mm f/4.0 MF

04/10/2007 12:07:20 AM · #8

Hint: I think looking at the diagram makes the story much easier to understand. Especially if you look at the ruler along the bottom edge.

shutterpuppy

Sony Alpha a7

04/10/2007 12:09:17 AM · #9

You can see from the diagram that the man can travel 3 sections in the time that it will take the train to reach point A. Subtract those three sections from the 5 sections that he will be able to travel until the train reaches point B. (You are subtracting because as the man is running to Point B, the train is traveling to Point A. Assuming that he runs as fast toward Point B as toward Point A, he can travel three sections of distance before the train reaches Point A, the start of the bridge.)

This means that while the train is traveling from Point A to Point B the man will be able to travel the last two sections. There are 8 sections total representing the distance of the bridge. The man is able to cover 2/8ths (1/4th) of the distance that the train can travel in the same time, therefore the man is 1/4th as fast as the train. 1/4th of 60mph is 15mph.

Do I get a gold star? ;)

Message edited by author 2007-04-10 00:10:23.

photokariangel

Canon EOS-5D Mark III

Canon EF 70-200mm f/4.0L IS USM

04/10/2007 12:10:30 AM · #10

*scratches head and mumbles a bit, drool emitting from the corner of her mouth*

04/10/2007 12:18:27 AM · #11

Originally posted by shutterpuppy:

Do I get a gold star? ;)

Canon FD 70-210mm f/4.0 MF

04/10/2007 12:19:53 AM · #12

Originally posted by photokariangel:

*scratches head and mumbles a bit, drool emitting from the corner of her mouth*

Hey, another Utahn! Welcome aboard, photokariangel. :-)

shutterpuppy

Sony Alpha a7

04/10/2007 12:21:09 AM · #13

Cool puzzle. Although it did give me a PTS flash back to the LSAT. ;)

zardoz

Nikon D200

Nikon AF-S DX NIKKOR 18-200mm f/3.5-5.6G ED VR II

04/10/2007 12:22:11 AM · #14

I typed it in but Shutterpuppy beat me to it lol. I may as well post it after all that work:

We know the man can run 3/8 of the length of the bridge in the time it takes for the train to reach point A because you say he can run back in that time. Therefore if he runs 3/8 of the length of the bridge towards point B he will get 6/8 (3/4) of the way along the bridge before the train reaches point A. He can then run the remaining 1/4 of the bridge in the same time as the train takes to travel the full length of the bridge. In other words he runs one 1/4 of the distance in the same length of time. Therefore he is running at 1/4 of the speed of the train i.e. 15mph.

edit typos

Message edited by author 2007-04-10 00:27:23.

photokariangel

Canon EOS-5D Mark III

Canon EF 70-200mm f/4.0L IS USM

04/10/2007 12:33:18 AM · #15

Originally posted by dwterry:

Originally posted by photokariangel:
*scratches head and mumbles a bit, drool emitting from the corner of her mouth*

Hey, another Utahn! Welcome aboard, photokariangel. :-)

Thank you David! (the way you said that made me laugh, in refrence to the question and my post) haha.