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Hey everyone, I was doing a chinese math problem set today for practice, and I came across a problem. Suppossedly, it is a really easy one, but then, I don't get it. I got everything after, but, just don't get that one...see if u can figure this damn thing out...

A team of people in a line stands a stretch of 100 M., when a person at the tail end of the team rushes up to deliver something to the captain(who is at the very front of the line), the team too, begins to move in a orderly fashion at the same time the messenger begins to walk up the line. After, the person walks back to the tail end, and all this time, the team has been moving at constant velocity for 100 M, ask, what is the total distance the messenger boy walked?

My thing with this problem is, it is confusing as hell, and, is there a number or two missing here???

Thanks...I'll post the answer after you guys give it a shot...

By Serene (Serene) on Wednesday, July 16, 2003 - 01:42 am: Edit

haha... doing this problem as practice for SAT??? you're overachieving.

i've seen this problem before, so allow me a few min.

but if i remember the answer had a sqrt(2) ... hmm

By Serene (Serene) on Wednesday, July 16, 2003 - 01:51 am: Edit

(1+sqrt(2)) * 100 meters

i'll give solution if it's right. tomorrow. =)

By Mattman (Mattman) on Wednesday, July 16, 2003 - 02:05 am: Edit

constant velocity of 100 meters by what? seconds, minutes, hours? You need some frame of time.

By Serene (Serene) on Wednesday, July 16, 2003 - 02:18 am: Edit

Mattman: that meant, with a constant velocity the team moved 100 meters away from where it was.

besides, this problem doesn't need any unit of time at all.

By Yoyo123 (Yoyo123) on Wednesday, July 16, 2003 - 12:53 pm: Edit

140m

By Flameball63 (Flameball63) on Wednesday, July 16, 2003 - 01:06 pm: Edit

No, the damn answer in the book is like 241 M....I am not sure...maybe the thing is wrong...urgh...

By Xiggi (Xiggi) on Wednesday, July 16, 2003 - 01:22 pm: Edit

241 meters - sounds right.

He ran from the back of the line to the front AND back. So that is 200 meters plus 41 meters to account for the movement of line.

By Serene (Serene) on Wednesday, July 16, 2003 - 01:35 pm: Edit

241 meters IS 100 * (1+sqrt(2))!!!!!!

sqrt(2) ~= 1.414...

*sniff* how dare you wrong my PERFECT answer...

now your book's answer is stupid. unless the problem asked for to the nearest meter. because as we all know EXACT ANSWERS is the way to go in math competitions ^_^

By Xiggi (Xiggi) on Wednesday, July 16, 2003 - 03:43 pm: Edit

Serene~
What do we get ... tomorrow?

By Serene (Serene) on Wednesday, July 16, 2003 - 04:30 pm: Edit

Xiggi~
I'm in AZ~~~ so "tomorrow" would have been today =)
But out of vanity I'm waiting for this guy to acknowledge that my answer is right first, lol...

j/k j/k

well... since you're really good at math too, i'll give you a hint, solve not for the speed, not for the time (since it's kind of impossible), but instead solve for x/y where x = messenger's speed and y = group's speed

By Flameball63 (Flameball63) on Wednesday, July 16, 2003 - 07:28 pm: Edit

Thank you guys very much...everything seems soooo clear now...haha. Well, thanks guys. And btw, I am NOT a guy...haha, I know the name sounds a lot like a guy's, but, naw...

By Flameball63 (Flameball63) on Wednesday, July 16, 2003 - 07:28 pm: Edit

And guess what serene...I am in AZ too...hahaha...

By Flameball63 (Flameball63) on Wednesday, July 16, 2003 - 07:29 pm: Edit

And one more thing to Serene...I am in AZ 2...hahaha...

By Jimjunior (Jimjunior) on Wednesday, July 16, 2003 - 07:36 pm: Edit

Serene never gave his solution. Here is how I solved it. Assume the line travels at speed of 1. Solve y= 100/(x+1) + 100/(x-1) for y=100. Multiply your answer by 100 because that is the length of time the line walks

By Serene (Serene) on Wednesday, July 16, 2003 - 07:56 pm: Edit

Jimjunior: Perfect =)

Flameball: How interesting. Do I know you? Or... what city are you in?

By Xiggi (Xiggi) on Wednesday, July 16, 2003 - 09:18 pm: Edit

Serene~

Neat. It is always so simple when someone shows you the solution. /wink /wink

Arizona! It always puzzled me that you knew I was from Texas the second I posted that question about dual-credit. For some reason, I thought you would live in a far away place!

By Serene (Serene) on Wednesday, July 16, 2003 - 09:19 pm: Edit

lol xiggi!!! because you had "Texas" in your profile ^^

By Flameball63 (Flameball63) on Wednesday, July 16, 2003 - 09:37 pm: Edit

HAHA, Serene, I live in Phoenix, over at Deer Valley(don't know why people call it that, since I never seen a deer runnin' around here in my life)...what city are u in?

By Serene (Serene) on Wednesday, July 16, 2003 - 09:47 pm: Edit

I'm in Gilbert =)

By Xiggi (Xiggi) on Wednesday, July 16, 2003 - 09:50 pm: Edit

Silly me! And you had AZ in yours. I never check the listed profiles since I assume most people leave them blank or fill them jokingly.

By Physicsstud (Physicsstud) on Thursday, July 17, 2003 - 12:15 pm: Edit

Jim, what does the x stand for in your equation? I"m an idiot i've been trying to figure that out for an hour. thanks.

By Physicsstud (Physicsstud) on Thursday, July 17, 2003 - 12:39 pm: Edit

bump

By Serene (Serene) on Thursday, July 17, 2003 - 03:08 pm: Edit

x=speed of messenger/speed of group

By Physicsstud (Physicsstud) on Thursday, July 17, 2003 - 03:30 pm: Edit

Then what does y =? Since jim said to plug 100 into it. How did u arrive at the formula? I totally lost, usually i can nail these problems. Also, why do u have 100/x+1 + 100/x-1. Thanks, this problem has been bothering me. I still can't give it up after 4 hours lol.

By Physicsstud (Physicsstud) on Thursday, July 17, 2003 - 03:54 pm: Edit

bump

By Idonotcare (Idonotcare) on Thursday, July 17, 2003 - 04:21 pm: Edit

i did the exact same problem in 5th grade!

By Physicsstud (Physicsstud) on Thursday, July 17, 2003 - 06:08 pm: Edit

bump

By Serene (Serene) on Thursday, July 17, 2003 - 06:10 pm: Edit

Idonotcare: anyone can say "I did that in 5th grade"... but not anyone would give the solution.


Physicsstud: it's good that you don't give up easily on a problem. =)

jim has good math but bad explanation. Here is my solution (variables are different from what jim used). Ask if you want more details on any step =)

let x = messenger's speed
let y = group's speed

100/y = how long it took the group to walk 100 meters

100/(x-y) = how long it took the messenger to go to the beginning of the line

100/(x+y) = how long it took the messenger to go from the beginning of the line back to the end of the line

since by the time the messenger got back, the group walked 100 meters, we can set these two time equal to each other... thus

100/y = 100/(x-y) + 100/(x+y)

a bit of simplifying get you
2xy = x^2 - y^2

divide both sides by y^2 we get

2(x/y) = (x/y)^2 - 1

treat x/y as one variable z
and solve for z in
2z = z^2 - 1

you'd get z = (1+sqrt(2))

Since z = x/y, z is the ratio messenger's speed/group's speed

so in the time group walked 100 meters, the messenger walked 100*z meters

so the final answer is 100 * (1+sqrt(2))

By Physicsstud (Physicsstud) on Thursday, July 17, 2003 - 06:15 pm: Edit

thanks serene, I see it now =) . You're always helpful. It's good to have a math genius on the board. If you don't mind me asking, where college are you attending next year?

By Serene (Serene) on Thursday, July 17, 2003 - 06:17 pm: Edit

lol thanks but i'm no genius at all ^^

hv.

By Physicsstud (Physicsstud) on Thursday, July 17, 2003 - 07:07 pm: Edit

I want to see if i understand the concept. You have x-y for on the way to the front because the group is moving, and that would increase the time to get there. Then you have x+y on the way back because the group is moving towards him which would decrease the time to get there.

By Jimjunior (Jimjunior) on Thursday, July 17, 2003 - 07:22 pm: Edit

Serene solved it like I did. Sorry for not providing an extensive solution. This problem will come up in a variety of forms. Often you will see it as someone in relation to a current,where someone has to go upstream or downstream. You will add the velocities when they are going in the same direction, and use the difference when they are opposites. I usually will use distance/rate = time to come up with my formula

By Serene (Serene) on Thursday, July 17, 2003 - 08:22 pm: Edit

just to clarify jimjunior's

---

Quote:

You will add the velocities when they are going in the same direction, and use the difference when they are opposites.

---

By Idonotcare (Idonotcare) on Wednesday, August 20, 2003 - 11:36 am: Edit

"Idonotcare: anyone can say "I did that in 5th grade"... but not anyone would give the solution."

Is there any one else did this problem in 5th grade?

By Digmedia (Digmedia) on Wednesday, August 20, 2003 - 12:03 pm: Edit

Well, if the line started moving at the same time the messenger started moving, and moved at the same speed as the messenger, then the messenger would walk 100 meters, the line would move 100 meters, and the "walk" back to the end of the line would be zero meters, because the end of the line would now be where the messenger was.

Thus, answer should be 100 meters...

BUT, suppose the line moved at a rate of 0.0000001 mm/year. Then then total distance would be 200 meters.

I'm also assuming that the messenger spent 0 seconds at the front of the line before turning back.

I disagree that the answer is 1+sqrt(2) without knowing the speed of the line movement.

Bad problem...

By Digmedia (Digmedia) on Wednesday, August 20, 2003 - 12:35 pm: Edit

ahhhh.... i reread the problem... didn't read that the line moved 100 m. So, ...AS USUAL... serene is right and i am wrong.


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"I thought I was wrong once,
but I was mistaken."