MAtH qUesTioN???~~

Discus: SAT/ACT Tests and Test Preparation: October 2003 Archive: MAtH qUesTioN???~~
 By Allosnake (Allosnake) on Sunday, October 05, 2003 - 06:12 am: Edit

1. A file of soldiers marching in a straight line one behind the other is one kilometer long. An inspecting officer starts at the rear, moves forward at a constant speed until he reaches the front, then turns aroudn and travels at the same speed until he reaches the last man in the year. by this time, the column, marching at a constant speed, has moved one kilometer forward so that the last soldier is now in the position the front soldier was when the whole movement started. How far did the inspecting officer travel?

i'll appreciate if u can help =)

 By Y17k (Y17k) on Sunday, October 05, 2003 - 09:23 am: Edit

3 kms

 By Allosnake (Allosnake) on Sunday, October 05, 2003 - 11:09 am: Edit

can tell me how u did it?

 By Sidis (Sidis) on Sunday, October 05, 2003 - 11:31 am: Edit

haha relative movement? I'd say 2kms.(or less)

1.5 km forward
.5 km backward

remember that the year is not completely 1km ahead when he starts coming back, and he takes much less time returning since the troop is still going forward so It would be close to 2kms

 By Y17k (Y17k) on Sunday, October 05, 2003 - 11:50 am: Edit

think of it as 2 lines following each other... so it becomes a 2k line.

you go 2ks to the front, but u come back only to the half way

2+1 = 3 ^^

 By Jszab (Jszab) on Sunday, October 05, 2003 - 01:24 pm: Edit

Hm. It depends, on relativistic effects. If the subtract between officer velocity, and the column is bigger it becomes more important. ;)

ok. 2 kms.
Imagine that the column is moving so fast, that commander even will not start. The column is moving, and reach one kilo forward the officer.

then commander is moving at the start. And column is repeating what has done before. But officer need not to ride. becuase he is at the end.

OKay. here you are mathematical explain:
We looking for relative velocity [speed, actualy will be enough].
Vo > Vk
V1 = Vofficer + Vkolumn
V2 = Vofficer - Vkolumn
Because it is relative velocity, we can assume, that the column is standing.

And then S1 = S2
or another
average officers speed: S1+S2/T1+T2 = (V1+Vk+V1-Vk)/2

2V1/2 = V1. If average is V1, and he is moving constant speed, he must travel with the V1 velocity. Easy. Isn't it?

 By Jszab (Jszab) on Sunday, October 05, 2003 - 01:29 pm: Edit

I would be very please if You would like to help me. I need some information about your education system [I am outside US]. Especially AP, and what is most valuable for Colleges.
Please write me if you would like to help me, so we would be able to find a day that fits us, for talk by ICQ, or name the topic on another discussion list. [I think this one is not appropriate.]
jszab@o2.pl

Please don't delete it. I am desperately looking for help.

 By Y17k (Y17k) on Sunday, October 05, 2003 - 07:54 pm: Edit

guys... the answer is 3km not 2km...

and u dont need all that complicated stuff, just common sense

 By Quarky (Quarky) on Sunday, October 05, 2003 - 09:04 pm: Edit

2km, I say.

 By Xiggi (Xiggi) on Sunday, October 05, 2003 - 09:20 pm: Edit

2,414 meters.

 By Xiggi (Xiggi) on Sunday, October 05, 2003 - 09:27 pm: Edit

let x = officer's speed
let y = soldier's speed

1000/y = how long it took the group to walk 1000 meters

1000/(x-y) = how long it took the officer to go to the beginning of the line

1000/(x+y) = how long it took the officer to go from the beginning of the line back to the end of the line

Since by the time the officer got back, the soldiers file walked 1000 meters, we can set these two time equal to each other... thus

1000/y = 1000/(x-y) + 1000/(x+y)

Simplify to
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 => z = (1+sqrt(2))

Since z = x/y, z is the ratio officer's speed/soldier file's speed

In the time the file walked 1000 meters, the officer walked 1000*z meters, or 1000 * (1+sqrt(2)).

 By Y17k (Y17k) on Monday, October 06, 2003 - 01:32 am: Edit

GUYS

jeez, the answer is 3 kilometres, coz the original poster asked for the answer, i replied, and he replied by saying "how did u do it" which implies that 3 kilometres is RIGHT

xiggi, you critisise people of not reading posts carefully, and in this situation you are guilty of the same crime :P

 By Xiggi (Xiggi) on Monday, October 06, 2003 - 02:07 am: Edit

Y17K~

First, are you so sure I criticize people for not reading post carefully? I know that I tell people to go look for past posts to find answers to questions that have been posted in the past.

As far the above problem, are you really so sure that the original poster has told you that the answer is correct? That would be a tad surprising but you are welcome to doubt my answer and cling to your answer of 3km.

Since you mentioned commom sense ... you may want to think about this statement: by this time, the column, marching at a constant speed, has moved one kilometer forward. It means that the column has moved ONLY 1 km in the time the officer has moved to the front and back to the end.

I hope you realize that he NEVER had to walk 2km to reach the front of the line. He started at the end and only had to travel a certain distance to reach the front, then he turned back and at the time he reached the end of the line, the whole line had only moved 1 km. That certain distance cannot be 2km, as you suggest, because that would give you zero time to travel the third km. The officer cannot be at the beginning and at the end of the line at the same time.

The reality is that the officer needs to run 1000+707 meters to the front of the line, and then runs back to the end of the line or 707 meters, for a total of 2,414 meters. During his run back, the line moves in the opposite direction for the remaining 293 meters. Surprisingly, the line moved 707+293 meters for 1000 meters.

But again, if you think that the distance is 3KM, that is OK by me

 By Nhlgoalie (Nhlgoalie) on Monday, October 06, 2003 - 02:17 am: Edit

This question is much more complex Y17k than you are realizing. I solved the problem exactly as Xiggi did and it makes sense.

Your basis for why your answer is correct is "oh, the poster implied it." Xiggi's is mathematically recreatable and I don't see any problems with it.

Before you assume your answer is right you may want to consider that the original poster may have seen this on a test, so he probably has no clue what the answer is. If it was multiple choice the answer "3" would certainly be there as a sucker answer. 2.414 would be my bet for the answer. Nice job Xiggi

 By Y17k (Y17k) on Monday, October 06, 2003 - 02:30 am: Edit

xiggi

firstly i have tremendous repsect for you, and my little criticism was supposed to be taken lightly. whether the criticism has any substance is up the original poster, but my point is, if my answer is not correct, why else would the original poster ask "can you tell me how you did it?"

 By Nhlgoalie (Nhlgoalie) on Monday, October 06, 2003 - 02:37 am: Edit

"Before you assume your answer is right you may want to consider that the original poster may have seen this on a test, so he probably has no clue what the answer is."

He would naturally ask you how you did it since he did not get the right answer and you didn't provide work

 By Quarky (Quarky) on Monday, October 06, 2003 - 03:08 am: Edit

Good job Xiggi!

 By Quarky (Quarky) on Monday, October 06, 2003 - 03:08 am: Edit

I was just using significant figures (that's my excuse) lol, so I am still right with my 2km.

 By Xiggi (Xiggi) on Monday, October 06, 2003 - 12:06 pm: Edit

Everybody is right ...
the answer is 3km > x > 2km

By the way, this will NEVER show up on a SAT1 test!

 By Jszab (Jszab) on Monday, October 06, 2003 - 04:05 pm: Edit

Maybe we should check it empirically?
Who has bike with velocity indicator?
We need, approximately, 1001 people...

 By Laniman (Laniman) on Tuesday, October 07, 2003 - 11:31 am: Edit

Let 'x' be speed of the soldiers, 'y' be the speed of the officer.
Now, total time taken (in terms of x)=1/x
When the officer travels in the direction of the soldiers, speed=y-x
Time= 1/y-x
Against the direction of the soldiers, 1/y+x
Now, to find out how much the officer travelled in the direction of the soldiers:
Distance=Speed*time
=y*(1/y-x)
Against the direction:
Distance=Speed*Time
=y*(1/y+x)
Total Distance = y*(1/y-x)+y*(1/y+x)
= y(2y/(y^2-x^2))
= 2(y^2)/(y^2-x^2)

 By Laniman (Laniman) on Tuesday, October 07, 2003 - 11:36 am: Edit

Y17K is an idiot

 By Xiggi (Xiggi) on Tuesday, October 07, 2003 - 12:03 pm: Edit

Laniman, you may want to show how you finish the problem after

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

 By Laniman (Laniman) on Tuesday, October 07, 2003 - 01:13 pm: Edit

Dude in my opinon, the answer differs based on the values for x and y.
eg: x=5 y=10
Distance=2*100/75=2.66 appx
x=3 y=1
Distance=18/8=2.25
The final answer cant be a fixed number IMO unless theyve given specific values for x and y for which to find the distance.

 By Xiggi (Xiggi) on Tuesday, October 07, 2003 - 01:40 pm: Edit

Dude Laniman~

Hehe!

The reason you can't solve it is because you use CIRCULAR references. You NEED to be able to write an equation that isolates an unknown value. Rewriting an equation that keeps tke same number of unknowns is not a SOLUTION.

This problem has an EXACT solution, you do not need to "offer" variables for x and y. And if you did, your values of x and y need to be in the appropriate ratio of 2.414/1.000.

The problem stated FIXED data that allows you to extrapolate a FIXED solution.

1. The distance the column moved
2. The fact that the officer was at the end of the line when the column had moved 1000 meters.

Do yourself a favor ... before calling people idiots, try to read and understand the posts that included the solution.

 By Laniman (Laniman) on Tuesday, October 07, 2003 - 02:16 pm: Edit

Hmm..youre right. I didn't equate the answers with one of the conditions. Good job with your solution though.
But the fact remains..yi7k is an idiot.

 By Sidis (Sidis) on Tuesday, October 07, 2003 - 10:55 pm: Edit

"haha relative movement? I'd say 2kms.(or less)"

My post... you don't have to beat your brains with a calculator in order to have a reasonable answer in a few seconds. Anyway since that is harder than any SAT q, I give a &&*&(\$ about it. You should try to solve other more interesting problems, we all know by now that the answer is app. 2kms (considering that men can march at a constant rate), but since they can't and the guy leading them does not move instantly it should be 2km +- 500m I agree with Xiggy after all..