[Guest]

A five mile long army was marching at a constant speed. The general at the rear of the army wanted to send a message to the general at the front. He sent Billy who galloped on his horse at a constant speed until he reached the front. Billy dropped the message and instantly turned around and galloped at the same speed until he got back to the rear of the army. When he reached the rear he was at the spot where the front of the army was when he originally set out with his message. How far did Billy travel?

Edited October 26, 2009 by Forcedhand

[Guest]

12.5 mi.

[Guest]

We know that in the time this takes to happen the army travels its five mile length. Since our rider travels to the front and back - two five mile trips - he covers a distance of 10 mi, or a displacement of 5 mi, relative to his starting point on the ground.

Edited October 26, 2009 by BrainInAVat

[Guest]

Say the army moved d miles in the time it took Billy to reach the front. Then the army traveled d miles in the time Billy traveled 5+d miles. On the other hand, when Billy got back (he had traveled 5+2d miles) the army had traveled 5 miles. This implies 5(5+d)=d(5+2d) and so Billy traveled 5+2d=5+5sqrt(2)=12.071 miles.

[Guest]

12.67mi

[Guest]

I totally agreed with palascosas. My answer is the same as palascosas's by approaching to the answer in different ways. Honestly, mine is not good as his, though .

12.071 . Instead of using the distance, I have used the velocity of the general and the army .

[Guest]

The army and the messenger travel at constant speeds during this. The messenger on the first trip has covered 5+d miles while on the return trip he has covered 5-d miles. Total miles traveled is 5+d+5-d = 10 miles...

[Guest]

10 miles! You don't need a masters in math which I have. 12.7? 12.3? You are all idiots. Idiots

[Guest]

10 miles

[Guest]

10 miles! You don't need a masters in math which I have. 12.7? 12.3? You are all idiots. Idiots

I, as one of the idiots, extend a hearty welcome to you, most esteemed new friend.....Why I do this is unfortunately an issue that is beyond my limited capacity.

[Guest]

The only thing that matters in this problem is how faster the messenger is than the army so that the two distances (first trip = 5miles plus d1 and back = 5miles minus d2) equal to 5 miles... With this in mind, the result is 12.07108 found with excel goalseek

[Guest]

10 miles! You don't need a masters in math which I have. 12.7? 12.3? You are all idiots. Idiots

Easy there Einstein.

By the way, welcome to the Idiot Den "Oh Light-Bearing Master of Math"

[Guest]

A five mile long army was marching at a constant speed. The general at the rear of the army wanted to send a message to the general at the front. He sent Billy who galloped on his horse at a constant speed until he reached the front. Billy dropped the message and instantly turned around and galloped at the same speed until he got back to the rear of the army. When he reached the rear he was at the spot where the front of the army was when he originally set out with his message. How far did Billy travel?

I am going out on a limb here and will guess 5 miles.

The question is how far did Billy travel? Not how much distance did Billy cover?

Billy started at the end of the line, and by the time he was done with his whole trip, he ended up where the front of the line was when he started. So he covered the length of the army which is 5 miles. The rest of the information is irrelevant.

Now the real question is why didn't the two generals use a cell phone?

Just because Billy is on horseback doesn't imply this is in a time period before cell phones.

[Guest]

I am going out on a limb here and will guess 5 miles.

The question is how far did Billy travel? Not how much distance did Billy cover?

Billy started at the end of the line, and by the time he was done with his whole trip, he ended up where the front of the line was when he started. So he covered the length of the army which is 5 miles. The rest of the information is irrelevant.

Now the real question is why didn't the two generals use a cell phone?

Just because Billy is on horseback doesn't imply this is in a time period before cell phones.

I think it is question on how much distance Billy covered... Simply because the problem can be solved with the above data, otherwise it wouldn't be solvable...

and BTW, Billy travels 141.42% faster than the army in order to manage

[Guest]

No I think he's right!!

Assume Y is the time and X is the distance. Then Y=X/V (V velocity)

Y(army) is Ya and Y(Billy) is Yb (Ya denotes the movement of the general in the back)

Ya=Xa/Va ; Yb=Xb/Vb=Xb/nVa (using the starting point of billy as origin)

when billy reaches the general in the front Ya=Yb and Xb-Xa=5

Xb=nXa and Xb=5+Xa => Xa=5/(n-1) and Xb=5n/(n-1) (of course n is not =1)

Now i'll take the point (Xa=5/(n-1) , Y1) as new origin where the general in the back is point (0,0) and billy is (5/n-1, Y1)

as Ya is still the same then Ya=Xa/Va

But Yb is now going the opposite way, then Yb=-Xb/nVa + c

We know that if Yb=0 then Xb=5, then we get c=5/nVa => Yb=(-Xb+5)/nVa

When Billy reaches the general in the back again Xa=Xb and Ya=Yb => X/Va=(-X+5)/nVa => X=5/(n-1)

Distance run by Billy is d= 5n/(n-1) + 5 - 5/(n-1) = (5n+5n-5-5)/(n-1) = 10 miles

[Guest]

It would be ten miles IF the army was stationary, but it's not....

I can follow palascosas when he says that Billy traveled 5+2d in the time the army traveled 5 mi. I can follow 5(5+d)=d(5+2d) becoming d=5/sqrt(2)=2.5sqrt(2), and plugging that in to 5+2d. But can someone please explain where he gets the original equation?

[Guest]

No I think he's right!!

Assume Y is the time and X is the distance. Then Y=X/V (V velocity)

Y(army) is Ya and Y(Billy) is Yb (Ya denotes the movement of the general in the back)

Ya=Xa/Va ; Yb=Xb/Vb=Xb/nVa (using the starting point of billy as origin)

when billy reaches the general in the front Ya=Yb and Xb-Xa=5

Xb=nXa and Xb=5+Xa => Xa=5/(n-1) and Xb=5n/(n-1) (of course n is not =1)

Now i'll take the point (Xa=5/(n-1) , Y1) as new origin where the general in the back is point (0,0) and billy is (5/n-1, Y1)

as Ya is still the same then Ya=Xa/Va

But Yb is now going the opposite way, then Yb=-Xb/nVa + c

We know that if Yb=0 then Xb=5, then we get c=5/nVa => Yb=(-Xb+5)/nVa

When Billy reaches the general in the back again Xa=Xb and Ya=Yb => X/Va=(-X+5)/nVa => X=5/(n-1)

Distance run by Billy is d= 5n/(n-1) + 5 - 5/(n-1) = (5n+5n-5-5)/(n-1) = 10 miles

but for a certain reason i'm still getting 13.33 hmm..

[Guest]

Being relatively ignorant mathematically, I get 10 miles with a more basic kind of analysis:

We know that when Billy finishes, the army has gone a total of 5 miles because the back of the line is where the front of the line was originally. So that means that halfway through his mission, the army had traveled 2 1/2 miles. And that means that Billy had traveled the original five mile line plus another 2 1/2 miles, or 7 1/2 miles. Right? And that means he needs to travel 2 1/2 miles to get back to the point where the front of the line originally had been (which is where the back of the line is now). So he travels 7 1/2 miles + 2 1/2 miles = 10 miles.

[Guest]

Being relatively ignorant mathematically, I get 10 miles with a more basic kind of analysis:

We know that when Billy finishes, the army has gone a total of 5 miles because the back of the line is where the front of the line was originally. So that means that halfway through his mission, the army had traveled 2 1/2 miles. And that means that Billy had traveled the original five mile line plus another 2 1/2 miles, or 7 1/2 miles. Right? And that means he needs to travel 2 1/2 miles to get back to the point where the front of the line originally had been (which is where the back of the line is now). So he travels 7 1/2 miles + 2 1/2 miles = 10 miles.

Wrong. Imagine this. Army travels at 10mph and Billy travels at 11mph. By the time he has reached the front line he will have covered 55 miles in total. To get back at the same speed he must travel (backwards) about 2.6 miles.

Now compare the distance needed if army travels at 10mph and Billy travels at 30 mph. He will reach the front line in only 5.5 miles and get to the back after 3.75 miles.

[Guest]

Wrong. Imagine this. Army travels at 10mph and Billy travels at 11mph. By the time he has reached the front line he will have covered 55 miles in total. To get back at the same speed he must travel (backwards) about 2.6 miles.

Now compare the distance needed if army travels at 10mph and Billy travels at 30 mph. He will reach the front line in only 5.5 miles and get to the back after 3.75 miles.

But in your examples, the back of the army won't be where the front of the army was originally, will it? I understand that at a faster rate or slower rate he'll cover less distance or more distance respectively to get to the front of the line and back, but there can only be one rate that he could travel in order to get to the front and back so that the line was exactly five miles further along. What am I missing?

[Guest]

Say the army moved d miles in the time it took Billy to reach the front. Then the army traveled d miles in the time Billy traveled 5+d miles. On the other hand, when Billy got back (he had traveled 5+2d miles) the army had traveled 5 miles. This implies 5(5+d)=d(5+2d) and so Billy traveled 5+2d=5+5sqrt(2)=12.071 miles.

Wouldn't the Billy's distance returned be less the distance covered to the front because his destination is moving toward him at a constant rate?

[Guest]

OK, here's what I am thinking...The army traveled 5 miles in the time the messenger traveled up and back. The distance the messenger traveled from the rear to the front was 5+d (where d=the distance the army traveled in the same time.) The distance he traveled back to the rear was 5+(5-d) (where 5-d is the remaining distance the army would have traveled.) So adding the two expressions together we get (5+d)+(5+(5-d)) (I know the parentheses are superfluous). So the total distance the messenger traveled is 15 miles, right?. Or am I an idiot? Where did I go wrong, if I am wrong?

[Guest]

If the question is a trick, then how "far" Bobby traveled is 5km

Otherwise, the total distance traveled by bobby is 12.07 as below:

Dividing the run into part (1) (where bobby run to the front) and part (2) (where bobby returns to his first general), and (a) for army, (b) for bobby and (d) for distance, we have db1+db2 = db (distance Bobby traveled ) and da1+da2 = da (distance traveled by the army)

Main equations:

xxxxxxxx

The ratio of the distances traveled for either bobby or the army, in part1 divided by the part2 are equal, since the times are the same (between bobby's t1 and the army's t1, etc.), and the speeds are constant. for example if the army moves at 5 mph for 2 hours (da1), then for 1 hour (da2) , its (5x2)/5 = 2 and bobby say moves at 20mph, for 2hours (db1) then for 1 hour (db2) = 20x2 / 20 = 2 also!!

so>>> db1/db2 = da1/da2

xxxxxxxx

When bobby travels to the Front, the distance he travels, is the same distance the army travels in the first part + the length of the army

so>>> db1=5+da1

xxxxxxxx

The second distance bobby travels, which is da2 is equal to the same distance the army traveled additionally in the first run to the front. The reason is bobby returns to the place where the army's Front was INITIALLY which is 5 miles off the starting point. so any distance travelled in the equation above of (db1=5+da1), he has to run back the same distance of da1, to get to the Initial location.

so>>> db2=da1

xxxxxxxx

da1+da2=5 as per the question, then da2 = 5-da1

Starting with the main equation: (say ^2 is a square)

db1/db2 = da1/da2, becomes:

db1/da1 = da1/(5-da1)

(5+da1)x(5-da1)= da1 x da1

5^2 - da1^2 = da1^2

25 = 2 x da1^2

da1^2= (25/2)

da1= 3.535533906

according to the equation above, db1=5+da1, so db1=8.535533906

and db2=da1 as bove, so db=db1+db2 = 8.535533906+3.535533906 = 12.07107

>>>>>>total distance traveled by bobby is = 12.07107

Edited October 26, 2009 by Ronin

[Guest]

If the question is a trick, then how "far" Bobby traveled is 5km

Sorry, thats Billy... and Miles not KM... but thats not the point right?

[Guest]

The army moves exactly 5 miles no matter how fast they are moving. In the time that the army moves 5 miles Billy makes it to the front them back to the back. Billy could not have traveled more than 10 mile due to the fact that the army only covers a total of 10 mile from the back at the start to the font at the finish.

On Billy’s way to the front the army should cover 2/3 the total distance traveled (5 miles) or twice what he would travel on the way back. On his way back he would only have to travel 1/2 the distance the army moved forward, or 1/3 the total distance.

He would travel 10 miles. This includes the 5 miles that the army traveled plus the distance for front to back of the army, or 5 miles, for a total of 10 miles.

[Guest]

The key is that the respective velocities of both the army and Billy are the same throughout.

Therefore, Va/Vb = C.

During the first part of the trip, the army travels Da₁ miles, and Billy travels Da₁ + 5 miles. Since each of these took the same amount of time,

the ratio of the velocities is Va₁/Vb₁ = Da₁/(Da₁+ 5).

For the total trip, the army travels 5 miles, and Billy travels 5 + 2Da₁. Similarly, Va_tot/Vb_tot = 5 / (5 + 2Da₁).

So,

Va/Vb = Va₁/Vb₁ = Va_tot/Vb_tot = Da₁/(Da₁+5) = 5 / (5 + 2Da₁).

Da₁ = SQRT (12.5) and Billy travels 5 + 2Da₁ or 12.071 miles.

Edited October 26, 2009 by smmills2