A tad around 8 days

(But remember, cows will not eat around places where they have 'staled' ...
therefore in the above, I have not calculated for the fact that in a week, 16 cows will have probably covered the whole 7 acres in S***.)

Problem solved!

---

ITI SAPIS
POTANDA
TINONE

Correct answer by Dominique!!!
Answer: 70 Days

1 Point to Dominique makes no difference to the Table Stats.... But It's always one step closer....

Standings Again:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 5 Points
4. Dominique - 3 1/2 Points
5. Sarge - 2 Points
6. John McLeod VII - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

A 1-Point Question to Finish off my Day....

Q24. A column of soldiers one mile long is marching forward at a constant rate. The soldier at the front of the column has to deliver a message to the soldier at the rear. He breaks rank and begins marching toward the rear at a constant rate while the column continues forward. The soldier reaches the rear, delivers the message and immediately turns to march forward at a constant rate. When he reaches the front of the column and drops back in rank, the column has moved one mile. Question- How far did the soldier delivering the message march?

Best Regards,
Luke.

---

- Luke.

I'm not letting this thread die anytime soon.... re-post...

Correct answer by Dominique!!!
Answer: 70 Days

1 Point to Dominique makes no difference to the Table Stats.... But It's always one step closer....

Standings Again:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 5 Points
4. Dominique - 3 1/2 Points
5. Sarge - 2 Points
6. John McLeod VII - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

A 1-Point Question to Finish off my Day....

Q24. A column of soldiers one mile long is marching forward at a constant rate. The soldier at the front of the column has to deliver a message to the soldier at the rear. He breaks rank and begins marching toward the rear at a constant rate while the column continues forward. The soldier reaches the rear, delivers the message and immediately turns to march forward at a constant rate. When he reaches the front of the column and drops back in rank, the column has moved one mile. Question- How far did the soldier delivering the message march?

Best Regards,
Luke.

---

- Luke.

I'm not letting this thread die anytime soon.... re-post...

Correct answer by Dominique!!!
Answer: 70 Days

1 Point to Dominique makes no difference to the Table Stats.... But It's always one step closer....

Standings Again:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 5 Points
4. Dominique - 3 1/2 Points
5. Sarge - 2 Points
6. John McLeod VII - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

A 1-Point Question to Finish off my Day....

Q24. A column of soldiers one mile long is marching forward at a constant rate. The soldier at the front of the column has to deliver a message to the soldier at the rear. He breaks rank and begins marching toward the rear at a constant rate while the column continues forward. The soldier reaches the rear, delivers the message and immediately turns to march forward at a constant rate. When he reaches the front of the column and drops back in rank, the column has moved one mile. Question- How far did the soldier delivering the message march?

Best Regards,
Luke.

too many unknowns.

Let d stand for the deliverer and c stand for the column. Also let b stand for back and f for forward.
r and t will, of course, stand for rate and time. rate * time = distance.
Consider the distances traveled over the time it takes for the courier to deliver the message to the back of the column and then during the return to the front of the column.
Since the column moves forward while the courier moves to the back of the line, the courier marches less than 1 mile. Call this amount x.

(1) r_d * t_b = x (the distance the courier travels back),
(2) r_d * t_f = 1 (the courier must cover the entire length of the column on the return trip),
(3) r_c * t_b = 1 - x,
and (4) r_c * t_f = x.

Substitute (4) into (3).

(5) r_c * t_b = 1 - r_c * t_f.

Combine the terms with r_c. (t_b + t_f) * r_c = r_c * t_b + r_c * t_f = 1.
Therefore, r_c = 1 / (t_b + t_f). (6) x = r_c * t_f = t_f / (t_b + t_f).

Isolate r_d in (2). r_d = 1 / t_f. Substitute this into (1). (7) x = t_b / t_f.

Therefore, by (6) and (7), (8) (t_b / t_f) = (t_f / (t_b + t_f)).

In (8), cross-multiply: t_b * t_f + t_b^2 = t_f^2. Solve for t_f. We obtain two possibilities.

t_f = ((1/2) + (1/2) * sqrt(5)) * t_b or t_f = ((1/2) - (1/2) * sqrt(5)) * t_b. (The first of these is the Golden Ratio, Phi.)

Substitute each of these into (7). Again, we obtain two possibilities, this time for x.

Either x = 2 / (1 + sqrt(5)), which is approximately x = .6180339886,
or x = 2 /(1 - sqrt(5)), which is approximately x = -1.618033988.

Since x is a distance and direction has already been accounted for from the start, x cannot be negative. The latter must be rejected.

Therefore, x = 2 / (1 + sqrt(5)), the distance the courier traveled back. Again, the courier then has to walk the entire length of the column, 1 mile, to return to the front.

1 + x = (3 + sqrt(5)) / (1 + sqrt(5)) = (-2 - 2 * sqrt(5)) / (-4) = (1 + sqrt(5)) / 2.

The total distance is the Golden Ratio, Phi, approximately 1.6180339886.

---

Capitalize on this good fortune, one word can bring you round ... changes.

The total distance is the Golden Ratio, Phi, approximately 1.6180339886

Is this just spooky or is there a reason for it ?

This will simplify the solution, perhaps, and maybe elaborate why it must be Phi.

1 / x = (distance courier forward) / (distance courier back) = (r_d * t_f) / (r_d * t_b) = t_f / t_b = (r_c / r_c) * (t_f / t_b) = (r_c * t_f) / (r_c * t_b) = (distance column while courier goes forward) / (distance column while courier goes back) = x / (1 - x).

The relationship here begins to give it away:
1 / x = x / (1 - x)

===> x^2 = 1 - x ===> x^2 + x - 1 = 0.
(x + 1)^2 + (x + 1) - 1 = (x^2 + 2x + 1) + (x + 1) - 1 = (x^ 2 + x - 1) + (2x + 1 + 1) = (x^2 + x - 1) + (2x + 2) = 0 + 2(x + 1) = 2(x + 1)

As does this version of the equation:
===> (x + 1)^2 - (x - 1) - 1 = 0. Solve for x + 1, not x.

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Capitalize on this good fortune, one word can bring you round ... changes.

(x + 1)^2 - (x - 1) - 1 = 0. Solve for x + 1, not x.

This should read as (x + 1)^2 - (x + 1) - 1 = 0.

---

Capitalize on this good fortune, one word can bring you round ... changes.

Let d stand for the deliverer and c stand for the column. Also let b stand for back and f for forward.
r and t will, of course, stand for rate and time. rate * time = distance.
Consider the distances traveled over the time it takes for the courier to deliver the message to the back of the column and then during the return to the front of the column.
Since the column moves forward while the courier moves to the back of the line, the courier marches less than 1 mile. Call this amount x.

(1) r_d * t_b = x (the distance the courier travels back),
(2) r_d * t_f = 1 (the courier must cover the entire length of the column on the return trip),
(3) r_c * t_b = 1 - x,
and (4) r_c * t_f = x.

Substitute (4) into (3).

(5) r_c * t_b = 1 - r_c * t_f.

Combine the terms with r_c. (t_b + t_f) * r_c = r_c * t_b + r_c * t_f = 1.
Therefore, r_c = 1 / (t_b + t_f). (6) x = r_c * t_f = t_f / (t_b + t_f).

Isolate r_d in (2). r_d = 1 / t_f. Substitute this into (1). (7) x = t_b / t_f.

Therefore, by (6) and (7), (8) (t_b / t_f) = (t_f / (t_b + t_f)).

In (8), cross-multiply: t_b * t_f + t_b^2 = t_f^2. Solve for t_f. We obtain two possibilities.

t_f = ((1/2) + (1/2) * sqrt(5)) * t_b or t_f = ((1/2) - (1/2) * sqrt(5)) * t_b. (The first of these is the Golden Ratio, Phi.)

Substitute each of these into (7). Again, we obtain two possibilities, this time for x.

Either x = 2 / (1 + sqrt(5)), which is approximately x = .6180339886,
or x = 2 /(1 - sqrt(5)), which is approximately x = -1.618033988.

Since x is a distance and direction has already been accounted for from the start, x cannot be negative. The latter must be rejected.

Therefore, x = 2 / (1 + sqrt(5)), the distance the courier traveled back. Again, the courier then has to walk the entire length of the column, 1 mile, to return to the front.

1 + x = (3 + sqrt(5)) / (1 + sqrt(5)) = (-2 - 2 * sqrt(5)) / (-4) = (1 + sqrt(5)) / 2.

The total distance is the Golden Ratio, Phi, approximately 1.6180339886.

The column is not standing still whilst the courier is returning to the front, so he must have to travel 1 mile plus the distance that the column has moved forward whilst he is making his return journey?

F.

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That is accounted for.
But, in case it is not, then in my second version:

1 / x = (distance courier forward) / (distance courier back) = (r_d * t_f) / (r_d * t_b) = t_f / t_b = (r_c / r_c) * (t_f / t_b) = (r_c * t_f) / (r_c * t_b) = (distance column while courier goes forward) / (distance column while courier goes back) = x / (1 - x).

The relationship here begins to give it away:
1 / x = x / (1 - x)

change it to:

(1 + x) / x = x / (1 - x).

x^2 = (1 + x)(1 - x) = 1 - x^2.
2 * x^2 = 1.
x^2 = 1 / 2.
x = sqrt(2) / 2, with obvious rejection of the negative branch of the square root.

The total distance in this case would be 1 + 2x = 1 + sqrt(2).

And, if Luke does not respond beforehand, then I will check both cases by determining all rates and times involved to check for consistency. I, for one, don't mind brain-storming and sharing an idea, with refinement / correction / improved explanation a little later. :)

---

Capitalize on this good fortune, one word can bring you round ... changes.

That is accounted for.
But, in case it is not, then in my second version:

1 / x = (distance courier forward) / (distance courier back) = (r_d * t_f) / (r_d * t_b) = t_f / t_b = (r_c / r_c) * (t_f / t_b) = (r_c * t_f) / (r_c * t_b) = (distance column while courier goes forward) / (distance column while courier goes back) = x / (1 - x).

The relationship here begins to give it away:
1 / x = x / (1 - x)

change it to:

(1 + x) / x = x / (1 - x).

x^2 = (1 + x)(1 - x) = 1 - x^2.
2 * x^2 = 1.
x^2 = 1 / 2.
x = sqrt(2) / 2, with obvious rejection of the negative branch of the square root.

The total distance in this case would be 1 + 2x = 1 + sqrt(2).

And, if Luke does not respond beforehand, then I will check both cases by determining all rates and times involved to check for consistency. I, for one, don't mind brain-storming and sharing an idea, with refinement / correction / improved explanation a little later. :)

Thank you. Somehow, your answer "felt" right so I was almost reluctant to highlight what appeared to a logical flaw in the argument.

F.

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That is accounted for.
But, in case it is not, then in my second version:

1 / x = (distance courier forward) / (distance courier back) = (r_d * t_f) / (r_d * t_b) = t_f / t_b = (r_c / r_c) * (t_f / t_b) = (r_c * t_f) / (r_c * t_b) = (distance column while courier goes forward) / (distance column while courier goes back) = x / (1 - x).

The relationship here begins to give it away:
1 / x = x / (1 - x)

change it to:

(1 + x) / x = x / (1 - x).

x^2 = (1 + x)(1 - x) = 1 - x^2.
2 * x^2 = 1.
x^2 = 1 / 2.
x = sqrt(2) / 2, with obvious rejection of the negative branch of the square root.

The total distance in this case would be 1 + 2x = 1 + sqrt(2).

And, if Luke does not respond beforehand, then I will check both cases by determining all rates and times involved to check for consistency. I, for one, don't mind brain-storming and sharing an idea, with refinement / correction / improved explanation a little later. :)

@Sarge
It is after midnight over here so forgive me if I seem particularly dense at this moment, but I don't see where your (4) comes from.

In English " r_c * t_f = x " is saying that the distance the column marches in the time that the courier is trying to return to his position at the front is the same as the distance that he had to go backwards originally.

Why??

F.

---

Fred: for a moment, you had me worried. See, here, it is mid-evening and the sky is beginning to darken. But, I've been spending much of the day working on grading papers from a different area of mathematics. So, my typing up those two solutions was a bit of multi-tasking and there could indeed be errors.

However, on that part, I am sure. total time = time courier is going back + time courier returns to the front = t_b + t_f. If the courier went distance x during t_b, since the column was a mile long, the column is marching 1 - x of the distance so that the courier meets the back of the column. If I am correct about that 1 - x, then the x the distance of the column moving forward during t_f is because the column moved a total of 1 mile during the total time, leaving 1 - (1 - x) = x during t_f.

So, since we are both tired, either I am either making more sense or making my self sound even more ridiculous. :) Sometimes multi-tasking = bad.

---

Capitalize on this good fortune, one word can bring you round ... changes.

Correct Answer goes to Sarge!! Congratulations!!!

Standings Again:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 5 Points
4. Dominique - 3 1/2 Points
5. Sarge - 3 Points
6. John McLeod VII - 1 Point
7. Scary Capitalist - 1/2 Point
8. TBA.......

I have now exhausted my current directory of Maths Problems... more easier ones coming soon....

Best Regards,
Luke.

---

- Luke.

OK, Luke, was it the first solution or the one I offered after Fred W asked what he asked?

If the first, which of the two solutions did you find easier to follow? More convincing? More elegant? Etc. ... .

---

Capitalize on this good fortune, one word can bring you round ... changes.

Fred:

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Capitalize on this good fortune, one word can bring you round ... changes.

OK, Luke, was it the first solution or the one I offered after Fred W asked what he asked?

If the first, which of the two solutions did you find easier to follow? More convincing? More elegant? Etc. ... .

I'll give Fred 1/2 a point for his help and directions.....
As for your solution Sarge, Your first one was the "beautiful" one.... but your second one convinced me more...

I've got one more also....
Q25:If a cylindrical water tank has a height of 10 and a diameter of 7 calculate the amount of sheet metal needed to create this tank. also Calculate the volume of water that this cylindrical tank will hold.

Best Regards,
Luke.

---

- Luke.

Fred: for a moment, you had me worried. See, here, it is mid-evening and the sky is beginning to darken. But, I've been spending much of the day working on grading papers from a different area of mathematics. So, my typing up those two solutions was a bit of multi-tasking and there could indeed be errors.

However, on that part, I am sure. total time = time courier is going back + time courier returns to the front = t_b + t_f. If the courier went distance x during t_b, since the column was a mile long, the column is marching 1 - x of the distance so that the courier meets the back of the column. If I am correct about that 1 - x, then the x the distance of the column moving forward during t_f is because the column moved a total of 1 mile during the total time, leaving 1 - (1 - x) = x during t_f.

So, since we are both tired, either I am either making more sense or making my self sound even more ridiculous. :) Sometimes multi-tasking = bad.

Ain't it wonderful, the difference a night's sleep can make??

Of course - the link that I was missing was that, since the column had travelled 1-x while the courier was going back they only had x left to do whilst he was going forward to reach the 1 mile mark.

We are, of course, making an assumption here that may not be explicit in the question (reference WinterKnight's post) and that is that the "constant speed" heading south is the same "constant speed" as when he is heading north. Any thoughts, Luke?

F.

---

OK, Luke, was it the first solution or the one I offered after Fred W asked what he asked?

If the first, which of the two solutions did you find easier to follow? More convincing? More elegant? Etc. ... .

I'll give Fred 1/2 a point for his help and directions.....
As for your solution Sarge, Your first one was the "beautiful" one.... but your second one convinced me more...

I've got one more also....
Q25:If a cylindrical water tank has a height of 10 and a diameter of 7 calculate the amount of sheet metal needed to create this tank. also Calculate the volume of water that this cylindrical tank will hold.

Best Regards,
Luke.

Without calculator and assuming laser cutting to minimise material loss:

The parts can be cut from sheet that is 28.99 x 10 = 289.9 Area.

The volume is 384.89.

F.

---

OK, Luke, was it the first solution or the one I offered after Fred W asked what he asked?

If the first, which of the two solutions did you find easier to follow? More convincing? More elegant? Etc. ... .

I'll give Fred 1/2 a point for his help and directions.....
As for your solution Sarge, Your first one was the "beautiful" one.... but your second one convinced me more...

I've got one more also....
Q25:If a cylindrical water tank has a height of 10 and a diameter of 7 calculate the amount of sheet metal needed to create this tank. also Calculate the volume of water that this cylindrical tank will hold.

Best Regards,
Luke.

Without calculator and assuming laser cutting to minimise material loss:

The parts can be cut from sheet that is 28.99 x 10 = 289.9 Area.

The volume is 384.89.

F.

Do we have rounding errors here?
2*Pi*3.5^2+2*Pi*3.5*10 is approximately 296.8805058 (referring to surface area, the amount of material needed for construction of the tank).

As for a night's worth of sleep, I only got 5 hours. :( It took me 30-45 minutes to wind down after I stopped doing work last night. And, in 30-60 minutes I'll be doing more work.

Luke: interesting, I figured the second solution would be the one you'd consider elegant.

---

Capitalize on this good fortune, one word can bring you round ... changes.

Fred: for a moment, you had me worried. See, here, it is mid-evening and the sky is beginning to darken. But, I've been spending much of the day working on grading papers from a different area of mathematics. So, my typing up those two solutions was a bit of multi-tasking and there could indeed be errors.

However, on that part, I am sure. total time = time courier is going back + time courier returns to the front = t_b + t_f. If the courier went distance x during t_b, since the column was a mile long, the column is marching 1 - x of the distance so that the courier meets the back of the column. If I am correct about that 1 - x, then the x the distance of the column moving forward during t_f is because the column moved a total of 1 mile during the total time, leaving 1 - (1 - x) = x during t_f.

So, since we are both tired, either I am either making more sense or making my self sound even more ridiculous. :) Sometimes multi-tasking = bad.

Ain't it wonderful, the difference a night's sleep can make??

Of course - the link that I was missing was that, since the column had travelled 1-x while the courier was going back they only had x left to do whilst he was going forward to reach the 1 mile mark.

We are, of course, making an assumption here that may not be explicit in the question (reference WinterKnight's post) and that is that the "constant speed" heading south is the same "constant speed" as when he is heading north. Any thoughts, Luke?

F.

I rejected the obvious answer, because with the squad marching at at any normal speed (3mph or greater) it would mean the courier would have to be a international grade athlete in running gear to reach the front of the column after it had completed that mile.