## The length of a river [solved]

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### The length of a river [solved]

Two ferry boats start at opposite sides of a river. They depart their stations and travel perpendicular to the shore at different but constant speeds until they reach the other station. They stop at the other station for exactly 20 minutes, and double back for their original station. The first time the boats "meet" eachother in the middle of the river, they are 720 yards from the first boat's station. The second time they "meet" eachother on the trip back, they are 400 yards from the other station. How wide is the river?
Edit: show reasoning
Edit: when I find a correct posting of the solution (with work shown), I will quote it and change the thread name.
Spoiler:
1760 yards.
Let the speed of the first ferry be u yards per minute, the speed of the second be v yards per minute, and the width of the river be L yards. Then the two ferries will meet after L/(u+v) minutes, at which point the first boat has traveled Lu/(u+v) yards from its station, which we are given to be 720 yards, so we find:

Lu/(u+v) = 720

Then each ferry continues across to the other station, waits at the station for 20 minutes, and returns to meet at the center. The first ferry leaves the opposite station at time L/u + 20; the second leaves at time L/v + 20. When will they meet (again)? They will meet at time T, where [T- (L/u + 20)] u + [T - (L/v + 20)] v = L, that is, when the total distance traveled by each ferry after they leave their rest stops is the width of the river. We can rewrite this as:

T (u+v) - 2L - 20(u+v) = L
T (u+v) = 3L + 20(u+v)
T = 3L/(u+v) + 20

An alternative way to produce T is as follows: Note that after meeting once already, the ferries must travel a total distance of 2L before they meet again, which they can accomplish in time 2L/(u+v). Add this to the time already spent reaching the first meeting point, L/(u+v), and add 20 minutes for the rest breaks they have, and you obtain the same result.

How far has the first ferry traveled from the second station to reach this point? The time it took to travel the distance is T - (L/u +20), or 3L/(u+v) - L/u, and its velocity is still u, so the distance is 3Lu/(u+v) - L, which we are given is 400 yards.

3Lu/(u+v) - L = 400

Use the relation we already know, that Lu/(u+v) = 720, and we obtain

3 * 720 - L = 400
L = 3 * 720 - 400 = 1760

If we want, we can continue to solve for the relative sizes of u and v. We know that Lu/(u+v) = 720, so (u+v)/u = 1760/720 = 22/9, so v/u = 13/9. Let us check our values for v = 130 yards per minute and u = 90 yards per minute. How long does it take the first boat to go 720 yards? It takes 720/90 = 8 minutes, in which time the second boat has gone 130*8 = 1040 yards, which is precisely the remaining distance to the opposite shore. The first boat continues to the opposite shore, which it reaches in a total amount of time 19+(5/9) minutes, and the second boat reaches its opposite shore at time 13+(7/13) minutes. Then the boats leave their respective shores when the clock reads 39+(5/9) minutes and 33+(7/13) minutes. How long does it take the first boat to travel 400 yards? Only 400/90 = 4+(4/9) minutes, bringing its clock up to 44 minutes even. The second boat then has 10+(6/13) minutes in the water, giving it time to travel 130 (10 + 6/13) = 1360 yards, which again is the exact distance it had to travel to meet the first boat. So the answer of 1760 yards passes the sanity check. Once you notice that the 20-minute wait has no ultimate bearing on the problem, then it becomes clear that only the ratio of the ferry speeds matters, making it plausible to solve a problem with three unknowns (L, u, and v) and only two pieces of information (the meeting points).

Well done Owehn, both on speed and completeness.
Last edited by headprogrammingczar on Sat Oct 27, 2007 1:12 pm UTC, edited 3 times in total.

Vengeful Donut
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### Re: The length of a river

Spoiler:
1280

Owehn
Posts: 479
Joined: Tue Oct 09, 2007 12:49 pm UTC
Location: Cambridge, UK

### Re: The length of a river [unsolved]

Spoiler:
1760 yards

Spoiler:
Let the speed of the first ferry be u yards per minute, the speed of the second be v yards per minute, and the width of the river be L yards. Then the two ferries will meet after L/(u+v) minutes, at which point the first boat has traveled Lu/(u+v) yards from its station, which we are given to be 720 yards, so we find:

Lu/(u+v) = 720

Then each ferry continues across to the other station, waits at the station for 20 minutes, and returns to meet at the center. The first ferry leaves the opposite station at time L/u + 20; the second leaves at time L/v + 20. When will they meet (again)? They will meet at time T, where [T- (L/u + 20)] u + [T - (L/v + 20)] v = L, that is, when the total distance traveled by each ferry after they leave their rest stops is the width of the river. We can rewrite this as:

T (u+v) - 2L - 20(u+v) = L
T (u+v) = 3L + 20(u+v)
T = 3L/(u+v) + 20

An alternative way to produce T is as follows: Note that after meeting once already, the ferries must travel a total distance of 2L before they meet again, which they can accomplish in time 2L/(u+v). Add this to the time already spent reaching the first meeting point, L/(u+v), and add 20 minutes for the rest breaks they have, and you obtain the same result.

How far has the first ferry traveled from the second station to reach this point? The time it took to travel the distance is T - (L/u +20), or 3L/(u+v) - L/u, and its velocity is still u, so the distance is 3Lu/(u+v) - L, which we are given is 400 yards.

3Lu/(u+v) - L = 400

Use the relation we already know, that Lu/(u+v) = 720, and we obtain

3 * 720 - L = 400
L = 3 * 720 - 400 = 1760

If we want, we can continue to solve for the relative sizes of u and v. We know that Lu/(u+v) = 720, so (u+v)/u = 1760/720 = 22/9, so v/u = 13/9. Let us check our values for v = 130 yards per minute and u = 90 yards per minute. How long does it take the first boat to go 720 yards? It takes 720/90 = 8 minutes, in which time the second boat has gone 130*8 = 1040 yards, which is precisely the remaining distance to the opposite shore. The first boat continues to the opposite shore, which it reaches in a total amount of time 19+(5/9) minutes, and the second boat reaches its opposite shore at time 13+(7/13) minutes. Then the boats leave their respective shores when the clock reads 39+(5/9) minutes and 33+(7/13) minutes. How long does it take the first boat to travel 400 yards? Only 400/90 = 4+(4/9) minutes, bringing its clock up to 44 minutes even. The second boat then has 10+(6/13) minutes in the water, giving it time to travel 130 (10 + 6/13) = 1360 yards, which again is the exact distance it had to travel to meet the first boat. So the answer of 1760 yards passes the sanity check. Once you notice that the 20-minute wait has no ultimate bearing on the problem, then it becomes clear that only the ratio of the ferry speeds matters, making it plausible to solve a problem with three unknowns (L, u, and v) and only two pieces of information (the meeting points).
[This space intentionally left blank.]

Meran
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Joined: Fri Mar 30, 2007 10:27 pm UTC

### Re: The length of a river [unsolved]

Spoiler:
1040 yards

Spoiler:
Suppose that the two boats are A and B, travelling at velocities Va and Vb respectively ( in yards per minute ) and that the width of the river is x.

At time t1 they both reach 720 yards from A's starting point (we'll reference that as 0).
(1) t1 = 720/Va = (x - 720)/Vb

At time t2 they both reach 400 yards from B's starting point (or x-400 yards from A's starting point).
(2) t2 = x/Va + 20 minutes + 400/Va = x/Vb + 20 minutes + (x-400)/Vb

Now time to do lots of maths... (I screwed this up the first time, so hopefully it's right now!)

From (1) we get...
720 + ( (720 * Vb) / Va ) = X

From (2) we get...
x * Vb + 400 * Vb = x * Va - 400 * Va
=> x = ( -400 * Va - 400 * Vb ) / ( Vb - Va ) = 720 + ( 720 * Vb ) / Va ( from (1) )

Solve for Va in terms of Vb...
... left as an exercise to the reader ... (apply quadratic formula yay!)
Va = 2.25 Vb

Subbing that back in to (1) we get...
x = 720 + ( (720 * Vb) / (2.25 * Vb) ) = 1040 yards

qed

Gwydion
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### Re: The length of a river [unsolved]

My answer jives with Owehn's, though I think my methods are a bit more brute-force-ish while his are probably more aesthetically pleasing. He gets all the credit, though if you didn't understand how he got where he got, try this:

Spoiler:
Here, let Va and Vb be the respective speeds of the boats, and let L be the width of the river. Therefore, at the first intersection 720/Va = (L-720)/Vb, since distance over velocity is time. At the second, by the same logic, 20 + (400+L)/Va = 20 + (2L-400)/Vb. The 20's cancel, leaving (400+L)/Va = (2L-400)/Vb.

From here, you can solve the first equation for L in terms of the other two, and plug into the second equation (do the math yourselves, I'm skipping a few steps here), which will eventually simplify into:
400Vb + 720(1+Vb/Va) = 1440Va*(1+Vb/Va) - 400Va
[Multiply through by Va, combine like terms]
720Vb^2 - 320VaVb - 1040Va^2 = 0
9Vb^2 - 4VaVb - 13Va^2 = 0
(Vb + Va) (9Vb - 13Va) = 0

Since neither speed can be negative, the first half drops out, leaving Vb/Va = 13/9. Plugging back into our original equation for L, we get L = 720*(22/9) = 1760 yards.

Mouffles
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Location: New Zealand

### Re: The length of a river [unsolved]

I much prefer the geometric solution, myself.
Spoiler: twoboats.JPG (23.13 KiB) Viewed 2314 times
The vertical axis is distance, where A and D are the two stations. The horizontal axis is time. The blue and red lines are the paths of the boats, intersecting at P1 and P2 (I left out the 20 minute stops because they don't change the answer, which you can see on the diagram: they would just mean moving the two right hand lines to the right, without changing the distance of P2 from the stations).

We know that CD = 720 and AB = 400. Let the width of the river, AD, be L; then we also know that AE = DF = L.

Because EFP2 and ADP1 are similar triangles, we have the following equation:
EB/BF = CD/AC
(L+400)/(2L-400) = 720/(L-720)
And solve this to get L = 1760.
In the spirit of taking things too far - the 5x5x5x5x5 Rubik's Cube.

Macbi
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### Re: The length of a river [unsolved]

Mouffles wrote:I much prefer the geometric solution, myself.

My method was geometric as well, but not nearly as pretty. (I worried too much about the stops)
Indigo is a lie.
Which idiot decided that websites can't go within 4cm of the edge of the screen?
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immute
Posts: 20
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### Re: The length of a river [unsolved]

Spoiler:
1760

I used basic algebra to solve.

immute
Posts: 20
Joined: Sun May 27, 2007 10:29 pm UTC

### Re: The length of a river [solved]

HEY! I should at least get credit for being still in high school and solving it all by myself! 