Four Rowers and a Swan
Four rowers wish to cross the river by means of a boat that can only hold two men. The rower R1 needs one minute to cross the river alone, and the rowers R2, R3, and R4 need 2, 6, and 9 minutes respectively. Since the boat will hold only two men at a time, the rowers planned an optimum strategy for crossing the river. At the moment the rowers implement their plan, a swan starts to swim across the river at a speed of 60 feet per minute. The swan reaches the other side precisely at the moment the rowers complete their plan of getting all across the river.
What was the plan? Who rows with who?
How long did it take to accomplish the plan?
How wide is the river?
What was the plan? Who rows with who?
How long did it take to accomplish the plan?
How wide is the river?
Labels: logic, mathschallenge
posted by Zaux at
5:24 PM
26 Comments:
I suspect there is a bug in the question. If not, R1 does all the rowing and simply ferries everyone across. Total crossings is 5 and takes 5 minutes. Assuming no current, the swan must travel 300 feet = width of the river.
Assume: two rowers in boat must row at slowest rowers speed
R1+R2 cross= 2 min
R1 returns= 1 min
R3+R4 cross=9 min
R2 returns= 2 min
R1+R2 cross= 2 min
Total=16 min
60ft/min*16 min= 960 feet
Cam
hey Cam-it seems to me that if R1 has someone else in the boat helping him,then they should be under a minute.
eg.R1+R2=45 sec...(thats a guess)
According to the problem source, 2 rowers will cross the river more quickly than a single rower.
Assume: two rowers have velocity added to eachother
v=L/T
V_R1=L/1
V_R2=L/2
V_R3=L/6
V_R4=L/9
R1+R2 cross= L/((1+1/2)L)=1/1.5=2/3 min
R1 returns= 1 min
R3+R4 cross= L/((1/6+1/9)L)=54/15 min
R2 returns= 2 min
R1+R2 cross= 2/3 min
Total=7.93 min
60ft/min*7.93 min= 475.8 feet
Cam
meh scratch that last one. the single rowers can't be optimal
Cam
i know i'm wrong but just going to throw this out there.
river is 213.3 feet wide
and freakin cold
Freakin' cold it is ... but not 213.3 ft.
I've nicked Cam's outline:
Assume: two rowers have velocity added to each other
R1+R2 => 1/(1 + 1/2) = 2/3 min
R1 returns => 1 min
R1+R3 cross => 1/(1 + 1/6) = 6/7 min
R1 returns => 1 min
R1+R4 cross => 1/(1 + 1/9) = 0.9 min
Total= 2/3 + 1 + 6/7 + 1 + 0.9 = 4.4238 mins (approx)
60ft/min * 4.43 min = 265.43 feet approx.
I see no ships!
i see all kinds of ships but that might be the mushroons.
not a correct answer yet ...
Got dragged away before I could repost. Anyhow, I confirm Chris's result.
And the water is cold!
Cam
According to the source:
Trips:
R1, R4 cross
R1 returns
R2, R3 cross
R2 returns
R1, R2 cross
Time:
91/15 minutes
River width:
Swan speed x time = 364 ft.
I don't understand why that is the answer.
Just to see if the solution is using added velocities:
9/10+1+12/8+2+2/3
=1/120*(108+120+180+240+80)
=1/120*(728)
=91/15~=6.1 min
So it appears to be...
Unless there is a a constraint on how many times R1 can cross I
have to stick with the 4.4 min solution.
Cam
Don't have time tonight... will try to post the lengthy solution tomorrow. Need a little help ... I've notice you guys indicate powers of a number n in this manner:
n^2 n^3 etc.
How is the best way to indicate subscripts?
Nsub1 Nsub2 etc.
My lack of mathematical prowess is revealing itself ... heh heh.
...oops. I accidentally deleted confirmation that the expected velocity addition rule seems to have been used.
I'm sure that the question hasn't been presented fully.
I presented the problem as fully as the source did ... heh heh.
Hi Zaux. Subscripts e.g.s: a_1, a_a is the safest. A1 and a1 are OK too, but Aa and aa are dodgy. It's usually clear from the context when suffices are being used. Array notation, A[1], A[a] might be suitable.
Perhaps your source wasn't good. These things happen. I frequently post problems that I haven't checked sufficiently. It probably made the problem more stimulating than it might have been.
Maybe so ... I will post "their" solution tomorrow sometime.
First, I will post the original wording of the problem ... and later I will post "their" solution:
Four rower-mathematicians wish to cross the river by means of a boat that can only hold two men. The rower R₁needs 1 minute to cross the river alone, R₂, R₃, R₄need
2, 6, and 9 minutes respectively. Being mathematicians, the rowers have planned an optimal strategy for crossing the river. They start to row toward the opposite river bank in a straight line. At the very moment they begin to row, a swan starts to swim over the river in a straight line with the speed of 60 feet per second. The swan reaches the opposite river bank at the exact moment when the rowers complete their transfer. How wide is the river?
I don't believe my wording of the problem violates any of the conditions presented in this original wording.
Isn't this a variation on the "Microsoft" question?
Here's the documented solution of the problem: (where possible, I will use actual subscripts, such as N₁... in situations where Character Map does not offer the needed subscript, I will use the form N_1, as suggested by Chris)
Here goes:
First, notice that a joined pair of rowers will cross the river faster than each of them individually. To find the speed of any pair of rowers, let us assume that an energy E is needed for one transfer. If F_k stands for the power of the rower R_k necessary for one crossing which lasts t_k minutes (k= 1,2,3,4), then
E=F₁t₁=F₂t₂= F₃t₃=F₄t₄(ref.A)
where t₁=1, t₂= 2, t₃=6, t₄=9
(expressed in miutes).
From ref.A (above)it follows that
F_k = E/t_k. The time t_ij for crossing the river by any two rowers R_i and R_j is equal to
t_ij = E/F_i+F_j = E/E/t_i+E/t_j = t_it_j/t_i+t_j.
Let -->R and <--R denote forward-trip and backward trip of a rower R over the river. The optimal strategy for crossing the river is as follows:(1-5 indicate trips)
(crossing time - min.)
1. -->R₁R4 t₁₄=t₁t₄/(t₁+t ₄)= 9/10 min.
2. <--R₁ t₁= 1 min.
3. -->R₂R₃ t ₂₃=(t₂t₃/(t₂+t₃)= 3/2 min.
4. <--R₂ t₂= 2 min.
5. -->R₁R₂ t₁₂= 2/3 min.
From the above scheme we find that the total time of all transfers of the rowers is:
t = t₁₄+ t₁+ t₂₃+ t₂+ t₁₂=
9/10 + 1 + 3/2 + 2 + 2/3 =
91/15 min.
The width of the river is equal to the distance traveled by the swan:
(60 ft./min.) x (91/15 min.)=
364 ft.
Answer: 364ft.
Hi Zaux. Thanks for that. Especial thanks for finding some subscript numbers - I've got a copy of subscript 1-4 now).
It seems that the velocity addition formula is the same as the one that Cam and I used. So the official solution is not optimal. I guess that rowers are being allowed to have a rest.
I concur with Chris on this one. The failing in the official solution is in its non-optimal selection of rowers. Steps 3 and 4 can clearly be shortenned by using R1 instead of R2.
Cam
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