Two ferry boats start moving at the same instant from opposite sides of a river.

One boat is faster than the other so that they meet (pass each other that is, not bash into each other) at a point 720 yards from the nearest bank.

After arriving at their destinations, each boat remains ten minutes in the slip to change passengers; then it starts on its return trip.

The boats meet again at a point 400 yards from the other bank.

What is the exact width of the river ?

Guest Dec 1, 2015

#1**+10 **

This problem comes from one of the greatest American puzzleists of all time.....Sam Lloyd.....it seems more difficult than it really is......

Let W = the river's width

When they first meet, they have traveled the same amout of time.....one boat has traveled 720 yards and the other has traveled W-720 yards......let the rate of the first boat = R1 and the rate of the second boat = R2 ....and we have that.....

T1 = T2 ....or......

[720]/R1 = [W - 720]/R2 → R2/ R1 = [W- 720] / [720] (1)

Wnen they next meet, they have again traveled the same amount of time......The first boat has traveled the width of the river, (W), plus 400 yards and the second boat has traveled 2W - 400....again we have that

T3 = T4 ....or.....

[W + 400] /R1 = [ 2W - 400] / R2 → R2/R1 = [2W - 400] / [W + 400] (2)

Equating (1) with (2), we have

[ W - 720] / 720= [2W - 400] [W + 400]

720[2W - 400] = [W - 720] [W + 400]

1440W - 288000 = W^2 - 320W - 288000

W^2 - 1760W = 0

W(W - 1760) = 0

Reject W = 0......and the solution is.....W = 1760 yards.....exactly 1 mile wide !!!!

CPhill Dec 1, 2015

#1**+10 **

Best Answer

This problem comes from one of the greatest American puzzleists of all time.....Sam Lloyd.....it seems more difficult than it really is......

Let W = the river's width

When they first meet, they have traveled the same amout of time.....one boat has traveled 720 yards and the other has traveled W-720 yards......let the rate of the first boat = R1 and the rate of the second boat = R2 ....and we have that.....

T1 = T2 ....or......

[720]/R1 = [W - 720]/R2 → R2/ R1 = [W- 720] / [720] (1)

Wnen they next meet, they have again traveled the same amount of time......The first boat has traveled the width of the river, (W), plus 400 yards and the second boat has traveled 2W - 400....again we have that

T3 = T4 ....or.....

[W + 400] /R1 = [ 2W - 400] / R2 → R2/R1 = [2W - 400] / [W + 400] (2)

Equating (1) with (2), we have

[ W - 720] / 720= [2W - 400] [W + 400]

720[2W - 400] = [W - 720] [W + 400]

1440W - 288000 = W^2 - 320W - 288000

W^2 - 1760W = 0

W(W - 1760) = 0

Reject W = 0......and the solution is.....W = 1760 yards.....exactly 1 mile wide !!!!

CPhill Dec 1, 2015

#2**+5 **

I like Sam Lloyd's solution to this problem.

Since the faster of the two ferry boats will cross the halfway point first, it follows that the 'nearest bank' is the one from which the slower boat has left and in which case the slower boat will have travelled 720 yards when they meet.

Also, the sum of the distances travelled by both boats (when they meet), will equal the width of the river.

When they next meet, their combined total distances travelled will be three times the width of the river, and in which case the slower of the boats will have travelled 3*720 = 2160 yards.

Since this is the width of the river plus 400 yards, it follows that the width of the river is

2160 - 400 = 1760 yards.

- Bertie

Guest Dec 2, 2015