### Riverboat Race

This is yet another kind of puzzle. Really it's just a math problem, with a twist (not a trick, just a legitimate technique that isn't immediately obvious to most). It does rate as a puzzle (to me) because it appears quite easy at first. But once into the problem, some will think that I have not provided enough information. Good luck!

Two boats on opposite shores of a river start moving toward each other at a constant but different speeds. Neglect all other factors, such as wind and current speed, etc. When they pass the first time they are 700 yards from one shoreline. They each continue to the opposite shore, turn around (instantly) and start moving toward each other again. When they pass the second time they are 300 yards from the other shoreline. How wide is the river?

If you give up, here is the solution:
http://blogs.sun.com/roller/resources/dcb/Boat.gif

```
The distance is a function of the speeds of the
two boats with boundaries at 700 and 1000 (non-inclusive)

distance = 400 + 600(1/(1+a/b))

a and b are the speeds of the boats such that b>a

```

Posted by whatsinaname on March 10, 2005 at 02:16 AM EST #

Nope. The river's width is fixed! There is a single correct answer. Hint, there are many combinations of boat speeds that work. But the river doesn't change.

Posted by Dave Brillhart on March 10, 2005 at 02:41 AM EST #

878.2329

Posted by whatsinaname on March 10, 2005 at 05:25 AM EST #

Nope. I can send you the answer off-line if you'd like. Or, try again. Hint #2 - the answer is less than a mile, and is a integer value of yards.

Posted by Dave Brillhart on March 10, 2005 at 05:40 AM EST #

2100.

Posted by Chad Mynhier on March 10, 2005 at 06:06 AM EST #

Posted by Chad Mynhier on March 10, 2005 at 06:15 AM EST #

Yep!! 1800 yards is exactly correct. I'll post my solution with the logic. I used a sequence of equations. There is also a clever way to solve this using ratios.

Posted by Dave Brillhart on March 10, 2005 at 06:22 AM EST #

```
Firstly 1800 is the right answer.

Secondly I had 4 scenarios in mind arising out
of the following 2 considerations:

1) When the boats initially meet 700 yards could
be the distance from them to either side of the
shore.
2) When they meet again they dont necessarily have
to be going in the opposite direction.

A reevaluation of the question invalidates the
latter 2 possibilities. Leaving only the first
two  which yield the same set of equations.

one of the solutions for the former is 878.2 which
is wrong for the question clearly states that
both the boats turn around and start back towards
each other.. Argh....(stupid me)

```

Posted by whatsinaname on March 10, 2005 at 01:59 PM EST #

Posted by whatsinaname on March 10, 2005 at 02:01 PM EST #

I just used basic algebra to get 1800. It's similar to Dave's solution, but don't need to solve for the time.

• W = width of river
• X = speed of first boat
• Y = speed of second boat
• T1 = time to first meeting (from a T0)
• T2 = time to second meeting (from a T0)
```Y\*T1 = 700
X\*T1 = W - 700

Since T1 is the same for both, solve each for T1
and then rearrange as ratio:

X/Y = (W - 700)/700

Now do the same for T2:

Y\*T2 = W + 300
X\*T2 = W + (W -300) = 2W -  300

solve for T2, etc., and we have

X/Y = (2W - 300)/(W + 300)

since the two ratios are even, solve for W:

(W+300)(W-700) = 700(2W - 300)
[...]
W\^2 - 1800W = 0
W(W-1800) = 0

W = either 0 or 1800.  I'll pick 1800.

```

Posted by Kevin on March 10, 2005 at 03:25 PM EST #

That's great Kevin... Here is a solution my dad sent to me:
1.At T1, combined distance traveled = x
2.When boats turn, combined distance = 2x
3.At T2, combined distance = 3x
4.At T2, each boat has traveled 3x more than at T1
5."B" traveled 700yds @ T1, and 2100yds @ T2
6."B" traveled 300yds from remote shore at T2
7.So, 2100yds must be 300yds greater than x
8.Therefore, width (x) 2100-300 = 1800yds!

Posted by Dave Brillhart on March 10, 2005 at 03:47 PM EST #

Comments are closed for this entry.

dcb

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