# Algebraic Solution to a Puzzle

1. Jun 12, 2012

### fishtail

On a recent Car Talk Show there recently was a puzzle from the "Hat in the River" series. The hosts of the show emphasized that the puzzle should be solved by thinking and not by using algebra which may confuse things.

So instead of thinking I tried to use algebra to try to solve the puzzle. I tried to get it down to two equations and two unknowns and it will be easy. So I thought, but I need help.
Here is the puzzle.
THE PROBLEM:
A vacationer decides to rent a rowboat and go for a ride on a river. He rows upstream for one mile and his hat falls in the river and the river's current carries the hat away from him towards the dock where he started. He continues rowing for 10 more minutes(0.1667 hours) in the same direction.
He instantly turns around after 10 minutes and rows to retrieve the hat which has flowed away from him. He rows with the same effort downstream as he rowed upstream. The hat and the boat arrive at the dock at the same time.
What was the speed of the current?

ATTEMPT AT A SOLUTION:

Vb=speed of boat(no current)
Vc= speed of current
Vb+Vc=speed of boat downstream
Vb-Vc=speed of boat upstrem

Some equations: T(h)=time for hat to reach the dock=1 mile /Vc or simply 1/Vc

T(b)=time for boat to reach the dock. T(b) is the sum of three parts of the trip

First part: 10 minutes(0.1667 hours)he continued rowing upstream after he lost his hat.

Second part: 0.1667x(Vb-Vc)/(Vb+Vc) is the time it takes the rower to row from the turnaround to where is hat fell in the water.This is downstream.

Third part: 1 mile/(Vb+Vc) = time it takes to complete the last mile to the dock.

We know: T(h)=T(b) so
1/Vc = 0.1667+.1667x(Vb-Vc)/(Vb+Vc) +1/(Vb+Vc)
Now I am stuck and I need help.I have one equation and two unknowns(Vb,Vc).
I need one more equation with these variables. Anybody have any ideas for the second equation.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 13, 2012

### azizlwl

Can you show how the hosts solved it?
By algebra:

(t2+t2)(Vman+Vriver)-10(Vman-Vriver)=1mile.

(10+t1+t2)vr=1mile

1 equation for 2 unknowns

http://img35.imageshack.us/img35/6119/upriver.jpg [Broken]

Last edited by a moderator: May 6, 2017
3. Jun 13, 2012

### Ratch

fishtail,

Forget the algebra and how fast the boat and stream are moving. No matter which direction the boat moves away from the bottle, it will take the same amount of time to get back to the bottle. So if it takes 20 mins to get back to the bottle, and the bottle has travelled 1 mile, the stream velocity is 3 mph.

Ratch

Last edited: Jun 13, 2012
4. Jun 13, 2012

### azizlwl

He continues rowing for 10 more minutes(0.1667 hours) in the same direction.
He instantly turns around after 10 minutes
--------
He is in 2 places 10 mins. after the hat dropped onto the river.
One picking his hat
The other making a u-turn.

5. Jun 13, 2012

### Ratch

azizlwl,

Yes, I made a stupid arithmetic error. It takes him 20 min to catch up to his hat, so the stream flow is 3 mph. Anyway, the thought is there. I edited my previous post to correct this error.

Ratch

6. Jun 13, 2012

### fishtail

Azizlwl. I dont understand your equations .You have two equations and four unknowns.

7. Oct 8, 2013

### jrhd00d

Awww seriously. I am trying to answer the same problem. People keep saying that answer is 3mph but they just don't understand the situation. Look he doesn't catch up to hat at the exact place where dropped it. He picks it up where the story began 1 mile behind where he dropped his at at the dock. And besides, he will be rowing faster going back for his hat because he is going DOWNSTREAM!!!! (HE ROWS WITH THE SAME EFFORT AS HE DID GOING UPSTREAM). This isn't a 10minutes, then 10 minutes, then another 10 minutes question.

8. Oct 8, 2013

### Staff: Mentor

Vb=speed of boat relative to current (miles/min)
Vc=speed of current (miles/min)
t = time for boat to return to dock and catch up with hat (min) after turning around

Boat: 1 + 10(Vb-Vc)-(Vb+Vc)t = 0
Hat: 1 - (t+10)Vc =0

Rearrange and refactor Boat equation:

1 - (t+10)Vc + Vb(10-t) =0

Subtract hat equation from this equation to get:

Vb(10-t) =0

Since Vb is not equal to 0, t = 10. Substitute this into Hat equation to get

Vb = 1/20 miles/min