A Tale of two ferries

[Badger]

Two ferry boats start moving at the same instant from opposite sides of the Hudson River, one boat going from New York to Jersey City, and the other from Jersey City to New York. One boat is faster than the other, so they meet at a point 720 yards from the nearest shore. After arriving at their destinations, each boat remains for 10 minutes to change passengers before starting on the return trip. The boats meet again at a point 400 yards from the other shore. What is the exact width of the river?

<hr width=100% size=1>

 

[jimi]

This scenario is impossible

<hr width=100% size=1>.. whit way roon should it be again ..

 

[Superstrath]

I don't think there's enough info here.

<hr width=100% size=1>

 

[tome]

Easy

1120 yds. They will meet at the same point.

Tom

<hr width=100% size=1>

 

[AndrewB]

While sympathising with Jimi, I think the theoretical answer is 1760 yds.

(The slower ferry covers 9 yds in the time the faster ferry does 13 yds, and their second meeting takes three times as long plus 10 minutes, as did their first. Exact speeds and times are indeterminate).

 

[Cactus]

According to my calculations, the answer is -400.

<hr width=100% size=1>

 

[BlueSkyNick]

No they won't because they start off at different times on their second trip., ie whatever time they arrived at the end of the first one (which are different) plus 10 minutes each.

<hr width=100% size=1>I'm average size, Its just that everybody else is short.

 

[tcm]

Re: 2160 yards ?

A is the speed of the fast one
B is the speed of the slow one
X is the width of the river

Speed = dist/time

So

Time = dist/speed, and the same after start when they meet each time.
--------------------------
T1 when they first meet is the same for each so

For the slow one (B) which will be nearest the ”nearest shore”

B= 720/T1

And for the fast one (farthest from that nearest shore)

A= (X- 720)/T1

Rearrange, equating both values of T1 for both ferries

T1= 720/B = (X-720)/A

B(X-720) = 720A

A= (B(X-720))/ 720

------------------
The second time they meet , at time = T2 mins after they set off

Speed = distance/ time

A = (X (X-400)) /T2
(note this is the fast one, completed almost two complete crossings, only 400yds short)

B= (X 400)/ T2

So , since T2 is the same for both

T2 = (2X –400)/A = (X 400)/B

Rearrange

B(2X-400) = A(X 400)

So A= B(2X-400)/(X 400) - another expression for A

So from above, using other exprn for A as well:
B(X-720)/720= B(2X-400)/(X 400)

Divide both sides by B :
(X-720)/720 = (2X-400)/(X 400)

So: 720(2X-400)= (X-720)(X 400)

Expand :
1440X – 720.400 = XX – 720X – 720.400 400X

Collect terms
XX – 720X –1440X = 0

so
XX = 2160 X

Divide by X and ignore zero values of X
X = 2160

So width of river = 2160 yds




<hr width=100% size=1>

 

[tome]

I reckoned on this being a trick question, because eventually the ferries would end up with both at one pier whereas we all know that they would leave at the same time.

<hr width=100% size=1>

 

[AndrewB]

Re: 2160 yards ?

Hey, a real sledgehammer approach! Possible slight error:

<blockquote><font size=1>In reply to:</font><hr>

<font size=1>Expand :
1440X – 720.400 = XX – 720X – 720.400 + 400X

Collect terms
XX – 720X –1440X = 0</font size=1>

<hr></blockquote>

Should be:
<blockquote><font size=1>In reply to:</font><hr>

<font size=1>
Expand :
1440X – 720.400 = XX – 720X – 720.400 + 400X

Collect terms
XX – 720X +400X –1440X = 0</font size=1>

<hr></blockquote>

Then we agree! /forums/images/icons/cool.gif



<hr width=100% size=1>

 

[tcm]

Re: Damn! ok 1760 yds

well spotted. :-(

<hr width=100% size=1>

 

[BlueSkyNick]

One Mile !!

I win, because everbody else has expressed their answer in small units of measure, instead of calculating the largest unit possible.

(I know there is a term for this but yet again, my memory is letting me down so far as school is concerned!)

<hr width=100% size=1>I'm average size, Its just that everybody else is short.

 

[Badger]

And the answer is ...

1760 Yards, which we all know is 1 statute mile... so congratulations to Mr Bebbington, BigNick and TCM.(he got there in the end)

<hr width=100% size=1>

 

[jimi]

Re: And the answer is ...

Well all that's fine but no one has taken the effect of the current into consideration .. so you're all wrong. And do'nt try and give me any crap with that algebra stuff .. I asked a simple question on relativity the other day and nae of yoose even bothered answering .. I guess TCM must hae done module 1 of basic maths as well as advanced nuclear fission whilst incarcerated in Goingforalongpee?

<hr width=100% size=1>.. whit way roon should it be again ..

 

[AndrewB]

My reasoning was ...

.. a good deal more basic than yours!

Since at the first meeting the combined distance travelled was equal to one crossing, and at the second meeting it was three crossings, at constant speed each ferry had travelled three times as far by the second meeting.

The slower ferry had done 720 yards up to the time of first meeting, and one crossing plus 400 yards up to the second meeting, so: 3 x 720 = a crossing + 400. Hence a crossing is 1760 yds.



<hr width=100% size=1>

 

[tcm]

Re: Maths as an artform

I bow to your more elegant approach. Mine is more more in keeping with my bicycle-clipped background

<hr width=100% size=1>