Answered

[Awkward_Turtle]

Thanks.

 

[soroban]

Re: Width of River

Hello, Awkward_Turtle!

This is a classic problem . . . and it is hard to set up.

Two passenger boats start moving at the same time from opposite sides of the river,
heading towards the other’s port. One boat is faster than the other, so they pass each
other at a point 720 yards from the nearest shore. After arriving, each boat remains
for 10 minutes to change passengers, then starts 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?

Note: since both ships stopped for the same length of time, we can disregard the stops.


Let \(\displaystyle x\) = width of the river (in yards).

Ship #1 starts at \(\displaystyle A\) moving at \(\displaystyle a\) yards/hour.
Ship #2 starts at \(\displaystyle B\) moving at \(\displaystyle b\) yards/hour.

At their first meeting, Ship #1 has gone \(\displaystyle 720\) yards to point \(\displaystyle P\).
. . At \(\displaystyle a\) yd/hr, this took: \(\displaystyle \,\frac{720}{a}\) hours.

Ship #2 has gone \(\displaystyle x\,-\,720\) yards to point \(\displaystyle P.\)
. . At \(\displaystyle b\) yd/hr, this took \(\displaystyle \,\frac{x-720}{b}\) hours.

                P
      * → → → → * ← ← ← ← ← ← ← ← ← *
          720          x-720

These times are equal: \(\displaystyle \:\frac{720}{a}\:=\:\frac{x-720}{b}\;\;\Rightarrow\;\;\L\frac{720}{x-720} \:=\:\frac{a}{b}\;\) [1]


At their second meeting, Ship #1 has crossed the river (\(\displaystyle x\) yards)
. . and has returned 400 yards.
At \(\displaystyle a\) yd/hr, this took \(\displaystyle \,\frac{x+400}{a}\) hours.

Ship #2 has crossed the river (\(\displaystyle x\) yards) and returned \(\displaystyle x-400\) yards.
At \(\displaystyle b\) yd/hr, this took \(\displaystyle \,\frac{2x-400}{b}\) hours.

                           Q
      * → → → → → → → → → → * ← ← ← *
            x-400             400

These two times are also equal: \(\displaystyle \:\frac{x+400}{a}\:=\:\frac{2x-400}{b} \;\;\Rightarrow\;\;\L\frac{x+400}{2x-400}\:=\:\frac{a}{b}\;\) [2]


Equate [1] and [2]: \(\displaystyle \L\:\frac{720}{x\,-\,720}\:=\:\frac{x\,+\,400}{2x\,-\,400}\)

. . which simplifies to: \(\displaystyle \:x^2\,-\,1760x\:=\:0\;\;\Rightarrow\;\;x(x\,-\,1760)\:=\:0\)


Therefore, the width of the river is: \(\displaystyle \,x\:=\:1760\) yards ... exactly one mile.

 

[Awkward_Turtle]

Thanks!

 

[TchrWill]

Two passenger boats start moving at the same time from opposite sides of the river, heading towards the other’s port. One boat is faster than the other, so they pass each other at a point 720 yards from the nearest shore. After arriving, each boat remains for 10 minutes to change passengers, then starts 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?

Letting x equal the river width:
After the first trip, the ratio of distances traveled by the twoo boats is (x - 720)/720.
On the second trip, the ratio of distances traveled is
2x - 400/x + 400.
Since these ratios are equal, equate and solve for x.