River Crossing

This is indeed a famous problem posed by Sam Loyd. When the boats meet the first time, together they cover one width of the river. Of that one boat that started from near shore has traveled 720 ft. When the two boats meet the second time, together they cover three times the width. So that boat has covered 720*3= 2160 feet. Now2160- 400 feet(from far shore) will give width of river=1760.

The first time they met (Point A), they traveled 1 width of the river in total.

The second time they met (Point B), they already traveled 3 widths of the river in total.

----------->A <------

<----------- ------> (turn around)

------> B<-----------

Thus, it took them three times the distance they traveled in the first encounter.

One of the boat traveled 720 feet in the first encounter.

The boat should have traveled 2160 feet (720 times 3) in the second encounter.

We know that the boat already traveled 1 width of the river plus another 400 feet.

So, let x be the width of the river.

x + 400 = 2160

x = 1760 feet

With absolute basics - A little lengthy solution provided below :P

Let w be the width of the river

At first point of meeting, Boat 1 has traveled 720 ft, and lets assume Boat 2 has traveled x ft [720 + x = w]

At second point of meeting, Boat 1 has covered an extra distance of x + 400 while Boat 2 has traveled a distance of 720 + (w-400) = 1040 + x

Equating the ratio of their speeds, we get (x+1040)/(x+400) = x/720 which leads to

x^2 - 320x - 1040 * 720 = 0

x^2 - 1040x + 720x - 1040 * 720 = 0

(x-1040)(x+720) = 0

x = 1040

w = 1040 + 720 = 1760 Hurraaahhh

Assuming the speeds of the two boats are v1 and v2 which they maintain throughout.

First time the boats meet at time t1, at which point we have

v1 (t1) = 720

v2 (t1) = d - 720

Second time the boats meet at time t2, at which point we have

v1 (t2) = d + 400

v2 (t2) = d + d - 400 = 2d - 400

Eliminating t1 from the first set of equations:

v2 (720/v1) = d - 720

which can be written as

(v2/v1) (720) = d - 720.......................(1)

Eliminating t2 from the second set of equations:

v2 ( d + 400 ) / v1 = 2d - 400..............(2)

Substituting (v2/v1) from (1) into (2)

(d + 400) (d - 720)/720 = 2d - 400

Expanding...

d^2 - 320 d - 288000 = 1440 d - 288000

d = 1440 + 320 = 1760

Let the wide of the river is w. Let the boat A covers the distant 720 ft, at first crossing.At the second crossing, boat A will cover 2(720). So 2(720) = w-720 + 400.Then w=1760 ft.

When they meet the 1st time Boat A crossed 720 ft Boat B crossed (W-720) ft

When they meet the 2nd time: Boat A has crossed (W-720) + 400 = W-320 ft Boat B has crossed 720 + (W-400) = W+320 ft

Assuming they travelled @ same speed then 720/(w-720) = (w-320)/(w+320)

on solving we get 1760

Let the speed of 1 boat = x. Width of river = z. When they first meet, distance from nearer shore = 720, distance from farthest shore = z-720. Time taken by both the boats to cover their respective distance is same. Hence speed of 2 boat = y = [z - 720]x/720. For second meeting, distance travelled by 1 boat = z - 320, and by 2 boat = z + 320 and time taken is also same. Hence, the expression comes, [z - 320]/x = [z + 320]720/[z - 320]x

solving, Z = 1760

Let the speed of the first (faster) boat be $v_1$ and the speed of the second (slower) boat be $v_2$ . Let the width of the river be $d$ .

We can equate the time that they meet for the first time because it is equal for both. The first (faster) boat travels a distance of $d - 720$ with speed $v_1$ and the second (slower) boat travels $720$ with speed $v_2$ .

Thus, $\Rightarrow \dfrac{720}{v_2} = \dfrac{d - 720}{v_1}$ or $\Rightarrow \dfrac{v_1}{v_2} = \dfrac{d-720}{720}$

Now, for the second condition.

At the time they meet for the second time, the first boat covers a distance of $d - 720 + 720 + d - 400 = 2d - 400$ with speed $v_1$ and the second boat covers a distance of $720 + d - 720 + 400 = d + 400$ with a speed of $v_2$ .

Thus,

$\Rightarrow \dfrac{d+400}{v_2} = \dfrac{2d - 400}{v_1}$

or

$\Rightarrow v_1(d+400) = v_2(2d - 400)$

or

$\Rightarrow \dfrac{v_1}{v_2} = \dfrac{2d - 400}{d+400}$

Now we already know what $\dfrac{v_1}{v_2}$ is from the first condition. We substitute that here which gives us

$\Rightarrow \dfrac{d-720}{720} = \dfrac{2d - 400}{d + 400}$

Solving this quadratic gives us $d = 0$ and $d = 1760$ . Since the distance is obviously $> 0$ , we get $\boxed{d = 1760}$

Let us say that the faster boat A travels a (ft/s), and boat B travels b (ft/s) Also, let the width of the river be x feet. The time it takes for the 2 boats to meet the first time is x/(a+b) The time it takes for the 2 boats to meet the second time is 3x(a+b) Hence, we get the two equations: {x/(a+b)} b = 720 {3x/(a+b)} b=x+400

When solved: bx=720(a+b)........................A 3bx=(400+x)(a+b) 3A = 3bx = 2160(a+b)=(x+400)(a+b)

Therefore, we can get the equation x+400=2160, and the answer: x=1760