+0

# hard!

0
64
3

Two ferries depart from opposite banks of a river simultaneously, and cross the river moving at constant speeds and perpendicularly to the river banks. The ferries meet each other at a point that is 700 meters away from the closest riverbank. When they reach the respective river banks toward which they are moving, they immediately turn around and travel back the other way, meeting again. This time they meet at a point that is 400 meters away from the second riverbank. How wide is the river, in meters?

Mar 24, 2020

#1
+23646
+1

Let 'x' be the distance from THE far bank   where 700 is the distance to the NEAR bank

boat one has travelled 700    (rate = 700/unit time)         boat two has travelled   x    rate = x / unit time

boat one then travels   x  + 400  more      and boat two travels   700 + (700+x -400) more    when they meet

The time is the same      rate x time = distance      distance/rate = time     equate the distances divided by the respective rates

(700 + x + 400)/700     =    (  x  +  700     +  (700+x-400) )/x

1100x + x^2 = 1400x + 700000

x^2-300x -700000 = 0                         quadratic formula yields  x = 1000

One boat travels 700   the other   1000 whe they first meet.....width of river =   700+ 1000 = 1700 m

Mar 24, 2020
#2
+1968
0

Nice, EP!

CalTheGreat  Mar 24, 2020
#3
0

When they meet the 2nd time at 400 meters, they had already travelled 3 times the width of the river, or 3 times their first meeting at 700 meters from the bank of the river.

Therefore the width of the river is: 3 x 700 meters - 400 meters =1,700 meters.

Note: This is a very famous puzzle by the great American "Puzzlist", Sam Lloyd. CPhill posed it as "challenge" about a year or two ago. Look it up.

Mar 24, 2020
edited by Guest  Mar 24, 2020