# A straight river that is 264 meters wide flows from west to east at a rate of 14 meters per minute. Malanie and Sherry sit on the south bank of the river, with Malanie a distance of D meters downstream from Sherry. Relative to the water, Malanie swims at 80 meters per minute, and Sherry swims at 60 meters per minute. At the same time, Malanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find D.

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Denote the common time as `T ` and the equidistant point as `A`. Each woman is under the influence of the current, so they have to direct to point `B`, `14T ` to the west of `A`. In the coordinate system connected with flowing water, they make `60T ` and `80T`, respectively.

We can write two equations using the Pythagorean theorem:

`sqrt( ( 60T )^2 - W^2 ) + 14T = D/2`, `sqrt( ( 80T )^2 - W^2 ) - 14T = D/2`, where `W ` means the river width. They are equivalent to `( 60T )^2 - W^2 = (D/2)^2 + (14T)^2 - 14TD`, `( 80T )^2 - W^2 = (D/2)^2 + (14T)^2 + 14TD`.

Add the two equations and obtain `( (60)^2 + (80)^2 - 2*(14)^2 ) T^2 - 2W^2 = D^2/2`.

Subtract them and obtain `((80)^2-(60)^2)T^2 = 28TD`, or `(40^2-30^2)T = 7D`, or `T = D/100`. Substitute this into the sum of equations:

`D^2 / 100^2 ((60)^2 + (80)^2 - 2*(14)^2 - 5000) = 2W^2`, or `D^2 = 100^2/48^2 W^2`, so `D = 100/48 W = 100*5.5 = 550 ` (meters), which is the answer.

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