Motorboat A leaves shore P as B leaves Q; they move across the lake at a constant speed.They meet first time 600 yards from P. Each returns from the opposite shore withouthalting, and they meet 200 yards from. How long is the lake?

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Aug 19th, 2006
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Archana.G
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Aug 21st, 2006

Let total distance be 'y'
When they first meet:    600:y-600---------eq1
when they meet for 2nd time ; y+200:2y-200----------eq2
Now from 1 and 2:    600:y-600::y+200:2y-200
   product of means = product of extremes
===> y=1600 yards

Bhoomika sharma
Profile Answers by Bhoomika sharma Questions by Bhoomika sharma

Dec 1st, 2009

Let the total distance be d metres.

From shore P when there is 600 yards thus the left distance from shore Q the distance is d-600 metres.

Now since they move with different velocities let's assume the velocity from shore P as x m/s and from shore Q as y m/s.

thus, 600/x=(d-600)/y -eqn1

And, when the ships reach the other ends they interchange the shores, thus now when they meet the time at which they again meet is:

d/y+(d-200)/y=d/x+(d+200)/x
Thus, on solving you get as:

d=1600 yards.