SOLUTION: 2 boats (B1 and B2) start off at opposite ends of a lake heading for the staring point of the other. They pass each other 800 yards from the starting point of boat 1. They contin

Question 266376: 2 boats (B1 and B2) start off at opposite ends of a lake heading for the staring point of the other. They pass each other 800 yards from the starting point of boat 1. They continue to the starting point of the other boat, turn around and return. This time they pass each other 300 yards from the starting point of boat 2. How wide is the lake? (Assume each boat travels at a fixed rate and ignore turn around times.)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
2 boats (B1 and B2) start off at opposite ends of a lake heading for the staring point of the other.
They pass each other 800 yards from the starting point of boat 1.
They continue to the starting point of the other boat, turn around and return.
This time they pass each other 300 yards from the starting point of boat 2.
How wide is the lake? (Assume each boat travels at a fixed rate and ignore turn around times.)
:
Let x = width of the lake
:
First meeting *
:
B1->-------800-------*---------------(x-800)---------------<-B2
:
2nd meeting
B2->------------------(x-300)-------------*--------300-----<-B1
:
The distance relationship of the two boats will be the same at 2nd meeting as it was with the 1st meeting.
Write an equation from this fact
:
=
:
=
Cross multiply, results
800x + 400000 = x^2 - 500x - 800x + 400000
:
800x = x^2 - 1300x; (subtracted 400000 from both sides)
:
x^2 - 1300x - 800x = 0
:
x^2 - 2100x = 0
Factor out x
x(x - 2100) = 0
:
x = +2100 yards is the width of the lake
:
:
See if this will make sense
1st boat went 800 and 2nd boat went 2100-800 = 1300 yds at the 1st meeting
Distance ratio: 800/1300
then
1st boat went 1300 + 300 = 1600 yds and 2nd boat went 800 + 1800 = 2600 yds to the 2nd meeting
Distance ratio: 1600/2600, the same as the 1st meeting
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