SOLUTION: Two campers left the camp at 8:00 AM to paddle their canoe 10 Km upstream to a fishing cove. They finished for 3.5 hours then paddled back to the campsite and arrived 6:00 PM. If
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Question 918875: Two campers left the camp at 8:00 AM to paddle their canoe 10 Km upstream to a fishing cove. They finished for 3.5 hours then paddled back to the campsite and arrived 6:00 PM. If the rate of the current was 3 km/hr. How fast could the camper paddle in still water?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Two campers left the camp at 8:00 AM to paddle their canoe 10 Km upstream to a fishing cove. They finished for 3.5 hours then paddled back to the campsite and arrived 6:00 PM. If the rate of the current was 3 km/hr. How fast could the camper paddle in still water?
let the rate of boat in still water be x
upstream time = 10/(x-3)
Time spent there = 3.5 hours
down stream time = 10/(x+3)
They left at 8.00 and reached at 6.00 pm
=10 hours
10/(x-3)+3.5+10/(x+3) = 10
10/(x-3)+10/(x+3) = 10-3.5
10/(x-3)+10/(x+3) = 6.5
take the LCM
10(x+3)+10(x-3)= 6.5(x+3)(x-3)
10x+30+10x-30=6.5(x^2-9)
20x=6.5x^2-58.5
6.5x^2-20x-58.5=0
Find the roots of the equation by quadratic formula
a= 6.5 , b= -20 , c= -58.5
b^2-4ac= 400 + 1521
b^2-4ac= 1921
x1=( 20 + 43.83 )/ 13
x1= 4.91
x2=( 20 -43.83 ) / 13
x2= -1.83
Ignore negative value
boat speed 4.910 km/h
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