SOLUTION: Hello! This is a very difficult question for me, considering the wording confuses me. Your help is greatly needed! And I would glady appreciate any help!
Question:
Jane can s
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: Hello! This is a very difficult question for me, considering the wording confuses me. Your help is greatly needed! And I would glady appreciate any help!
Question:
Jane can s
Log On
Question 775819: Hello! This is a very difficult question for me, considering the wording confuses me. Your help is greatly needed! And I would glady appreciate any help!
Question:
Jane can sail upstream in a river at an average rate of 4 miles per hour, and downstream at an average rate of 6 miles per hour. If she starts at 10:00 A.M, within what period of time must she turn around if she is to return to her point of depature between 6:00 P.M and 8:00 P.M?
Again I would really appreciate anyone that could help me! Thanks! Found 2 solutions by Alan3354, mananth:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Jane can sail upstream in a river at an average rate of 4 miles per hour, and downstream at an average rate of 6 miles per hour. If she starts at 10:00 A.M, within what period of time must she turn around if she is to return to her point of depature between 6:00 P.M and 8:00 P.M?
------------
Find the avg speed of a round-trip.
It's 2*4*6/(4+6) = 4.8 mi/hr regardless of the distance.
-----------
From 10 to 6 = 8 hours --> RT distance = 38.4 miles, 19.2 miles each way.
19.2 mi/4 mi/hr = 4.8 hours = 4:48
1000 + 4:48 = 1448
-------------
From 10 to 8 = 10 hours --> RT distance = 48 miles, 24 miles each way.
24 mi/4 mi/hr = 6 hours
1000 + 6 = 1600
----------------
She has to turnaround between 1448 and 1600, or 2:48 and 4:00
You can put this solution on YOUR website! Jane can sail upstream in a river at an average rate of 4 miles per hour, and downstream at an average rate of 6 miles per hour. If she starts at 10:00 A.M, within what period of time must she turn around if she is to return to her point of depature between 6:00 P.M and 8:00 P.M?
let the distance she has to go one way be x
time upstream = x/4
time downstream = x/6
Total time should be less than 10 hours or greater than 8 hours
8<
8<
multiply by 12
/5
19.2
She should return after 19.2 miles or before 24 miles
speed upstream is 4 mph
Earliest time to turn = 19.2/4 = 4.8 hours after starting ( 4 hours 48 min)
Latest time to turn 24/4 = 6 hours
She should turn after 2.48 pm
OR
She should turn before 4.00 pm
