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A tugboat goes 180 miles upstream in 10 hours, the return trip downstream takes 5 hours. Find the speed of the tugboat without the current, and the speed of the current

 Mar 13, 2020
 #1
avatar+272 
+1

D=RT
180=(R-C)10
180=10R-10C DIVIDE BY 2 & ADD.
180=(R+C)5
180=5R+5C
90=5R-5C
---------------
270=10R
R=270/10
R=27 MPH. ANS. FOR THE RATE OF THE BOAT.
180=(27-C)10
180=270-10C
10C=270-180
10C=90
C=90/10
C=9 MPH. ANS.FOR THE CURRENT.
PROOF:
180(27+9)5
180=36*5
180=180

credited to checkley79

 Mar 13, 2020
 #3
avatar+272 
+1

27 for the rate of the boat

C=9 for the current

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.735649.html

credited to checkley79

 Mar 13, 2020
 #4
avatar+21827 
+2

180 m / 10 hr   = 18 m/hr

180 m / 5 hr = 36 m/hr            ( 18 +36 )/2 = 27    speed of boat         9 = current

 Mar 13, 2020

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