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
+298 D=RT
180=(R-C)10
180=10R-10C DIVIDE BY 2 & ADD.
180=(R+C)5
180=5R+5C
90=5R-5C
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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
+298 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
+36915 180 m / 10 hr = 18 m/hr
180 m / 5 hr = 36 m/hr ( 18 +36 )/2 = 27 speed of boat 9 = current
