SOLUTION: the speed of a boat in still water is 15 km/hr. it needs four more hours to travel 63 km against the current of a river than it needs to travel down the river. Determine the spee

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Question 1149799: the speed of a boat in still water is 15 km/hr. it needs four
more hours to travel 63 km against the current of a river
than it needs to travel down the river. Determine the speed
of the current of the river.






Found 2 solutions by ikleyn, Boreal:
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

The time equation is


     -  = 4   hours.


It is literal translation English to Math.


    63*(15+x) - 63*(15-x) = 4*(15^2-x^2)


Reduce this quadratic equation to the standard form and solve it by any method you know.


Here x is the speed of the current.



    //  Without any complicated calculations, I see the answer by my non-armed eye: x= 6 miles per hour for the rate of the current.

        You may confirm it by solving quadratic equation formally.

-------------------

To see many other similar solved problems, see the lessons
    - Wind and Current problems
    - More problems on upstream and downstream round trips
    - Wind and Current problems solvable by quadratic equations
    - Unpowered raft floating downstream along a river
    - Selected problems from the archive on the boat floating Upstream and Downstream
in this site, where you will find other similar solved problems with detailed explanations.

Read them attentively and learn how to solve this type of problems once and for all.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the section "Word problems",  the topic "Travel and Distance problems".


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to your archive and use it when it is needed.



Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
d=v*t
63=(15+c)*t where c is current speed
t=63/(15+c)
and
63=(15-c)(t+4)
and t=[63/(15-c)]-4, which is (4c+3)/(15-c), from putting -4 over 15-c and getting -60+4c
those t s are equal
cross-multiply and get 2c^2+63c-450=0
and this is (2c+75)(c-6)=0
positive root is c=6 kph current
this would be 21 kph downstream and 3 hours to do 63 km and 9 kph upstream, which takes 7 hours.

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