SOLUTION: A tourist traveled on a motorboat against the current for 25 km. And then returned back on a raft. In the boat the tourist traveled for 10 hours less than on the raft. Find the spe

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Question 1075485: A tourist traveled on a motorboat against the current for 25 km. And then returned back on a raft. In the boat the tourist traveled for 10 hours less than on the raft. Find the speed of the current if the speed of the motorboat in still water is 12 km/hour. Show your work!
Found 3 solutions by ikleyn, ankor@dixie-net.com, jorel1380:
Answer by ikleyn(52772)   (Show Source):
You can put this solution on YOUR website!
.
A tourist traveled on a motorboat against the current for 25 km. And then returned back on a raft.
In the boat the tourist traveled for 10 hours less than on the raft.
Find the speed of the current if the speed of the motorboat in still water is 12 km/hour.
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Let x be the speed of the current, in km/h units. Then the speed of the boat traveling against the current is (12-x) km/h. The time spent traveling 25 miles against the current is hours. The time spent rafting 25 miles (with the current) is hours. The equation is = 10. To solve it, multiply both sides by x*(12-x). You will have 25*(12-x) - 25x = 10x*(12-x), 300 - 25x - 25x = , = 0, = 0, (x-2)*(x-15) = 0. The roots are x= 2 and x= 15. Since 12-x must be positive, the only root x = 2 survives. Answer. The current speed is 2 km/h. Check. The time traveling against the current = 2.5 hours. The time rafting is = 12.5 hours. 12.5 - 2.5 = 10 hours. Correct !

It is a typical and standard Upstream and Downstream round trip word problem.
You can find many similar fully solved problems on upstream and downstream round trips with detailed solutions in 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.

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".

Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website!
A tourist traveled on a motorboat against the current for 25 km.
And then returned back on a raft.
In the boat the tourist traveled for 10 hours less than on the raft.
Find the speed of the current if the speed of the motorboat in still water is 12 km/hour.
:
let c = the speed of the current (also the speed of the raft)
then
(12-c) = boat speed against the current
:
Write a time equation; time = dist/speed
raft time - boat time = 10 hrs
-
multiply equation by c(12-c), cancel the denominators
25(12-c) - 25c = 10c(12-c)
300 - 25c - 25c = 120c - 10c^2
300 - 50c = 120c - 10c^2
10c^2 - 50c - 120c + 300 = 0
10c^2 - 170c + 300 = 0
simplify, divide by 10
c^2 - 17c + 30 = 0
Factors to
(c-15)(c-2) = 0
Two solutions, but only one is reasonable
c = 2 km/hr is the rate of the current
:
;
Check this by finding the time each way (boat speed: 12 - 2 = 10 km/hr
25/2 = 12.5 hrs
25/10 = 2.5 hrs
-----------------
time dif: 10 hrs as given

Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website!
Let n be the amount of time it takes to travel the river, and c be the speed of the current. Then:
25/(12-c)=25/c -10
25c=25(12-c)-10(c)(12-c)
25c=300-25c-120c+10c�
10c�-170c+300=0
c�-17c+30=0
(c-15)(c-2)=0
c=15 or 2
Since a value of 15 would give a negative result for the motorboat trip, we get a logical value of 2 kph as the speed of the current. ☺☺☺☺