Question
A tugboat goes upstream 120 miles in 20 hours. The return trip downstream takes 10 hours. Find the speed of the tugboat without the current and the speed of the current.
Answers
Set-up:
Let v=speed of boat
c=speed of current
Distance = speed*time
Upstream:
120=(v-c)*20 ...(1)
Downstream:
120=(v+c)*10 ...(2)
Solve for v and c.
Let v=speed of boat
c=speed of current
Distance = speed*time
Upstream:
120=(v-c)*20 ...(1)
Downstream:
120=(v+c)*10 ...(2)
Solve for v and c.
Okay, I got:
v+6-c and c=c-6 for 120=(v-c)*20
AND:
c=12-v and v=6/5-c for 120=(v+c)*10
Now what do I do?
v+6-c and c=c-6 for 120=(v-c)*20
AND:
c=12-v and v=6/5-c for 120=(v+c)*10
Now what do I do?
What you really want to do is to reduce the variables in simpler terms by dividing both sides by 20:
v-c=120/20=6 ...(1a)
v+c=120/10=12...(1b)
This is a sum and difference form where the solution can be done mentally:
v=(sum+difference)/2=9
c=(sum-difference)/2=3
v-c=120/20=6 ...(1a)
v+c=120/10=12...(1b)
This is a sum and difference form where the solution can be done mentally:
v=(sum+difference)/2=9
c=(sum-difference)/2=3