Question 125513
a boat goes upstream 30 miles and downstream 44 miles in 10 hours.the next day it goes upstream 40 miles and downstream 55 miles in 13 hours.what is the average speed of the boat and the stream?
:
Let x = speed of the boat in still water
Let y = speed of the current:
then
(x-y) = speed upstream
(x+y) = speed downstream
:
Write a time equation for the 1st trip: time = {{{distance/speed}}}
:
Time upstream + time down stream = 10 hrs
{{{30/((x-y))}}} + {{{44/((x+y))}}} = 10
:
Do the same for the second trip:
{{{40/((x-y))}}} + {{{55/((x+y))}}} = 13
:
We can use elimination; multiply the 1st eq by 4, and the 2nd eq by 3:
{{{120/((x-y))}}} + {{{176/((x+y))}}} = 40
{{{120/((x-y))}}} + {{{165/((x+y))}}} = 39
---------------------------subtraction eliminates {{{120/(x-y)}}}leaving us with
{{{11/((x+y))}}} = 1
which is:
x + y = 11
:
We can replace(x+y) with 11 in the 1st trip equation
{{{30/((x-y))}}} + {{{44/11}}} = 10
{{{30/((x-y))}}} + 4 = 10
{{{30/((x-y))}}} = 10 - 4
{{{30/((x-y))}}} = 6
Cross multiply
6(x-y) = 30
x - y = 5; divided equation by 6
:
Using these two simple equation eliminate y
x + y = 11
x - y =  5
-------------adding eliminates y
2x = 16
x = 8 mph is the boat in still water
then
8 + y = 11
y = 3 mph is the current
:
:
Check our solution in the 2nd trip equation:
{{{40/5}}} + {{{55/11}}} = 13
:
:
Seems like there should be a simpler way, but I can't seem to come up with tonight.