It took a boat 7 hours to travel 231 miles with the current but 11 hours to travel the same distance against that same current. Find the speed of the boat in the still water and the speed of the current.
2007-12-03 22:41:50 · 4 answers · asked by journey 1 in Science & Mathematics ➔ Mathematics
Boat speed = x mph
River speed = y mph
________time(h)__speed(mph)
With______7______x + y
Against___11______x - y
Distance = 231 miles
231 = 7x + 7y
231 = 11x - 11y
77x + 77y = 2541
77x - 77y = 1617
154x = 4158
x = 27
231 = 189 + 7y
7y = 42
y = 6
Boat speed = 27 mph
River speed = 6 mph
2007-12-04 03:24:57 · answer #1 · answered by Como 7 · 3⤊ 1⤋
x1=231 , t1=7, x2=231 , t2=11
v1:boat speed in current
v2:boat speed against current
x1=v1t1
231=7*v1 ; v1=231/7=33
231=v1t1=v2t2=11v2 ; v2=21
boat speed in still water=(33+21)/2=27
current speed=27-21=33-27=6
2007-12-03 22:57:34 · answer #2 · answered by goodi 3 · 0⤊ 0⤋
(b-- c) / (b + c) = 7/11
=> 11(b -- c) = 7(b + c)
=> 4b = 18c
=> 2b = 9c
=> b = 9c/2
then 231/(b + c) = 7
=> 7(b + c) = 231
=> b + c = 33
=> 9c/2 + c = 33 => 11c/2 = 33 => c = 6
then b = 9/2(6) = 27
boat b = 27 mph, current c = 6 mph
2007-12-03 22:54:17 · answer #3 · answered by sv 7 · 1⤊ 0⤋
youre going 33mi/h with current and 21 mi/h against the curent, therefore, assume the speed of the current and boat are constant
x (speed boat) + current =33.......................(eq1)
x - current = 21................................................(eq2)
subtract e1-e2 and we have 2*current=12
Therefore the speed of the current is 6 mi/h, and the speed of the boat is 27 mi/h
2007-12-03 23:00:14 · answer #4 · answered by Mikey D.S. 2 · 0⤊ 0⤋