SOLUTION: I would very much appreciate if you could answer this question I do not know what category this is:
The speed of a boat in still water is 10km/h. The boat travels 12km upstream
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The speed of a boat in still water is 10km/h. The boat travels 12km upstream
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Question 138079This question is from textbook Prentice Hall Algebra 1
: I would very much appreciate if you could answer this question I do not know what category this is:
The speed of a boat in still water is 10km/h. The boat travels 12km upstream and 28km downstream in a total time of 4 hours. What is the speed of the stream?
You can put this solution on YOUR website! The speed of a boat in still water is 10km/h. The boat travels 12km upstream and 28km downstream in a total time of 4 hours. What is the speed of the stream?
:
Let x = speed of the current:
then
(10-x) = speed upstream
and
(10-x) = speed downstream
;
Write a time equation: Time = Dist/speed
:
Upstream time + downstream time = 4 hrs + = 4
:
Multiply equation by (10-x)(10+x), get rid of the denominator, you have:
12(10+x) + 28(10-x) = 4(10-x)(10+x)
:
120 + 12x + 280 - 28x = 4(100 - x^2)
:
-16x + 400 = 400 - 4x^2
:
+4x^2 - 16x + 400 - 400 = 0
:
4x^2 - 16x = 0
:
4x(x - 4) = 0
:
x = +4 km/h is the speed of the current
:
:
Check solution by finding the times (Speed up = 6 km/h; Speed down = 14 km/h): + = 4 hrs
You can put this solution on YOUR website! The speed of a boat in still water is 10km/h. The boat travels 12km upstream and 28km downstream in a total time of 4 hours. What is the speed of the stream?
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Let the speed of the stream be "s".
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Upstream DATA:
Distance = 12 km ; Rate = 10-s kph ; Time = d/r = 12/(10-s) hrs.
-----------------------------------
Downstream DATA:
Distance = 28 km ; Rate = 10+s kph ; Time = d/r = 28/(10+s) hrs
-----------------------------------
EQUATION:
time up + time down = 4 hrs
12/(10-s) + 28/(10+s) = 4
12(10+s) + 28(10-s) = 4(100-s^2)
3(10+s) + 7(10-s) = 100-s^2
30+3s + 70-7s = 100-s^2
s^2-4s= 0
s(s-4) = 0
s = 4 kph (speed of the stream)
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Cheers,
Stan H.
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