Question 508847: pls answer my word problems..............
two pipes together can fill a large tank in 2 hours. one of the pipes, used alone, takes 3 hours longer than the other to fill the tank. how long would each pipe take to fill the tank alone.
FASTER PIPE: RATE:1/x IN 2 HRS.
SLOWER PIPE: RATE:????? IN 2 hrs.
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! With work problems like this, each device does a fraction of the whole job.
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One pipe has an unknown rate, so it does 1/x job per hr.
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One pipe does the job in 3 more hr than the other one, so it does 1/(x+3) job per hr.
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Working together they get the whole job done in 2 hrs.
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( 1/x + 1/(x+3) ) * 2 = 1 whole job
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(x+3 +x ) / x(x+3) = 1/2
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cross multiply
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2(2x+3) = x(x+3)
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4x +6 = x^2 + 3x
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x^2 +3x = 4x +6
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x^2 +3x -4x = 6
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x^2 -x -6 = 0
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factor
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(x-3)(x+2) = 0
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so x=3 or x=-2
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-2 is nonsense, so the only realistic answer is x=3
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One pipe does the work in 3 hr, so it does 1/3 job per hr.
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The other takes 3 hr longer, so it takes 6 hr, or 1/6 job per hr.
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Check to see if they can do the job in 2 hr working together.
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(1/3 + 1/6) *2 = ?
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(2/6 + 1/6) = 3/6
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3/6*2 = 6/6 = 1
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Correct. the two can do 1 whole job in 2 hrs working together.
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Answer: Working alone, the faster pipe can do the work in 3 hrs; the slower pipe can do the work in 6 hrs.
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Done.
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