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February 27th, 2014, 09:18 AM   #1
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problem on windmills; algebra

Two windmills working simultaneously and jointly can fill a reservoir in 12 hours
One takes longer than the other to fill the reservoir
How long does each take working separately to fill the reservoir ?
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February 27th, 2014, 10:21 AM   #2
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Can you clarify "One takes longer than the other to fill the reservoir"? Does that apply when they work together or only when they work individually? Are you expecting an exact answer even though you haven't stated how much longer?
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February 27th, 2014, 11:01 AM   #3
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Re: problem on windmills; algebra

Hello, bongantedd!

Quote:
Two windmills working simultaneously and jointly can fill a reservoir in 12 hours
One takes longer than the other to fill the reservoir.[color=beige] . [/color][color=blue]This is no help . . . More information, please![/color]
How long does each take working separately to fill the reservoir?
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March 2nd, 2014, 07:59 PM   #4
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Re: problem on windmills; algebra

missing info:
One takes 10hrs longer than the other to fill the reservoir.
how long does each take working separately to fill the reservoir?
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March 3rd, 2014, 05:23 AM   #5
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Re: problem on windmills; algebra

The crucial point here is that when two machines or people, etc. work together their rates of work add. Let "A" be the time it takes for the first wind mill to fill the reservoir. Then its rate of work is . Let "B" be the time it takes for the second wind mill to fill the reservoir. Then its rate of work is . Working together, their rate of work is 'reservoir per hour'.

Now that we know that "one takes 10 hours longer than the other to fill the reservoir", letting A be the time for that slower wind mill, A= B+ 10.

So we must solve and A= B+ 10. Of course, the first step might be to write the first equation as . And the second might be to multiply through by the denominators, 12B(B+ 10). That gives a quadratic equation to solve for B.
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March 3rd, 2014, 09:10 AM   #6
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Re: problem on windmills; algebra

I find that students have a difficult time "picturing" what goes on with these problems.
Pretend what you have is 1 mile being travelled at 2 speeds;
then you can use ye old distance = speed * time formula; like:
Code:
[1]@(a+b)..........................1........................>12 hours
[2]@a..............................1........................>x hours
[3]@b..............................1........................>x+10 hours
a = faster speed, b = slower speed, x = time @ a; then:
[1]: 12(a+b) = 1 ; a+b = 1/12
[2]: ax = 1 ; a = 1/x
[3]: b(x+10) = 1 ; b = 1/(x+10)

So now you have:
1/x + 1/(x+10) = 1/12 ; which simplifies to:
x^2 - 14x - 120 = 0

Which is same as what Halls was telling you...solve for x
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