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February 27th, 2014, 09:18 AM  #1 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  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 ? 
February 27th, 2014, 10:21 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 
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?

February 27th, 2014, 11:01 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: problem on windmills; algebra Hello, bongantedd! Quote:
 
March 2nd, 2014, 07:59 PM  #4 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  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? 
March 3rd, 2014, 05:23 AM  #5 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  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. 
March 3rd, 2014, 09:10 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,590 Thanks: 1038  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 [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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