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Incinerators

 
 
fdrhs
 
Reply Wed 20 Jul, 2005 05:32 pm
Each of two incinerators can process a day's trash in 20 hours. Together
with a third incinerator, they process the trash in 6 hours. In what time
can the third incinerator do the job alone?

I just want someone to set up the proper equation needed to solve this question.
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raprap
 
  1  
Reply Wed 20 Jul, 2005 06:33 pm
How much electrical circuit analysis do you know? If you any, think of this as an equivalent parallel resistor problem.

Let T1 be the time incinerator one can process the days trash (20 hours) & T2 be the time incinerator one can process the days trash (20 hours).

Working together they can process the trash in T12 hrs. This is calculated by
1/T1+1/T2=1/T12
or
1/20+1/20=1/T12
combining and reducing
1/T12=1/20+1/20=2/20=1/10
inverting this T12=10 hrs, so working together incinerator 1 and 2 can process the days trash in 10 hrs.

Now throw in a third incinerator that can incinerate the days trash in T3 hrs and the resistance ladder becomes
1/T1+1/T2+1/T3=1/T123

where T123 is the time all three process the days garbage together (6hrs)

& plugging in knowns this becomes

1/20+1/20+1/T3=1/6

now you can rearrange and solve for T3

Remember this routine, it is the same as the "Bill can dig a hole in 2 hrs and Ted takes 3 hrs, if Bill and Ted work together how long will it take for them to dig a hole?" style problems.

Rap
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fdrhs
 
  1  
Reply Wed 20 Jul, 2005 08:28 pm
okay
rap,

Your notes are very helpful. I can now solve for T3.
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