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Dumb question

  1. Dec 11, 2008 #1
    I feel really blond at the moment...

    Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?

    Can anyone guide me to the solution of solving the problem. I know the answer but I just cannot seem to be able to get to it...

    I have had a few other questions like this and I would like to know how to do them.
     
  2. jcsd
  3. Dec 11, 2008 #2

    mgb_phys

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    Would it be easier to picture if you put some numbers in?
    Suppose the pool held 3 buckets of water (or 30 or 300 or any other number - it doesn't matter)
    How many buckets/hour does each carry?

    Now if they were all carrying that number how many buckets per hour would you have in total? How many pools/hour can you fill? So what fraction of an hour does it take for one?
     
  4. Dec 11, 2008 #3
    I still am not figuring out how to write it onto paper. What answer do you get for that problem and how? I have the answer just not the how.
     
  5. Dec 11, 2008 #4

    mgb_phys

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    Write it here using those numbers and show what you get.
     
  6. Dec 11, 2008 #5
    I am using a notepad right now...and I do not think I can write it here cause I cannot even write it on the notepad...every which way I have tried to write it does not come out to the answer therefore my equations are wrong and the problem I am having is translating the word problem to an equation...

    Let us use 30 buckets to fill the pool...

    Jim is going to do 1 bucket a minute...
    Sue is going to do 1.5 buckets a minute...
    Tony is going to do 3 buckets a minute...
     
  7. Dec 11, 2008 #6

    mgb_phys

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    So the total number of buckets/min for all of them together is?
    How many minutes will it take this number of buckets to fill the pool?

    Then just rewrite this with 'n' buckets to fill the pool and you have the equation.
     
  8. Dec 11, 2008 #7
    Total number of buckets per minute...

    5.5

    30/5.5 = 5.45

    Which does not equal the answer...
     
  9. Dec 11, 2008 #8
    Think you have something upside down, even tho the approach is correct. Tony is the slowest of the threee, correct?
     
  10. Dec 11, 2008 #9
    You are right and now I have to see if I can do the buckets per minute correctly lol.
     
  11. Dec 11, 2008 #10
    Thank you for your attempt to help...I just got to walk away for now as I cannot do this. I cannot get the right answer no matter how I am writing it. lol I cannot even figure out how to do buckets/minute for each because I am so frustrated.
     
  12. Dec 11, 2008 #11
    Make sure you put Tony's work rate in terms of minutes. Always make sure all your information is in the same units to avoid confusion. 1.5 hours = 90 minutes.

    Therefore tony does 30 buckets/90 minutes = .3333 buckets minute
     
  13. Dec 11, 2008 #12

    symbolipoint

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    The rates of all three working together are additive.
    This type of problem usually is most easily analyzed using columns for RATE(as jobs per unit time), TIME, JOBS. Those three columns. You set up rows named with each person who does a job. Not really certain if all the chart is necessary since you know the rates will here be additive and you want the combined rate,... not sure unless I actually try to solve the problem myself.
     
  14. Dec 11, 2008 #13
    30/30 = 1
    30/45 = .6666
    30/90 = .3333

    Total - 1.9999

    30/1.9999 = 15.00007

    Is there a simpler way to do this as an equation for a timed test? Potentially it would be easier if I kept them fractions since I will not have the aid of a calculator...

    So...in these word problems you have to go the extra step and decide how long it takes for a pool to be filled up?...

    Like we did 30 buckets...

    So Jim = 30/30 = 3/3
    Sue = 30/45 = 2/3
    Tony = 30/90 = 1/3

    6/3 = 2 and 30/2 = 15...

    So the easiest and quickest way for these word problems like this one is to designate the amount so you can get a fraction or decimal to work off of?

    Thank you greatly for helping me out of confusion.
     
  15. Dec 11, 2008 #14

    symbolipoint

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    ...In fact, you will not need to fill out a chart on this one. The rates should be additive. If done in jobs per HOUR, your rate for the combined team will be 2+[tex]\frac{4}{3}+\frac{2}{3}[/tex] jobs per hour.
     
  16. Dec 11, 2008 #15
    No, not necessarily. But sometimes it is easier to think in terms of everyday terms.

    I did it in terms of time/job, much like symboli did it.
     
  17. Dec 11, 2008 #16

    symbolipoint

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    Actually, I used the numbers in the dimensions of jobs per time. This is a carry-over from the first critical exposure to applied "rate-time-distance" problems from year-1-algebra. Some people deal with the inverted forms for work-rate applied problems but that form is more difficult for some people (and for me, too which is why I use rate-time-jobs, where "rate" means number of jobs per unit of time.)
     
  18. Dec 11, 2008 #17
    Well well...seems like I was going through some more word problems similar and had ok time with them...but another one has me stuck...

    If Sam can do a job in 4 days that Lisa can do in 6 days and tom can do it in 2 days, how long would the job take if Sam, Lisa, and Tom worked together to compete it?

    I do not know what it is...but even the few I did earlier...it is taking me forever to set them up...I am having a hard time understanding.

    Ok...so this would be

    1/4+1/6+1/2 = 3/12 +2/12 + 6/12 = 11/12

    Then 1*12/11 = 12/11 = 1 1/11 = 1.09

    So... Job/Rate and add them together (if additive)...and then Job/Rate the answer?
     
    Last edited: Dec 11, 2008
  19. Dec 11, 2008 #18
    We did these like this:
    You always consider the units of time that you are faced with. Here it is minutes.
    In one minute, Jim can do 1/30 th of the job. In one minute Sue can do 1/45 th of the job, and in one minute, Tony can do 1/{3\2} of the job. We set it up as follows:

    [tex]\frac{1}{30}x+\frac{1}{45}x+\frac{2}{3}x=1[/tex] one being one job completed...the x is the time....rate x time = work completed.
    Find a common denominator and solve.
    CC
     
  20. Dec 11, 2008 #19
    Sam can do the job in 4 days..so he gets 1/4 th of the job done in a day. Lisa can get 1/6th of the job done in a day, and tom can get 1/2 of the job done in a day.

    Use the work completed=rate x time to get the equation.
    [tex]\frac{1}{4}x+\frac{1}{6}x+\frac{1}{2}x=1[/tex]...one job completed...x is the time and the rate is "job completed per day"....

    CC
     
    Last edited: Dec 11, 2008
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