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Jack can paint a house in 5 days, and Richard can paint the same house in 7 days. Working together, how long will it take them to finish the job?


A second method for such problems, often taught in school, is to change from using the days it takes to do a job to the amount of the job one can finish in a day. In other words, we want to translate

           can do a job in N units of time
into
           can do 1/N of a job in 1 unit of time


Why would we do that? Well, now we can picture what happens in a single unit of time:

  +------+------+------+------+------+
  |######|      |      |      |      |  
  +------+------+------+------+------+

     Jack: 1/5 of the job in 1 day

  +----+----+----+----+----+----+----+
  |    |    |####|    |    |    |    |  
  +----+----+----+----+----+----+----+

     Richard: 1/7 of the job in 1 day

  +----------+-----------------------+
  |##########|                       |   
  +----------+-----------------------+
  
     Together: 1/5 + 1/7 of the job in 1 day

Of course, now we have to add 1/5 and 1/7 to find out how much of the job they can do together in one day. To do that, we need to find a common denominator; the easiest way to do that is to multiply each fraction by d/d, where d is the denominator of the other fraction:

  1   1   1   7     1   5
  - + - = - * -  +  - * -
  5   7   5   7     7   5

            7       5
        = ----- + -----
          5 * 7   7 * 5

          5 + 7
        = -----   
          5 * 7

        = 12/35 of the job in 1 day

Now we do the inverse of the previous translation, from

           can do a/b of a job in 1 unit of time
into
           can do a job in b/a units of time

In this case, we translate from

           can paint 12/35 of a house in 1 day
to
           can paint a house in 35/12 of a day

You may have noticed that when we began our explanation (not solving by adding fractions), instead of finding a common denominator for two fractions, we found an interval during which we knew that both Jack and Richard would be able to paint whole numbers of houses - and that this interval was the same as the denominator of the two fractions.

This is no coincidence. In the fractional method shown on this page, we're adding fractional completions in a single unit of time; whereas when we began, we were adding whole completions in multiple units of time.

That is, with the fractional method we're asking: how finely do we have to slice up a day in order to do this addition? In the other method, we're asking: how many days do I have to let this stretch out in order to do this addition? The number of slices is the number of days. They're really just two ways of looking at the same thing, except one is expressed as a fraction and the other isn't.

So we still haven't really found the answer to why one learns the fractional method.

The ability to add fractions with different denominators will be extremely important later on when you get to algebra, e.g.,

    3       2       3(x-2)       2(x+1)     3(x-2) + 2(x+1)
  ----- + ----- = ---------- + ---------- = ---------------
  (x+1)   (x-2)   (x+1)(x-2)   (x-2)(x+1)      (x+1)(x-2)

    ^       ^          ^           ^      
    |       |          |           |
    +---+---+          +-----+-----+
        |                    |
    Different             Same 
    denominators          denominator
        |                    |
    +---+---+          +-----+-----+
    |       |          |           |
    v       v          v           v

    3       2         3(5)        2(7)      3(5) + 2(7)
    -   +   -   =     ----   +    ----    = ----------- 
    7       5         7(5)        5(7)         (5)(7)

The technique is the same in each case, so if you learn to do it using numbers, you should have no trouble later on learning to do it using expressions involving variables.

Thus the answer to the question is pedagogical, rather than mathematical: by converting the rates from time per unit job to jobs per unit time, you're forced to add two fractions.

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