Question

The OfficeJet printer can copy Maria's dissertation in 16 min. The LaserJet printer can copy the same document in 18 min. If the two machines work​ together, how long would they take to copy the​ dissertation?

Answer 1

If the two printers divide the work, it will take them about 8.47 minutes (= 8 minutes 28 seconds) to complete the job.

Explanation:

Let the number of pages in Maria's dissertation = `n`.

Let's assume that we will split her dissertation into two parts. One part, we will have printed by the Office Jet, and the remaining part we will have printed by the Laser Jet. Let

`x` = the number of pages that we will have printed by the Office Jet

This means that we will have `n-x` pages printed by the Laser Jet.

The time it takes the Office Jet to print a page is `16/n` minutes per page.

The time it takes the Laser Jet to print a page is `18/n` minutes per page.

The time it takes the Office Jet to print `x` pages is `16/nx` minutes.

The time it takes the Laser Jet to print `n-x` pages is `18/n(n-x)` minutes.

We want to divide the job between the two printers in such a way that they each take the same time to print the pages assigned to them. Therefore, we can write

`16/nx=18/n(n-x)`

`16x=18(n-x)`

`16x=18n-18x`

`34x-18n`

`x/n=18/34=9/17`

As we noted above, the time it takes for the Office Jet to print its pages is

`16/nx=16(x/n)=16(9/17)=144/17~~8.47` minutes

This is about 8 minutes and 28 seconds.

Note that this is the same amount of time it takes the Laser Jet to print its pages. As we noted above, the time it takes for the Laser Jet to print its pages is

`18/n(n-x)=18(1-x/n)=18(1-9/17)=18(8/17)=144/17`.

Answer 2

`8.47`min.

Explanation:

The combined time will be slightly less that the arithmetic mean of the 'half' time of the two (8.50) because the faster printer will print more than one-half of the document.

Taking an arbitrary length of 100 pages to avoid too many variables (it works out the same either way), we have the first rate as:
`R_1 = 100/16 = 6.25`
And the second rate as:
`R_2 = 100/18 = 5.55`

The combined rate is thus 11.75, and the time to print 100 pages would be:
`100/11.75 = 8.47`min.

In general then,
`R_1 = P/T_1` ; `R_2 = P/T_2` ; `P/(R_1 + R_2) = T_3`

We can remove the arbitrary "P" with either original expression.
`R_1 = P/T_1` ; `P = R_1xxT_1`

`(R_1xxT_1)/(R_1 + R_2) = T_3 = (R_2xxT_2)/(R_1 + R_2)`
But, that only works when you know the rate in the first place, and that is scalable over any range, so picking an arbitrary number of pages works well.