Word Problem

[KCarney10]

I'm studying for a practice test I have to take for nursing school and I keep coming across word problems that look like this but I can't remember how to set it up to get the answer. Help?
Jonathan can type a 20 page document in 40 minutes, Susan can type it in 30 minutes, and Jack can type it in 24 minutes. Working together, how much time will it take them to type the same document?

 

[mmm4444bot]

Here's one approach.

Express each person's work in terms of their fractional amount of the total job done per unit of time.

We can ignore the number 20; it does not matter how many pages comprise "the job". We don't even care what they're actually doing; it suffices to know that the task is the same for each of them.

We're more concerned with the time unit -- which is one minute -- and how much of the job gets done in that amount of time, by each person.

Then, we simply sum up these three fractional amounts of work completed per minute because that sum represents how much work gets done in one minute when all three work on "the job" together.

The reciprocal of the sum represents the required minutes.

Jonathan needs 40 minutes to complete the job; therefore, Jonathon does 1/40th of the job per minute.

Likewise, Susan completes 1/30th of the job per minute.

Jack completes 1/24th of the job per minute.

With all three working at it at once, the fractional amount of the job that gets done each minute is:

1/40 + 1/30 + 1/24

The reciprocal of this number is the time needed to complete the job.

I mean, for example, if the three contributions per minute add up to 1/4, then it takes 4 minutes to complete the entire job working together.

There are other approaches to these types of exercises.

Cheers ~ Mark

 

[tkhunny]

The ONLY difficulty with such problems is that you have incompatible information. pages / minute just doesn't add. The secret is that minutes / page DOES add nicely.

Thinking logically, if they don't trip over each other, the right answer should be less than 24 minutes. That is the fastest single time. Working together, they must do it faster than that. If we get ANYTHING greater than 24 minutes, we'll have to go back to the drawing board.

I like to think in terms of "jobs". If the job is a 20 page document, how much of it can Jon tackle in 1 minute? 1/40. How much for Sue in 1 min? 1/30. And Jack? 1/24. So, how much of the JOB can be accomplished in 1 minute? 1/40 + 1/30 + 1/24 = 1/10

Finally, how long does it take to complete the entire JOB at this rate: (1/10 document) / minute?

This section is for the VERY interested student.
Caveats: It is unlikely that this rate will be achieved. Remember that "trip over" assumption? Well, unless they all finish a page at EXACTLY the same time, the rate will slow substantially somewhere toward the end. Check it out. If we think one page comes out every 40 seconds, we'll be wrong. Each must start a different page and a page does not actually appear until someone finishes one. The moment of the first page is when Jack prints page one at the 72 second mark. The second page is by Sue at 90 seconds and the 3rd by Jon at 120 seconds.
The next three are at 144, 180, and 216, by Jack, Sue, and Jack respectively. Now we have six done.
The next six are at 240, 270, 288, by Jack, Sue and Jon THEN all three finish a page at 360 seeconds. We have 12 done.
13 is at 432 by Jack
14 is at 450 by Sue
15 is at 480 by Jon
We're getting close to the exciting part!
16 is at 504 by Jack
17 is at 540 - this is Sue's 6th page.
18 is at 576 - this is Jack's 8th page - But john must now stop, since Jon and Sue are working on the last two pages. There is no more work for Jack.
19 is at 600 by Jon - when we thought we were done, but Sue is only 2/3 done with the last page. We must wait another 30 seconds for Sue to finish.

Of course, what we SHOULD have done was get rid of Jon when Jack was finished with #8 If we lose Jon and substitute Jack, we get a slightly different result.
Jack takes over and punches out Jon's last page at just under the 593 mark. Oddly, this realy doesn't help, since we still have to wait for Sue, who won't finish until 630.

Clearly, then, our only hope is to have Jack sub for Sue when he is done with Jon's last partial page. This saves only a little time, finishing at just under 623, still nowhere near 600. Why? Jon was idle after 576. Sue was idle after 593.

That "trip over" assumption is VERY important in practice.

 

[Denis]

Ok ok TK...
Need a li'l "program/strategy" to assign complete pages; theoretical = 10 minutes:

Jonathan can type a 20 page document in 40 minutes ; 5 pages = 10 min

Susan can type it in 30 minutes ; 7 pages = 10.5 min

and Jack can type it in 24 minutes ; 8 pages = 9.6 min

In other words, closest possible to 10 min each :idea: :shock:

 

[BigGlenntheHeavy]

\(\displaystyle In \ doing \ a \ certain \ job, \ in \ this \ case \ typing \ a \ 20 \ word \ document, \ Jonathan\)

\(\displaystyle can \ do \ it \ in \ 40 \ minutes, \ Susan \ can \ do \ it \ in \ 30 \ minutes, \ and \ Jack \ can \ do \ it \ in\)

\(\displaystyle 24 \ minutes. \ So \ Jonathan's \ rate \ is \ 1/40, \ Susan's \ rate \ is \ 1/30 \ and \ Jack's \ rate \ is \ 1/24.\)

\(\displaystyle Now, \ all \ working \ together, \ how \ long \ will \ it \ take \ them \ to \ complete \ the \ job?\)

\(\displaystyle Ergo, \ (1/40+1/30+1/24)t \ = \ 1 \ (unity, \ the \ completion \ of \ the \ job, \ t \ = \ time \ required).\)

\(\displaystyle Therefore \ t \ = \ 10 \ minutes.\)

 

[soroban]

Hello, KCarney10!

Here's another explanation . . .

Jonathan can type a document in 40 minutes, Susan can type it in 30 minutes, and Jack can type it in 24 minutes.
Working together, how much time will it take them to type the same document?

\(\displaystyle \text{Jonathan can do the job in 40 minutes.}\)
. . \(\displaystyle \text{In one minute, he can do }\frac{1}{40}\text{ of the job.}\)
. . \(\displaystyle \text{In }t\text{ minutes, he can do }\frac{t}{40}\text{ of the job.}\)

\(\displaystyle \text{Susan can do the job in 30 minutes.}\)
. . \(\displaystyle \text{In one minute, she can do }\frac{1}{30}\text{ of the job.}\)
. . \(\displaystyle \text{In }t\text{ minutes, she can do }\frac{t}{30}\text{ of the job.}\)

\(\displaystyle \text{Jack can do the job in 24 minutes.}\)
. . \(\displaystyle \text{In one minute, he can do }\frac{1}{24}\text{ of the job.}\)
. . \(\displaystyle \text{In }t\text{ minute, he can do }\frac{t}{24}\text{ of the job.}\)

\(\displaystyle \text{Working together for }t\text{ minutes, they can do }\frac{t}{40} + \frac{t}{30} + \frac{t}{24}\text{ of the job.}\)


\(\displaystyle \text{But in }t\text{ minutes, we expect them to do }the\:whole\:job.\)

\(\displaystyle \text{There is our equation . . . }\;\frac{t}{40} + \frac{t}{30} + \frac{t}{24} \;=\;1\)


\(\displaystyle \text{Multiply by 120: }\;3t + 45 + 5t \:=\:120 \quad\Rightarrow\quad 12t \:=\:120\)

. . . . . . . . . . . . \(\displaystyle \boxed{t \:=\:10\text{ minutes}}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{Check}\)

\(\displaystyle \text{In 10 minutes, Jonathan will do }\frac{10}{40} \,=\,\frac{1}{4}\text{ of the job.}\)

\(\displaystyle \text{In 10 minutes, Susan will do }\frac{10}{30} \,=\,\frac{1}{3}\text{ of the job.}\)

\(\displaystyle \text{In 10 minutes, Jack will do }\frac{10}{24} \,=\,\frac{5}{12}\text{ of the job.}\)


\(\displaystyle \text{Together, they will do: }\:\frac{1}{4} + \frac{1}{3} + \frac{5}{12} \;=\;1\text{ job . . . }\;\text{ta-}D\!AA!\)