Candy's rate = route/70
Tim's rate = route/80
combined rate = route/80 + route/70
= 3(route)/112
so time to do route with combined rate
= route/[3route/112]
= 112/3
= 37 1/3 minutes
and of course 1/3 minute is 20 seconds,
so 37 minutes, 20 seconds
Candy and Tim share a paper route. It takes Candy 70min to deliver all the papers, and it takes Tim 80min. How long does it take the two when they work together?
The answer is 37min and 20s. Here's what I have. 7/6t + 8/6t = 15/6
I have t = 5/2. I don't think that's 37min and 20s. Is it?
4 answers
how did you get 3(route)/112 ?
You may want to consider a slightly different approach. It makes it easier for me to understand.
Assume the paper route is 100 papers.
Then Candy can deliver 100 papers/70 min = 1.42857 papers/min.
Tim can deliver 100 papers/80 min = 1.25 papers/min.
Together they can deliver 1.42857/min + 1.25/min = 2.67857 papers/min.
So if there are 100 papers, it will take 100 papers x (1 min/2.67857) = 37.3333 minutes. This method has the disadvantage that the decimal equivalent of 1/3 is a repeating fraction; however, the answer is 37 min and 19.999999 seconds which for all practical purposes is 37 min and 20 seconds.
Assume the paper route is 100 papers.
Then Candy can deliver 100 papers/70 min = 1.42857 papers/min.
Tim can deliver 100 papers/80 min = 1.25 papers/min.
Together they can deliver 1.42857/min + 1.25/min = 2.67857 papers/min.
So if there are 100 papers, it will take 100 papers x (1 min/2.67857) = 37.3333 minutes. This method has the disadvantage that the decimal equivalent of 1/3 is a repeating fraction; however, the answer is 37 min and 19.999999 seconds which for all practical purposes is 37 min and 20 seconds.
Sean, if in DrBob's way you replace "100 papers" with my "route" the solutions are the same, except Bob changed his fractions to decimals.
in my case, "route" canceled, in Bob's case "100 papers" canceled.
In these "working together" kind of problems, if one person can do the job in a hours, another in b hours, c hours for the next and so on,
then when all are working together, the time taken will be
1/(1/a + 1/b + 1/c + ...)
in my case, "route" canceled, in Bob's case "100 papers" canceled.
In these "working together" kind of problems, if one person can do the job in a hours, another in b hours, c hours for the next and so on,
then when all are working together, the time taken will be
1/(1/a + 1/b + 1/c + ...)