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# Rational equations word problem: combined rates

Sal solves a word problem about the combined leaf-bagging rates of Ian and Kyandre, by creating a rational equation that models the situation. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• I have the gut feeling to start out with their rates in hours/lawn. Thus their combined rate (K + I) would be 3hours/lawn + 5hours/lawn= t/lawn or total hours per lawn. But this is obviously incorrect. What am I doing wrong? Sal starts out correctly with lawn/hour. Why is my intuition to start out with hour/lawn instead of lawn/hour is wrong. I just can't figure out why one rate (hour/lawn) is incorrect to solve the problem while going with another rate (lawn/hour) is correct. •  Hours per lawn is not an unreasonable intuition to start off with, but we can quickly see that adding their times together to get 8 hours per lawn doesn't make sense, as the time should get shorter when they are working together.

What you really want to ask yourself is "How much can each person do in a given time?" When the amount of time is the same for both people, we can get a much better comparison of their efforts.
This is why Sal talks in terms of one third of a lawn per hour, and one fifth of a lawn per hour.
It is in this way that you can add their efforts together to see how much of one lawn they can complete in one hour.

That's the way I see it at least. Hope it helps.
• It is not so intuitive for me.
Say Ian takes 5 hrs for 1 lawn & Kyandre takes 3 hrs for 1 lawn.
So they take 8 hours for 2 lawns and hence 4 hours for a lawn ?
what is wrong in above assumption ? •  They don't take 8 hours for 2 lawn.
Here is why. After 3 hours Kyandre has finished 1 lawn and Ian has finished 3/5 of his lawn. At this point, Kyandre, helps Ian with the second lawn and they finish faster than Ian would alone.
• At , why can Sal flip the left and right sides and just use the reciprocal of both sides? I can't remember the rule or property that justifies it. Help would be greatly appreciated, thanks. Also, I tried to do it by cross multiplying but couldn't figure out how the units get changed from lawns per hour to hours per lawn. • One way to see it is that Sal is doing a few steps at once. He starts with 8/15=1 (lawn)/t (hours). Next lets multiply both sides by t (hours) to cancel the t (hours) on the right and get (8/15)t (hours)=1 (lawn). Next multiply both sides by 15/8 to cancel the 8/15 on the left and get t (hours)=15/8 (lawn). As a last step we could divide both sides by 1(lawn) to get t (hours/lawn)=15/8. That is how he did it.

You could also justify taking the reciprocal by the following reasoning. If A=B then 1/A=1/B (assuming A and B are not 0). This is true since we can directly substitute B in for A because of the equality. In general if you have an equality you can do whatever you want to both sides as long as you do it to both and you don't do something like dividing by 0.

As a side note, a real example of reciprocal units is period and frequency of waves. Period measures seconds per cycle and frequency measures cycles per second. If the frequency is 30 beats per minute than the period is 2 seconds per beat.
• at , why does Sal multiply 60 by 7/8? What happened to the 1 7/8? • The time needed if they both worked together (t) was 1 7/8 HOURS.
So Sal kept the 1 hour separate and then converted 7/8 of an hour to minutes by multiplying by 60 min/hr.
After he got the answer of 52.5 minutes, he then combined that with the 1 hour to get the final answer of 1 hour 52.5 min.
• When combining the rates my intuition tells me to take the average of the 2 rates (8/30) rather than summing them (8/15). Can anyone explain the logic as to why you sum them? I cant quite get my head around it. • The product over sum formula is one of the most useful tools in applied math imho.
(3*5)/(3+5) = 15/8 • What is the formula for this? • By now, i'm quite familiar with work rate problems, but I have hit a particular snag that no one on the internet seems to have addressed...
How do I algebraically represent one of the two workers leaving the job before it's complete? Here is the problem:

" A man can lay a concrete sidewalk in 5 days; his assistant can lay the same sidewalk in 8 days. After working together for 2 days, the man is called away. How long will it take the assistant to finish the work? "

I have gotten so far as to determine that it would take the two workers 40/13 days to lay the sidewalk if they were to work together the whole way through, but after that, I'm completely stuck. I'm sure i'm not the only one who has seen a problem like this, but as i have scoured the internet for examples of such problems, it seems that either no one else has seen such a problem, or I'm missing something painfully obvious. can somebody clue me in?
(1 vote) • In order to treat such problems more easily you first need to convert (in mind or on paper) the rates into (y/x part of the work done / (per) time interval notation) which in your particular problem would be 13/40 (combined efforts) / (per) day. It may seem not that intuitive but 13/40 represents the same daily rate but in a more disguised way.
Personally I find very confusing multiplying 13/40 (sidewalks/days) by 2 days (different frame of reference) but I have no problem when I know that per day 13/40 of sidewalk is done and when 2 days pass I just double the rate.
Thence, when you start solving a problem it is much more keep track of what you are doing when you keep in mind the same frame of reference.

I would apply similar thought process to the second part.

Cheers.  