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February 2nd, 2012, 04:32 PM  #1  
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  "Workload" (rate) problems Quote:
For a simpler problem, such as: Sarah takes 10 minutes to mow a single lawn and Jessica takes 15 minutes to mow that same lawn, apparently the algebra is: (1/10) + (1/15) = 1/t For the solution to how long it will take both to mow the same lawn if they do it together. I don't understand why it's adding the two efforts; though I understand that the '1' is that they can mow 1 lawn every number of minutes which is why it's a fraction. Isn't adding them just increasing the total time it takes to mow the lawn, by adding both Sarah's time to mow one lawn and Jessica's time to mow one lawn while doing it at the same time? So back to the original question, I'll attempt to just plug in numbers without even understanding why and how the formula even works: A (Andy) = 75/20 B (Ben) = 100/30 C = 75/25 D = 70/15 So I guess the first step is to subtract the square feet that Andy already mowed at his pace for 5 minutes: 500/t  [(75/20)*5] 500/t  75/4 Actually, I'm not even sure. My head hurts right now so I'll probably attempt it tomorrow after I sleep, so I'll just leave this here for tomorrow...  
February 2nd, 2012, 06:31 PM  #2  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: "Workload" (rate) problems Quote:
Suppose we define the task to be traveling 100 miles. If we know it takes a person 2 hours to complete this task, then we know his speed, or rate of accomplishing the task is: Suppose we define the task to be filling a 100 gallon tank with water from a hose. If we find it takes 1 hour 15 minutes to fill the tank, then we know the rate of flow from the hose is: Sometimes, if we are given the rate and the amount of time, to find the amount of the task done, we may wish to arrange our definition as: Suppose we are told a person can paint 200 square feet per hour, and we wish to know how much will be painted in 2.5 hours. We can then find this by stating: Or, suppose we find 900 square feet have been painted at that rate, and we wish to know how long the person was painting, then we arrange the definition as: Quote:
But, let's look at this from the perspective of combining or adding what portion of the task has been completed in 1 hour. The first person will have done 20/100 = 2/10 of the task while the second person will have done 30/100 = 3/10 of the task. If t represents the total time it will take them to accomplish the task, then we also know that after 1 hour, 1/t of the task has been done. So, their combined portions will equal the total portion: Now, let's look at Sarah and Jessica. In 1 minute, Sarah will mow 1/10 of the lawn and Jessica will mow 1/15 of the lawn, and working together for that 1 minute, they will mow 1/t of the lawn, where t is the number of minutes it takes them to complete the lawn. So we add their individual efforts to find the total portion done after 1 minute: So, we find that working together, Sarah and Jessica can mow the lawn in 6 minutes. For a quick and easy method for solving a problem of this type, simply compute the ratio of the product of their individual times over the sum of their individual times: I recommend this only after you understand how the formula above actually works. Here's another way to look at the problem. Suppose we let t be the time it takes them to mow the lawn. If we add the portion Sarah does to the portion Jessica does, then we will have 1 job done. Recall we found that the total is rate times time, so we have: And here's another method I thought clever that I observed [color=#00BF00]skipjack[/color] using before: In 30 minutes, Sarah can mow 3 lawns and Jessica can mow 2 lawns, so working together they can mow 5 lawns in 30 minutes, or 1 lawn in 6 minutes. To address your observation that adding the efforts increases the time, observe that we are adding to get 1/t. When we make 1/t larger, we are making t smaller, as this is an inverse relationship. If we make the denominator of a fraction smaller while holding the numerator fixed, the fraction gets larger, do you see what I mean? Now, let's look at the problem you posted: Quote:
For the next 5 minutes, Andy continues mowing, so we add another , and we add Ben's contribution during that five minutes. Ben mows at a rate of so during that 5 minutes, he mows So, after ten minutes, the total mowed is . Now, for the next 8 minutes, Andy is on break, Ben continues mowing and Charlie and Dave join in. During this time Ben will mow , Charlie will mow and Dave will mow . So the total for this 8 minute period is So, for the first 18 minutes, we have a total of . For the next 2 minutes, Ben continues mowing alone, adding another for a total for the first 20 minutes of . For the remainder of the 500 square feet, Andy and Ben work together. They have left to do. Combining their rates, we find: and adding this to the previous 20 minutes, we find the total time is approximately 69.58 min.  

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