
Jonathan F. answered 09/20/14
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Math and Computer Tutor for All Ages
This is actually a two-part question.
1. If a team of 10 can do the job in 36 hours, then how large a team is required to get the job done in just 30 hours?
2. At a labor rate of $28 per hour, ignoring the additional men on the team (from the problem description, they don't affect the cost of the job), how much money is saved by decreasing the number of hours from 36 to 30?
Part 1:
This is a proportion: x * y = a * b. Three of these values are given, so adjusting any one of them would change the value of the unknown (the one that isn't given). In the computer world, to avoid confusion with the letter, "x", the * symbol is used to denote multiplication.
10 men * 36 hours (for the whole job) = 360 "man-hours" of work to be done.
On the other side of the proportion, you have an unknown ("a") number of men, and 30 hours.
10 * 36 = a * 30
Solve for "a", by isolating it on one side of the equation.
10 * 36 / 30 = a [Given how ÷ requires you to type Alt+0247 on the computer, / is usually used to denote division there.]
360 / 30 = a
12 = a
So, it will take 12 men 30 hours to do the job, which is still 360 man-hours' worth of work.
A little aside: This isn't usually how it works in the real world! Adding additional people to a project introduces various inefficiencies (getting the new people up to speed, communication between them, logistics), increases costs (they all have to get paid, and be managed effectively), and may even cause the project to take LONGER! See http://en.wikipedia.org/wiki/The_Mythical_Man-Month and http://en.wikipedia.org/wiki/Brooks%27s_law.
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Part 2:
This is just figuring.
$28 per hour * 36 hours - $28 per hour * 30 hours = amount of money saved.
This can be simplified to: $28 per hour * (36 hours - 30 hours).
28 * (36 - 30) = 28 * 6 = $168 saved.
Phillip R.
09/18/14