|
|
|
|
||
|
|
|
|
|
"Work" Word Problems (page 4 of 4)
Well, Bill can do it in "b" hours, so he does 1/b per hour. Tom does it in "t" hours, so he does 1/t per hour. Then, working together, they can do 1/b + 1/t = 1/8 per hour, since they took 8 hours. I also know that Tom takes 12 hours longer than Bill, so t = b + 12. Then I get 1/b + 1/(b + 12) = 1/8. That is: hours to complete job: completed per hour: adding their labor: 1/b + 1/(b + 12) = 1/8 To solve for "b", I'll multiply through by 8b(b + 12) to get rid of all of the denominators: 8(b + 12) + 8(b) = (b)(b
+ 12) (See Factoring Quadratics, if you're not sure how I just got to that last line.) Then b = 12 and b = –8. Of course, for the purposes of our problem, b must be positive, so we'll ignore the "b = –8" solution; it's "extraneous" (pronounced "ek-STRAY-nee-uss", meaning "valid mathematically, but pointless as far as our situation is concerned"). Then Bill takes 12 hours, so Tom takes 24 hours. This next one is a bit different: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
Ben can do 250 dishes per hour, and Frank can do 150 dishes per hour. Working together, they can do 250 + 150 = 400 dishes an hour. That is: Ben: 500
dishes / 2
hours = 250
dishes / hour Now that I have their hourly rate, I have to find the number of hours that it takes to wash 1000 dishes. Set things up so units cancel: (1000
dishes) × (1
hour / 400
dishes) It will take two and a half hours for the two of them to wash 1000 dishes.
Convert this to man-hours, or, in this case, man-days. If it takes six guys fourteen days, that is: (6 men) × (14 days) = 84 man-days Now you're expanding the time allowed from fourteen days to twenty-one days. How many guys do you need? The job itself still takes the same number of man-days, but, since you're spreading the work over more days, you need fewer guys. But how many guys, exactly? (x guys) × (21 days) = 84 man-days ...or, in algebra: 21x = 84 x = 4 So you'll need four guys to do the job in twenty-one days. You may have noticed that each of these problems used the "how much can be done per time unit" construction, but other than that, each problem was done differently. That's how "work" problems are. You'll have to be alert and clever to do these. But as you saw above, if you do things neatly and orderly, you'll probably find your way to the solution. << Previous Top | 1 | 2 | 3 | 4 | Return to Index
|
|
|
|
Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
|
|
|
|
|
|