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Work problem # 2

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by oks » Fri Jun 12, 2009 4:59 pm
I must be missing a point with the work problems, b/c I keep missing all of them.

Five farm workers can pick 10 acres of blueberries in 20 hours. How many hours would it take 8 workers to pick the same amount of blueberries?

A. 12.5
B. 14
C. 16
D. 23
E. 32

OA is A

Since the amount is the same, we need to find how long does it take 1 worker: 5 workers - 20 hours, so 1 worker - 4 hours. Then it would take 32 hours for 8 workers?

Please explain. Thanks.
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by ssmiles08 » Fri Jun 12, 2009 5:17 pm
5 workers pick 10 acres of blueberry fields in 20 hours

1 worker can pick 2 acres of blueberry fields in 20 hrs.

1 worker can pick 1/10 (2/20) acres of blueberry fields in 1 hr.

8 workers can pick 4/5 (8/10) acres of blueberry fields in 1 hr.

8 workers can pick 10 (4/5)*(25/2) acres of blueberry fields in 25/2 hrs.

So total time it takes to pick for 8 workers in 10 acres is: 25/2 = 12.5hrs.
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Re: Work problem # 2

by Naruto » Fri Jun 12, 2009 7:02 pm
oks wrote:I must be missing a point with the work problems, b/c I keep missing all of them.

Five farm workers can pick 10 acres of blueberries in 20 hours. How many hours would it take 8 workers to pick the same amount of blueberries?

A. 12.5
B. 14
C. 16
D. 23
E. 32

OA is A

Since the amount is the same, we need to find how long does it take 1 worker: 5 workers - 20 hours, so 1 worker - 4 hours. Then it would take 32 hours for 8 workers?

Please explain. Thanks.
You know i used to face the same confusion in work, but if u like i can just clear a basic funda of yours which would always help you. Just remember in work its usually inverse variation instead of direct. You can always check it with logic, like here since 5 people take 20 hrs so it would be obvious that more ppl will take less than 20 hours right?
so now just plug in the equation but inverse variation.
i.e
lets assume the no. of hrs required to be x
so,
5/8=x/20
now solving for x you'll get x=12.5
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by Naruto » Fri Jun 12, 2009 7:04 pm
oh and one more thing , here it formed a simple inverse equation, because 10 acres was constant, else you would have to first calculate for a constant amount and then vary, had 10 acres changed to something else.
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by vpmba2009 » Sat Jun 13, 2009 8:57 am
Work problem is my concern, too.
I found an easier way to approach this kind of problem.
If it takes M workers H hours to finish a job, then each worker does 1/MH of the job in an hour. So the rate of each worker is 1/MH.
----------------------
Apply this to your problem: Five farm workers can pick 10 acres of blueberries in 20 hours. So the rate of each worker is 1/(5*20)=1/100.
8 workers will do 8*1/100=8/100 of the job in an hour.
Call T is the time it takes 8 workers to complete the job (1 job).
Then (8/100)(T)=1 => T=100/8=12.5
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Re: Work problem # 2

by Ian Stewart » Sat Jun 13, 2009 11:09 am
oks wrote: Five farm workers can pick 10 acres of blueberries in 20 hours. How many hours would it take 8 workers to pick the same amount of blueberries?

A. 12.5
B. 14
C. 16
D. 23
E. 32
Alternatively, we can use the management principle of 'person-hours' to solve problems when all of our workers are identical. The '10 acres' is irrelevant here; we know that 5 workers can finish the job in 20 hours. So the job requires a total of 5*20 = 100 hours of work. If we have 8 workers, they will each therefore need to work for 100/8 = 12.5 hours.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by tohellandback » Sat Jun 13, 2009 11:42 am
hey u can use the allegation method too:

5 x


8 20

5*20=8*x
x=12.5
The powers of two are bloody impolite!!
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by Umar82 » Fri Jun 19, 2009 3:23 pm
since you are keeping the "10 acres" constant . . . there is an easier way to do this.

5 workers__________10 acres___________20hours

1 worker___________10 acres___________100 hours

so . . . if it takes 1 worker 100 hours, it would take eight workers, 100/8 = 12.5
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