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Work Problem: Re-opening a thread

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Work Problem: Re-opening a thread

by kanha81 » Thu Apr 23, 2009 12:32 pm
Is there an efficient way to solve these type of problems?

Working alone, Maria can complete a task in 100 minutes. Shaniqua can complete the same task in two hours. They work together for 30 minutes when Liu, the new employee, joins and begins helping. They finish the task 20 minutes later. How long (in minutes) would it take Liu to complete the task alone?

A. 180
B. 240
C. 200
D. 160
E. 220

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Re: Work Problem: Re-opening a thread

by Vemuri » Thu Apr 23, 2009 7:37 pm
Given, M can do the work in 100mins & S can do the same work in 120mins.

Remember the standard rate formula ==> rate = work/time

M can do the work alone in 100mins. This means that she can do 1/100th of the work in 1 minute. Similarly, S can do 1/120th of the work in 1 minute. So, in 1 minute both M & S can do (1/100 + 1/120)th of the work. The expression reduces to 11/600.

If M & S work for 30 minutes together, then 30*11/600 ==> 11/20th of the work is completed. Hence, the remaining work is 1-(11/20) ==> 9/20.

The question stem says that Liu joins at this stage & the remaining work is completed in 20mins. So, 20 * (1/100 + 1/120 + 1/L) = 9/20
==> 1/5 + 1/6 + 20/L = 9/20
==> 20/L = 9/20-1/5-1/6
==> 20/L = 1/12
==> L = 240. So, L alone can do the work in 240mins.

Hope this helps.
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by vittalgmat » Fri Apr 24, 2009 12:45 am
Awesome explanation Vemuri !!!!!.
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Re: Work Problem: Re-opening a thread

by kanha81 » Fri Apr 24, 2009 8:11 am
Vemuri wrote:Given, M can do the work in 100mins & S can do the same work in 120mins.

Remember the standard rate formula ==> rate = work/time

M can do the work alone in 100mins. This means that she can do 1/100th of the work in 1 minute. Similarly, S can do 1/120th of the work in 1 minute. So, in 1 minute both M & S can do (1/100 + 1/120)th of the work. The expression reduces to 11/600.

If M & S work for 30 minutes together, then 30*11/600 ==> 11/20th of the work is completed. Hence, the remaining work is 1-(11/20) ==> 9/20.

The question stem says that Liu joins at this stage & the remaining work is completed in 20mins. So, 20 * (1/100 + 1/120 + 1/L) = 9/20
==> 1/5 + 1/6 + 20/L = 9/20
==> 20/L = 9/20-1/5-1/6
==> 20/L = 1/12
==> L = 240. So, L alone can do the work in 240mins.

Hope this helps.
Excellent explanation, mate! Thanks a bunch. It really makes it easier to understand.
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