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Age Problem Part 2 - Little Tougher

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Age Problem Part 2 - Little Tougher

by jnellaz » Thu Nov 20, 2008 6:45 pm
Using the knowledge you have imparted in my last Age Problem post, can you show me how it would apply to this problem? Thanks all!


Deb can do a job alone in 6 hours. After working alone for 2 hours, she is joined by Sue who can do the job alone in 8 hours. They work together and finish the job. How many hours did Deb work?

[spoiler]Answer: 4 2/7[/spoiler]
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by cramya » Thu Nov 20, 2008 7:44 pm
RATE OF DEB = 1/6
RATE OF SUE = 1/8

Deb works two hours she completes 1/6*2 =1/3 of the job
(Rate*Time = Work)

Remaining work 2/3

When Deb ans Sue work together u add their rates

1/6+1/8

Use the formula rate * time = work again to find time

(1/6+1/8)*T = 2/3 (why its not 1 since Deb has completed 1/3 rd of job working alone so only 2/3rds remain)

7/24*T = 2/3

T=16/7

Deb has worked 2+16/7 = 4 2/7 hours

Jnellaz, hope this helps; let me know if u still hv questions.
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by Neo2000 » Thu Nov 20, 2008 7:47 pm
After 2 hrs Sue has done 2/6 = 1/3 of the work

Remaining work = 2/3

New Efficiency = (1/6) + (1/8)

Time taken to complete work = [spoiler](2/3)/((1/6) + (1/8))[/spoiler]
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by jnellaz » Thu Nov 20, 2008 8:16 pm
Much appreciated! By the way...is it safe to assume that 8) = 8?
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by Neo2000 » Thu Nov 20, 2008 8:33 pm
jnellaz wrote:Much appreciated! By the way...is it safe to assume that 8) = 8?
LOL! yes "pretty safe" i should say :)
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by Neo2000 » Thu Nov 20, 2008 8:36 pm
Also this is not so much an age problem as it is a "Work" or specifically "Time and Work" problem.
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by jnellaz » Fri Nov 21, 2008 6:41 am
Bad wording on my part. I meant to say Work. (You see what happens when you go 2 days without sleep.)
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