Work problem

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Work problem

by nadib002 » Mon May 09, 2011 2:27 pm
A, B and C can finish a piece of work in 30, 40, and 60 days respectively. 10 days after they started to work together B leaves. 4 days after B left, A leaves and C completes the remaining work. Find how many days C had worked altogether.

This is how i did it: Please let me know if there is another easier way of solving it:

rate of A = 1/30

rate of B = 1/40

rate of C = 1/60

Work completed by all 3 in 10 days is W = rate * time


(1/30 + 1/40 + 1/60) * 10 = 3/4


Work left to complete = 1-3/4 = 1/4


Since B left after 10 days, the remaining work will have to be finished by A + C

rate of A + C together = (1/30 + 1/60)

Work done by A+C = (1/30 + 1/60) * 4 = 1/5

Work Left for C to complete alone = 1/4 - 1/5 = 1/20

Time required for C to complete 1/20 work alone = rate of C * time = 1/60 * time

Therefore time required for C to complete the remaining , 1/20 of the work @ 1/60 = 3 days

Total days worked by C = 10 + 4 + 3 = 17 days
Last edited by nadib002 on Mon May 09, 2011 2:53 pm, edited 1 time in total.

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by djiddish98 » Mon May 09, 2011 2:46 pm
1/30 + 1/40 + 1/60 = Completion in 1 day when working all 3 together = 8/240 + 6/240 + 4/240 = 18/240 = 9/120 = 3/40.

They all work together for 10 days - 3/40 * 10 = 30/40 = 3/4 done before B leaves.

B leaves - now working at 1/30(2/60) + 1/60 = 3/60 = 1/20 per day.

Do that for 4 days so 4*(1/20) = 1/5 completed before A leaves.

3/4 + 1/5 = 19/20 completed. All that's left is C now, working 1/60 per day. Needs to complete 1/20 or 3/60 units, so add 3 more days.

I'm getting a total of 10 days + 4 days + 3 days = 17 days.

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by GMATGuruNY » Mon May 09, 2011 2:53 pm
nadib002 wrote:A, B and C can finish a piece of work in 30, 40, and 60 days respectively. 10 days after they started to work together B leaves. 4 days after B left, A leaves and C completes the remaining work. Find how many days C had worked altogether.
Let job = 120 units.

Rate for A = w/t = 120/30 = 4 units per day.
Rate for B = w/t = 120/40 = 3 units per day.
Rate for C = w/t = 120/60 = 2 units per day.

Combined rate for A+B+C = 4+3+2 = 9 units per day.
Over 10 days, work completed by A+B+C = r*t = 9*10 = 90 units.

Combined rate for A+C = 4+2 = 6 units per day.
Over 4 days, work completed by A+C = 6*4 = 24 units.

Remaining work = 120-90-24 = 6 units.
Time for C to finish = w/r = 6/2 = 3 days.

Total days for C = 10+4+3 = 17 days.
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by nadib002 » Mon May 09, 2011 3:08 pm
Thank you all for replying

The answer is 17 days.

@Mitch

I saw that you plugged in numbers. I did it using algebra. When should one plug in numbers vs algebra?

Thank you

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by GMATGuruNY » Mon May 09, 2011 3:21 pm
nadib002 wrote:Thank you all for replying

The answer is 17 days.

@Mitch

I saw that you plugged in numbers. I did it using algebra. When should one plug in numbers vs algebra?

Thank you
The approaches are the same: each determines respective rates for A, B and C.
The key difference: by plugging in a value for the job, we can solve using only integer values. No messy fractions.
For most rate problems, I plug in a value for the job.
Use whichever approach you prefer.
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