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Work And Time

by neya » Sat Oct 08, 2011 6:29 pm
A completes the job in 30 days, B in 20 days and C demolishes the construction in 60 days, each person working alone. When will the construction be complete if they work on alternate day starting with A, then B and then C??
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by Anurag@Gurome » Sat Oct 08, 2011 7:33 pm
neya wrote:A completes the job in 30 days, B in 20 days and C demolishes the construction in 60 days, each person working alone. When will the construction be complete if they work on alternate day starting with A, then B and then C?
In the first 3 days, they complete (1/30 + 1/20 - 1/60) = 4/60 = 1/15 of the complete work.

Hence, in the first 42 days, they will complete (42/3)*(4/60) = 14/15 of the complete work.

Thus, work left after 42nd day = 1/15 of the complete work.

In 43rd day, A will finish 1/30 of the work.

And B will complete the rest of the (1/15 - 1/30) = 1/30 of the work in 20/30 = 2/3 day.

Hence, total time required = (43 + 2/3) days.
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by fcabanski » Sat Oct 08, 2011 7:55 pm
Work problems are just like distance, rate, time problems. D=rt (I call it dirt, distance is rate times time...Dirt)

Job = work rate * time. Generally the job is 1, so 1=rate*time.

A problem like this comes down to how much of the job each person completes in a day. If it isn't obvious, use the formula for each.

A:

1 = rate*30 ... 1=30r...solve for r by dividing both sides by 30, so the rate is r=1/30. A completes 1/30th of the job per day.


B :

1 = 20r so r=1/20. B completes 1/20th of the job per day.

C:

C is destroying instead of building. That's a negative job.

-1=60r so r = -1/60. C destroys (or un-builds) 1/60th of the building per day.

They work on consecutive days, ABC. So every three days they complete 1/30 + 1/20 + (-1/60) = 4/60 = 1/15

In 15 cycles of 3 days the job will be complete. That's 45 days. But C un-builds each 3rd day.

So the job is complete before the 45th day.

Day 42 14 cycles of 3 days have been completed, so 14*1/15 = 14/15 is complete.

Day 43 A completes 1/30th more. 14/15 + 1/30 = (28+1)/30 = 29/30

Day 44 B completes 1/20th more. 29/30 + 1/20 = (58+3)60 = 61/60. So the job is complete on the 44th day. Specifically it's 2/3 of the way through the 44th day. 2/3 * 1/20 = 2/60. 2/60 + 58/60 = 60/60 = 1.
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by parul9 » Sat Oct 08, 2011 8:18 pm
Anurag@Gurome wrote:
neya wrote:A completes the job in 30 days, B in 20 days and C demolishes the construction in 60 days, each person working alone. When will the construction be complete if they work on alternate day starting with A, then B and then C?
In the first 3 days, they complete (1/30 + 1/20 - 1/60) = 4/60 = 1/15 of the complete work.

Hence, in the first 42 days, they will complete (42/3)*(4/60) = 14/15 of the complete work.

Thus, work left after 42nd day = 1/15 of the complete work.

In 43rd day, A will finish 1/30 of the work.

And B will complete the rest of the (1/15 - 1/30) = 1/30 of the work in 20/30 = 2/3 day.

Hence, total time required = (43 + 2/3) days.
Hi Anurag,

I got 45 as the answer.
Please see my approach here, can you please tell me where I went wrong?
Let total work = W.
A would do W/30 work in 1 day.
B - W/20
C - W/60

Total work done in 3 days - W/30 + W/20 - W/60 = W/15.

Then applying the unitary method -

W/15 work is done in 3 days.
So, W will be done in - 3 * 15/W * W = 45.
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by GMATGuruNY » Sat Oct 08, 2011 8:20 pm
neya wrote:A completes the job in 30 days, B in 20 days and C demolishes the construction in 60 days, each person working alone. When will the construction be complete if they work on alternate day starting with A, then B and then C??
Let the job = 60 units.
Rate for A = w/t = 60/30 = 2 units per day.
Rate for B = w/t = 60/20 = 3 units per day.
Rate for C = w/t = 60/60 = 1 unit of demolition per day.
Work completed every 3 days = A+B-C = 2+3-1 = 4 units.

Thus, after 42 days -- equal to 14 3-day cycles -- the amount of work completed = 14*4 = 56 units.
Work completed by A on the 43rd day = 2 units.
Time for B to complete the remaining 2 units = w/r = 2/3 of a day.

Total time = 43 2/3 days.
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by Anurag@Gurome » Sat Oct 08, 2011 8:33 pm
parul9 wrote:...W/15 work is done in 3 days.
So, W will be done in - 3 * 15/W * W = 45.
There lies the trap.
As I've already shown, the work will be done in 43 2/3 days.

Now on the remaining 1/3 time of the 44th day, B will do (1/3)*(1/20) = 1/60 of a new work.
And on the 45th day, C will demolish 1/60 of the new work.

Hence, by the end of the 45th day, the status is same as that of at the end of 43 2/3 days.

Blind application of unitary method completely misses the intermediate status and leads us to the end of 45th day.

Hope that helps.
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by fcabanski » Sat Oct 08, 2011 9:05 pm
What C should do is use his tools to attack and kill A and B. Then they won't finish the project.

Parul, you neglected to note that every three days is a build, build, un-build cycle. They complete 1/15th of the project, as a unit, every three days. But every third day is a step backwards. The project finishes before one of those 3rd day un-build days.

A simpler example is if A builds a home in a day, and B destroys 1/2 home a day. They work consecutive days.

How many days, working on alternating days, does it take them to build a home?

Only one. A builds it in the first day.
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