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Re: 40 skilled employees need 60 days to complete a project while working [#permalink]
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pushpitkc wrote:
40 skilled employees need 60 days to complete a project while working 8 hours a day. They begin working as per schedule but one month into the project, 30 employees need to shift to a different project. These 30 employees are replaced by 40 new employees who are 50% less productive. By approximately how many days will the project be delayed if the new team decides to work 10 hours a day?

A. No delay
B. 2 days
C. 5 days
D. 10 days
E. More than 10 days


The rate for the 40 skilled employees is 1/(60x8) = 1/480 job/hour.

So after 30 days half of the job has been completed.

Since 30 of them left after 30 days, the remaining 10 skilled employees will work at a rate that is 1/4 that of the 40 skilled employees, i.e., at a rate of 1/4 x 1/480 = 1/1,920 job/hour.

The 40 new less-productive employees who join in will work at a rate that is 1/2 of the 40 skilled employees, i.e., at at a rate of 1/2 x 1/480 = 1/960 job/hour.

Thus, the combined rate of the 10 original skilled employees and the 40 new less-productive employees is 1/1,920 + 1/960 = 3/1,920 = 1/640 job/hour.

We can let x = the number of days needed for the 50 (10 original and 40 new) employees to complete the remaining half of the job given that they are working 10 hours a day now. So we have:

(1/640)(10)(x) = 1/2

10x/640 = 1/2

x/64 = 1/2

x = 32

Recall that after half the job has been completed in 30 days, originally it would have taken 30 more days to finish the job, but now they need 32 more days. Thus, the project completion has been delayed by 2 days.

Alternate Solution:

40 skilled employees complete one month of work, which leaves another 60 - 30 = 30 days of work to be completed. Since they work for 8 hours a day, this makes 30 x 8 = 240 hours.

Since the new addition of workers are only half as productive as the skilled workers, we can think of these 40 new workers as equivalent to 20 skilled workers. Thus, after one month, we have (40 - 30) + 20 = 30 skilled workers.

Let’s set up an inverse proportion to determine the number of hours, call it t, needed for 30 skilled workers to complete a job that can be completed in 240 hours by 40 skilled workers.

(240)(40) = (t)(30)

t = (8)(40) = 320

Thus, the 30 skilled workers (which are in reality 10 skilled workers and 40 workers that are only half as productive) can complete the job in 320 hours. At 10 hours per day, this makes 320/10 = 32 days. Since the job would have been completed in 30 days by the 40 skilled workers, this substitution produces a delay of 32 - 30 = 2 days.

Answer: B
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Re: 40 skilled employees need 60 days to complete a project while working [#permalink]
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pushpitkc wrote:
40 skilled employees need 60 days to complete a project while working 8 hours a day. They begin working as per schedule but one month into the project, 30 employees need to shift to a different project. These 30 employees are replaced by 40 new employees who are 50% less productive. By approximately how many days will the project be delayed if the new team decides to work 10 hours a day?

A. No delay
B. 2 days
C. 5 days
D. 10 days
E. More than 10 days

Source: Experts Global

hi...

you can solve this qus. within 1 min. or less if you good in basics and if u have deep understanding of work time.

given that 40 man can do work in 60 days ...

given after days 30 of workers shift to different project and new 40 works joined with 50% less productive
so
we say 2 person will do work of one i.e. we say 20 workers joined with same productivity

now we have to find work per unit time per unit person i.e. speed
w/40*60*8
we multiply these with now 30 *10*x (now total person = 10 +20 , hours worked = 10 , days = x)
and equate with w/2 (because now w/2 work left)

(w/40*60*8) *30*10*x = w/2
we get x = 32
extra day = 32 -30 =2
ans B

above three lines is only require two solve rest are information only..
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Re: 40 skilled employees need 60 days to complete a project while working [#permalink]
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Re: 40 skilled employees need 60 days to complete a project while working [#permalink]
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