Buttercup3 wrote:
Sangeeta2018 wrote:
40 employees take 30 days,working at 8 hrs per day,to complete a task. 40 employees start the work but after 10 days,20 leave and are replaced by employees who are 1/2 as productive.How many hours per day should the new team work if the work has to be completed in the scheduled timeline?
1. 12
2. 20
3. 10.6
4. 6
5. 30
Source:
Experts' Global generis please help with this question using your method.
Buttercup3 - Sure. The approach, though, works when all workers are equally productive.
When all the workers are not equally productive, we have to finesse the formula a bit.*
These employees are all going to be men. Time is in hours.
Rewrite the standard work formula RT=W by adding (# of workers) to LHS:
(# of workers) * (Rate) * (Time) = WorkScenario 1, FIRST 10 DAYS - equally productive workers
1) Find the rate of each man
# * R * T = W# = 40
R = ??
T = 240 (30 days * 8 hrs per day)
W = 1
40 * R * 240 = 1
R =
\(\frac{1}{(40)(240)}=\frac{1}{9600}\) = rate of individual man
2) How much work is finished?**
W = # * R * T# \(= 40\)\(R = \frac{1}{9600}\)\(T = 80\) (10 days * 8 hrs per day)
\(W = 40 * \frac{1}{9600} * 80\)\(W = \frac{3200}{9600}=\frac{1}{3}\)Work is \(\frac{1}{3}\) finished, \(\frac{2}{3}W\) remains
Scenario 2: NEXT 20 DAYS, workers not equally productive
20 men leave. Replaced by 20 men who are half as productive
Adding individual rates (fast and slow) and simply multiplying by number of men (40) will not work because the workers are not equally productive.
chetan2u shows one way to handle the different levels of productivity.
Pair one fast man with one slow man, add their rates, then multiply by the number of pairs (20 pairs).
You can also find each group's rate (or each "set's" rate), then add them.
Group rate in this context is simply
# (of men) *
\(R\) (individual man) -- i.e. two of the three variables on LHS
Find each group's / set's rate, then add• FAST set's rate: (
# \(* R) = (20 * \frac{1}{9600}) = \frac{20}{9600}\)• SLOW set's rate: (
# \(* R * \frac{1}{2}) =\) \((20*\frac{1}{9600}*\frac{1}{2})= \frac{20}{(9600)(2)} = \frac{20}{19200}\)FAST set + SLOW set = new rate =
\(R_2\) \(R_{2}= (\frac{20}{9600}+\frac{20}{19200})=(\frac{40}{19200} + \frac{20}{19200})=\)\(\frac{60}{19200}=\frac{6}{1920}= \frac{1}{320} = R_2\)With
\(R_2\) , we have calculated (
#\(* R)\) on LHS.
We can substitute
\(R_{2}\) for those two variables, thus
\(R_2 * T = W\)Number of hours per day? Total hours divided by number of days
Total hours,
\(R_2 * T = W\), so
\(T = \frac{W}{R_{2}}\)\(W = \frac{2}{3}\)
\(R_{2} = \frac{1}{320}\)\(T = \frac{\frac{2}{3}}{\frac{1}{320}}= \frac{2}{3} * 320 = \frac{640}{3}\) total hours
Hours per day: \(\frac{TotalHours}{NumberOfDays}\)
\(\frac{\frac{640}{3}}{20}= \frac{640}{60} = \frac{64}{6} = \frac{32}{3} \approx{10.66}\approx{10.6}\) hours per day
Answer C
Hope that helps.
* If all 40 workers were replaced by workers half as productive, you would stay with an unmodified (# * R * T) = W
You would: 1) Find "normal" rate 2) Find work finished after 10 days; 3) Find the slower rate (1/2 * R); 4) use the formula straightforwardly. No modifications
**We know this already, from direct proportionality between time and work when rate is constant.