Work and time problem . People leaving work one by one and at the same time reducing their rate of doing work. 
Question: A group of $30$ people can complete a job working for $10$ hours a day in
  $15$ days. The group starts a work. But at the end of every day, starting
  from the first day, one person leaves the group and the remaining people
  work for $20$ minutes less on the next day.
  On which day will the work be completed?
Answer: $29$th day.

Attempted solution:

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Please help me solve this problem.Please tell me where I have gone wrong. Whether in concept or in calculations....Thank you..
 A: The job will require $30\times 15\times 10=4500$ person hours. 
The number of person hours worked on day $k$ is $(31-k)\times\frac{31-k}{3}$; this is the number of people working times the number of hours worked.  
Thus the total work done by day $n$, in person hours is:
$$W(n)=\sum_{k=1}^{n}\frac{(31-k)^2}{3}$$. 
Your problem is to find the least $n$ such that $W(n)\geq 4500$.  
A: Needed amount of work: 
$$
W 
= 30 \cdot 10 \frac{\text{h}}{\text{d}} \cdot 15 \text{d}
= 4500 \text{h}
$$
Rate of work per person over time: 
$$
r(t) = 10 \frac{\text{h}}{\text{d}}
$$
Active people:
\begin{align}
n(1) &= 30 \\
n(t) &= 30 - (t - 1) = 31 - t
\end{align}
Work done until day $t$, where the group in total works $1/3$ hour less each day:
\begin{align}
W(t) 
&= \sum_{k=1}^t 
\left( 
n(k) \, r(k)- \frac{k-1}{3} \frac{\text{h}}{\text{d}} 
\right) \, 1 \text{d} \\
&= \sum_{k=1}^t 
\left( (31 - k) 10 - \frac{1}{3} k + \frac{1}{3} \right) \text{h} \\
&= \sum_{k=1}^t 
\left(
\frac{931}{3} - \frac{31}{3} k \right) \text{h} \\
&= 
\left(
\frac{931}{3} \sum_{k=1}^t 1 - \frac{31}{3} \sum_{k=1}^t k \right) \text{h} \\
&= 
\left(
\frac{931}{3} t - \frac{31}{3} \frac{t^2+t}{2} \right) \text{h} \\
&= 
\left( \frac{1831}{6} t - \frac{31}{6} t^2 \right) \text{h} \\
\end{align}
We can now solve the quadratic equation:
$$
-\frac{4500 \cdot 6}{31} 
= t^2 - \frac{1831}{31} t
= \left( t - \frac{1831}{62} \right)^2 - \left( \frac{1831}{62} \right)^2 \iff \\
t = \frac{1831 \pm\sqrt{1831^2 - 4500 \cdot 6 \cdot 4 \cdot 31}}{62} \\
= \frac{1831 \pm \sqrt{4561}}{62} \in \{ 28.44, 30.62  \}
$$
So only $t= 28.44$ makes sense (we have $30$ people, which are all gone after $30$ days), thus we need $29$ days to achieve the $4500$ hours of work.
A: Use $20$ minutes as time unit. The total work load then is $30\cdot 10\cdot15\cdot 3=13\,500$ units. On day${}_k$ there are $31-k$ people at work, and each of them works during $31-k$ units. After day${}_{30}$ there is nobody left to do any work. The total number of units performed by then is
$$\sum_{k=1}^{30}(31-k)^2=9455<13\,500\ .$$
It follows that the work will not be completed at all.
