Work and time problem . People leaving work one by one and at the same time reducing their rate of doing work.

Question

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..

Answer 1

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.

Answer 2

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 $W(n)\geq 4500$.

Answer 3

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.