Question

150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped the second day. Four more workers dropped by the third day and so on. It now takes 8 more days to finish the work. Find the number of days in which the work was completed.

Answer 1

Hint: Assume that a worker completes ${{\left( \dfrac{1}{x} \right)}^{th}}$ fraction of work in a day. Assume that the work takes n days to finish. Using the fact that the sum of total of all the works done by all the workers over the period of time till the job is completed should be equal to the total work, form two equations in x and n. Eliminate the variable x from the equations and hence form a quadratic equation in n. Using the fact that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ find the two roots of the so formed quadratic equation in n. Remove the roots which are not natural numbers and hence find the value of n and hence find the number of days in which the work was completed.

Complete step-by-step answer:
Let a worker’s work for one day is equal to ${{\left( \dfrac{1}{x} \right)}^{th}}$ fraction of the total work and let it takes the n days to complete the work.
So initially it would have taken $n-8$ days to complete the work.
The work will be completed when the fractional contribution of all the workers for all the days adds up to 1.
Hence we have $150\dfrac{\left( n-8 \right)}{x}=1$
Multiplying both sides by n , we get
$150(n-8) = x$
i.e. $150n-1200 = x $....................(i).
Also, we have
Fractional Contribution of all the workers on first day $=150\times \dfrac{1}{x}=\dfrac{150}{x}$
Fractional Contribution of all the workers on the second day $=\dfrac{1}{x}\times \left( 150-4 \right)=\dfrac{150-4}{x}$
Continuing in this way
Fractional contribution of all the workers on $n^{th}$ day $=\dfrac{150-\left( n-1 \right)4}{x}$
Since sum of fractional contribution of all the workers on all the days till the jobs is concluded is 1, we have
$\dfrac{150}{x}+\dfrac{150-4}{x}+\dfrac{150-8}{x}+\cdots +\dfrac{150-\left( n-1 \right)4}{x}=1$
Multiplying both sides by x, we get
$150+146+\cdots +\left( 150-\left( n-1 \right)4 \right)=x$, which is sum up to $n^{th}$ term of the A.P 150,146,142,…
Here $a = 150, d = -4$
We know that in A.P sum of n terms is given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
Using the above formula, we get
$\begin{align}
  & \dfrac{n}{2}\left( 2\times 150-\left( n-1 \right)4 \right)=x \\
 & \Rightarrow n\left( 150-2n+2 \right)=x \\
\end{align}$
Substituting the value of x from equation (i), we get
$\begin{align}
  & n\left( 152-2n \right)=150n-1200 \\
 & \Rightarrow 152n-2{{n}^{2}}=150n-1200 \\
\end{align}$
Subtracting $152n$ form both sides, we get
$\begin{align}
  & 152n-2{{n}^{2}}-152n=150n-1200-152n \\
 & -2{{n}^{2}}=-1200-2n \\
 & \Rightarrow -{{n}^{2}}=-600-n \\
\end{align}$
Adding ${{n}^{2}}$ on both sides, we get
${{n}^{2}}-n-600=0$
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are given $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here $a = 1, b = -1$ and $c = -600$.
Hence $n=\dfrac{1\pm \sqrt{1+2400}}{2}=\dfrac{1\pm \sqrt{2401}}{2}=\dfrac{1\pm 49}{2}=25,-24$
Since the number of days is a natural number $n = -24$ is rejected.
Hence $n = 25$ and hence the number of days in which the work was completed = 25.

Note: [1] Do not assume one days work to be ${{\left( \dfrac{p}{q} \right)}^{th}}$ of total. The assumption that one days’ work of a man is $\dfrac{1}{x}$ is valid since every fraction $\dfrac{p}{q}$ can be converted $\dfrac{1}{\dfrac{q}{p}}=\dfrac{1}{n}$ form. This leads to the formation of a system of two equations and not in three equations.