Question

150 workers were engaged to finish a piece of work in a certain number of days. 4 workers dropped the second day, 4 more workers dropped the third day, and so on. It takes eight more days to finish work now. The number of days in which the work was completed is
(a) 15
(b) 20
(c) 25
(d) 30

Answer 1

Hint: To solve this question we will first analyze that numbers of workers form an AP (Arithmetic Progression) with the common difference as – 4 as 4 workers get reduced each day. And the sum of AP is given by \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] where d is the common difference, n is the number of terms and a is the first term. Finally, we will equate the work done when the workers are not reduced and when the workers are reduced to get the result.

Complete step-by-step solution:
Let us assume x as work is done by each worker per day. Initially, we had 150 workers. 4 dropped the second day. Therefore, workers left is \[150-4=146.\]
Again 4 dropped on the third day.
\[\Rightarrow \text{Workers left now}=146-4=142\]
Similarly, this goes on. So we have a sequence obtained as
\[150,146,142......\left( i \right)\]
Now the work done will be calculated by adding work done each day. Therefore, we get,
\[\text{Work Done each day}=\text{Number of workers}\times x\]
\[\Rightarrow \text{Work Done total}=150\times x+146\times x+142\times x+.....\]
\[\Rightarrow \text{Work Done}=x\left( 150+146+142+..... \right).....\left( ii \right)\]
So we have obtained a sequence as in equation (i). Hence the common difference d will be
\[d=146-150=-4\]
It is the same so it is an Arithmetic Progression.
The first term, a = 150, and the common difference d = – 4.
Now, the equation (ii) can be obtained by the sum of n terms of the given AP. The sum of n terms of AP of the first term ‘a’ and common difference ‘d’ is given by
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
Applying this in the above formula, we get,
\[Sum=\dfrac{n}{2}\left[ 2\times 150+\left( n-1 \right)\left( -4 \right) \right]\]
\[\Rightarrow Sum=n\left[ 150+\left( -2 \right)\left( n-1 \right) \right]......\left( iii \right)\]
As it is given in the question that it took 8 more days to finish a task. If 150 workers worked till the end it would have taken n – 8 days to complete the work. Then the work done would be calculated as
\[150\times x\times \left( n-8 \right)......\left( iv \right)\]
(As 150 workers with x, each worker work and in (n – 8) days)
Now, the work done is the same in both cases.
Equation (ii) = Equation (iv)
Using (iv) in (ii), we get,
\[\left( ii \right)=x\left[ n\left( 150+\left( -2 \right)\left( n-1 \right) \right) \right]\]
Equating (ii) and (iii), we get,
\[\Rightarrow 150x\left( n-8 \right)=x\left( n\left( 150-2\left( n-1 \right) \right) \right)\]
\[\Rightarrow 150\left( n-8 \right)=n\left( 150 \right)-2n\left( n-1 \right)\]
\[\Rightarrow 150n-1200=150n-2{{n}^{2}}+2n\]
\[\Rightarrow 2{{n}^{2}}-2n-1200=0\]
\[\Rightarrow {{n}^{2}}-n-600=0\]
Using the splitting by the middle term, we have,
\[{{n}^{2}}-25n+24n-600=0\]
\[\Rightarrow n\left( n-25 \right)+24\left( n-25 \right)=0\]
\[\Rightarrow n-25=0;\left( n+24 \right)=0\]
\[\Rightarrow n=25;n=-24\]
Now, as the number of days can’t be negative, n = 25.
Therefore, the number of days in which work was completed is 25.
Hence, the option (c) is the right answer.

Note: Because the task was certain, means, for example, the worker had to build a 5 storey building then if whether 5 workers had work or 7 workers had work, the amount of work is the same. The days of working can differ with the addition of the number of workers. So we have used them to solve this question.