Question

A man saves $Rs200$ in each of the first three months of his service. In each of the subsequent months, his saving increases by $Rs40$ more than the saving of immediately previous month. His total saving from the start of service will be $Rs11040$ after

• A. $19$ months
• B. $20$ months
• C. $21$months
• D. $18$months

Answer 1

The correct option is C

$21$months

Explanation for the correct option:

Step 1. Finding the time taken to save $Rs11040$:

Saving for first 3 months is

$$\begin{gathered} =3\times 200 \\ =600 \end{gathered}$$

Step 2. Finding the time taken

Let the time taken to save $Rs11040$ be $n+3$months.

Here $d=40,a=240$ [As it is in AP]

In $n+3$ months, savings will be:

$\begin{array}{rcl}600+\frac{n}{2}(2\times 240+(n-1\left)40\right)& =& 11040\end{array}$ [Using the formula of sum of ‘n’ terms of an AP]

$\Rightarrow$ $\begin{array}{rcl}\frac{n}{2}(480+40n-40)& =& 10440\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(440+40n)n& =& 20880\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{n}^2+11n-522& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(n-18)(n+29)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}n& =& 18or-29\end{array}$

$n$ cannot be negative.

$\begin{array}{rcl}\therefore n+3& =& 18+3\\ & =& \mathbf{21}\end{array}$

Hence option C is the answer.