Question

A person is to count $4500$ currency notes. Let an denote the number of notes he counts in the nth minute. If $a1=a2=...=a10=150anda10,a11,....$are in AP with common difference $-2$ , then the time taken by him to count all notes, is

• A. $24$ min
• B. $34$ min
• C. $125$ min
• D. $135$ min

Answer 1

The correct option is B

$34$ min

Explanation for the correct option:

Step 1. Find the time taken to cont all notes:

The number of notes counted in first $10$ m$=150\times 10$ $=1500$
Suppose, the person counts the remaining $3000$ currency notes in $n$ minutes, then

Step 2. $3000=$ Sum of $n$ terms of an A.P. with first term $148$ and common difference $-2$

$\Rightarrow$ $3000=\frac{n}{2}\left[2\times 148+(n-1)\times (-2)\right]$ $\left[\mathbf{\because }{\mathit{S}}_{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)d\right]\right]$

$\Rightarrow$ $\begin{array}{rcl}3000& =& n(149-n)\end{array}$

$\Rightarrow$$\begin{array}{rcl}{n}^2-149n+3000& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(n-125)(n-24)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}n& =& 125,24\end{array}$

Clearly, $n=125$ is not possible.

$\therefore$Total time taken $=(10+24)$

$=\mathbf{34}$ min

Hence, Option ‘B’ is Correct.