Question

A person is to count $4500$ currency notes. Let $a_n$ denotes the number of notes he counts in the nth minute. If $a_1=a_2=.....=a_{10}=150$ and $a_{10},a_{11}...$ are in AP with common difference $-2$, then the time taken by him to count all notes is?

• A. $12.5$ min
• B. $135$ min
• C. $34$ min
• D. $24$ min

Answer 1

The correct option is C $34$ min

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

Then, $3000=$ Sum of $n$ terms of an A.P. with first term $148$ and common difference $-2$
$\Rightarrow 3000=\frac{n}{2}\{2\times 148+(n-1)\times (-2\left)\right\}$
$\Rightarrow 3000=n(149-n)$
$\Rightarrow n^2-149n+3000=0$
$\Rightarrow (n-125)(n-24)=0$
$\Rightarrow n=125,24$
Clearly, $n=125$ is not possible.
Total, time taken$=(10+24)=34$ min.