Question

A person is to count 4500 currency notes. Let $a_n$ denotes the number of notes he counts in the $n^{\text {th }} \min$. If $a_1=a_2=\ldots .=a_{10}=150$ and $a_{10}, a_{11}, \ldots$. are in A.P. 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

Answer: (B)

Number of notes that the person counts in $10 \min$
$=10 \times 150=1500$
Since $a_{10}, a_{11}, a_{12}, \ldots$ are in A.P. with common difference $-2$. Let $n$ be the time taken to count remaining 3000 notes, then
$ \frac{n}{2}[2 \times 148+(n-1) \times-2] =3000 $
$\Rightarrow n^2-149 n+3000 =0$
$\Rightarrow (n-24)(n-125) =0 $
$\Rightarrow n =24,125$
Then, the total time taken by the person to count all notes
$ =10+24=34 \min $
$(\because n \neq 125 $ as $a_{125}=-100,$ which is not possible $)$