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 $a_1=a_2=\dots .=a_{10}=150$ and $a_{10},a_{11},\dots$. 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 $a_{10},a_{11},a_{12},\dots$ are in A.P. with common difference $-2$. Let $n$ be the time taken to count remaining 3000 notes, then
$\frac{n}{2}[2\times 148+(n-1)\times -2]=3000$
$\Rightarrow n^2-149n+3000=0$
$\Rightarrow (n-24)(n-125)=0$
$\Rightarrow n=24,125$
Then, the total time taken by the person to count all notes
$=10+24=34\min$
$(\because n\ne 125$ as $a_{125}=-100,$ which is not possible $)$