Question

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

• A. 34 minutes
• B. 125 minutes
• C. 135 minutes
• D. 24 minutes

Answer 1

The correct option is A 34 minutes

Given

$a_1=a_2=a_3=.......a_{10}=150$

$a_{10}=150a_{11}=148a_{12}=146$

Total currency notes $=4500$

Counted notes in 9 minutes $=150\times 9=1350$

Remaining notes $=4500-1350$

$=3150$

Let $n$ be the remaining minutes of the count

$3150=\frac{n}{2}[2a_{10}+(n-1\left)\right(-2\left)\right]$

$3150=n[a_{10}+(n-1\left)\right(-1\left)\right]$

$3150=(150+1-n)n$

$3150=151n-n^2$

$n^2-151n+3150=0$

We have a formula for solving quadratic equation $a{x}^2+bx+c=0$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\therefore n=\frac{151-\sqrt{{\left(151\right)}^2-4\times 3150}}{2}=25$

Total time taken $=9+25=34$ minutes