Question

A person is to count $4500$ notes. Let $a_n$ denote the number of notes he counts in the $n^{th}$ minute. If $a_1=a_2=a_3=...=a_{10}=150$, and $a_{10},a_{11},a_{12}$,... are in $AP$ with the common difference $-2$, then the time taken by him to count all the notes is

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

Answer 1

The correct option is B $34$ minutes

Let us assume that total minutes $=x$
For first nine minutes $=150\times 9$

Let the remaining minutes $=y=x-9$

Now,

$(150\times 9)+\frac{y}{2}[2\times 150+(y-1\left)\right(-2\left)\right]=4500$

$\Rightarrow \frac{y}{2}(302-2y)=3150$

$\Rightarrow y^2-151y+3150=0$

$\Rightarrow (y-126)(y-25)=0$

$\Rightarrow y=25$ or $126\left(Rejected\right)$

So, $x=y+9\Rightarrow x=25+9=34$

Therefore, the answer is $34$ minutes.