JEE MAIN - Mathematics (2010 - No. 18)
A person is to count 4500 currency notes. Let $${a_n}$$ denote the number of notes he counts in the $${n^{th}}$$ minute. If $${a_1}$$ = $${a_2}$$ = ....= $${a_{10}}$$= 150 and $${a_{10}}$$, $${a_{11}}$$,.... are in an AP with common difference - 2, then the time taken by him to count all notes is
34 minutes
125 minutes
135 minutes
24 minutes
Explanation
Till $$10$$th minute number of counted notes $$ = 1500$$
$$3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = n\left[ {148 - n + 1} \right]$$
$$ \Rightarrow $$$${n^2} - 149n + 3000 = 0$$
$$ \Rightarrow n = 125,24$$
But $$n=125$$ is not possible
$$\therefore$$ total time $$ = 24 + 10 = 34$$ minutes.
$$3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = n\left[ {148 - n + 1} \right]$$
$$ \Rightarrow $$$${n^2} - 149n + 3000 = 0$$
$$ \Rightarrow n = 125,24$$
But $$n=125$$ is not possible
$$\therefore$$ total time $$ = 24 + 10 = 34$$ minutes.
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