A person is to count 4500 currency notes. Let $a_{n}$ denote the number of notes he counts in the n^thminute. 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

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135 minutes

24 minutes

34 minutes

125 minutes

answer is C.

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Detailed Solution

Till 10^th minute number of counted notes = 1500

$3000 = \frac{n}{2} [2 \times 148 + (n - 1) (- 2)] = n [148 - n + 1]$

$\Rightarrow n^{2} - 149 n + 3000 = 0$

$\Rightarrow n = 125, 24$

$\Rightarrow n = 125$ is not possible

$\Rightarrow$ Total time $= 24 + 10 = 34 \min$

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