Math miscellaneousHard
Question
A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ...... = a10 = 150 and a10, a11, ...... are in A.P. with common difference - 2, then the time taken by him to count all notes is
Options
A.34 minutes
B.125 minutes
C.135 minutes
D.24 minutes
Solution
Till 10th minute number of counted notes = 1500
3000 =
[2 × 148 + (n - 1)(-2)] = n[148 - n + 1]
n2 - 149n + 3000 = 0
n = 125, 24
n = 125 is not possible.
Total time = 24 + 10 = 34 minutes.
3000 =
[2 × 148 + (n - 1)(-2)] = n[148 - n + 1]n2 - 149n + 3000 = 0
n = 125, 24
n = 125 is not possible.
Total time = 24 + 10 = 34 minutes.
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