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A person is to count 4500 currency notes...

A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If `a_1=""a_2="". . . . . .""=""a_(10)=""150` and `a_(10),""a_(11),"". . . . . .` are in A.P. with common difference 2, then the time taken by him to count all notes is

A

24 min

B

34 min

C

125 min

D

135 min

Text Solution

Verified by Experts

The correct Answer is:
b
Number of notes that the person counts in 0 min
` = 10 xx150 xx = 1500`
Since , `a_(10),a_(11),a_(12),…` are in AP with common difeerence -2 Let n be the time taken to count remaining 3000 notes , then
` b/a[2xx 148 +(n-1)xx-2] = 3000 " " [ :' a_(11) = 148]`
`rArr n^(2) - 149 n + 3000=0`
` rArr (n-24)(n-125) = 0`
`rArr n = 24,125`
Then, the total time taken by the person to count all notes
` 10 + 24 = 34 "min " [ :' n != 125, as a_(125)lt 0`
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