InterviewSolution
Saved Bookmarks
| 1. |
A person is to count 4500 currency notes. Let an denote the numberof 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 commondifference 2, then the time taken by him to count all notes is(1) 34 minutes(2) 125 minutes (3) 135minutes (4) 24 minutesA. 125 minutesB. 135 minutesC. 24 mintutesD. 34 minutes |
|
Answer» Correct Answer - D The number of notes counted in first 10 minutes `=150xx10=1500` Suppose the person counts the remaining 3000 currency notes in n minutes. Then, 3000=Sum of terms of an AP with first term 148 and common difference -2 `rArr" "3000=(n)/(2){2xx148+(n-1)xx(-2)}` `rArr" "3000=n(149-n)` `rArr" "n^(2)-149n+3000=0` `rArr" "(n-125)(n-24)=0rArrn=125,24` Clearly, n-125 is not possible. `:.` Total time taken =(10+42)mintutes =34 minutes. |
|