# Numbers- Aptitude Questions and Answers

 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 an AP with common difference – 2, then the time taken by him to count all notes is ? A) 24 minutes B) 34 minutes C) 125 minutes D) 135 minutes Correct Answer :34 minutes Explanation :Suppose he takes n minutes to count 4500 notes. Number of notes he counts in the 10 minute => a1+a2+........+a10 = 150*10 =1500 (a1 = a2 = .... = a10 = 150) As is given a1+a2+........+an=4500 Hence, a11+a12+.....+an = 4500-1500 = 3000 a10 , a11 ,... are in an AP with common difference −2 then a10=150 and a11=148 Now, 3000 = Sum of n terms of an A.P.(first term=148 , difference= −2) Sum of first n terms of an AP: S =(n/2)[2a + (n- 1)d] ⇒3000 = (n/2)[2*148+(n−1)*(−2)] ⇒3000 = n(148−(n−1)) ⇒3000=n(149−n) ⇒n2−149n+3000=0 ⇒(n−125)(n−24)=0 ⇒n=125,24 n=125 is not possible because it is negative[ a125= 148+(125-1)*-2 =148 - 252 < 0 ] Hence, time taken=(10+24)=34 min. Post/View Answer Post comment Cancel Thanks for your comment.! Write a comment(Click here) ...