Q170 / 270 · NDA 2017 (Phase 2) · Mathematics

A person is to count 4500 notes. Let $a_n$ denote the number of notes he counts in the nth minute. If $a_1=a_2=a_3=\ldots$. $=a_{10}=150$, and $a_{10}, a_{11}, a_{12}, \ldots$. are in AP with the common difference -2 , then the time taken by him to count all the notes is

  • A
    24 minutes
  • Correct answer
    B
    34 minutes
  • C
    125 minutes
  • D
    135 minutes

Solution

Given, $\mathrm{a}_1=\mathrm{a}_2=\mathrm{a}_3=$ $\qquad$ $=a_{10}=150$ Also, $\mathrm{a}_{10}, \mathrm{a}_{11}, \mathrm{a}_{12}, \ldots$. are in A.P. and $\mathrm{d}=-2$ Since, $\mathrm{a}_{10}=150, \mathrm{~A} . \mathrm{P}$. is $150,148,146, \ldots$. For the first 10 minutes, he has counted $150 \times 10=1500$ notes. Time taken to count remaining 3000 notes $\begin{aligned} & \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ & \Rightarrow 3000=\frac{\mathrm{n}}{2}[2 \times 148+(\mathrm{n}-1)(-2)]\end{aligned}$ $\begin{aligned} & \Rightarrow 3000=\frac{\mathrm{n}}{2} \times 2(148-\mathrm{n}+1) \\ & \Rightarrow 3000=148 \mathrm{n}-\mathrm{n}^2+\mathrm{n} \\ & \Rightarrow \mathrm{n}^2-149 \mathrm{n}+3000=0\end{aligned}$ $\begin{aligned}& \Rightarrow(\mathrm{n}-24)(\mathrm{n}-125) \\ & \Rightarrow \mathrm{n}=24, \text { or } \mathrm{n}=125 . \end{aligned}$ Since he has taken 10 minutes to count 1500 notes, he will not take 125 min to count 3000 notes. So, $\mathrm{n}=24$. $\therefore$ Total time $=10+24=34$ minutes.

Hence, the correct answer is option: 'B'.

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