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Step 1: Understand the Problem
The person needs to count 4500 currency notes. For the first 10 minutes, he counts 150 notes each minute. After that, the number of notes counted each minute follows an arithmetic progression (AP) with a common difference of −2.
Step 2: Determine the Notes Counted in the First 10 Minutes
Since each minute for the first 10 minutes he counts 150 notes, the total counted in these 10 minutes is:
$10 \times 150 = 1500$.
Step 3: Set Up the Remaining Count with an AP
After the first 10 minutes, the person still needs to count the remaining notes:
$4500 - 1500 = 3000.$
In this period, the first term of the AP is:
$a_{11} = a_{10} + d = 150 + (-2) = 148.$
The common difference $d = -2.$
Step 4: Use the Sum Formula for an Arithmetic Progression
Let $n$ be the number of minutes (terms) after the 10th minute. The sum of these $n$ terms must be 3000:
\[
S_n = \frac{n}{2}\bigl(2a + (n - 1)d\bigr).
\]
Here, $a = 148$, $d = -2$, and $S_n = 3000.$ Substituting these values gives:
\[
3000 = \frac{n}{2}\Bigl(2 \times 148 + (n - 1)\times (-2)\Bigr).
\]
Step 5: Simplify and Solve the Quadratic Equation
Simplify inside the brackets:
\[
2 \times 148 = 296,\quad
(n - 1)\times (-2) = -2n + 2.
\]
Hence:
\[
3000 = \frac{n}{2} \left( 296 - 2n + 2 \right) = \frac{n}{2} \left(298 - 2n \right).
\]
This simplifies to:
\[
3000 = n(149 - n).
\]
\[
\implies n(149 - n) = 3000.
\]
\[
\implies n^2 - 149\,n + 3000 = 0.
\]
Solving this quadratic gives $n = 24$ or $n = 125.$
Step 6: Choose the Feasible Solution
If $n = 125,$ the number of notes snapped each minute eventually becomes non-sensical (going very low or possibly negative), so we discard $n=125.$ The physically meaningful choice is $n = 24.$
Step 7: Compute the Total Time
Total time = (the first 10 minutes) + (the next 24 minutes)
\[
= 10 + 24 = 34 \text{ minutes.}
\]
Hence, the person takes 34 minutes to count all 4500 notes.
