Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence of numbers such that the difference between consecutive terms is constant.
  • The $n$-th term of an AP is given by $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
  • The sum of the first $n$ terms of an AP is given by $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}[a + l]$, where $l$ is the last term.
• Constant Rate: When a quantity changes at a constant rate, the total amount is simply the rate multiplied by the time.

Step-by-Step Solution

The problem requires us to find the total time taken to count 4500 currency notes, given a changing rate of counting. We can break this into two distinct phases based on the counting rate.

Phase 1: Initial Constant Counting Rate

• Step 1: Determine notes counted at a constant rate. The problem states that for the first 10 minutes, the person counts 150 notes per minute. This is a period of constant rate. The number of notes counted in the first 10 minutes is: $\text{Notes counted in first 10 minutes} = \text{(Rate per minute)} \times \text{(Number of minutes)}$ $\text{Notes counted in first 10 minutes} = 150 \text{ notes/minute} \times 10 \text{ minutes} = 1500 \text{ notes}$ Reasoning: This step establishes the baseline of notes counted before the counting rate begins to change.

• Step 2: Calculate the remaining notes. The total number of notes to be counted is 4500. After the first 10 minutes, we subtract the notes already counted. $\text{Remaining notes} = \text{Total notes} - \text{Notes counted in Phase 1}$ $\text{Remaining notes} = 4500 - 1500 = 3000 \text{ notes}$ Reasoning: This step identifies the quantity of work that still needs to be done in the subsequent phase.

Phase 2: Counting with a Decreasing Arithmetic Progression

• Step 3: Identify the parameters of the AP for the remaining notes. The problem states that $a_{10}, a_{11}, \dots$ are in an AP with a common difference $d = -2$. We know $a_{10} = 150$. The counting in the 11th minute will be $a_{11}$. $a_{11} = a_{10} + d = 150 + (-2) = 148$ The remaining 3000 notes are counted starting from the 11th minute, with the rate decreasing by 2 notes each minute. We need to find how many additional minutes ($N$) it takes to count these 3000 notes. For this AP segment (starting from the 11th minute):

  • The first term ($A$) is $a_{11} = 148$.
  • The common difference ($d$) is $-2$.
  • The sum of these $N$ terms ($S_N$) must equal the remaining notes, which is 3000.

• Step 4: Set up the sum of AP formula for the remaining notes. Using the formula for the sum of $N$ terms of an AP, $S_N = \frac{N}{2}[2A + (N-1)d]$: $3000 = \frac{N}{2}[2(148) + (N-1)(-2)]$ Reasoning: This equation represents the total number of notes counted during the AP phase, which must equal the remaining notes.

• Step 5: Solve the quadratic equation for the number of additional minutes ($N$). Simplify the equation: $3000 = \frac{N}{2}[296 - 2N + 2]$ $3000 = \frac{N}{2}[298 - 2N]$ Divide the bracket by 2: $3000 = N[149 - N]$ $3000 = 149N - N^2$ Rearrange into a standard quadratic form $N^2 + BN + C = 0$: $N^2 - 149N + 3000 = 0$ Factor the quadratic equation. We need two numbers that multiply to 3000 and add up to -149. These numbers are -125 and -24. $(N - 125)(N - 24) = 0$ This yields two possible values for $N$: $N = 125 \quad \text{or} \quad N = 24$ Reasoning: Solving this algebraic equation provides potential durations for the second counting phase.

• Step 6: Validate the solutions for $N$ based on physical constraints. The number of notes counted per minute cannot be negative. Let's check the counting rate in the last minute for each potential value of $N$. The rate in the $k$-th minute of this AP segment is $A + (k-1)d$.

  • Case 1: $N=125$ The counting rate in the $125^{th}$ minute of this phase (which is the $10+125 = 135^{th}$ minute overall) would be: $\text{Rate} = 148 + (125-1)(-2) = 148 + (124)(-2) = 148 - 248 = -100$ A negative counting rate is physically impossible. This solution implies the person would have stopped counting or the rate would have become zero before reaching this point. Thus, $N=125$ is not a valid solution.

  • Case 2: $N=24$ The counting rate in the $24^{th}$ minute of this phase (which is the $10+24 = 34^{th}$ minute overall) would be: $\text{Rate} = 148 + (24-1)(-2) = 148 + (23)(-2) = 148 - 46 = 102$ This rate is positive (102 notes/minute), which is physically possible. Therefore, $N=24$ is the valid number of additional minutes. Reasoning: This step is crucial for applying mathematical solutions to real-world problems. We must discard solutions that violate the physical constraints of the scenario.

• Step 7: Calculate the total time taken. The total time is the sum of the time from Phase 1 and the valid time from Phase 2. $\text{Total time} = (\text{Time in Phase 1}) + (\text{Valid time in Phase 2})$ $\text{Total time} = 10 \text{ minutes} + 24 \text{ minutes} = 34 \text{ minutes}$ Reasoning: This step combines the durations of both phases to arrive at the total time required to complete the task.

Common Mistakes & Tips

• Incorrectly defining the first term of the AP: When a sequence rule changes, be precise about what the "first term" of the new sequence is. Here, the AP for the remaining notes starts with the count in the 11th minute, not the 10th.
• Ignoring physical constraints: Always check if mathematical solutions are realistic. A negative rate of counting notes is impossible.
• Confusing total minutes with additional minutes: Ensure you are clear whether a variable like $N$ represents the total time elapsed or the duration of a specific phase.

Summary

The problem was solved by dividing the counting process into two phases: an initial period of constant counting and a subsequent period where the counting rate followed an arithmetic progression. We calculated the notes counted in the first phase, determined the remaining notes, and then used the sum of an arithmetic progression formula to find the duration of the second phase. Critically, we validated the mathematical solutions to ensure they were physically plausible, discarding any solution that implied a negative counting rate. The total time was obtained by summing the durations of both phases.

The final answer is $\boxed{34 \text{ minutes}}$ which corresponds to option (A).