Understanding the Note Counting Problem

This problem involves calculating the total time taken to count 4500 notes. The counting process is described in two phases:

• Phase 1: The first 10 minutes, where the number of notes counted per minute is constant.
• Phase 2: After the first 10 minutes, where the number of notes counted per minute follows an Arithmetic Progression (AP) with a decreasing rate.

Notes Counted in the First 10 Minutes

The number of notes counted in the nth minute is denoted by \(a_n\). We are given that \(a_1 = a_2 = \dots = a_{10} = 150\).

The total number of notes counted in the first 10 minutes is the sum of notes counted in each of these minutes:

\( \text{Notes in first 10 minutes} = a_1 + a_2 + \dots + a_{10} \)

Since each term is 150, this sum is:

\( \text{Notes in first 10 minutes} = 10 \times 150 = 1500 \)

Calculating Remaining Notes to Count

The total number of notes to be counted is 4500.

After the first 10 minutes, the remaining notes are:

\( \text{Remaining notes} = \text{Total notes} - \text{Notes in first 10 minutes} \)

\( \text{Remaining notes} = 4500 - 1500 = 3000 \)

These 3000 notes must be counted in the minutes following the first 10 minutes, where the counting rate changes according to the given AP.

Analyzing the Counting Rate After 10 Minutes (Arithmetic Progression)

We are told that \(a_{10}, a_{11}, a_{12}, \dots\) are in an AP with \(a_{10} = 150\) and a common difference \(d = -2\).

The number of notes counted in the nth minute for \(n \ge 10\) can be given by the formula for the nth term of an AP:

\( a_n = a_{10} + (n - 10)d \)

Substituting the given values:

\( a_n = 150 + (n - 10)(-2) \)

\( a_n = 150 - 2n + 20 \)

\( a_n = 170 - 2n \quad \text{for } n \ge 10 \)

The counting rates for minutes 11, 12, 13, ... are:

• \(a_{11} = 170 - 2(11) = 170 - 22 = 148\)
• \(a_{12} = 170 - 2(12) = 170 - 24 = 146\)
• \(a_{13} = 170 - 2(13) = 170 - 26 = 144\)
• ... and so on.

Note that the counting rate must be non-negative. \(a_n = 170 - 2n \ge 0 \implies 170 \ge 2n \implies n \le 85\). So, the counting rate becomes zero at minute 85.

Calculating Time Taken for Remaining 3000 Notes

Let \(k\) be the number of additional minutes required after the first 10 minutes to count the remaining 3000 notes. These \(k\) minutes correspond to minutes \(11, 12, \dots, 10+k\).

The total notes counted in these \(k\) minutes is the sum of the terms \(a_{11}, a_{12}, \dots, a_{10+k}\).

This is an AP starting with the first term \(a_{11} = 148\) and having a common difference \(d = -2\). There are \(k\) terms in this sum.

The last term in this sum is \(a_{10+k}\).

\( a_{10+k} = 170 - 2(10+k) = 170 - 20 - 2k = 150 - 2k \)

Alternatively, using the first term of this sub-sequence \(a_{11}=148\):

\( a_{10+k} = a_{11} + (k-1)d = 148 + (k-1)(-2) = 148 - 2k + 2 = 150 - 2k \)

The sum of an AP is given by \( S_k = \frac{k}{2} (\text{first term} + \text{last term}) \) or \( S_k = \frac{k}{2} [2 \times \text{first term} + (k-1)d] \).

Using the first formula with \(a_{11}=148\) and \(a_{10+k} = 150-2k\):

\( S_k = \frac{k}{2} (148 + 150 - 2k) \)

\( S_k = \frac{k}{2} (298 - 2k) \)

\( S_k = k (149 - k) = 149k - k^2 \)

We need this sum \(S_k\) to be equal to the remaining notes, which is 3000.

\( 149k - k^2 = 3000 \)

Rearranging into a quadratic equation:

\( k^2 - 149k + 3000 = 0 \)

We can solve this quadratic equation for \(k\) using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a=1\), \(b=-149\), and \(c=3000\).

\( k = \frac{-(-149) \pm \sqrt{(-149)^2 - 4(1)(3000)}}{2(1)} \)

\( k = \frac{149 \pm \sqrt{22201 - 12000}}{2} \)

\( k = \frac{149 \pm \sqrt{10201}}{2} \)

The square root of 10201 is 101.

\( k = \frac{149 \pm 101}{2} \)

This gives two possible values for \(k\):

• \( k_1 = \frac{149 + 101}{2} = \frac{250}{2} = 125 \)
• \( k_2 = \frac{149 - 101}{2} = \frac{48}{2} = 24 \)

We must consider which value of \(k\) makes sense in the context of the problem. The counting rate decreases by 2 notes each minute after the 10th minute. The rate in minute \(n\) is \(a_n = 170 - 2n\). If \(k=125\), the counting would continue up to minute \(10+125 = 135\). However, the rate \(a_n\) becomes 0 at minute 85 and negative thereafter (\(a_{85} = 0, a_{86} = -2\), etc.). Counting negative notes doesn't fit the problem. The sum formula for an AP assumes all terms contribute positively or according to the sequence. In a practical counting scenario with decreasing rate, the person would finish counting when the *total sum* reaches 4500, which would occur *before* the rate becomes negative or zero if 3000 notes are finished before minute 85.

The smaller value of \(k=24\) means the additional counting takes 24 minutes. This finishes at minute \(10+24=34\). The rate in minute 34 is \(a_{34} = 170 - 2(34) = 170 - 68 = 102\), which is positive. The sum of rates from minute 11 to minute 34 would be \(a_{11} + \dots + a_{34}\), which is the sum of an AP with \(k=24\) terms, starting at 148 and ending at 102. As calculated earlier, this sum is indeed 3000.

The larger value \(k=125\) would imply counting for 125 minutes after the first 10, resulting in 135 total minutes. This would involve summing terms that are negative, which is not logical for counting notes. The smaller positive solution for \(k\) typically represents the time taken to reach the target sum in problems involving sums of sequences where terms might become non-positive or the sum might exceed the target after that point.

Therefore, the correct number of minutes taken after the first 10 minutes is \(k = 24\).

Total Time Taken

The total time taken to count all 4500 notes is the sum of the time in the first phase and the time in the second phase.

\( \text{Total time} = \text{Time in Phase 1} + \text{Time in Phase 2} \)

\( \text{Total time} = 10 \text{ minutes} + k \text{ minutes} \)

\( \text{Total time} = 10 + 24 = 34 \text{ minutes} \)

The time taken by him to count all the notes is 34 minutes.

Revision Table: Key Calculations

Additional Information: Arithmetic Progressions (AP)

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\).

The general form of an AP is \(a, a+d, a+2d, a+3d, \dots\).

The nth term of an AP is given by the formula:

\( a_n = a_1 + (n-1)d \)

where \(a_1\) is the first term and \(d\) is the common difference.

The sum of the first \(n\) terms of an AP is given by the formula:

\( S_n = \frac{n}{2} (a_1 + a_n) \)

or

\( S_n = \frac{n}{2} [2a_1 + (n-1)d] \)

In this problem, the sequence of counting rates for \(n \ge 10\) forms an AP where \(a_{10}\) is the term corresponding to \(n=10\). The sequence of notes counted in minutes 11, 12, ... is also an AP starting from \(a_{11}\).