Problem 153 - A Complete Resource Book in Mathematics for JEE Main

Question

A person is to count 4500 currency notes. Let \(a_{n}\) denote the number of notes he counts in the \(n^{\text {th }}\) minute. If \(a_{1}=a_{2}=\ldots \ldots=a_{10}=150\) and \(a_{10}, a_{11} \ldots\) are in A.P. with common difference \(-2\), then the time taken by him to count all notes is (A) 34 minutes (B) 125 minutes (C) 135 minutes (D) 24 minutes

Step-by-Step Solution

Answer

The person takes 34 minutes to count all the notes.

Step 1: Initial Counting Setup

The person counts 4500 currency notes in total. For the first 10 minutes, they count 150 notes per minute, which means they count a total of \( 10 \times 150 = 1500 \) notes in the first 10 minutes.

Step 2: Counting Notes from 11th Minute

From the 11th minute onward, the number of notes counted forms an arithmetic progression (AP) with a common difference of -2. Therefore, the number of notes counted in the 11th minute, \(a_{11}\), is 148 notes (since \(a_{10} = 150\)).

Step 3: Setting Up the Arithmetic Progression

For minutes starting from the 11th, the progression is \(a_{11} = 148, a_{12} = 146, a_{13} = 144\), and so on. The nth term formula for an AP is given by \( a_n = a_{11} + (n-1)d \), where \(a_{11} = 148\) and \(d = -2\).

Step 4: Determine Total Notes Counted by AP Terms

The sum of the notes counted in the AP for \(n\) terms is given by the formula \(S_n = \frac{n}{2}(a_1 + a_n)\). Since the total number of notes must add up to 4500, and 1500 have been counted already, we need 3000 more notes counted from the AP sequence.

Step 5: Calculate the Number of Minutes

We need the sum \(S_n = 3000\) using the initial term \(a_1 = a_{11} = 148\) and common difference \(d = -2\). We solve for \(n\) by setting the equation \( n/2 (2 \times 148 + (n-1)(-2)) = 3000 \). Simplifying \(148n - n(n-1) = 3000\) leads to solving the quadratic equation \(n^2 - 149n + 3000 = 0\).

Step 6: Solve the Quadratic Equation

Solving \(n^2 - 149n + 3000 = 0\) using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we have: \(n = \frac{-(-149) \pm \sqrt{(-149)^2 - 4\times 1\times 3000}}{2\times 1}\). Calculating gives \(n = 24 \).

Step 7: Total Time Calculation

Add the minutes from the initial setup (10 minutes) to the result from the quadratic equation (24 minutes) to find the total time taken: \(10 + 24 = 34\) minutes.

Key Concepts

Counting Techniques

Counting techniques are crucial for solving problems involving sequences or progressions. In this exercise, the person is tasked with counting 4500 currency notes. To efficiently manage the counting, they initially count a fixed amount for the first 10 minutes. Calculating the number of notes is quite simple: for each of those minutes, 150 notes are counted, totaling 1500 notes after 10 minutes (since 10 x 150 = 1500). This straightforward technique helps us establish a baseline for how many notes have been counted before a different pattern begins.

These techniques not only apply to currency notes but to any situation where items are systematic and follow a specific pattern. Utilizing counting techniques allows us to break down complex counting tasks into manageable units and can often involve:

• Setting up a repeatable process for initial counts.
• Calculating totals for a set period, then switching to another counting pattern.
• Adjusting counting strategies when sequences change, like moving into an arithmetic progression.

By understanding this systematic approach, we develop various strategies that come in handy in quantitative comparisons and problem-solving.

Quadratic Equations

Quadratic equations are fundamental in many mathematical scenarios, especially when dealing with sequences like an arithmetic progression. Here, after the initial 10 minutes of fixed counting, the person switches to an arithmetic progression with a common difference of -2. This decrease means the number of notes counted per minute dwindles by 2 with each passing minute.

In order to determine how many minutes are needed for the remaining 3000 notes, we rely on forming and solving a quadratic equation. The sum of the arithmetic sequence is given by \( S_n = \frac{n}{2} (2a + (n-1)d) \), where \( a \) is the first term, and \( d \) is the common difference. The problem then becomes solving for \( n \), where \( a = 148 \) and \( d = -2 \). Simplifying this results in the quadratic equation \( n^2 - 149n + 3000 = 0 \).

To solve this equation, we use the quadratic formula: \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plug in the coefficients \( a = 1 \), \( b = -149 \), and \( c = 3000 \) to find the solution. Understanding quadratic equations can vastly expand one's ability to solve real-world problems involving patterns and progressions.

Time Calculation in Sequences

Time calculations are integral components when dealing with sequences, especially when trying to determine how long it takes to reach a specific count or output. In this scenario, the person must continuously adjust their counting strategy over time to reach the overall target of 4500 notes. After counting for 10 minutes at a rate of 150 notes per minute, and switching to a pattern where they count fewer notes each subsequent minute, the task becomes calculating total time.

The ability to measure time in sequential tasks is valuable in numerous applications. Here, by solving the quadratic equation, we discover that after the initial phase, 24 more minutes are needed (from step 6) to reach the remaining count of 3000 notes. Thus, the total time taken, combining both counting phases, is 34 minutes.

When dealing with such sequences, consider:

• Breaking down the question into smaller phases and tasks.
• Utilizing arithmetic sequences to predict time or counts accurately.
• Solving relevant equations to bridge the gap between different phases.

This calculated approach ensures precision in estimating time while managing sequential tasks efficiently.