Problem 153 - A Complete Resource Book in Mathematics for JEE Main 2017

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 [2010] (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: Calculate Notes Counted in First 10 Minutes

The person counts 150 notes each minute for the first 10 minutes, so the total notes counted in this period is \(150 \times 10 = 1500\).

Step 2: Set Up Arithmetic Progression for Subsequent Minutes

After 10 minutes, the number of notes counted per minute forms an arithmetic progression (A.P.) starting from the 11th minute, where \(a_{11} = 150 - 2 = 148\), with a common difference of \(-2\).

Step 3: Formulate the Total Counting Equation

Let the total time taken be \(n\) minutes. The notes counted in the remaining \(n - 10\) minutes can be represented as the sum of an arithmetic sequence: \(S_{n-10} = (n - 10)/2 \times (a_{11} + a_{n})\).The notes counted from the 11th to the last minute must sum up to \(4500 - 1500 = 3000\).

Step 4: Calculate the Sum of Notes from 11th Minute Onwards

The sum of an arithmetic progression is given by: \(S_{n-10} = \frac{(n-10)}{2} \times (148 + a_{n})\). We know \(a_{n} = 150 - 2(n-10) = 150 - 2n + 20 = 170 - 2n\). Hence, \((n-10) \times (318 - 2n)/2 = 3000\).

Step 5: Solve the Equation for Total Time \(n\)

Substitute \((n-10) \times (318 - 2n)/2 = 3000\) and simplify: \((n-10)(159-n) = 3000\). Expanding gives: \(-n^2 + 169n - 1590 = 3000\). Rearranging gives the quadratic equation: \(-n^2 + 169n - 4590 = 0\).

Step 6: Solve the Quadratic Equation

The quadratic equation \(-n^2 + 169n - 4590 = 0\) can be solved using the quadratic formula: \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -1\), \(b = 169\), \(c = -4590\). Calculate the discriminant: \(b^2 - 4ac = 169^2 - 4(-1)(-4590) = 28561 - 18360 = 10201\). \(\sqrt{10201} = 101\). Then, \(n = \frac{-169 \pm 101}{-2}\).

Step 7: Calculate the Positive Root

The positive root of the equation is \(n = \frac{-169 + 101}{-2} = \frac{-68}{-2} = 34\). Since time cannot be negative, \(n = 34\) minutes is the total time to count all the notes.

Key Concepts

Quadratic Equation

A quadratic equation is an important concept in algebra that helps in finding unknown variables when given polynomial equations of degree two. Such equations have the standard form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \).To solve a quadratic equation, you can use various methods such as factoring, completing the square, or the quadratic formula. The quadratic formula is particularly useful, especially when dealing with non-factorable equations. It is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula provides solutions by calculating the values of \( x \) that satisfy the equation. Here, \( b^2 - 4ac \) is known as the discriminant, which determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (double root).
• If it is negative, the roots are complex or imaginary.

In the exercise, solving the quadratic equation helped in determining the time taken to count all the notes, which was crucial for reaching the solution.

Sum of Arithmetic Sequence

An arithmetic sequence is a list of numbers where each term is obtained by adding a constant difference to the previous term. This constant is known as the common difference, denoted as \( d \). If the first term of the sequence is \( a_1 \), the \( n^{\text{th}} \) term, \( a_n \), can be found using:\[ a_n = a_1 + (n-1)d \]When you need to find the sum of the first \( n \) terms of an arithmetic sequence, you use the formula:\[ S_n = \frac{n}{2} (a_1 + a_n) \]This formula calculates the sum by averaging the first and last terms, then multiplying by the number of terms, \( n \). In this exercise, after counting a steady number of notes per minute initially, a diminishing count per minute followed an arithmetic progression. Thus, the sum formula was used to determine how many notes were counted during this phase and eventually helped deduce the total time taken.

Time Calculation

Calculating time accurately is essential in many practical problems, especially when rates or sequences are involved. In this context, time calculation was used to determine how long it took to count a specific number of currency notes at given rates.Initially, the person involved was counting for a fixed period at a constant rate, making it straightforward to calculate the number of notes counted during this time. Subsequently, the rate of counting changed and followed an arithmetic progression, introducing a need for calculations using the sum of an arithmetic sequence.The time taken overall had to satisfy the total number of notes counted, both in the initial fixed-rate period and during the variable rate period. By appropriately setting up equations and solving for the total time \( n \), it became possible to find that counting all 4500 notes required 34 minutes. Such calculations can be essential when planning or analyzing time-dependent activities.