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Maharashtra Board Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.4 Solutions

Welcome to the complete Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we provide a detailed, step-by-step conceptual guide to solving Practice Set 3.4 from Chapter 3: Arithmetic Progression. These word problems are crucial for board exams, so we have structured every solution with clear Given data, Formulas, and Calculations to help you maximize your scores!

Practice Set 3.4 Questions and Answers

Question 1

On \( 1^{\text{st}} \) Jan 2016, Sanika decides to save ₹10, ₹11 on the second day, ₹12 on the third day. If she decides to save like this, then on \( 31^{\text{st}} \) Dec 2016 what would be her total saving?

Given: Sanika's daily savings form an Arithmetic Progression (A.P.): \( 10, 11, 12, \dots \)
First term \( a = 10 \)
Common difference \( d = 11 - 10 = 1 \)
Total number of days \( n = 366 \) (since 2016 is a leap year).
Formula:
Key Formula (Sum of an A.P.):
\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: \[ \begin{aligned} S_{366} &= \frac{366}{2}[2(10) + (366 - 1)1] \\ &= 183(20 + 365 \times 1) \\ &= 183(20 + 365) \\ &= 183 \times 385 \\ S_{366} &= 70455 \end{aligned} \]
Final Answer: Sanika's total saving on \( 31^{\text{st}} \) December 2016 would be ₹70,455.

Question 2

A man borrows ₹8000 and agrees to repay with a total interest of ₹1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹40. Find the amount of the first and last instalment.

Given: The instalments are in an A.P.
Total amount repaid in 12 instalments (\( S_{12} \)) = Amount borrowed + Total interest = \( 8000 + 1360 = 9360 \).
Number of instalments \( n = 12 \).
Each instalment is less than the preceding one by ₹40, therefore \( d = -40 \).
Formula:
Formulas Used:
Sum formula: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( n^{\text{th}} \) term formula: \( t_{n} = a + (n-1)d \)
Substitution & Calculation: Finding the first instalment (\( a \)): \[ \begin{aligned} S_{12} &= \frac{12}{2}[2a + (12 - 1)(-40)] \\ 9360 &= 6[2a + (11)(-40)] \\ 9360 &= 6(2a - 440) \\ \frac{9360}{6} &= 2a - 440 \\ 1560 &= 2a - 440 \\ 1560 + 440 &= 2a \\ 2000 &= 2a \\ a &= \frac{2000}{2} \\ a &= 1000 \end{aligned} \] Finding the last instalment (\( t_{12} \)): \[ \begin{aligned} t_{12} &= 1000 + (12 - 1)(-40) \\ &= 1000 + 11(-40) \\ &= 1000 - 440 \\ t_{12} &= 560 \end{aligned} \]
Final Answer: The amount of the first instalment is ₹1000 and that of the last instalment is ₹560.

Question 3

Sachin invested in a national saving certificate scheme. In the first year he invested ₹5000, in the second year ₹7000, in the third year ₹9000 and so on. Find the total amount that he invested in 12 years.

Given: Amounts invested by Sachin each year: \( 5000, 7000, 9000, \dots \)
This sequence is an A.P.
First term \( a = 5000 \)
Common difference \( d = 7000 - 5000 = 2000 \)
Number of years \( n = 12 \).
Formula: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: \[ \begin{aligned} S_{12} &= \frac{12}{2}[2(5000) + (12 - 1)2000] \\ &= 6(10000 + 11 \times 2000) \\ &= 6(10000 + 22000) \\ &= 6(32000) \\ S_{12} &= 192000 \end{aligned} \]
Final Answer: The total amount invested by Sachin in 12 years is ₹1,92,000.

Question 4

There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the \( 15^{\text{th}} \) row and also find how many total seats are there in the auditorium?

Given: The number of seats arranged row-wise are: \( 20, 22, 24, \dots \)
This is an A.P. where:
First term \( a = 20 \)
Common difference \( d = 22 - 20 = 2 \)
Total rows \( n = 27 \).
Formula:
\( n^{\text{th}} \) term: \( t_{n} = a + (n-1)d \)
Sum of n terms: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: Finding seats in the \( 15^{\text{th}} \) row (\( t_{15} \)): \[ \begin{aligned} t_{15} &= 20 + (15 - 1)2 \\ &= 20 + 14 \times 2 \\ &= 20 + 28 \\ t_{15} &= 48 \end{aligned} \] Finding total seats (\( S_{27} \)): \[ \begin{aligned} S_{27} &= \frac{27}{2}[2(20) + (27 - 1)2] \\ &= \frac{27}{2}(40 + 26 \times 2) \\ &= \frac{27}{2}(40 + 52) \\ &= \frac{27}{2} \times 92 \\ &= 27 \times 46 \\ S_{27} &= 1242 \end{aligned} \]
Final Answer: The number of seats in the \( 15^{\text{th}} \) row is 48 and the total seats in the auditorium are 1242.

Question 5

Kargil's temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was \( 5^{\circ}\text{C} \) more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was \( -30^{\circ}\text{C} \) then find the temperature on the other five days.

Given: Let the temperatures from Monday to Saturday in A.P. be:
\( a,\ a+d,\ a+2d,\ a+3d,\ a+4d,\ a+5d \)
Condition 1: Sum of Monday & Saturday = Sum of Tuesday & Saturday + \( 5^{\circ} \)
Condition 2: Temperature on Wednesday = \( -30^{\circ}\text{C} \)
Substitution & Calculation: From the first condition: \[ \begin{aligned} (a) + (a+5d) &= (a+d) + (a+5d) + 5 \\ 2a + 5d &= 2a + 6d + 5 \\ 5d - 6d &= 5 \\ -d &= 5 \\ d &= -5^{\circ} \end{aligned} \] From the second condition (Wednesday is the \( 3^{\text{rd}} \) term): \[ \begin{aligned} a + 2d &= -30^{\circ} \\ a + 2(-5^{\circ}) &= -30^{\circ} \\ a - 10^{\circ} &= -30^{\circ} \\ a &= -30^{\circ} + 10^{\circ} \\ a &= -20^{\circ} \end{aligned} \] Finding the temperatures for all days:
- Monday (\( a \)): \( -20^{\circ}\text{C} \)
- Tuesday (\( a+d \)): \( -20^{\circ} - 5^{\circ} = -25^{\circ}\text{C} \)
- Wednesday (\( a+2d \)): \( -30^{\circ}\text{C} \) (Given)
- Thursday (\( a+3d \)): \( -20^{\circ} + 3(-5^{\circ}) = -20^{\circ} - 15^{\circ} = -35^{\circ}\text{C} \)
- Friday (\( a+4d \)): \( -20^{\circ} + 4(-5^{\circ}) = -20^{\circ} - 20^{\circ} = -40^{\circ}\text{C} \)
- Saturday (\( a+5d \)): \( -20^{\circ} + 5(-5^{\circ}) = -20^{\circ} - 25^{\circ} = -45^{\circ}\text{C} \)
Final Answer: The temperatures on the other five days are \( -20^{\circ}\text{C} \), \( -25^{\circ}\text{C} \), \( -35^{\circ}\text{C} \), \( -40^{\circ}\text{C} \), and \( -45^{\circ}\text{C} \).

Question 6

On the world environment day, a tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Given: The number of trees planted row-wise are: \( 1, 2, 3, \dots \)
This sequence is an A.P.
First term \( a = 1 \)
Common difference \( d = 2 - 1 = 1 \)
Total number of rows \( n = 25 \).
Formula: \( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
Substitution & Calculation: \[ \begin{aligned} S_{25} &= \frac{25}{2}[2(1) + (25 - 1)1] \\ &= \frac{25}{2}(2 + 24) \\ &= \frac{25}{2} \times 26 \\ &= 25 \times 13 \\ S_{25} &= 325 \end{aligned} \]
Final Answer: The total number of trees planted in the 25 rows is 325.

Class 10 Maths Part 1 Chapter 2 Practice Set 2.3 - Quadratic Equations Solutions (Maharashtra Board)

Welcome to the comprehensive Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we will cover Chapter 2: Quadratic Equations, specifically focusing on Practice Set 2.3. This exercise is essential as it deals with solving quadratic equations using the completing the square method. Our step-by-step guide will help you understand the concepts clearly and score full marks in your board exams!

Exam Tip: When using the "Completing the Square" method, always ensure the coefficient of \( x^2 \) (or the leading variable) is exactly 1 before proceeding. The formula for the third term (k) is \( \left( \frac{1}{2} \times \text{coefficient of } x \right)^2 \). Add and subtract this term to form a perfect square binomial.

Question 1: Solve the following quadratic equations by completing the square method.

1. \( x^{2} + x - 20 = 0 \)

Solution & Calculation:

\( x^{2} + x - 20 = 0 \)
Let \( x^{2} + x + k = (x + a)^{2} \)
\( x^{2} + x + k = x^{2} + 2ax + a^{2} \)
Comparing the coefficients, we get:
\( 1 = 2a \) and \( k = a^{2} \)
\( a = \frac{1}{2} \) and \( k = \left(\frac{1}{2}\right)^{2} = \frac{1}{4} \)
Now, substitute \( k \) back into the original equation context:
\( x^{2} + x + \frac{1}{4} - \frac{1}{4} - 20 = 0 \)
\( \left(x + \frac{1}{2}\right)^{2} - \left(\frac{1 + 80}{4}\right) = 0 \)
\( \left(x + \frac{1}{2}\right)^{2} - \frac{81}{4} = 0 \)
\( \left(x + \frac{1}{2}\right)^{2} = \frac{81}{4} \)
Taking the square root of both sides, we get:
\( x + \frac{1}{2} = \pm \frac{9}{2} \)
\( x + \frac{1}{2} = \frac{9}{2} \) or \( x + \frac{1}{2} = -\frac{9}{2} \)
\( x = \frac{9}{2} - \frac{1}{2} \) or \( x = -\frac{9}{2} - \frac{1}{2} \)
\( x = \frac{8}{2} = 4 \) or \( x = -\frac{10}{2} = -5 \)

Final Answer: The roots of the given quadratic equation are 4 and -5.

2. \( x^{2} + 2x - 5 = 0 \)

Solution & Calculation:

\( x^{2} + 2x - 5 = 0 \)
Let \( x^{2} + 2x + k = (x + a)^{2} \)
\( x^{2} + 2x + k = x^{2} + 2ax + a^{2} \)
Comparing the coefficients, we get:
\( 2 = 2a \) and \( k = a^{2} \)
\( a = 1 \) and \( k = (1)^{2} = 1 \)
Now, substitute \( k \):
\( x^{2} + 2x + 1 - 1 - 5 = 0 \)
\( (x + 1)^{2} - 6 = 0 \)
\( (x + 1)^{2} = 6 \)
Taking the square root of both sides, we get:
\( x + 1 = \pm \sqrt{6} \)
\( x + 1 = \sqrt{6} \) or \( x + 1 = -\sqrt{6} \)
\( x = \sqrt{6} - 1 \) or \( x = -\sqrt{6} - 1 \)

Final Answer: The roots of the given quadratic equation are \( \sqrt{6} - 1 \) and \( -\sqrt{6} - 1 \).

3. \( m^{2} - 5m = -3 \)

Solution & Calculation:

\( m^{2} - 5m = -3 \)
\( m^{2} - 5m + 3 = 0 \)
Let \( m^{2} - 5m + k = (m + a)^{2} \)
\( m^{2} - 5m + k = m^{2} + 2am + a^{2} \)
Comparing the coefficients, we get:
\( -5 = 2a \) and \( k = a^{2} \)
\( a = \frac{-5}{2} \) and \( k = \left(\frac{-5}{2}\right)^{2} = \frac{25}{4} \)
Now, substitute \( k \):
\( m^{2} - 5m + \frac{25}{4} - \frac{25}{4} + 3 = 0 \)
\( \left(m - \frac{5}{2}\right)^{2} + \left(\frac{-25 + 12}{4}\right) = 0 \)
\( \left(m - \frac{5}{2}\right)^{2} - \frac{13}{4} = 0 \)
\( \left(m - \frac{5}{2}\right)^{2} = \frac{13}{4} \)
Taking the square root of both sides, we get:
\( m - \frac{5}{2} = \pm \frac{\sqrt{13}}{2} \)
\( m - \frac{5}{2} = \frac{\sqrt{13}}{2} \) or \( m - \frac{5}{2} = -\frac{\sqrt{13}}{2} \)
\( m = \frac{\sqrt{13} + 5}{2} \) or \( m = \frac{-\sqrt{13} + 5}{2} \)

Final Answer: The roots of the given quadratic equation are \( \frac{\sqrt{13} + 5}{2} \) and \( \frac{-\sqrt{13} + 5}{2} \).

4. \( 9y^{2} - 12y + 2 = 0 \)

Solution & Calculation:

\( y^{2} - \frac{4}{3}y + \frac{2}{9} = 0 \) (Dividing both sides by 9)
Let \( y^{2} - \frac{4}{3}y + k = (y + a)^{2} \)
\( y^{2} - \frac{4}{3}y + k = y^{2} + 2ay + a^{2} \)
Comparing the coefficients, we get:
\( -\frac{4}{3} = 2a \) and \( k = a^{2} \)
\( a = -\frac{2}{3} \) and \( k = \left(-\frac{2}{3}\right)^{2} = \frac{4}{9} \)
Now, substitute \( k \):
\( y^{2} - \frac{4}{3}y + \frac{4}{9} - \frac{4}{9} + \frac{2}{9} = 0 \)
\( \left(y - \frac{2}{3}\right)^{2} - \frac{2}{9} = 0 \)
\( \left(y - \frac{2}{3}\right)^{2} = \frac{2}{9} \)
Taking the square root of both sides, we get:
\( y - \frac{2}{3} = \pm \frac{\sqrt{2}}{3} \)
\( y - \frac{2}{3} = \frac{\sqrt{2}}{3} \) or \( y - \frac{2}{3} = -\frac{\sqrt{2}}{3} \)
\( y = \frac{\sqrt{2} + 2}{3} \) or \( y = \frac{-\sqrt{2} + 2}{3} \)

Final Answer: The roots of the given quadratic equation are \( \frac{\sqrt{2} + 2}{3} \) and \( \frac{-\sqrt{2} + 2}{3} \).

5. \( 2y^{2} + 9y + 10 = 0 \)

Solution & Calculation:

\( y^{2} + \frac{9}{2}y + 5 = 0 \) (Dividing both sides by 2)
Let \( y^{2} + \frac{9}{2}y + k = (y + a)^{2} \)
\( y^{2} + \frac{9}{2}y + k = y^{2} + 2ay + a^{2} \)
Comparing the coefficients, we get:
\( \frac{9}{2} = 2a \) and \( k = a^{2} \)
\( a = \frac{9}{4} \) and \( k = \left(\frac{9}{4}\right)^{2} = \frac{81}{16} \)
Now, substitute \( k \):
\( y^{2} + \frac{9}{2}y + \frac{81}{16} - \frac{81}{16} + 5 = 0 \)
\( \left(y + \frac{9}{4}\right)^{2} + \left(\frac{-81 + 80}{16}\right) = 0 \)
\( \left(y + \frac{9}{4}\right)^{2} - \frac{1}{16} = 0 \)
\( \left(y + \frac{9}{4}\right)^{2} = \frac{1}{16} \)
Taking the square root of both sides, we get:
\( y + \frac{9}{4} = \pm \frac{1}{4} \)
\( y + \frac{9}{4} = \frac{1}{4} \) or \( y + \frac{9}{4} = -\frac{1}{4} \)
\( y = \frac{1}{4} - \frac{9}{4} \) or \( y = -\frac{1}{4} - \frac{9}{4} \)
\( y = \frac{-8}{4} = -2 \) or \( y = -\frac{10}{4} = -\frac{5}{2} \)

Final Answer: The roots of the given quadratic equation are -2 and \( -\frac{5}{2} \).

6. \( 5x^{2} = 4x + 7 \)

Solution & Calculation:

\( 5x^{2} - 4x - 7 = 0 \)
\( x^{2} - \frac{4}{5}x - \frac{7}{5} = 0 \) (Dividing both sides by 5)
Let \( x^{2} - \frac{4}{5}x + k = (x + a)^{2} \)
\( x^{2} - \frac{4}{5}x + k = x^{2} + 2ax + a^{2} \)
Comparing the coefficients, we get:
\( -\frac{4}{5} = 2a \) and \( k = a^{2} \)
\( a = -\frac{2}{5} \) and \( k = \left(-\frac{2}{5}\right)^{2} = \frac{4}{25} \)
Now, substitute \( k \):
\( x^{2} - \frac{4}{5}x + \frac{4}{25} - \frac{4}{25} - \frac{7}{5} = 0 \)
\( \left(x - \frac{2}{5}\right)^{2} - \left(\frac{4 + 35}{25}\right) = 0 \)
\( \left(x - \frac{2}{5}\right)^{2} - \frac{39}{25} = 0 \)
\( \left(x - \frac{2}{5}\right)^{2} = \frac{39}{25} \)
Taking the square root of both sides, we get:
\( x - \frac{2}{5} = \pm \frac{\sqrt{39}}{5} \)
\( x - \frac{2}{5} = \frac{\sqrt{39}}{5} \) or \( x - \frac{2}{5} = -\frac{\sqrt{39}}{5} \)
\( x = \frac{2 + \sqrt{39}}{5} \) or \( x = \frac{2 - \sqrt{39}}{5} \)

Final Answer: The roots of the given quadratic equation are \( \frac{2 + \sqrt{39}}{5} \) and \( \frac{2 - \sqrt{39}}{5} \).

Class 10 Maths Part 1 Practice Set 2.2 Solutions - Maharashtra Board

Welcome to the complete, step-by-step Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we will cover Chapter 2: Quadratic Equations, focusing entirely on Practice Set 2.2. These Chapter Notes and solutions are designed to help you master solving quadratic equations by factorization, ensuring you score full marks in your board exams!

Exam Tip: The Zero Product Property
The factorization method relies heavily on this core mathematical rule: If the product of two numbers is zero, then at least one of them must be zero. In exam terms, if \( (a)(b) = 0 \), then either \( a = 0 \) or \( b = 0 \). Keep this principle in mind for every factorization sum!

Practice Set 2.2

Question 1. Solve the following quadratic equations by factorisation.

Solution i.

Given:
\[ x^2 - 15x + 54 = 0 \]
Calculation: \[ x^2 - 9x - 6x + 54 = 0 \] \[ x(x - 9) - 6(x - 9) = 0 \] \[ (x - 9)(x - 6) = 0 \] By using the zero product property: \[ x - 9 = 0 \quad \text{or} \quad x - 6 = 0 \] \[ x = 9 \quad \text{or} \quad x = 6 \]
Final Answer:
The roots of the given quadratic equation are 9 and 6.

Solution ii.

Given:
\[ x^2 + x - 20 = 0 \]
Calculation: \[ x^2 + 5x - 4x - 20 = 0 \] \[ x(x + 5) - 4(x + 5) = 0 \] \[ (x + 5)(x - 4) = 0 \] By using the zero product property: \[ x + 5 = 0 \quad \text{or} \quad x - 4 = 0 \] \[ x = -5 \quad \text{or} \quad x = 4 \]
Final Answer:
The roots of the given quadratic equation are -5 and 4.

Solution iii.

Given:
\[ 2y^2 + 27y + 13 = 0 \]
Calculation: \[ 2y^2 + 26y + y + 13 = 0 \] \[ 2y(y + 13) + 1(y + 13) = 0 \] \[ (y + 13)(2y + 1) = 0 \] By using the zero product property: \[ y + 13 = 0 \quad \text{or} \quad 2y + 1 = 0 \] \[ y = -13 \quad \text{or} \quad 2y = -1 \] \[ y = -13 \quad \text{or} \quad y = -\frac{1}{2} \]
Final Answer:
The roots of the given quadratic equation are -13 and \( -\frac{1}{2} \).

Solution iv.

Given:
\[ 5m^2 = 22m + 15 \]
Calculation:
First, bring all terms to one side to form the standard quadratic equation \( ax^2 + bx + c = 0 \): \[ 5m^2 - 22m - 15 = 0 \] \[ 5m^2 - 25m + 3m - 15 = 0 \] \[ 5m(m - 5) + 3(m - 5) = 0 \] \[ (m - 5)(5m + 3) = 0 \] By using the zero product property: \[ m - 5 = 0 \quad \text{or} \quad 5m + 3 = 0 \] \[ m = 5 \quad \text{or} \quad 5m = -3 \] \[ m = 5 \quad \text{or} \quad m = -\frac{3}{5} \]
Final Answer:
The roots of the given quadratic equation are 5 and \( -\frac{3}{5} \).

Solution v.

Given:
\[ 2x^2 - 2x + \frac{1}{2} = 0 \]
Calculation:
Multiplying both sides by 2 to remove the fraction: \[ 4x^2 - 4x + 1 = 0 \] \[ 4x^2 - 2x - 2x + 1 = 0 \] \[ 2x(2x - 1) - 1(2x - 1) = 0 \] \[ (2x - 1)(2x - 1) = 0 \] By using the zero product property: \[ 2x - 1 = 0 \quad \text{or} \quad 2x - 1 = 0 \] \[ 2x = 1 \quad \text{or} \quad 2x = 1 \] \[ x = \frac{1}{2} \quad \text{or} \quad x = \frac{1}{2} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{1}{2} \) and \( \frac{1}{2} \).

Solution vi.

Given:
\[ 6x - \frac{2}{x} = 1 \]
Calculation:
Multiplying both sides by \( x \) to remove the denominator: \[ 6x^2 - 2 = x \] Rearranging to standard form: \[ 6x^2 - x - 2 = 0 \] \[ 6x^2 - 4x + 3x - 2 = 0 \] \[ 2x(3x - 2) + 1(3x - 2) = 0 \] \[ (3x - 2)(2x + 1) = 0 \] By using the zero product property: \[ 3x - 2 = 0 \quad \text{or} \quad 2x + 1 = 0 \] \[ 3x = 2 \quad \text{or} \quad 2x = -1 \] \[ x = \frac{2}{3} \quad \text{or} \quad x = -\frac{1}{2} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{2}{3} \) and \( -\frac{1}{2} \).

Solution vii.

Given:
\[ \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \]
Calculation: \[ \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \] \[ x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \] \[ (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \] By using the zero product property: \[ \sqrt{2}x + 5 = 0 \quad \text{or} \quad x + \sqrt{2} = 0 \] \[ \sqrt{2}x = -5 \quad \text{or} \quad x = -\sqrt{2} \] \[ x = \frac{-5}{\sqrt{2}} \quad \text{or} \quad x = -\sqrt{2} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{-5}{\sqrt{2}} \) and \( -\sqrt{2} \).

Solution viii.

Given:
\[ 3x^2 - 2\sqrt{6}x + 2 = 0 \]
Calculation: \[ 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \] Taking \( \sqrt{3}x \) common from the first two terms, and \( \sqrt{2} \) from the last two: \[ \sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0 \] \[ (\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0 \] By using the zero product property: \[ \sqrt{3}x - \sqrt{2} = 0 \quad \text{or} \quad \sqrt{3}x - \sqrt{2} = 0 \] \[ \sqrt{3}x = \sqrt{2} \quad \text{or} \quad \sqrt{3}x = \sqrt{2} \] \[ x = \frac{\sqrt{2}}{\sqrt{3}} \quad \text{or} \quad x = \frac{\sqrt{2}}{\sqrt{3}} \]
Final Answer:
The roots of the given quadratic equation are \( \frac{\sqrt{2}}{\sqrt{3}} \) and \( \frac{\sqrt{2}}{\sqrt{3}} \).

Solution ix.

Given:
\[ 2m(m - 24) = 50 \]
Calculation: \[ 2m^2 - 48m = 50 \] \[ 2m^2 - 48m - 50 = 0 \] Dividing both sides by 2 to simplify: \[ m^2 - 24m - 25 = 0 \] \[ m^2 - 25m + m - 25 = 0 \] \[ m(m - 25) + 1(m - 25) = 0 \] \[ (m - 25)(m + 1) = 0 \] By using the zero product property: \[ m - 25 = 0 \quad \text{or} \quad m + 1 = 0 \] \[ m = 25 \quad \text{or} \quad m = -1 \]
Final Answer:
The roots of the given quadratic equation are 25 and -1.

Solution x.

Given:
\[ 25m^2 = 9 \]
Calculation: \[ 25m^2 - 9 = 0 \] Expressing the equation in the form of \( a^2 - b^2 \): \[ (5m)^2 - (3)^2 = 0 \] Using the identity \( a^2 - b^2 = (a + b)(a - b) \): \[ (5m + 3)(5m - 3) = 0 \] By using the zero product property: \[ 5m + 3 = 0 \quad \text{or} \quad 5m - 3 = 0 \] \[ 5m = -3 \quad \text{or} \quad 5m = 3 \] \[ m = -\frac{3}{5} \quad \text{or} \quad m = \frac{3}{5} \]
Final Answer:
The roots of the given quadratic equation are \( -\frac{3}{5} \) and \( \frac{3}{5} \).

Solution xi.

Given:
\[ 7m^2 = 21m \]
Calculation: \[ 7m^2 - 21m = 0 \] Dividing both sides by 7 to simplify: \[ m^2 - 3m = 0 \] Taking \( m \) common: \[ m(m - 3) = 0 \] By using the zero product property: \[ m = 0 \quad \text{or} \quad m - 3 = 0 \] \[ m = 0 \quad \text{or} \quad m = 3 \]
Final Answer:
The roots of the given quadratic equation are 0 and 3.

Solution xii.

Given:
\[ m^2 - 11 = 0 \]
Calculation: Expressing 11 as a square to apply the difference of squares identity: \[ m^2 - (\sqrt{11})^2 = 0 \] Using the identity \( a^2 - b^2 = (a + b)(a - b) \): \[ (m + \sqrt{11})(m - \sqrt{11}) = 0 \] By using the zero product property: \[ m + \sqrt{11} = 0 \quad \text{or} \quad m - \sqrt{11} = 0 \] \[ m = -\sqrt{11} \quad \text{or} \quad m = \sqrt{11} \]
Final Answer:
The roots of the given quadratic equation are \( -\sqrt{11} \) and \( \sqrt{11} \).

Maharashtra Board Class 10 Maths Part 1 Practice Set 2.1 Solutions (Quadratic Equations)

Welcome to our comprehensive study guide for Class 10 Maths Part 1 of the Maharashtra Board. In this post, we provide complete, step-by-step Textbook Exercise Solutions for Chapter 2: Quadratic Equations (Practice Set 2.1). Whether you are revising for your upcoming board exams or working through daily assignments, these verified answers will help you grasp the core concepts, identify the correct standard forms, and easily find the roots of equations. Let's master these numericals together!

Exam Tip: A standard quadratic equation is always written in the form \( ax^2 + bx + c = 0 \). To check if an equation is quadratic, ensure it has only one variable and the highest index (power) of that variable is exactly 2. (Also, \( a \neq 0 \)).

Practice Set 2.1 Solutions

Question 1. Write any two quadratic equations.

Solution:

\( y^2 - 7y + 12 = 0 \)
\( x^2 - 8 = 0 \)

(Note: Students can write any two valid quadratic equations of their choice.)

Question 2. Decide which of the following are quadratic equations.

i. \( x^2 + 5x - 2 = 0 \)

Solution:
The given equation is \( x^2 + 5x - 2 = 0 \)
Here, \( x \) is the only variable and the maximum index of the variable is 2.
Comparing with standard form, \( a = 1, b = 5, c = -2 \) are real numbers and \( a \neq 0 \).
Final Answer: Therefore, the given equation is a quadratic equation.

ii. \( y^2 = 5y - 10 \)

Solution:
The given equation is \( y^2 = 5y - 10 \)
\[ \therefore y^2 - 5y + 10 = 0 \] Here, \( y \) is the only variable and the maximum index of the variable is 2.
\( a = 1, b = -5, c = 10 \) are real numbers and \( a \neq 0 \).
Final Answer: Therefore, the given equation is a quadratic equation.

iii. \( y^2 + \frac{1}{y} = 2 \)

Solution:
The given equation is \( y^2 + \frac{1}{y} = 2 \)
Multiplying both sides by \( y \), we get:
\[ y^3 + 1 = 2y \] \[ \therefore y^3 - 2y + 1 = 0 \] Here, \( y \) is the only variable, but the maximum index of the variable is 3 (not 2).
Final Answer: Therefore, the given equation is not a quadratic equation.

iv. \( x + \frac{1}{x} = -2 \)

Solution:
The given equation is \( x + \frac{1}{x} = -2 \)
Multiplying both sides by \( x \), we get:
\[ x^2 + 1 = -2x \] \[ \therefore x^2 + 2x + 1 = 0 \] Here, \( x \) is the only variable and the maximum index of the variable is 2.
\( a = 1, b = 2, c = 1 \) are real numbers and \( a \neq 0 \).
Final Answer: Therefore, the given equation is a quadratic equation.

v. \( (m + 2)(m - 5) = 0 \)

Solution:
The given equation is \( (m + 2)(m - 5) = 0 \)
\[ m(m - 5) + 2(m - 5) = 0 \] \[ m^2 - 5m + 2m - 10 = 0 \] \[ \therefore m^2 - 3m - 10 = 0 \] Here, \( m \) is the only variable and the maximum index of the variable is 2.
\( a = 1, b = -3, c = -10 \) are real numbers and \( a \neq 0 \).
Final Answer: Therefore, the given equation is a quadratic equation.

vi. \( m^3 + 3m^2 - 2 = 3m^3 \)

Solution:
The given equation is \( m^3 + 3m^2 - 2 = 3m^3 \)
\[ 3m^3 - m^3 - 3m^2 + 2 = 0 \] \[ \therefore 2m^3 - 3m^2 + 2 = 0 \] Here, \( m \) is the only variable, but the maximum index of the variable is 3 (not 2).
Final Answer: Therefore, the given equation is not a quadratic equation.

Question 3. Write the following equations in the form \( ax^2 + bx + c = 0 \), then write the values of a, b, c for each equation.

i. \( 2y = 10 - y^2 \)
Solution:
\[ y^2 + 2y - 10 = 0 \] Comparing the above equation with \( ay^2 + by + c = 0 \), we get:
\( a = 1 \), \( b = 2 \), \( c = -10 \)

ii. \( (x - 1)^2 = 2x + 3 \)
Solution:
Using the formula \( (a-b)^2 = a^2 - 2ab + b^2 \):
\[ x^2 - 2x + 1 = 2x + 3 \] \[ x^2 - 2x - 2x + 1 - 3 = 0 \] \[ \therefore x^2 - 4x - 2 = 0 \] Comparing the above equation with \( ax^2 + bx + c = 0 \), we get:
\( a = 1 \), \( b = -4 \), \( c = -2 \)

iii. \( x^2 + 5x = -(3 - x) \)
Solution:
\[ x^2 + 5x = -3 + x \] \[ x^2 + 5x - x + 3 = 0 \] \[ \therefore x^2 + 4x + 3 = 0 \] Comparing the above equation with \( ax^2 + bx + c = 0 \), we get:
\( a = 1 \), \( b = 4 \), \( c = 3 \)

iv. \( 3m^2 = 2m^2 - 9 \)
Solution:
\[ 3m^2 - 2m^2 + 9 = 0 \] \[ m^2 + 9 = 0 \] Writing in standard form:
\[ \therefore m^2 + 0m + 9 = 0 \] Comparing the above equation with \( am^2 + bm + c = 0 \), we get:
\( a = 1 \), \( b = 0 \), \( c = 9 \)

v. \( p(3 + 6p) = -5 \)
Solution:
\[ 3p + 6p^2 = -5 \] \[ \therefore 6p^2 + 3p + 5 = 0 \] Comparing the above equation with \( ap^2 + bp + c = 0 \), we get:
\( a = 6 \), \( b = 3 \), \( c = 5 \)

vi. \( x^2 - 9 = 13 \)
Solution:
\[ x^2 - 9 - 13 = 0 \] \[ x^2 - 22 = 0 \] Writing in standard form:
\[ \therefore x^2 + 0x - 22 = 0 \] Comparing the above equation with \( ax^2 + bx + c = 0 \), we get:
\( a = 1 \), \( b = 0 \), \( c = -22 \)

Question 4. Determine whether the values given against each of the quadratic equations are the roots of the equation.

i. \( x^2 + 4x - 5 = 0 \); \( x = 1, -1 \)

Solution:
The given equation is \( x^2 + 4x - 5 = 0 \) ... (i)
Checking for \( x = 1 \):
Putting \( x = 1 \) in L.H.S. of equation (i), we get:
L.H.S. \( = (1)^2 + 4(1) - 5 = 1 + 4 - 5 = 0 \)
\( \therefore \text{L.H.S.} = \text{R.H.S.} \)
Final Answer: \( x = 1 \) is a root of the given quadratic equation.

Checking for \( x = -1 \):
Putting \( x = -1 \) in L.H.S. of equation (i), we get:
L.H.S. \( = (-1)^2 + 4(-1) - 5 = 1 - 4 - 5 = -8 \)
\( \therefore \text{L.H.S.} \neq \text{R.H.S.} \)
Final Answer: \( x = -1 \) is not a root of the given quadratic equation.

ii. \( 2m^2 - 5m = 0 \); \( m = 2, \frac{5}{2} \)

Solution:
The given equation is \( 2m^2 - 5m = 0 \) ... (i)
Checking for \( m = 2 \):
Putting \( m = 2 \) in L.H.S. of equation (i), we get:
L.H.S. \( = 2(2)^2 - 5(2) = 2(4) - 10 = 8 - 10 = -2 \)
\( \therefore \text{L.H.S.} \neq \text{R.H.S.} \)
Final Answer: \( m = 2 \) is not a root of the given quadratic equation.

Checking for \( m = \frac{5}{2} \):
Putting \( m = \frac{5}{2} \) in L.H.S. of equation (i), we get:
L.H.S. \( = 2\left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) \)
\( = 2\left(\frac{25}{4}\right) - \frac{25}{2} \)
\( = \frac{25}{2} - \frac{25}{2} = 0 \)
\( \therefore \text{L.H.S.} = \text{R.H.S.} \)
Final Answer: \( m = \frac{5}{2} \) is a root of the given quadratic equation.

Question 5. Find k if \( x = 3 \) is a root of equation \( kx^2 - 10x + 3 = 0 \)

Solution:

Given: \( x = 3 \) is the root of the equation \( kx^2 - 10x + 3 = 0 \).
Substitution & Calculation:
Putting \( x = 3 \) in the given equation:
\[ k(3)^2 - 10(3) + 3 = 0 \] \[ 9k - 30 + 3 = 0 \] \[ 9k - 27 = 0 \] \[ 9k = 27 \] \[ k = \frac{27}{9} \] \[ k = 3 \]
Final Answer: The value of \( k \) is 3.

Question 6. One of the roots of equation \( 5m^2 + 2m + k = 0 \) is \( -\frac{7}{5} \). Complete the following activity to find the value of 'k'.

Solution (Completed Activity):

\( -\frac{7}{5} \) is a root of quadratic equation \( 5m^2 + 2m + k = 0 \).
Put \( m = \) \( -\frac{7}{5} \) in the equation.

\[ 5 \times \left( \mathbf{-\frac{7}{5}} \right)^2 + 2 \times \left( \mathbf{-\frac{7}{5}} \right) + k = 0 \] \[ \mathbf{\frac{49}{5}} + \mathbf{\frac{-14}{5}} + k = 0 \] \[ \frac{35}{5} + k = 0 \] \[ 7 + k = 0 \] \[ k = \mathbf{-7} \]

Additional Textbook Activities (Pg. 31 & 33)

Activity 1: Write the polynomials in the index form. Observe the coefficients and fill in the boxes. (Textbook pg. no. 31)

Polynomials: \( x^2 + 3x - 5 \), \( 3x^2 - 5x \), \( 5x^2 \)

Answer (Completed Index Form):

\( x^2 + 3x - 5 \)
\( 3x^2 - 5x + 0 \)
\( 5x^2 + 0x + 0 \)

Coefficients of \( x^2 \) are 1, 3, and 5 respectively, and these coefficients are non-zero.
Coefficients of \( x \) are 3, -5, and 0 respectively.
Constant terms are -5, 0, and 0 respectively.

(Here, constant terms of the second and third polynomial are zero.)

Activity 2: Complete the following table (Textbook pg. no. 31)

Quadratic Equation	General Form (\( ax^2 + bx + c = 0 \))	a	b	c
\( x^2 - 4 = 0 \)	\( x^2 + 0x - 4 = 0 \)	1	0	-4
\( y^2 = 2y - 7 \)	\( y^2 - 2y + 7 = 0 \)	1	-2	7
\( x^2 + 2x = 0 \)	\( x^2 + 2x + 0 = 0 \)	1	2	0

Activity 3: Decide which of the following are quadratic equations? (Textbook pg. no. 31)

Solution (Completed Activity):

i. \( 9y^2 + 5 = 0 \)
In the equation \( 9y^2 + 5 = 0 \), y is the only variable and maximum index of the variable is 2.
\( \therefore \) It is a quadratic equation.

ii. \( m^3 - 5m^2 + 4 = 0 \)
In the equation \( m^3 - 5m^2 + 4 = 0 \), m is the only variable and maximum index of the variable is not 2.
\( \therefore \) It is not a quadratic equation.

iii. \( (l + 2)(l - 5) = 0 \)
\[ l(l - 5) + 2(l - 5) = 0 \] \[ l^2 - 5l + 2l - 10 = 0 \] \[ \therefore l^2 - 3l - 10 = 0 \] In this equation l is the only variable and maximum index of the variable is 2.
\( \therefore \) It is a quadratic equation.

Activity 4: Find the value of k (Textbook pg. no. 33)

If \( x = 5 \) is a root of equation \( kx^2 - 14x - 5 = 0 \), then find the value of k by completing the following activity.

Solution (Completed Activity):

One of the roots of equation \( kx^2 - 14x - 5 = 0 \) is 5.
Put \( x = \) 5 in the given equation.
\[ k(\mathbf{5})^2 - 14(\mathbf{5}) - 5 = 0 \] \[ 25k - 70 - 5 = 0 \] \[ 25k - 75 = 0 \] \[ 25k = \mathbf{75} \] \[ k = \frac{\mathbf{75}}{\mathbf{25}} = \mathbf{3} \]