Open App

Class 6 Exam > Class 6 Notes > Mathematics Class 6 (Maharashtra Board) > Textbook Solutions: Equations

Textbook Solutions: Equations | Mathematics Class 6 (Maharashtra Board) PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


Chapter 10: Equations 
 
 
PRACTICE SET 26 [PAGE 51] 
Practice Set 26 | Q 1 | Page 51 
Different mathematical operations are given in the two rows below. Find out the number 
you get in each operation and make equations. 
16 ÷ 2; 5 × 2; 9 + 4; 72 ÷ 3; 4 + 5; 8 × 3; 19 - 10; 10 - 2; 37 - 27; 6 + 7 
 
SOLUTION 
16 ÷ 2 = 8 
 
5 × 2 = 10 
 
9 + 4 = 13 
 
72 ÷ 3 = 24 
 
4 + 5 = 9 
 
8 × 3 = 24 
 
19 - 10 = 9 
 
10 - 2 = 8 
 
37 - 27 = 10 
 
6 + 7 = 13 
 
So, 
16 ÷ 2 = 10 - 2 
 
5 × 2 = 37 - 27 
 
9 + 4 = 6 + 7 
 
72 ÷ 3 = 8 × 3 
 
4 + 5 = 19 - 10 
PRACTICE SET 27 [PAGE 55] 
 
Practice Set 27 | Q 1.1 | Page 55 
Rewrite the following using a letter. 
Page 2


Chapter 10: Equations 
 
 
PRACTICE SET 26 [PAGE 51] 
Practice Set 26 | Q 1 | Page 51 
Different mathematical operations are given in the two rows below. Find out the number 
you get in each operation and make equations. 
16 ÷ 2; 5 × 2; 9 + 4; 72 ÷ 3; 4 + 5; 8 × 3; 19 - 10; 10 - 2; 37 - 27; 6 + 7 
 
SOLUTION 
16 ÷ 2 = 8 
 
5 × 2 = 10 
 
9 + 4 = 13 
 
72 ÷ 3 = 24 
 
4 + 5 = 9 
 
8 × 3 = 24 
 
19 - 10 = 9 
 
10 - 2 = 8 
 
37 - 27 = 10 
 
6 + 7 = 13 
 
So, 
16 ÷ 2 = 10 - 2 
 
5 × 2 = 37 - 27 
 
9 + 4 = 6 + 7 
 
72 ÷ 3 = 8 × 3 
 
4 + 5 = 19 - 10 
PRACTICE SET 27 [PAGE 55] 
 
Practice Set 27 | Q 1.1 | Page 55 
Rewrite the following using a letter. 
The sum of a certain number and 3. 
 
SOLUTION 
Let a certain number be x.  
 
? Sum of a certain number and 3 = x + 3 
Practice Set 27 | Q 1.2 | Page 55 
Rewrite the following using a letter. 
The difference obtained by subtracting 11 from another number. 
 
SOLUTION 
Let another number be x. 
? Difference obtained by subtracting 11 from another number = x – 11 
Practice Set 27 | Q 1.3 | Page 55 
Rewrite the following using a letter. 
The product of 15 and another number. 
 
SOLUTION 
Let another number be x.  
? Product of 15 and another number = 15 × x = 15x  
Practice Set 27 | Q 1.4 | Page 55 
Rewrite the following using a letter. 
Four times a number is 24. 
 
SOLUTION 
Let the number be x. 
 
Four time a number = 24 
 
? 4 × x = 24 
 
? 4x = 24  
Practice Set 27 | Q 2.1 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
x + 9 = 11 
 
SOLUTION 
Subtract 9 from both sides 
Page 3


Chapter 10: Equations 
 
 
PRACTICE SET 26 [PAGE 51] 
Practice Set 26 | Q 1 | Page 51 
Different mathematical operations are given in the two rows below. Find out the number 
you get in each operation and make equations. 
16 ÷ 2; 5 × 2; 9 + 4; 72 ÷ 3; 4 + 5; 8 × 3; 19 - 10; 10 - 2; 37 - 27; 6 + 7 
 
SOLUTION 
16 ÷ 2 = 8 
 
5 × 2 = 10 
 
9 + 4 = 13 
 
72 ÷ 3 = 24 
 
4 + 5 = 9 
 
8 × 3 = 24 
 
19 - 10 = 9 
 
10 - 2 = 8 
 
37 - 27 = 10 
 
6 + 7 = 13 
 
So, 
16 ÷ 2 = 10 - 2 
 
5 × 2 = 37 - 27 
 
9 + 4 = 6 + 7 
 
72 ÷ 3 = 8 × 3 
 
4 + 5 = 19 - 10 
PRACTICE SET 27 [PAGE 55] 
 
Practice Set 27 | Q 1.1 | Page 55 
Rewrite the following using a letter. 
The sum of a certain number and 3. 
 
SOLUTION 
Let a certain number be x.  
 
? Sum of a certain number and 3 = x + 3 
Practice Set 27 | Q 1.2 | Page 55 
Rewrite the following using a letter. 
The difference obtained by subtracting 11 from another number. 
 
SOLUTION 
Let another number be x. 
? Difference obtained by subtracting 11 from another number = x – 11 
Practice Set 27 | Q 1.3 | Page 55 
Rewrite the following using a letter. 
The product of 15 and another number. 
 
SOLUTION 
Let another number be x.  
? Product of 15 and another number = 15 × x = 15x  
Practice Set 27 | Q 1.4 | Page 55 
Rewrite the following using a letter. 
Four times a number is 24. 
 
SOLUTION 
Let the number be x. 
 
Four time a number = 24 
 
? 4 × x = 24 
 
? 4x = 24  
Practice Set 27 | Q 2.1 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
x + 9 = 11 
 
SOLUTION 
Subtract 9 from both sides 
x + 9 = 11 
? x + 9 - 9 = 11 - 9 (Subtract 9 from both sides) 
? x + 0 = 2 
? x = 2  
Practice Set 27 | Q 2.2 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
x - 4 = 9 
 
SOLUTION 
Add 4 to both sides 
x - 4 = 9  
? x - 4 + 4 = 9 + 4 (Add 4 to both sides) 
? x + 0 = 13 
? x = 13  
Practice Set 27 | Q 2.3 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
8x = 24 
 
SOLUTION 
 
Practice Set 27 | Q 2.4 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
 
Page 4


Chapter 10: Equations 
 
 
PRACTICE SET 26 [PAGE 51] 
Practice Set 26 | Q 1 | Page 51 
Different mathematical operations are given in the two rows below. Find out the number 
you get in each operation and make equations. 
16 ÷ 2; 5 × 2; 9 + 4; 72 ÷ 3; 4 + 5; 8 × 3; 19 - 10; 10 - 2; 37 - 27; 6 + 7 
 
SOLUTION 
16 ÷ 2 = 8 
 
5 × 2 = 10 
 
9 + 4 = 13 
 
72 ÷ 3 = 24 
 
4 + 5 = 9 
 
8 × 3 = 24 
 
19 - 10 = 9 
 
10 - 2 = 8 
 
37 - 27 = 10 
 
6 + 7 = 13 
 
So, 
16 ÷ 2 = 10 - 2 
 
5 × 2 = 37 - 27 
 
9 + 4 = 6 + 7 
 
72 ÷ 3 = 8 × 3 
 
4 + 5 = 19 - 10 
PRACTICE SET 27 [PAGE 55] 
 
Practice Set 27 | Q 1.1 | Page 55 
Rewrite the following using a letter. 
The sum of a certain number and 3. 
 
SOLUTION 
Let a certain number be x.  
 
? Sum of a certain number and 3 = x + 3 
Practice Set 27 | Q 1.2 | Page 55 
Rewrite the following using a letter. 
The difference obtained by subtracting 11 from another number. 
 
SOLUTION 
Let another number be x. 
? Difference obtained by subtracting 11 from another number = x – 11 
Practice Set 27 | Q 1.3 | Page 55 
Rewrite the following using a letter. 
The product of 15 and another number. 
 
SOLUTION 
Let another number be x.  
? Product of 15 and another number = 15 × x = 15x  
Practice Set 27 | Q 1.4 | Page 55 
Rewrite the following using a letter. 
Four times a number is 24. 
 
SOLUTION 
Let the number be x. 
 
Four time a number = 24 
 
? 4 × x = 24 
 
? 4x = 24  
Practice Set 27 | Q 2.1 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
x + 9 = 11 
 
SOLUTION 
Subtract 9 from both sides 
x + 9 = 11 
? x + 9 - 9 = 11 - 9 (Subtract 9 from both sides) 
? x + 0 = 2 
? x = 2  
Practice Set 27 | Q 2.2 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
x - 4 = 9 
 
SOLUTION 
Add 4 to both sides 
x - 4 = 9  
? x - 4 + 4 = 9 + 4 (Add 4 to both sides) 
? x + 0 = 13 
? x = 13  
Practice Set 27 | Q 2.3 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
8x = 24 
 
SOLUTION 
 
Practice Set 27 | Q 2.4 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
 
SOLUTION 
 
Practice Set 27 | Q 3 | Page 55 
Given below are some equations and the values of the variables. Are these values the 
solutions to those equations? 
 
No Equation Value of the 
variable 
Solution 
(Yes/No) 
1 y - 3 = 11 y = 3 No 
2 17 = n + 7 n = 10   
3 30 = 5x x = 6   
4 m/2 = 14 m = 7   
 
SOLUTION 
 
No Equation  Value of the variable  
Solution(Yes / No) 
1 y - 3 = 11 y = 3 No 
2 17 = n + 7 n = 10 Yes 
3 30 = 5x x = 6 Yes 
4 m
2
 =14m
2
 = 14 m = 7   No 
Explanation: 
(1) When y = 3, 
LHS = y - 3 = 3 - 3 = 0 
RHS = 11 
Since LHS ? RHS, so y = 3 is not a solution of equation 
y - 3 = 11.  
(2) When n = 10, 
Page 5


Chapter 10: Equations 
 
 
PRACTICE SET 26 [PAGE 51] 
Practice Set 26 | Q 1 | Page 51 
Different mathematical operations are given in the two rows below. Find out the number 
you get in each operation and make equations. 
16 ÷ 2; 5 × 2; 9 + 4; 72 ÷ 3; 4 + 5; 8 × 3; 19 - 10; 10 - 2; 37 - 27; 6 + 7 
 
SOLUTION 
16 ÷ 2 = 8 
 
5 × 2 = 10 
 
9 + 4 = 13 
 
72 ÷ 3 = 24 
 
4 + 5 = 9 
 
8 × 3 = 24 
 
19 - 10 = 9 
 
10 - 2 = 8 
 
37 - 27 = 10 
 
6 + 7 = 13 
 
So, 
16 ÷ 2 = 10 - 2 
 
5 × 2 = 37 - 27 
 
9 + 4 = 6 + 7 
 
72 ÷ 3 = 8 × 3 
 
4 + 5 = 19 - 10 
PRACTICE SET 27 [PAGE 55] 
 
Practice Set 27 | Q 1.1 | Page 55 
Rewrite the following using a letter. 
The sum of a certain number and 3. 
 
SOLUTION 
Let a certain number be x.  
 
? Sum of a certain number and 3 = x + 3 
Practice Set 27 | Q 1.2 | Page 55 
Rewrite the following using a letter. 
The difference obtained by subtracting 11 from another number. 
 
SOLUTION 
Let another number be x. 
? Difference obtained by subtracting 11 from another number = x – 11 
Practice Set 27 | Q 1.3 | Page 55 
Rewrite the following using a letter. 
The product of 15 and another number. 
 
SOLUTION 
Let another number be x.  
? Product of 15 and another number = 15 × x = 15x  
Practice Set 27 | Q 1.4 | Page 55 
Rewrite the following using a letter. 
Four times a number is 24. 
 
SOLUTION 
Let the number be x. 
 
Four time a number = 24 
 
? 4 × x = 24 
 
? 4x = 24  
Practice Set 27 | Q 2.1 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
x + 9 = 11 
 
SOLUTION 
Subtract 9 from both sides 
x + 9 = 11 
? x + 9 - 9 = 11 - 9 (Subtract 9 from both sides) 
? x + 0 = 2 
? x = 2  
Practice Set 27 | Q 2.2 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
x - 4 = 9 
 
SOLUTION 
Add 4 to both sides 
x - 4 = 9  
? x - 4 + 4 = 9 + 4 (Add 4 to both sides) 
? x + 0 = 13 
? x = 13  
Practice Set 27 | Q 2.3 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
8x = 24 
 
SOLUTION 
 
Practice Set 27 | Q 2.4 | Page 55 
Find out which operation must be done on both sides of these equation in order to solve 
them. 
 
SOLUTION 
 
Practice Set 27 | Q 3 | Page 55 
Given below are some equations and the values of the variables. Are these values the 
solutions to those equations? 
 
No Equation Value of the 
variable 
Solution 
(Yes/No) 
1 y - 3 = 11 y = 3 No 
2 17 = n + 7 n = 10   
3 30 = 5x x = 6   
4 m/2 = 14 m = 7   
 
SOLUTION 
 
No Equation  Value of the variable  
Solution(Yes / No) 
1 y - 3 = 11 y = 3 No 
2 17 = n + 7 n = 10 Yes 
3 30 = 5x x = 6 Yes 
4 m
2
 =14m
2
 = 14 m = 7   No 
Explanation: 
(1) When y = 3, 
LHS = y - 3 = 3 - 3 = 0 
RHS = 11 
Since LHS ? RHS, so y = 3 is not a solution of equation 
y - 3 = 11.  
(2) When n = 10, 
RHS = n + 7 = 10 + 7 = 17 
LHS = 17 
Since LHS = RHS, so n = 10 is a solution of equation 
17 = n + 7.  
(3)  When x = 6,  
RHS = 5x = 5 × 6 = 30 
LHS = 30 
Since LHS = RHS, so x = 6 is a solution of equation 
30 = 5x. 
(4) When m = 7, 
 
Since LHS ? RHS, so m = 7 is not a solution of equation m/2 = 14 
Practice Set 27 | Q 4.1 | Page 55 
Solve the following equation: y - 5 = 1 
 
SOLUTION 
y - 5 = 1 
 
? y - 5 + 5 = 1 + 5 (Add 5 to both sides) 
 
? y + 0 = 6 
 
? y = 6 
 
Thus, the solution of the given equation is y = 6.  
Practice Set 27 | Q 4.2 | Page 55 
Solve the following equation: 8 = t + 5 
 
SOLUTION 
8 = t + 5 
? 8 - 5 = t + 5 - 5 (Subtract 5 from both sides) 
? 3 = t + 0

	Mathematics Class 6 (Maharashtra Board) 30 videos\|112 docs\|15 tests

Mathematics Class 6 (Maharashtra Board)

30 videos|112 docs|15 tests

Join Course for Free

FAQs on Textbook Solutions: Equations - Mathematics Class 6 (Maharashtra Board)

1. What are equations, and why are they important in mathematics?

Ans. Equations are mathematical statements that assert the equality of two expressions, typically involving variables and constants. They are important because they help us understand relationships between different quantities and solve problems in various fields, including science, engineering, and finance.

2. How can I solve simple equations in Class 6?

Ans. To solve simple equations, you can follow these steps: first, isolate the variable by performing the same operation on both sides of the equation. For example, if you have x + 5 = 12, subtract 5 from both sides to find x = 7. Always check your solution by substituting it back into the original equation.

3. What are some common types of equations that students learn in Class 6?

Ans. In Class 6, students typically learn about linear equations, which involve variables raised to the first power. Common types include one-step equations (like x + 3 = 7) and two-step equations (like 2x - 4 = 10). Understanding these basics is crucial for tackling more complex equations later on.

4. How do I check if my solution to an equation is correct?

Ans. To check if your solution is correct, substitute the value of the variable back into the original equation. If both sides of the equation are equal after substitution, then your solution is correct. This process helps verify your work and ensures accuracy.

5. Can you explain the difference between equations and expressions?

Ans. Yes, equations and expressions are different. An expression is a combination of numbers, variables, and operations without an equality sign (e.g., 3x + 5). An equation, on the other hand, includes an equality sign, indicating that two expressions are equal (e.g., 3x + 5 = 11). Understanding this difference is essential for solving mathematical problems.

About this Document

Sep 23, 2025 Last updated

Related Exams

Class 6

Document Description: Textbook Solutions: Equations for Class 6 2025 is part of Mathematics Class 6 (Maharashtra Board) preparation. The notes and questions for Textbook Solutions: Equations have been prepared according to the Class 6 exam syllabus. Information about Textbook Solutions: Equations covers topics like and Textbook Solutions: Equations Example, for Class 6 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Textbook Solutions: Equations.

Introduction of Textbook Solutions: Equations in English is available as part of our Mathematics Class 6 (Maharashtra Board) for Class 6 & Textbook Solutions: Equations in Hindi for Mathematics Class 6 (Maharashtra Board) course. Download more important topics related with notes, lectures and mock test series for Class 6 Exam by signing up for free. Class 6: Textbook Solutions: Equations | Mathematics Class 6 (Maharashtra Board)

Description

Full syllabus notes, lecture & questions for Textbook Solutions: Equations | Mathematics Class 6 (Maharashtra Board) - Class 6 | Plus excerises question with solution to help you revise complete syllabus for Mathematics Class 6 (Maharashtra Board) | Best notes, free PDF download

Information about Textbook Solutions: Equations

In this doc you can find the meaning of Textbook Solutions: Equations defined & explained in the simplest way possible. Besides explaining types of Textbook Solutions: Equations theory, EduRev gives you an ample number of questions to practice Textbook Solutions: Equations tests, examples and also practice Class 6 tests

	Mathematics Class 6 (Maharashtra Board) 30 videos\|112 docs\|15 tests

Mathematics Class 6 (Maharashtra Board)

30 videos|112 docs|15 tests

Join Course for Free

Download as PDF

Explore Courses for Class 6 exam

Free

past year papers

Objective type Questions

MCQs

Exam

Textbook Solutions: Equations | Mathematics Class 6 (Maharashtra Board)

video lectures

study material

ppt

Textbook Solutions: Equations | Mathematics Class 6 (Maharashtra Board)

Semester Notes

Textbook Solutions: Equations | Mathematics Class 6 (Maharashtra Board)

pdf

shortcuts and tricks

practice quizzes

Important questions

Sample Paper

Extra Questions

Previous Year Questions with Solutions

Summary

Viva Questions

mock tests for examination

;

Additional Information about Textbook Solutions: Equations for Class 6 Preparation

Textbook Solutions: Equations Free PDF Download

The Textbook Solutions: Equations is an invaluable resource that delves deep into the core of the Class 6 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the Textbook Solutions: Equations now and kickstart your journey towards success in the Class 6 exam.

Importance of Textbook Solutions: Equations

The importance of Textbook Solutions: Equations cannot be overstated, especially for Class 6 aspirants. This document holds the key to success in the Class 6 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

Textbook Solutions: Equations Notes

Textbook Solutions: Equations Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to Textbook Solutions: Equations. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, Textbook Solutions: Equations Notes on EduRev are your ultimate resource for success.

Textbook Solutions: Equations Class 6 Questions

The "Textbook Solutions: Equations Class 6 Questions" guide is a valuable resource for all aspiring students preparing for the Class 6 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study Textbook Solutions: Equations on the App

Students of Class 6 can study Textbook Solutions: Equations alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the Textbook Solutions: Equations, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of Textbook Solutions: Equations is prepared as per the latest Class 6 syllabus.

Education Revolution