NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (EX 4.3) Exercise 4.3
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Mathematics has always been considered one of the most difficult subjects to learn and understand. It is also one of the most important subjects for students of all grades. It plays a very important role in improving a student’s overall score. The curriculum is so extensive that many students find it difficult to follow the steps and concepts of mathematics. It causes confusion among students. The challenge for students is also to identify the correct approach and method to apply in order to respond to each question asked. Selfstudying this subject can be very timeconsuming. Students need a deeper understanding of concepts and formulas. To address their concerns, Extramarks offers curative and diligently prepared solutions. Each chapter in the NCERT textbook is carefully divided into different segments, each equipped with comprehensive exercises. Each exercise allows students to take a holistic approach while solving its questions and achieving the highest marks.
Each step in every solution is clearly explained, giving students a clear understanding of the concepts they are dealing with. Through intensive study of the NCERT solutions, students gain insight into each concept and method, enabling them to solve any question posed on any topic of the subject matter. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 introduce concepts such as Integers, Fraction Multiplication and Division, Geometry, Algebra, Rational Numbers, Simple Equations, and more. These solutions lay the foundation for higher Mathematics. Therefore, students should focus carefully on individual solutions given in NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3. Extramarks facilitates comprehension by framing details in plain language.The NCERT Solutions for Class 7 Maths, Chapter 4 Exercise 4.3, not only helped students with their school annual exams, but they can also be used to excel in various competitive exams.The Class 7 Mathematics NCERT solutions provide students with an easy way to comprehend Mathematics.
Students can find NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 in PDF format. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 pertain to Exercise 4.3 of Chapter 4 of Class 7 Mathematics. Extramarks encourages students to download and practise the NCERT Solutions For Class 7 Mathematics Chapter 4 Exercise 4.3. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 PDF versions consist of 15 short chapters. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 are prepared by efficient mentors of Extramarks, so the chances of inconsistencies are negligible.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (EX 4.3) Exercise 4.3
Students can download PDF versions of NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 and solutions to all chapters’ exercises in one place, created by experienced teachers according to NCERT guidelines. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 include questions with answers to help students revise the entire syllabus and gain more points. Students can register on Extramarks to receive access to all the solutions. Each NCERT solution is provided to make learning with the assistance of Extramarks easy and interesting. Subjects such as Science, Mathematics, and English are easier to learn with access to Extramarks.
Access NCERT Solutions for Class 7 Maths Chapter 4 – Simple Equation
Choosing NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 could be considered the best choice for students who want to prepare for the exam. Extramarks provides NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 in PDF format. Students can download these solutions at their convenience or access them directly online via the Extramarks website or mobile application. Extramarks’ subject matter experts have engineered solutions to various kinds of questions in line with all of the CBSE guidelines. A student in Class 7 who knowsall the concepts in the Mathematics textbook and has practised all the problems in the exercises in the book will easily get the highest score in the final exam. With the help of the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3, students can easily understand the types of questions that may appear during the examination, related to this chapter and familiarise themselves with the topic weightage pertaining to the chapter. In addition to Exercise 4.3, to which the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 pertain, this chapter also has many exercises with different types of questions. As mentioned, the solutions to all of these questions have been produced by the subject specialists of Mathematics of Extramarks. Hence, all of these solutions are of the highest quality and can be read by students while preparing for the exam.
Understanding all the concepts in the textbook and solving the adjacent exercises is very important to getting good marks in exams. Students must download the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 from the Extramarks website and get prepared for the examination. The most remarkable feature of the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 is that they are accessible both online and offline.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3
The Mathematics textbook for Class 7 includes chapters on Integrals, Fractions and Decimals, Simple Equations, Data Handling, Lines and Angles, Triangles and their Properties, Comparing Quantities, Congruence in Triangles, etc. Understanding the concepts embedded in these chapters requires regular practice. In order to master the problemsolving skills necessary for this purpose, the use of NCERT solutions to make learning easier, faster, and better is highly recommended. Students looking for credible answers to questions in NCERT books are suggested to access the Extramarks website. These NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 are created after extensive research that sets Extramarks apart. Extramarks NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 are also important reference materials for students with regard to the completion of their homework assignments. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 offer a straightforward and innovative approach to building a strong conceptual foundation for a variety of academic themes taught to students in Class 7. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 facilitate an engaging unhindered learning experience and can also be used for revision just before exams. The easily downloadable format of NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 ensures that stepbystep explanations of all the issues and concepts covered in the Class 7 textbooks required by the CBSE board are provided.
Q.1
$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}:\\ \left(\mathrm{a}\right)2\mathrm{y}+\frac{5}{2}=\frac{37}{2}\left(\mathrm{b}\right)5\mathrm{t}+28=10\left(\mathrm{c}\right)\frac{\mathrm{a}}{5}+3=2\\ \left(\mathrm{d}\right)\frac{\mathrm{q}}{4}+7=5\left(\mathrm{e}\right)\frac{5}{2}\mathrm{x}=10\left(\mathrm{f}\right)\frac{5}{2}\mathrm{x}=\frac{25}{4}\\ \left(\mathrm{g}\right)7\mathrm{m}+\frac{19}{2}=13\left(\mathrm{h}\right)6\mathrm{z}+10=2\left(\mathrm{i}\right)\frac{3\mathrm{l}}{2}=\frac{2}{3}\\ \left(\mathrm{j}\right)\frac{2\mathrm{b}}{3}5=3\end{array}$
Ans.
\begin{array}{l}\left(\text{a}\right)\text{2}y+\frac{5}{2}=\frac{37}{2}\\ 2y=\frac{37}{2}\frac{5}{2}=\frac{375}{2}=\frac{32}{2}=16\\ 2y=16\\ y=\frac{16}{2}=8\\ Thus,\text{\hspace{0.17em}}\overline{)y=8}\\ \left(\text{b}\right)\text{5t}+\text{28}=\text{1}0\\ 5t=1028\\ 5t=18\\ \overline{)t=\frac{18}{5}}\\ \left(\text{c}\right)\frac{\text{a}}{5}+3=2\\ \frac{a}{5}=23\\ \frac{a}{5}=1\\ \overline{)a=5}\\ \left(\text{d}\right)\frac{q}{4}+7=5\\ \frac{q}{4}=57\\ \frac{q}{4}=2\\ \overline{)q=8}\\ \left(\text{e}\right)\frac{5}{2}x=10\\ 5x=10\times 2\\ x=\frac{10\times 2}{5}=\frac{\overline{)5}\times 2\times 2}{\overline{)5}}=4\\ Thus,\text{\hspace{0.17em}}\overline{)x=4}\\ \left(\text{f}\right)\frac{5}{2}x=\frac{25}{4}\\ 5x=\frac{25\times 2}{4}\\ x=\frac{\overline{)5}\times 5\times \overline{)2}}{\overline{)2}\times 2\times \overline{)5}}\\ Thus,\text{\hspace{0.17em}}\overline{)x=\frac{5}{2}}\\ \left(\text{g}\right)7m+\frac{19}{2}=13\\ 7m=13\frac{19}{2}=\frac{2619}{2}=\frac{7}{2}\\ 7m=\frac{7}{2}\\ m=\frac{\overline{)7}}{2\times \overline{)7}}\\ Thus,\text{\hspace{0.17em}}\overline{)m=\frac{1}{2}}\\ \left(\text{h}\right)\text{6z}+\text{1}0=\text{2}\\ 6z=210\\ 6z=12\\ z=\frac{12}{6}=\frac{2\times \overline{)6}}{\overline{)6}}0\\ Thus,\text{\hspace{0.17em}}\overline{)z=2}\\ \left(\text{i}\right)\frac{3l}{2}=\frac{2}{3}\\ 3l=\frac{4}{3}\\ \overline{)l=\frac{4}{9}}\\ \left(\text{j}\right)\frac{2b}{3}5=3\\ \frac{2b}{3}=8\\ 2b=24\\ b=\frac{24}{2}=\frac{\overline{)2}\times 12}{\overline{)2}}\\ Thus,\text{\hspace{0.17em}}\overline{)b=12}\end{array}
Q.2
$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}:\\ \left(\mathrm{a}\right)2(\mathrm{x}+4)=12\left(\mathrm{b}\right)3(\mathrm{n}5)=21\left(\mathrm{c}\right)3(\mathrm{n}5)=21\\ \left(\mathrm{d}\right)32(2\mathrm{y})=7\left(\mathrm{e}\right)4(2\mathrm{x})=9\left(\mathrm{f}\right)4(2\mathrm{x})=9\\ \left(\mathrm{g}\right)4+5(\mathrm{p}1)=34\left(\mathrm{h}\right)345(\mathrm{p}1)=4\end{array}$
Ans.
\begin{array}{l}\text{(a})\text{}2(x+\text{}4)=12\\ \text{Divide both sides by 2 to get}\\ x+4=6\\ \text{Thus,}\text{\hspace{0.17em}}\overline{)x=2}\\ \left(\text{b}\right)\text{}3(n5)=21\\ \text{Divide both sides by 3 to get}\\ n5=\text{7}\\ \text{Thus,}\overline{)\text{}n=2}\\ \left(\text{c}\right)\text{}3(n5)=\text{21}\\ \text{Divide both sides by 2 to get}\\ n5=7\\ Thus,\text{\hspace{0.17em}}\overline{)n=12}\\ \left(\text{d}\right)\text{3}\text{2}(\text{2}y)=\text{7}\\ 2\left(2y\right)=4\\ \text{Divide both sides by}\text{2 to get}\\ 2y=2\\ \text{Multiply both sides by}\text{1 to get}\\ y2=2\\ \text{Thus,}\overline{)y=4}\\ \left(\text{e}\right)\text{}4(2x)=\text{9}\\ \text{Divide both sides by}\text{4 to get}\\ \text{2}x=\frac{9}{4}\\ \text{Multiply both sides by}\text{1 to get}\\ x2=\frac{9}{4}\\ x=\frac{9}{4}+2=\frac{9+8}{4}\\ Thus,\text{\hspace{0.17em}}\overline{)x=\frac{17}{4}}\\ \left(\text{f}\right)\text{}4(2x)=\text{9}\\ \text{Divide both sides by 4 to get}\\ \text{2}x=\frac{9}{4}\\ \text{Multiply both sides by 1 to get}\\ \text{x}2=\frac{9}{4}\\ x=\frac{9}{4}+2=\frac{9+8}{4}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)x=\frac{1}{4}}\\ \left(\text{g}\right)\text{}4+5\text{}(p1)=34\\ 5\left(p1\right)=30\\ \text{Divide both sides by 5 to get}\\ \text{p}1=6\\ p=6+1=7\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=7}\\ \left(\text{h}\right)\text{}345(p1)=4\\ 5\left(p1\right)=30\\ \text{Divide both sides by}\text{5 to get}\\ p1=6\\ p=6+1=7\\ \text{Thus,}\text{\hspace{0.17em}}\overline{)p=7}\end{array}
Q.3
$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}.\\ \left(\mathrm{a}\right)4=5(\mathrm{p}2)\left(\mathrm{b}\right)4=5(\mathrm{p}2)\left(\mathrm{c}\right)16=5(2\mathrm{p})\\ \left(\mathrm{d}\right)10=4+3(\mathrm{t}+2)\left(\mathrm{e}\right)28=4+3(\mathrm{t}+5)\\ \left(\mathrm{f}\right)0=16+4(\mathrm{m}6)\end{array}$
Ans.
\begin{array}{l}\left(\text{a}\right)\text{4}=\text{5}(p\text{2})\\ \text{Divide both sides by 5 to get}\\ \frac{4}{5}=p2\\ p=\frac{4}{5}+2=\frac{4+10}{5}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=\frac{14}{5}}\\ \left(\text{b}\right)\text{4}=\text{5}(p2)\\ \text{Divide both sides by 5 to get}\\ \frac{4}{5}=p2\\ p=\frac{4}{5}+2=\frac{4+10}{5}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=\frac{6}{5}}\\ \left(\text{c}\right)16=5\text{}(2p)\\ \text{Divide both sides by}\text{5 to get}\\ \frac{16}{5}=2p\\ \text{Multiply both sides by}\text{1 to get}\\ \frac{\text{16}}{5}=p2\\ p=\frac{16}{5}+2=\frac{16+10}{5}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=\frac{6}{5}}\\ \left(\text{d}\right)\text{1}0=\text{4}+\text{3}(t+2)\\ 6=3\left(t+2\right)\\ \text{Divide both sides by 3 to get}\\ 2=t+2\\ \text{Thus,}\overline{)t=0}\\ \left(\text{e}\right)\text{28}=\text{4}+\text{3}(t+\text{5})\\ 24=3\left(t+5\right)\\ \text{Divide both sides by 3 to get}\\ 8=t+5\\ \text{Thus,}\overline{)t=3}\\ \left(\text{f}\right)0=\text{16}+\text{4}(m\text{6})\\ 16=4\left(m6\right)\\ \text{Divide both sides by 4 to get}\\ 4=m6\\ \text{Thus,}\overline{)m=2}\end{array}
Q.4
$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=2\\ \left(\mathrm{b}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=2\end{array}$
Ans.
\begin{array}{l}\text{(a) We have}\\ x=2\\ \text{Multiply both sides by 4 to get}\\ \overline{)4x=8}\dots \dots \dots \dots \left(\text{i}\right)\\ \text{Subtract 2 from both sides to get}\\ \overline{)4x2=6}\dots \dots \left(\text{ii}\right)\\ \text{Divide both sides by 2 to get}\\ \overline{)2x1=3}\dots \dots \text{(iii)}\\ \text{(b)}\text{\hspace{0.17em}}\text{We have}\\ x=2\\ \text{Adding 4 to both sides to get}\\ \overline{)x+4=2}\dots \dots \dots \left(\text{i}\right)\\ \text{Multiply 2 from both sides to get}\\ \overline{)2x+8=4}\dots \dots \left(\text{ii}\right)\\ \text{Divide both sides by 2 to get}\\ \overline{)x+4=2}\dots \dots \left(\text{iii}\right)\end{array}
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FAQs (Frequently Asked Questions)
1. How many questions are there in Chapter 4 to which solutions to Class 7 Maths Exercise 4.3 pertain?
NCERT solutions for Chapter 4 of Class 7 Mathematics contain approximately 18 questions divided into 4 exercises. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3 provide detailed answers to each question so students can understand the concepts step by step. These questions require a lot of attention and practice as each step is important for the exam. Students are better able to understand the chapter if they spend a lot of time solving each problem along with the example.
2. Should students practice all the problems provided in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.3?
Yes, all problems provided by the Class 7 Maths Chapter 4 Exercise 4.3 Solutions should be solved. Simple Equation is a chapter that requires a lot of practice to fully understand the topic and learn how to apply them to solve a problem. Students can use it as a reference guide for preparing for Chapter 4. This process can be made easy with the help of the Class 7 Mathematics Chapter 4 NCERT solutions.
3. What does a simple equation explained in Class 7 Maths Chapter 4 Exercise 4.3 mean?
The representation of the relationship between two expressions on both sides of the equal to sign is known as a simple equation. These equations can have 2 or more than 2 variables. Students can get indepth knowledge about simple equations in the NCERT solutions for Chapter 4 of Class 7 Mathematics. The NCERT solutions provide students with accurate methods to solve each question in a manner that helps students comprehend them at the same time.
4. How do students write an equation according to NCERT Class 7 Maths Chapter 4 Exercise 4.3?
Students can learn how to write equations and all of the simple equations in Chapter 4 through NCERT solutions. Learning the theory and applying all the fundamentals gets easier with the aid and assistance of NCERT solutions. These solutions are also valuable aid for selfassessment and for reference purposes. With the right preparation, students can be ready for the exam.