CBSE Class 8 Maths Chapter 2 Linear Equation in One Variable Exercise 2.5
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Class 8 is considered a critical year in which students learn and explore a variety of new academic themes. Students discover their interests and hobbies. The academic curriculum of Class 8 gives students insight into new subjects. The academic curriculum of Class 8 is very important and also necessary to get good results in semester exams. Students follow different strategies for examination preparation in order to perform well on their annual exams. In addition to these strategies, it is also necessary to prepare a suitable schedule. If students adhere to the timetable, they can perform well in exams.
Students should be familiar with the respective syllabus before starting to prepare for the exam. They should solve past years’ papers and sample papers regularly to help students do better in exams. Students should have a clear understanding of the concepts and topics being taught in a particular class. Their strategy for examination preparation should comply with the syllabus specified with regard to each prescribed textbook. They should study concepts thoroughly and be familiar with the appropriate model for applying them while writing answers in the examination. For the latest information on CBSE/ICSE/State Board exams and competitive exams, they should keep learning and stay up to date with the help of Extramarks. They can also download the Extramarks Learning App to watch interactive learning videos.
CBSE Class 8 Maths NCERT Solutions Linear Equation in One Variable
The academic curriculum of Class 8 Mathematics is very important to get good grades; therefore, students need to focus and learn concepts properly. Class 8 Mathematics helps build a good conceptual foundation necessary for the learning of Mathematics in senior classes. The Class 9 Mathematics syllabus has 16 chapters, and each chapter should be studied thoroughly. Students have to practice every day and solve as many problems as possible. Students are advised to solve the practice problems and practice the examples given in the NCERT textbook to better understand the concepts. Students are recommended to solve various sample papers, and they could refer to the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, for additional help. Students should read and retain information about various concepts and learn how to apply them in order to solve problems. Another important thing is that students need to study the chapters and related topics carefully in order to attain proficiency in the subject. Students are expected to understand the topic and concepts in their entirety. They can use NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, to gain credible answers to the key questions in the Class 8 Mathematics textbook.
A PDF download option for NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, which are created by expert teachers in accordance with the format of the latest edition of the CBSE (NCERT) book, is available on the Extramarks website. Students can get excellent scores in the examination by practising important questions from the Class 8 Mathematics textbooks with the assistance of NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5. Students could secure more points on their exams by registering on the Extramarks educational portal for accessing NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5. Extramarks is a platform that provides students with reliable NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, along with a multitude of other learning materials. Students of Mathematics looking for trustworthy solutions can download the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 to efficiently revise the entire syllabus and score more points on the exam. Along with practising with the aid of NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, here are some key tips for students that could be helpful in improving their Mathematics grades.
 Students are advised to create a memory card for formulas. This allows students to use formulas precisely, so they are easy to apply.
 Doubts tend to get in the way of subsequent concepts, so students should dispel them as quickly as possible with the assistance of their teachers as well as with the aid of resources like NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5.
 In addition to practising the required questions from the exercises provided in the prescribed NCERT book, students should not forget to refer to CBSE sample papers and past years’ papers.
Solving the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, is very important for doing well on the exam. Many small but important issues and concepts are mentioned in the NCERT book exercises and revision notes. Problems based on these concepts have been succinctly solved in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, which comprises detailed solutions for Exercise 2.5 of Chapter 2 of the prescribed textbook of Mathematics for Class 8. Students can download the PDF versions of these solutions and incorporate NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, into their learning process.
Access NCERT solutions for Maths Chapter 2 – Linear Equations in One Variable Exercise 2.5
A questionwise analysis of the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 has been done as follows:
With regard to the chapter on linear equations in one variable, there are four primary types of questions that could be a part of the examination. The first type of question is called a Very Short Answer type question. Each question has one mark allotted to it. The problems in this section look easy but can be very challenging. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, would equip students with an adequate understanding of these types of questions so that they can adequately answer them in the examination. The different types of very shortanswer questions that are asked are:
 Fill in the blanks – Here students have to complete the sentence so that it makes sense. In most cases, the question statements are part of different theories of Mathematics that must be completed.
Longanswer type questions may also be considered important components of the question paper of Mathematics. Every question of this type has 5 points. So it is hoped that the answer will be more detailed. The answers here are meant to provide very detailed, stepbystep solutions to given problems. These questions have been systematically solved as a part of the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5.
Why is the NCERT Class 8 Maths Chapter 2 on Linear Equation Important?
Students begin learning the basic concepts of advanced Mathematics as a part of the syllabus of Mathematics of Class 8. The chapter on Linear Equations in One Variable is very important for understanding almost all of the mathematics curricula in senior classes. Linear Equations are the basis for solving almost all Mathematics problems. From Algebra and Analysis to Geometry and Arithmetic, knowledge of Linear Equations plays an important role. Students can study NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, on the Extramarks educational portal. These solutions are also available as a PDF. Knowing how to solve Linear Equations in One Variable will help students solve more complex problems in the future. The chapter on Linear Equations in One Variable in Class 8 is very interesting, but the nature of this academic theme of Mathematics requires consistent practice. Gaining proficiency in Mathematics requires constant effort and diligence. Extramarks’ NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 makes learning Mathematics an engaging and rewarding experience.
These solutions include completed exercises, descriptive solutions along with comprehensive explanations which can equip students with all the quintessential knowledge pertaining to Exercise 2.5 of Linear Equations in One Variable. Students can download NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 as a PDF file to make preparation for the Mathematics examination convenient.
What Will You Learn in Cbse Class 8 Maths Chapter 2 Linear Equation 2.5?
Chapter 2 of Linear Equations in One Variable shows how to find unknown variables using one or more real numbers. In this chapter, students will learn how to solve some complex problems consisting of Linear Equations in One Variable. If students study this chapter with reference to Extramarks’s NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, and notes, they can easily comprehend other chapters of the Mathematics textbook for Class 8. Students are likely to find it easy to solve the problem and determine the value of the unknown variable. It is highly recommended that students download the PDF versions of NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, from the Extramarks learning platform.
NCERT Solutions for Class 8
The chapter on Linear Equations in One Variable in the prescribed textbook of Mathematics for Class 8 contains all the important questions, and the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, is centered around Exercise 2.5 of this chapter. Before the exam, students should be familiar with all the concepts, including the scheme of mark distribution and the pattern of the question paper.
When it comes to catering to students’ academic requirements, Extramarks’ learning assets could prove to be indispensable. These learning assets would simplify the most difficult concepts for better understanding. These are made for their convenience. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, are written in straightforward language and are easy to understand. Students would have to peruse the solutions on a routine basis in order to understand how to solve the problem intelligently. Extramarks understands students’ needs and will provide them with the services they require. PDF versions of NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5, are available on Extramarks. Students can conveniently download these and study for the exam.
Q.1
$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{linear}\mathrm{equations}.\\ 1.\frac{\mathrm{x}}{2}\frac{1}{5}=\frac{\mathrm{x}}{3}+\frac{1}{4}\text{\hspace{0.17em}}\\ 2.\frac{\mathrm{n}}{2}\frac{3\mathrm{n}}{4}+\frac{5\mathrm{n}}{6}=21\\ 3.\mathrm{x}+7\frac{8\mathrm{x}}{3}=\frac{17}{6}\frac{5\mathrm{x}}{2}\text{\hspace{0.17em}}\end{array}$
$\begin{array}{l}4.\frac{\mathrm{x}5}{3}=\frac{\mathrm{x}3}{5}\text{\hspace{0.17em}}\\ 5.\frac{3\mathrm{t}2}{4}\frac{2\mathrm{t}+3}{3}=\frac{2}{3}\mathrm{t}\text{\hspace{0.17em}}\\ 6.\mathrm{m}\frac{\mathrm{m}1}{2}=1\frac{\mathrm{m}2}{3}\\ \mathrm{Simplify}\mathrm{and}\mathrm{solve}\mathrm{the}\mathrm{following}\mathrm{linear}\mathrm{equations}.\\ 7.3(\mathrm{t}3)=5(2\mathrm{t}+1)\\ 8.15(\mathrm{y}4)2(\mathrm{y}9)+5(\mathrm{y}+6)=0\\ 9.3(5\mathrm{z}7)2(9\mathrm{z}11)=4(8\mathrm{z}13)17\\ 10.0.25(4\mathrm{f}3)=0.05(10\mathrm{f}9)\end{array}$
Ans
$\begin{array}{l}1.\frac{\mathrm{x}}{2}\frac{1}{5}=\frac{\mathrm{x}}{3}+\frac{1}{4}\text{\hspace{0.17em}}\\ \text{Taking L.C.M on both the sides ,we get}\\ \Rightarrow \frac{5\mathrm{x}2}{10}=\frac{4\mathrm{x}+3}{12}\\ \text{Multiplying both the sides by10,we get}\\ \Rightarrow \frac{5\mathrm{x}2}{10}\times 10=\frac{4\mathrm{x}+3}{12}\times 10\\ \Rightarrow (5\mathrm{x}2)=\frac{4\mathrm{x}+3}{12}\times 10\end{array}$
$\begin{array}{l}\text{Multiplying both the sides by12,we get}\\ \Rightarrow (5\mathrm{x}2)\times 12=\frac{4\mathrm{x}+3}{12}\times 10\times 12\\ \Rightarrow 60\mathrm{x}24=40\mathrm{x}+30\\ \text{Transposing 40x to L.H.S and 24 to R.H.S,we get}\\ \Rightarrow 60\mathrm{x}40\mathrm{x}=30+24\\ \Rightarrow 20\mathrm{x}=54\\ \text{Dividing both sides by 20 ,we get}\\ \Rightarrow \frac{20\mathrm{x}}{20}=\frac{54}{20}\\ \Rightarrow \mathrm{x}=2.7\end{array}$ \begin{array}{l}\mathrm{}\end{array}
$\begin{array}{l}2.\frac{\mathrm{n}}{2}\frac{3\mathrm{n}}{4}+\frac{5\mathrm{n}}{6}=21\\ \text{Taking L.C.M of 2,4 and 6,we get}\\ \Rightarrow \frac{6\mathrm{n}9\mathrm{n}+10\mathrm{n}}{12}=21\\ \Rightarrow \frac{7\mathrm{n}}{12}=21\end{array}$ \begin{array}{l}\text{}\end{array}
$\begin{array}{l}\text{Multiplying bith sides by 12,}\text{we get}\\ \Rightarrow \frac{7\mathrm{n}}{12}\times 12=21\times 12\\ \Rightarrow 7\mathrm{n}=21\times 12\\ \Rightarrow 7\mathrm{n}=252\\ \text{Dividing both sides by 7,}\text{we get}\\ \Rightarrow \frac{7\mathrm{n}}{7}=\frac{252}{7}\\ \Rightarrow \mathrm{n}=\frac{252}{7}\\ \Rightarrow \mathrm{n}=36\end{array}$
$\begin{array}{l}3.\mathrm{x}+7\frac{8\mathrm{x}}{3}=\frac{17}{6}\frac{5\mathrm{x}}{2}\text{\hspace{0.17em}}\\ \text{Transposing}\frac{5\mathrm{x}}{2}\text{\hspace{0.17em}to L.H.S and 7 to R.H.S,}\text{we get}\\ \Rightarrow \mathrm{x}\frac{8\mathrm{x}}{3}+\frac{5\mathrm{x}}{2}\text{\hspace{0.17em}}=\frac{17}{6}7\\ \text{Taking LCM both the sides, we get}\end{array}$
$\begin{array}{l}\Rightarrow \frac{6\mathrm{x}16\mathrm{x}+15\mathrm{x}}{6}=\frac{1742}{6}\\ \Rightarrow \frac{5\mathrm{x}}{6}=\frac{25}{6}\\ \text{Multiplying both the sides by 6}\\ \Rightarrow \frac{5\mathrm{x}}{6}\times 6=\frac{25}{6}\times 6\\ \Rightarrow 5\mathrm{x}=25\\ \text{Dividing both the sides by 5}\\ \Rightarrow \frac{5\mathrm{x}}{5}=\frac{25}{5}\\ \Rightarrow \mathrm{x}=5\end{array}$ \begin{array}{l}\mathrm{}\end{array}
$\begin{array}{l}4.\frac{\mathrm{x}5}{3}=\frac{\mathrm{x}3}{5}\text{\hspace{0.17em}}\\ \text{Multiplying both the sides by 3,we get}\\ \Rightarrow \frac{\mathrm{x}5}{3}\times 3=\frac{\mathrm{x}3}{5}\times 3\\ \Rightarrow \mathrm{x}5=\frac{\mathrm{x}3}{5}\times 3\end{array}$
$\begin{array}{l}\text{Multiplying both the sides by 5,we get\hspace{0.17em}}\\ \Rightarrow (\mathrm{x}5)\times 5=\frac{\mathrm{x}3}{5}\times 3\times 5\\ \Rightarrow 5(\mathrm{x}5)=3(\mathrm{x}3)\\ \Rightarrow 5\mathrm{x}25=3\mathrm{x}9\\ \Rightarrow 5\mathrm{x}3\mathrm{x}=9+25\\ \Rightarrow 2\mathrm{x}=16\\ \Rightarrow \mathrm{x}=8\end{array}$
$\begin{array}{l}5.\frac{3\mathrm{t}2}{4}\frac{2\mathrm{t}+3}{3}=\frac{2}{3}\mathrm{t}\text{\hspace{0.17em}}\\ \Rightarrow \frac{3(3\mathrm{t}2)4(2\mathrm{t}+3)}{12}=\frac{23\mathrm{t}}{3}\\ \Rightarrow \frac{9\mathrm{t}68\mathrm{t}12}{12}=\frac{23\mathrm{t}}{3}\\ \Rightarrow \frac{\mathrm{t}18}{12}=\frac{23\mathrm{t}}{3}\\ \text{Multiplying both the sides by 12,we get}\\ \Rightarrow \frac{\mathrm{t}18}{12}\times 12=\frac{23\mathrm{t}}{3}\times 12\\ \Rightarrow \mathrm{t}18=(23\mathrm{t})4\\ \Rightarrow \mathrm{t}18=812\mathrm{t}\end{array}$ \begin{array}{l}\text{}\end{array}
$\begin{array}{l}\text{Transposing 12t to L.H.S and 18 to R.H.S ,we get}\\ \Rightarrow \mathrm{t}+12\mathrm{t}=8+18\\ \Rightarrow 13\mathrm{t}=26\\ \Rightarrow \mathrm{t}=\frac{26}{13}=2\end{array}$
$\begin{array}{l}6.\mathrm{m}\frac{\mathrm{m}1}{2}=1\frac{\mathrm{m}2}{3}\\ \Rightarrow \frac{2\mathrm{m}\mathrm{m}+1}{2}=\frac{3\mathrm{m}+2}{3}\\ \Rightarrow \frac{\mathrm{m}+1}{2}=\frac{5\mathrm{m}}{3}\\ \text{Multiplying both the sides by 2,we get}\\ \frac{\mathrm{m}+1}{2}\times 2=\frac{5\mathrm{m}}{3}\times 2\\ \Rightarrow \mathrm{m}+1=\frac{5\mathrm{m}}{3}\times 2\end{array}$
$\begin{array}{l}\text{Multiplying both the sides by 3,we get \u21d2}\\ (\mathrm{m}+1)\times 3=\frac{5\mathrm{m}}{3}\times 2\times 3\\ \Rightarrow 3\mathrm{m}+3=102\mathrm{m}\\ \Rightarrow 3\mathrm{m}+2\mathrm{m}=103\\ \Rightarrow 5\mathrm{m}=7\\ \Rightarrow \mathrm{m}=\frac{7}{5}\end{array}$ \begin{array}{l}\mathrm{}\end{array}
$\begin{array}{l}7.3(\mathrm{t}3)=5(2\mathrm{t}+1)\\ \Rightarrow 3\mathrm{t}9=10\mathrm{t}+5\end{array}$
$\begin{array}{l}\mathrm{Transposing}\text{}10\mathrm{t}\text{}\mathrm{to}\text{}\mathrm{L}.\mathrm{H}.\mathrm{S}\text{}\mathrm{and}\text{}9\text{}\mathrm{to}\text{}\mathrm{R}.\mathrm{H}.\mathrm{S},\mathrm{we}\text{}\mathrm{get}\\ \Rightarrow 3\mathrm{t}10\mathrm{t}=5+9\\ \Rightarrow 7\mathrm{t}=14\\ \Rightarrow \mathrm{t}=2\end{array}$
$\begin{array}{l}\begin{array}{l}8.15(\mathrm{y}4)2(\mathrm{y}9)+5(\mathrm{y}+6)=0\\ \Rightarrow 15\mathrm{y}602\mathrm{y}+18+5\mathrm{y}+30=0\end{array}\\ \Rightarrow 15\mathrm{y}2\mathrm{y}+5\mathrm{y}60+18+30=0\\ \Rightarrow 18\mathrm{y}12=0\\ \Rightarrow 18\mathrm{y}=12\\ \Rightarrow \mathrm{y}=\frac{12}{18}\\ \Rightarrow \mathrm{y}=\frac{2}{3}\end{array}$ \begin{array}{l}\mathrm{}\end{array}
$\begin{array}{l}9.3(5\mathrm{z}7)2(9\mathrm{z}11)=4(8\mathrm{z}13)17\\ \Rightarrow 15\mathrm{z}2118\mathrm{z}+22=32\mathrm{z}5217\\ \Rightarrow 3\mathrm{z}+1=32\mathrm{z}69\\ \text{Transposing 32z to L.H.S and 1 to R.H.S,we get}\\ \Rightarrow 3\mathrm{z}32\mathrm{z}=691\\ \Rightarrow 35\mathrm{z}=70\\ \Rightarrow \mathrm{z}=\frac{70}{35}=2\\ \Rightarrow \mathrm{z}=2\end{array}$
$\begin{array}{l}10.0.25(4\mathrm{f}3)=0.05(10\mathrm{f}9)\\ \Rightarrow 1\mathrm{f}0.75=0.5\mathrm{f}0.45\\ \text{Transposing}0.5\mathrm{f}\text{to L.H.S and}0.75\text{to R.H.S,we get}\\ \Rightarrow 1\mathrm{f}0.5\mathrm{f}=0.750.45\\ \Rightarrow 0.5\mathrm{f}=0.3\\ \Rightarrow \mathrm{f}=\frac{0.3}{0.5}=0.6\\ \Rightarrow \mathrm{f}=0.6\end{array}$
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FAQs (Frequently Asked Questions)
1. What are the most complex kinds of questions encapsulated in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5?
With regard to the academic discipline of Mathematics, students should pay adequate attention to all kinds of problems. However, questions 5 and 6 of Exercise 2.5 of Chapter 2 of Class 8 Mathematics could be perceived as a little challenging for students. Extramarks offers NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 so students can practice difficult problems and revise them for the exam. Students can procure the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 to understand all the concepts. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 are created by highly qualified instructors with decades of experience and are available for download.
2. Are the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 reliable reference materials for Exercise 2.5 of Class 8 Mathematics?
The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 are the best learning resources for students to ace the CBSE curriculum of Mathematics. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.5 cover all the questions of Exercise 2.5 of Chapter 2 of Class 8 Mathematics so that students can understand the concepts well. The solutions to Class 8 Maths Chapter 2 Exercise 2.5 on Extramarks are developed by experienced teachers to help students become familiar with all the concepts. Students should check Class 8 Maths Chapter 2 Exercise 2.5 Solutions on the Extramarks learning portal. Students can download these solutions and save them to their computers. Students can always refer to these solutions while preparing for exams.
3. What is Class 8 Maths Chapter 2.5?
Chapter 2 of Class 8 Mathematics focuses on Linear Equations in One Variable. NCERT Solutions from Extramarks will help students prepare for their exams. Students can visit the Extramarks educational website to access solutions created by subject matter specialists.
4. Why should students refer to Extramarks's NCERT Class 8 Maths Chapter 2 Exercise 2.5?
Rational Numbers, Square Roots, Cube Roots, Data Manipulation and Factorisation are some of the subjects included in the CBSE Class 8 Mathematics syllabus. On Extramarks, students can find chapterbychapter solutions, key questions and revision notes for all chapters covered in the Class 8 Mathematics syllabus. These key questions have been prepared by subject matter experts to give students a questionbyquestion analysis of each chapter. This will give students a clear understanding of all the concepts introduced in CBSE Class 8 Mathematics. Students will also learn appropriate problemsolving techniques for each academic theme.