NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.1) Exercise 2.1

Fractions and Decimals are covered in the Class 7 Maths Chapter 2 Exercise 2.1 Solutions. This very basic but important chapter in Mathematics, presented in Class 7, has four main divisions or subjects in Class 7 Maths Exercise 2.1. The key points discussed in the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, are explained in a detailed manner on the Extramarks website. To properly comprehend the concepts associated with the chapter, Extramarks suggests students carefully read through each topic. This will undoubtedly enable students to fully make use of Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 for understanding the crucial chapter on Fractions and Decimals.

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Topics relating to the introduction of Fractions are covered in NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. Through this assignment, students can recall the study of fractions, including their addition and subtraction, as well as Proper, Improper, and Mixed fractions. The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, is an essential resource for covering the complete Math curriculum because it offers a variety of questions that measure students’ conceptual comprehension.

Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.1

The key concepts of Fractions and Decimals covered in the chapter are reviewed in the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. The fundamentals of fractions and decimals, as well as addition and subtraction of fractions and decimals, can all be learned by students using the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. There are eight questions in the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, four of which are short answers and four of which are longer in nature.

Students can deepen their comprehension of fundamental fractional and decimal themes and subtopics as well as other important concepts like the addition and subtraction of positive and negative fractions and decimals, among other things, with the aid of the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. Class 7 Maths Chapter 2 Exercise 2.1, which can boost their preparation for the examination.

It is crucial for students to reinforce their fundamental knowledge of fractions and decimals with the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. Students will be comfortable working with the addition and subtraction of fractions and decimals owing to the constant practice of representing fractions and decimals on a number line and other concepts based on fractions and decimals arithmetic provided in Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. With the help of NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, they can also evaluate how to represent fractions and decimals as proper and improper fractions.

Important Topics under NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals, Exercise 2.1

Expert teachers who are experienced and capable of writing the answers simply created Extramarks’ NCERT Class 7 Maths Chapter 2 Exercise 2.1 solutions. Students will benefit from this as they learn about the basic ideas behind fractions. The examiners will be impressed by their thorough approach to problem-solving, and they can use the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 to get high scores.

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The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 was created in accordance with the NCERT curriculum. Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 PDF is written in straightforward language, making it easier for students to comprehend the solutions. The experts have not answered all the solutions in one line; instead, they have provided enough explanation. Students will therefore find it simple to comprehend the strategies employed in the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1.

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Access NCERT Solutions for Class 7 Maths Chapter 2 – Fractions and Decimals

Students should refer to the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1

to prepare for their examination. In order to aid students in their test preparation, Extramarks provides a quick explanation of the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1.  A student-focused digital learning platform such as Extramarks provides the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 to students who are registered to its website so that they can understand every concept clearly. Students are allowed to choose the type of education that is best for them. The CBSE curriculum is followed by the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, which is a relatively simple study guide.

Extramarks provides all the necessary study materials for the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 in order to help students fully comprehend the concept. With the aid of the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, students must learn and memorise concepts for the Mathematics examination. For students to fully understand numerous other subjects, Extramarks provides solutions. Students learn more effectively by using the in-depth solutions to the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. The Extramarks website provides assistance for students practising the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1.

One of the key benefits of using the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, for the examination is that students do not need to refer to any other resources. The CBSE curriculum and grading system are followed by the NCERT textbooks and their solutions. The credibility of NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, is an additional advantage. Since the published solutions were created by experts, students need not worry about the accuracy of the data. Extramarks publishes the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. It is a reputable website that offers students trustworthy solutions. Subject-matter specialists simplify the content so that students can comprehend the key ideas in NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1.

Consistency and practice are essential for students to succeed in examinations. Students should refer to the study materials for the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 on the Extramarks website in order to better understand all the concepts. With the aid of Extramarks, students have the opportunity to communicate with the best educators in the nation from the convenience of their homes. Extramarks’ solutions provide a thorough grasp of the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. With the help of the Extramarks website, students may learn at their own pace while keeping track of their advancement and generating in-depth analyses and reports to aid in their understanding.

Specialists help the students understand the concepts they are having trouble with after analysing the report and identifying their weak points. They may easily get ready with the help of the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, which are available on the Extramarks website. Most students find Mathematics to be challenging, but with Extramarks’ expert instruction, they may rapidly master the principles. The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1, are accessible on the Extramarks website. A variety of educational programmes are offered by Extramarks to promote students’ academic growth. For students to have a better learning experience and increase their comprehension, Extramarks provides the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1.

NCERT Solutions for Class 7 Maths Chapter 2 Exercise 2.1

Extramarks specialises in providing a variety of instructional courses to help students effectively learn, practice, and perform. When they combine their practice with Extramarks’ assistance, students will be able to reach excellent accomplishments. Extramarks provides the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 so that students may organise their thoughts and avoid having difficulty with the topics. The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 can be used by students to better understand the concepts behind each question. Students study more quickly and effectively when they utilise the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 to guide their studies.

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The detailed and step-by-step NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 are available at Extramarks to help students better their conceptual comprehension and exam performance. Special care is taken to ensure that all CBSE regulations are observed when releasing the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. For practice questions with a detailed explanation, students can get the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1. This gives students the chance to rehearse and review their solutions in order to better prepare for the examination.

NCERT Solutions for Class 7 Maths

The basis for higher courses is laid around Class 7. Therefore, it is a crucial class in students’ careers. Only studying for classes is insufficient in today’s fast-paced, fiercely competitive world. Students should conduct independent research, learn all the concepts and theories, and be able to apply the theories in real-world situations. Students who use Extramarks’ NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 are better able to learn and achieve their goals. All of the solutions to the NCERT textbook are included on the Extramarks website and were created by Extramarks’ experts. These solutions have been written so that the students can study and comprehend all the concepts more effectively and efficiently. The language is basic and suitable for students.

Students who perform well on their examinations must properly understand the concepts. They may revise topics rapidly if they want to comprehend them clearly. Understanding and acquiring concepts, as well as solving problems based on them, are the foundations of Mathematics. They can have trouble answering the questions if they do not have access to NCERT solutions. As a result, Extramarks provides students with NCERT solutions in a chapter-by-chapter format. Students can cover all the material in this manner and work through the problems to improve their comprehension. The provided solutions are written in straightforward terms that students can quickly comprehend and retain.

Highly skilled and knowledgeable topic specialists create the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 PDF at the Extramarks website. The answers are intended to clear up any confusion students could have when attempting to solve problems from the NCERT Math textbook. They can increase their examination grades with this method. Every topic from every chapter has solutions available to students, written in plain language for everyone to grasp.

Proper Fractions

A proper fraction is one in which the numerator (the higher portion of the fraction) is less than the denominator (the lower portion). The component of a whole is represented by a correct fraction.

For example- A person must take 2/3 of the cake when they divide a cake into 3 pieces and want 2 pieces from each of them. The cake is divided into three sections, and they are taking two of those parts.

Improper Fraction

An improper fraction is one in which the numerator exceeds the denominator. A cake cannot be cut into 3 pieces and then cut into 4 pieces. The result would be – 4/3 in a mathematical sense.

A person needs to get a brand-new 3-slice cake in order to fix the issue of being unable to remove 4 pieces out of 3. They just need to take one slice from the original 3-piece cake because they already have 3 slices from the new entire. Mathematically, it would appear as

1 whole ⅓ or 1⅓.

A mixed fraction is an improper fraction with this altered appearance. 5/2, 11/9, 800/700, and it includes more improper fractions. Examples of mixed fractions are 1⅓, 100⅔, 11¾ etc.

The NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 offer students the confidence they need to perform better on the CBSE examination along with a selection of activities and problem sets.

Addition and subtraction of fractions

Students only need to add the numerators of fractions with the same denominator and can leave the denominator alone.

For instance, 5/2 + 6/2 = 12/2.

If the fractions are, for example, 5/2 and 4/6, they must first get the LCM of the denominators before dividing each one by the LCM. The appropriate numerator must now be multiplied by the quotient of the previous division. For each fraction involved, the procedure must be repeated. The sum of the outcomes of the multiplications would be the resulting numerator, and the resultant LCM would be the resulting denominator.

The same is true of subtraction.

Students can access the NCERT Solutions For Class 7 Maths Chapter 2 Exercise 2.1 are recognised as being immensely beneficial for preparing for the CBSE Class 7 examination.

To Find the Ascending Order or Descending Order

Students must remember this before they proceed: The fraction with the larger numerator is thought to be the greatest when two fractions have the same denominator. And the numerator fraction with the smallest value is regarded as the lowest. Now, if they want to determine the order in ascending –

5/2, ¾, ⅛, 8/4

First, they must equalise the denominator. These fractions can be multiplied with other fractions as long as the denominators are the same in each instance. These fractions can be multiplied by other fractions. As a result, 5/2 multiplied by 4/4 will equal 20/8 (which finally equals 5/2, keeping the number constant).

NCERT Solutions for Class 7

Mathematical chapters cannot be fully grasped by only viewing the PDF files. Students will undoubtedly have questions about any chapter. The Extramarks website offers online lessons where students can ask questions right away in order to erase these doubts from their minds. The learning process can be made enjoyable and interesting by the experts. Students’ understanding of the chapters will improve with the help of this visual learning method, as Extramarks adds graphics and animations to its videos to enhance students’ learning. They can use live classes to get ready for their examinations. In the live classes, students can rediscover their concepts of Mathematics. The instructors in the live classes will help them develop their capacity for logical reasoning in addition to teaching students Mathematics, its numerous formulas, and procedures.

Extramarks aims to assist students in every way possible, including providing answers to each question. According to the curriculum, NCERT textbooks are regarded as the finest for CBSE. Many students find Mathematics to be a frightening subject, which causes them to become anxious and perform poorly on examinations. However, if a student studies carefully and makes use of the right tools, they will undoubtedly perform well.

Extramarks’ NCERT solutions are created in a way that clearly explains the fundamentals of each concept as well as many approaches to answering problem questions. The solutions should be familiar to students in order to give them an idea of the kinds of questions that have appeared in the past years’ papers. Therefore, Extramarks advises students to study a number of sample papers and exams from past years before taking their examinations. Understanding the ideas presented in the NCERT solutions will help students advance in their preparation. Although the chapter’s material might sometimes be complex, students can easily understand it with the help of Extramarks’ NCERT solutions.


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(i)2 3 5 = 103 5 = 7 5 (ii)4+ 7 8 = 32+7 8 = 39 8 = 4 7 8 (iii) 3 5 + 2 7 = 21+10 35 = 31 35 (iv) 9 11 4 15 = 135+44 165 = 179 165 = 1 14 165 (v) 7 10 + 2 5 + 3 2 = 7+4+15 10 = 26 10 = 13 5 =2 3 5 (vi)2 2 3 +3 1 2 = 8 3 + 7 2 = 16+21 6 = 37 6 =6 1 6 (vii)8 1 2 3 5 8 = 17 2 29 8 = 6829 8 = 39 8 =4 7 8 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqaaeaacaGGOaGaamyAaiaacMcacaaMe8UaaGOm aiabgkHiTmaalaaabaGaaG4maaqaaiaaiwdaaaGaeyypa0ZaaSaaae aacaaIXaGaaGimaiabgkHiTiaaiodaaeaacaaI1aaaaiabg2da9maa L4babaWaaSaaaeaacaaI3aaabaGaaGynaaaaaaaabaGaaiikaiaadM gacaWGPbGaaiykaiaaysW7caaI0aGaey4kaSYaaSaaaeaacaaI3aaa baGaaGioaaaacqGH9aqpdaWcaaqaaiaaiodacaaIYaGaey4kaSIaaG 4naaqaaiaaiIdaaaGaeyypa0ZaaSaaaeaacaaIZaGaaGyoaaqaaiaa iIdaaaGaeyypa0ZaauIhaeaacaaI0aWaaSaaaeaacaaI3aaabaGaaG ioaaaaaaaabaGaaiikaiaadMgacaWGPbGaamyAaiaacMcacaaMe8+a aSaaaeaacaaIZaaabaGaaGynaaaacqGHRaWkdaWcaaqaaiaaikdaae aacaaI3aaaaiabg2da9maalaaabaGaaGOmaiaaigdacqGHRaWkcaaI XaGaaGimaaqaaiaaiodacaaI1aaaaiabg2da9maaL4babaWaaSaaae aacaaIZaGaaGymaaqaaiaaiodacaaI1aaaaaaaaeaacaGGOaGaamyA aiaadAhacaGGPaGaaGjbVpaalaaabaGaaGyoaaqaaiaaigdacaaIXa aaaiabgkHiTmaalaaabaGaaGinaaqaaiaaigdacaaI1aaaaiabg2da 9maalaaabaGaaGymaiaaiodacaaI1aGaey4kaSIaaGinaiaaisdaae aacaaIXaGaaGOnaiaaiwdaaaGaeyypa0ZaaSaaaeaacaaIXaGaaG4n aiaaiMdaaeaacaaIXaGaaGOnaiaaiwdaaaGaeyypa0ZaauIhaeaaca aIXaWaaSaaaeaacaaIXaGaaGinaaqaaiaaigdacaaI2aGaaGynaaaa aaaabaGaaiikaiaadAhacaGGPaGaaGjbVpaalaaabaGaaG4naaqaai aaigdacaaIWaaaaiabgUcaRmaalaaabaGaaGOmaaqaaiaaiwdaaaGa ey4kaSYaaSaaaeaacaaIZaaabaGaaGOmaaaacqGH9aqpdaWcaaqaai aaiEdacqGHRaWkcaaI0aGaey4kaSIaaGymaiaaiwdaaeaacaaIXaGa aGimaaaacqGH9aqpdaWcaaqaaiaaikdacaaI2aaabaGaaGymaiaaic daaaaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjb VlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8 UaaGjbVlaaysW7cqGH9aqpdaWcaaqaaiaaigdacaaIZaaabaGaaGyn aaaaaeaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8 UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7 caaMe8UaaGjbVlabg2da9iaaikdadaWcaaqaaiaaiodaaeaacaaI1a aaaaqaaiaacIcacaWG2bGaamyAaiaacMcacaaMe8UaaGOmamaalaaa baGaaGOmaaqaaiaaiodaaaGaey4kaSIaaG4mamaalaaabaGaaGymaa qaaiaaikdaaaGaeyypa0ZaaSaaaeaacaaI4aaabaGaaG4maaaacqGH RaWkdaWcaaqaaiaaiEdaaeaacaaIYaaaaiabg2da9maalaaabaGaaG ymaiaaiAdacqGHRaWkcaaIYaGaaGymaaqaaiaaiAdaaaaabaGaaGjb VlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8 UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7 caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVl abg2da9maalaaabaGaaG4maiaaiEdaaeaacaaI2aaaaaqaaiaaysW7 caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVl aaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Ua aGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7cq GH9aqpcaaI2aWaaSaaaeaacaaIXaaabaGaaGOnaaaaaeaacaGGOaGa amODaiaadMgacaWGPbGaaiykaiaaysW7caaI4aWaaSaaaeaacaaIXa aabaGaaGOmaaaacqGHsislcaaIZaWaaSaaaeaacaaI1aaabaGaaGio aaaacqGH9aqpdaWcaaqaaiaaigdacaaI3aaabaGaaGOmaaaacqGHsi sldaWcaaqaaiaaikdacaaI5aaabaGaaGioaaaaaeaacaaMe8UaaGjb VlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8 UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpdaWcaaqaaiaa iAdacaaI4aGaeyOeI0IaaGOmaiaaiMdaaeaacaaI4aaaaaqaaiaays W7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjb VlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0ZaaSaaaeaaca aIZaGaaGyoaaqaaiaaiIdaaaaabaGaaGjbVlaaysW7caaMe8UaaGjb VlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8 UaaGjbVlaaysW7cqGH9aqpcaaI0aWaaSaaaeaacaaI3aaabaGaaGio aaaaaaaa@AD9D@


Arrange the following in descending order.(i) 29, 23, 821 (ii) 15, 37, 710


First, we find the LCM of 9, 3 and 21. LCM of 9, 3 and 21 = 63 = 2×7 9×7 , 2×21 3×21 , 8×3 21×3 = 14 63 , 42 63 , 24 63 Since, 42 > 24 > 14. Therefore, 42 63 > 24 63 > 14 63 2 3 > 8 21 > 2 9 (ii) 1 5 , 3 7 , 7 10 First, find the LCM of 5, 7 and 10 LCM of denominators 5, 7 and 10 is 70. So, 1 5 , 3 7 , 7 10 1×14 5×14 , 3×10 7×10 , 7×7 10×7 14 70 , 30 70 , 49 70 Since, 49 > 30 > 14. Therefore, 49 70 > 30 70 > 14 70 7 10 > 3 7 > 1 5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiFz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGgbGaaeyAaiaabkhacaqGZbGaaeiD aiaacYcacaqGGaGaae4DaiaabwgacaqGGaGaaeOzaiaabMgacaqGUb GaaeizaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeitaiaaboea caqGnbGaaeiiaiaab+gacaqGMbGaaeiiaiaabMdacaGGSaGaaeiiai aabodacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaabkdacaqGXaGa aiOlaaqaaiaabYeacaqGdbGaaeytaiaabccacaqGVbGaaeOzaiaabc cacaqG5aGaaiilaiaabccacaqGZaGaaeiiaiaabggacaqGUbGaaeiz aiaabccacaqGYaGaaeymaiaabccacqGH9aqpcaqGGaGaaeOnaiaabo daaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8Ua eyypa0ZaaSaaaeaacaaIYaGaey41aqRaaG4naaqaaiaaiMdacqGHxd aTcaaI3aaaaiaacYcacaaMe8+aaSaaaeaacaaIYaGaey41aqRaaGOm aiaaigdaaeaacaaIZaGaey41aqRaaGOmaiaaigdaaaGaaiilaiaays W7daWcaaqaaiaaiIdacqGHxdaTcaaIZaaabaGaaGOmaiaaigdacqGH xdaTcaaIZaaaaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMc8UaaG PaVlaaykW7cqGH9aqpdaWcaaqaaiaaigdacaaI0aaabaGaaGOnaiaa iodaaaGaaiilaiaaysW7daWcaaqaaiaaisdacaaIYaaabaGaaGOnai aaiodaaaGaaiilamaalaaabaGaaGOmaiaaisdaaeaacaaI2aGaaG4m aaaaaeaacaqGtbGaaeyAaiaab6gacaqGJbGaaeyzaiaacYcacaqGGa GaaeinaiaabkdacaqGGaGaeyOpa4JaaeiiaiaabkdacaqG0aGaaeii aiabg6da+iaabccacaqGXaGaaeinaiaac6caaeaacaqGubGaaeiAai aabwgacaqGYbGaaeyzaiaabAgacaqGVbGaaeOCaiaabwgacaGGSaaa baWaaSaaaeaacaaI0aGaaGOmaaqaaiaaiAdacaaIZaaaaiabg6da+i aaysW7daWcaaqaaiaaikdacaaI0aaabaGaaGOnaiaaiodaaaGaeyOp a4ZaaSaaaeaacaaIXaGaaGinaaqaaiaaiAdacaaIZaaaaaqaaiabgk DiEpaaL4babaWaaSaaaeaacaaIYaaabaGaaG4maaaacqGH+aGpdaWc aaqaaiaaiIdaaeaacaaIYaGaaGymaaaacqGH+aGpdaWcaaqaaiaaik daaeaacaaI5aaaaaaaaeaacaGGOaGaamyAaiaadMgacaGGPaGaaGjb VpaalaaabaGaaGymaaqaaiaaiwdaaaGaaiilaiaaysW7daWcaaqaai aaiodaaeaacaaI3aaaaiaacYcacaaMe8+aaSaaaeaacaaI3aaabaGa aGymaiaaicdaaaaabaGaaeOraiaabMgacaqGYbGaae4Caiaabshaca GGSaGaaeiiaiaabAgacaqGPbGaaeOBaiaabsgacaqGGaGaaeiDaiaa bIgacaqGLbGaaeiiaiaabYeacaqGdbGaaeytaiaabccacaqGVbGaae OzaiaabccacaqG1aGaaiilaiaabccacaqG3aGaaeiiaiaabggacaqG UbGaaeizaiaabccacaqGXaGaaGimaaqaaiaabYeacaqGdbGaaeytai aabccacaqGVbGaaeOzaiaabccacaqGKbGaaeyzaiaab6gacaqGVbGa aeyBaiaabMgacaqGUbGaaeyyaiaabshacaqGVbGaaeOCaiaabohaca qGGaGaaeynaiaacYcacaqGGaGaae4naiaabccacaqGHbGaaeOBaiaa bsgacaqGGaGaaeymaiaaicdacaqGGaGaaeyAaiaabohacaqGGaGaae 4naiaaicdacaGGUaaabaGaae4uaiaab+gacaGGSaGaaGjbVdqaamaa laaabaGaaGymaaqaaiaaiwdaaaGaaiilaiaaysW7daWcaaqaaiaaio daaeaacaaI3aaaaiaacYcacaaMe8+aaSaaaeaacaaI3aaabaGaaGym aiaaicdaaaaabaGaeyO0H49aaSaaaeaacaaIXaGaey41aqRaaGymai aaisdaaeaacaaI1aGaey41aqRaaGymaiaaisdaaaGaaiilamaalaaa baGaaG4maiabgEna0kaaigdacaaIWaaabaGaaG4naiabgEna0kaaig dacaaIWaaaaiaacYcadaWcaaqaaiaaiEdacqGHxdaTcaaI3aaabaGa aGymaiaaicdacqGHxdaTcaaI3aaaaaqaaiabgkDiEpaalaaabaGaaG ymaiaaisdaaeaacaaI3aGaaGimaaaacaGGSaWaaSaaaeaacaaIZaGa aGimaaqaaiaaiEdacaaIWaaaaiaacYcadaWcaaqaaiaaisdacaaI5a aabaGaaG4naiaaicdaaaaabaGaae4uaiaabMgacaqGUbGaae4yaiaa bwgacaGGSaGaaeiiaiaabsdacaqG5aGaaeiiaiabg6da+iaabccaca qGZaGaaGimaiaabccacqGH+aGpcaqGGaGaaeymaiaabsdacaGGUaaa baGaaeivaiaabIgacaqGLbGaaeOCaiaabwgacaqGMbGaae4Baiaabk hacaqGLbGaaiilaiaaysW7aeaadaWcaaqaaiaaisdacaaI5aaabaGa aG4naiaaicdaaaGaeyOpa4JaaGjbVpaalaaabaGaaG4maiaaicdaae aacaaI3aGaaGimaaaacqGH+aGpcaaMe8+aaSaaaeaacaaIXaGaaGin aaqaaiaaiEdacaaIWaaaaaqaaiabgkDiEpaaL4babaWaaSaaaeaaca aI3aaabaGaaGymaiaaicdaaaGaeyOpa4ZaaSaaaeaacaaIZaaabaGa aG4naaaacqGH+aGpdaWcaaqaaiaaigdaaeaacaaI1aaaaaaaaaaa@87DC@

Q.3 In a magic square, the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?



Finding the sum of first row: Sumof1strow= 4 11 + 9 11 + 2 11 = 4+9+2 11 = 15 11 Finding the sum of second row: Sum of 2nd row= 3 11 + 5 11 + 7 11 = 15 11 Finding the sum of third row: Sum of 3rd row= 8 11 + 1 11 + 6 11 = 8+1+6 11 = 15 11 Finding the sum of first column: Sum of 1st column = 4 11 + 3 11 + 8 11 = 4+3+8 11 = 15 11 Finding the sum of second column: Sum of 2nd column = 9 11 + 5 11 + 1 11 = 9+5+1 11 = 15 11 Finding the sum of third column: Sum of 3rd column = 2 11 + 7 11 + 6 11 = 2+7+6 11 = 15 11 Now, we must find diagonal sum. Sum of diagonalfromleftbottomtorighttop = 8 11 + 5 11 + 2 11 = 8+5+2 11 = 15 11 Sum of diagonalfromleftbottomtorighttop = 6 11 + 5 11 + 4 11 = 6+5+4 11 = 15 11 Since the sum of the numbers in each row, in each column along the diagonal is the same, which is equal to 15 11 . Therefore, given square is magical. 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A rectangular sheet of paper is 1212cmlong and1023cm wide. Find its perimeter.


We are given: Lengthofrectangularsheetofpaper=12 1 2 cm= 25 2 cm Widthofrectangularsheetofpaper=10 2 3 cm= 32 3 cm Since, Perimeter of a rectangle = 2×( length+width ) Therefore, Perimeter of the given rectangular sheet of paper =2×( 25 2 + 32 3 )cm =2×( 75+64 6 )cm =2× 139 6 cm = 139 3 cm= 46 1 3 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqaaeaacaqGxbGaaeyzaiaabccacaqGHbGaaeOC aiaabwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaGG6a aabaGaaeitaiaabwgacaqGUbGaae4zaiaabshacaqGObGaaGjbVlaa b+gacaqGMbGaaGjbVlaabkhacaqGLbGaae4yaiaabshacaqGHbGaae OBaiaabEgacaqG1bGaaeiBaiaabggacaqGYbGaaGjbVlaabohacaqG ObGaaeyzaiaabwgacaqG0bGaaGjbVlaab+gacaqGMbGaaGjbVlaabc hacaqGHbGaaeiCaiaabwgacaqGYbGaeyypa0JaaGymaiaaikdadaWc aaqaaiaaigdaaeaacaaIYaaaaiaabogacaqGTbGaeyypa0ZaaSaaae aacaaIYaGaaGynaaqaaiaaikdaaaGaae4yaiaab2gaaeaacaqGxbGa aeyAaiaabsgacaqG0bGaaeiAaiaaysW7caqGVbGaaeOzaiaaysW7ca qGYbGaaeyzaiaabogacaqG0bGaaeyyaiaab6gacaqGNbGaaeyDaiaa bYgacaqGHbGaaeOCaiaaysW7caqGZbGaaeiAaiaabwgacaqGLbGaae iDaiaaysW7caqGVbGaaeOzaiaaysW7caqGWbGaaeyyaiaabchacaqG LbGaaeOCaiaab2dacaqGXaGaaeimamaalaaabaGaaGOmaaqaaiaaio daaaGaae4yaiaab2gacqGH9aqpdaWcaaqaaiaaiodacaaIYaaabaGa aG4maaaacaqGJbGaaeyBaaqaaiaabofacaqGPbGaaeOBaiaabogaca qGLbGaaiilaiaaysW7daqjEaqaaiaabcfacaqGLbGaaeOCaiaabMga caqGTbGaaeyzaiaabshacaqGLbGaaeOCaiaabccacaqGVbGaaeOzai aabccacaqGHbGaaeiiaiaabkhacaqGLbGaae4yaiaabshacaqGHbGa aeOBaiaabEgacaqGSbGaaeyzaiaabccacqGH9aqpcaqGGaGaaeOmai abgEna0oaabmaabaGaaeiBaiaabwgacaqGUbGaae4zaiaabshacaqG ObGaey4kaSIaae4DaiaabMgacaqGKbGaaeiDaiaabIgaaiaawIcaca GLPaaaaaaabaGaaeivaiaabIgacaqGLbGaaeOCaiaabwgacaqGMbGa ae4BaiaabkhacaqGLbGaaiilaaqaaiaabccacaqGqbGaaeyzaiaabk hacaqGPbGaaeyBaiaabwgacaqG0bGaaeyzaiaabkhacaqGGaGaae4B aiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabEgacaqGPb GaaeODaiaabwgacaqGUbGaaeiiaiaabkhacaqGLbGaae4yaiaabsha caqGHbGaaeOBaiaabEgacaqG1bGaaeiBaiaabggacaqGYbGaaeiiai aabohacaqGObGaaeyzaiaabwgacaqG0bGaaeiiaiaab+gacaqGMbGa aeiiaiaabchacaqGHbGaaeiCaiaabwgacaqGYbaabaGaeyypa0JaaG OmaiabgEna0oaabmaabaWaaSaaaeaacaaIYaGaaGynaaqaaiaaikda aaGaey4kaSYaaSaaaeaacaaIZaGaaGOmaaqaaiaaiodaaaaacaGLOa GaayzkaaGaae4yaiaab2gaaeaacqGH9aqpcaaIYaGaey41aq7aaeWa aeaadaWcaaqaaiaaiEdacaaI1aGaey4kaSIaaGOnaiaaisdaaeaaca aI2aaaaaGaayjkaiaawMcaaiaabogacaqGTbaabaGaeyypa0JaaGOm aiabgEna0oaalaaabaGaaGymaiaaiodacaaI5aaabaGaaGOnaaaaca qGJbGaaeyBaaqaaiabg2da9maalaaabaGaaGymaiaaiodacaaI5aaa baGaaG4maaaacaqGJbGaaeyBaiabg2da9maaL4babaGaaGinaiaaiA dadaWcaaqaaiaaigdaaeaacaaIZaaaaiaabogacaqGTbaaaaaaaa@3682@


Find the perimeters of (i)ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?


Since, the Perimeter of a triangle = side + side +side Therefore, Perimeter of triangle ABE = AB + BE + AE = 5 2 +2 3 4 +3 3 5 cm = 5 2 + 11 4 + 18 5 cm = 50+55+72 20 cm = 177 20 cm In rectangle BCDE, length BE=2 2 4 cm= 11 4 cm Width ED= 7 6 cm Since, Perimeter of a rectangle = 2×( length+width ) Therefore, perimeter of rectangle BCDE=2×( BE+ED ) =2×( 11 4 + 7 6 )cm =2×( 33+14 12 )cm =2× 47 12 cm = 47 6 cm Hence, the perimeter ofΔABC= 177 20 cm and The perimeter of rectangle BCDE= 47 6 cm Now, comparing both the perimeter to find which is greater. Since, The LCM of denominators of both the perimeter ( 20 and 6 ) is equal to 60. Therefore, 177 20 , 47 6 177×3 20×3 , 47×10 6×10 531 60 , 470 10 Since, 531 is greater than 470. Therefore, 531 60 > 470 60 177 20 > 47 6 Thus, the perimeter of triangle ABE is greater than the perimeter of rectangle BCDE. 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Sali wants to put a picture in a frame. The picture is 735 cmwide. To fit in the frame the picture cannot be more than7310cm wide. How much should the picture be trimmed?


We are given: Thewidthofthepicture=7 3 5 cm Requiredwidthof picture to be fit in frame=7 3 10 cm Therefore, picture to be trim=7 3 5 7 3 10 cm = 38 5 73 10 cm = 3 10 cm Thus picture should be trimmed by 3 10 cm MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqaaeaacaqGxbGaaeyzaiaabccacaqGHbGaaeOC aiaabwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaGG6a aabaGaaeivaiaabIgacaqGLbGaaGjbVlaabEhacaqGPbGaaeizaiaa bshacaqGObGaaGjbVlaab+gacaqGMbGaaGjbVlaabshacaqGObGaae yzaiaaysW7caqGWbGaaeyAaiaabogacaqG0bGaaeyDaiaabkhacaqG LbGaaeypaiaabEdadaWcaaqaaiaabodaaeaacaqG1aaaaiaabogaca qGTbaabaGaaeOuaiaabwgacaqGXbGaaeyDaiaabMgacaqGYbGaaeyz aiaabsgacaaMe8Uaae4DaiaabMgacaqGKbGaaeiDaiaabIgacaaMe8 Uaae4BaiaabAgacaqGGaGaaeiCaiaabMgacaqGJbGaaeiDaiaabwha caqGYbGaaeyzaiaabccacaqG0bGaae4BaiaabccacaqGIbGaaeyzai aabccacaqGMbGaaeyAaiaabshacaqGGaGaaeyAaiaab6gacaqGGaGa aeOzaiaabkhacaqGHbGaaeyBaiaabwgacaqG9aGaae4namaalaaaba GaaG4maaqaaiaaigdacaaIWaaaaiaabogacaqGTbaabaGaaeivaiaa bIgacaqGLbGaaeOCaiaabwgacaqGMbGaae4BaiaabkhacaqGLbGaai ilaiaabccacaqGWbGaaeyAaiaabogacaqG0bGaaeyDaiaabkhacaqG LbGaaeiiaiaabshacaqGVbGaaeiiaiaabkgacaqGLbGaaeiiaiaabs hacaqGYbGaaeyAaiaab2gacqGH9aqpcaaI3aWaaSaaaeaacaaIZaaa baGaaGynaaaacqGHsislcaaI3aWaaSaaaeaacaaIZaaabaGaaGymai aaicdaaaGaae4yaiaab2gaaeaacqGH9aqpdaWcaaqaaiaaiodacaaI 4aaabaGaaGynaaaacqGHsisldaWcaaqaaiaaiEdacaaIZaaabaGaaG ymaiaaicdaaaGaae4yaiaab2gaaeaacqGH9aqpdaWcaaqaaiaaioda aeaacaaIXaGaaGimaaaacaqGJbGaaeyBaaqaaiaabsfacaqGObGaae yDaiaabohacaqGGaGaaeiCaiaabMgacaqGJbGaaeiDaiaabwhacaqG YbGaaeyzaiaabccacaqGZbGaaeiAaiaab+gacaqG1bGaaeiBaiaabs gacaqGGaGaaeOyaiaabwgacaqGGaGaaeiDaiaabkhacaqGPbGaaeyB aiaab2gacaqGLbGaaeizaiaabccacaqGIbGaaeyEaiaaysW7daqjEa qaamaalaaabaGaaG4maaqaaiaaigdacaaIWaaaaiaabogacaqGTbaa aaaaaa@E802@


Ritu ate35part of an apple and the remaining apple waseaten by her brother Somu. How much part of the appledid Somu eat? Who had the larger share? By how much?


Since, Ritu ate 35 part of an apple and her brother eaten therest part of the apple.So, the part of apple left after eaten by Ritu=135=535=25Therefore, Somu ate 25 parts of apple.Now comparing the part eaten by them 35, ​25Since, 3>2.Therefore35>25That means Ritu ate larger part.The difference in both parts=3525                                 =325                                 =15Thus,Somu ate 25 part of apple.Ritu ate the larger part of appleRitu ate part more apple than her brother Somu.


Michael finished colouring a picture in712 hour. Vaibhav finished colouring the same picture in34hour.Who worked longer? By what fraction was it longer?


We are given:Michael worked for 712hourVaibhavworkedfor34hourTherefore,In order to find the longer hour of work we have to compare both the fractions 712,34The LCM of 12 and 4=12So, 712, 34 =712,912Therefore,912>712That means Vaibhav worked for longer hour. Difference in their working hour  =912712                                       =9712                                       =212                                       =16Thus, Vaibhav worked for 16 hour more than Michael.

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1. Which chapters in Mathematics for Class 7 are crucial in terms of exams?

Every chapter is significant for achieving higher examination scores. However, students may choose which chapters are more significant, and they should concentrate on those according to the PDF. The chapters on Fractions and Decimals, Data Handling, Simple Equations, and Lines And Angles have the highest weightage and are therefore more significant, according to the marks distribution. Geometry, Mensuration, Quantity Comparison, and Integers are also crucial.

2. How do students get good grades in Mathematics in Class 7?

The subject that necessitates frequent practice is Mathematics. No matter how skilled students are, if they do not personally tackle the difficulties, the questions get hard and time-consuming. Every day, students must schedule at least an hour to spend answering the questions. They should complete all of the NCERT exercises and verify the solutions with Extramarks’ NCERT solutions.

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