NCERT Solutions Class 7 Mathematics Chapter 2

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals

Fractions and decimals are the focus of NCERT Class 7 Mathematics Chapter 2. We all understand the difference between a fraction and a decimal. We even use fractions and decimals in our daily lives, whether we realise it or not. However , we might wonder  how  it helps the students to learn these things? 

All the concepts like fixed fractions, proper fractions, and improper fractions with their addition and subtraction, equivalent fractions, fraction comparisons, fraction ordering, and fraction representation on a number line are studied under fractions. In previous classes, we learned fractions and decimals, as well as how to add and subtract them. We will be learning how to solve questions related to multiplication and division of fractions and decimals in NCERT Class 7 Chapter 2 and going through our NCERT Solutions by Extramarks will assist you to learn these topics in a systematic and organised manner to excel in academics.

NCERT Solutions for Class 7 Mathematics Chapter 2 Extramarks have detailed answers to every question in the NCERT textbook. They will aid in scoring better marks in exams. These solutions are prepared by subject-matter experts, thus you can count on them as they are reliable and genuine resources.

NCERT Solutions for Class 7 Mathematics Chapter 2 Fractions and Decimals 

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

NCERT Solutions for Class 7 Chapter 2 Mathematics 

Students can refer to the NCERT Class 7 Mathematics Chapter 2 solutions here at Extramarks. All the questions are solved in such a way so that  students can  understand them well. 

2.1 Introduction 

Students have already learned about fraction and its types in previous  classes. They will learn about multiplication and division of fractions in Class 7 Chapter 2. The concept of fractions focuses on the ratios and proportions, their distribution etc. Students will learn all these Mathematical concepts  in this chapter.

2.2 Recollect

How well have you learnt about fractions? While solving questions of NCERT Class 7 Mathematics Chapter 2, students will recall what they learned in their previous classes. 

2.3 Multiplication of Fractions

Students will learn how to multiply two fractions in this section. For example, if they have values like x and y, they can say xy is the product of x and y. If the values are a/b, how will we multiply? In this scenario, there are two methods - by using a whole number and by using a fraction. .

2.3.1 Multiplication of Fractions Using the Whole Number

This section explains the parts of a fraction. The lower part of a fraction is a whole number (except zero) and the upper part is an integer. The lower part of a fraction is called the denominator, while the upper part is called the numerator. If the fractions are the same, we can multiply them with a whole number. Let's say we have a/b as an example. Then we can multiply by 3×a/b since 3 is a whole number. A fraction can also be multiplied with improper or mixed fractions. However, before multiplying, students must reduce them to their simplest forms. 

2.3.1 Multiplication of Fractions Using the Fraction

This section deals with how to multiply two fractions that are dissimilar to each other. To multiply them, students need to use this formula:

(Product of the numerators) / (Product of the denominators)

However, it must be noted that when two proper fractions are multiplied, the resultant product is less than the two fractions. When two improper fractions are multiplied, the result is greater than the two fractions. 

2.4 Division of Fractions

It is possible to divide a whole number by a fraction and a fraction by a whole number.

Similarly, when mixed fractions are divided with a whole number, mixed fractions should be changed to improper fractions. It makes division easy. Students also learn how to divide a fraction with another fraction by changing one of the fractions to its reciprocal form.

2.5 Recall of Decimals

In this section, with the help of a quick recap, students will be forced to recall their decimal knowledge. Height, distance, weight, measuring values, interest rates, shares, and fractions can all be expressed with decimals. We can multiply with 10,100, etc. to change the place value of a point. 

2.6 Multiplication of Decimals

The purpose of this section is for students to practise  multiplication of decimals. Even though multiplication is simple, students may have doubts. To do so, count the number of values after a decimal point in both numbers and keep the point before that number of places in the result. Another variation of decimal multiplication is changing the decimal point's place value by multiplying it by 10 multiples.  

2.7 Division of Decimals

  • Students will learn how to divide decimals and its variations. 
  • If a whole number is divided with 10 multiples, we derive decimals. 
  • If decimals are divided, we get whole numbers
  • Students also learn the division of decimals with a whole number. In this case, the place value of the decimal point will not change in the result as well. 

Key Features of NCERT Solutions for Class 7 Mathematics Chapter 2

Fractions and decimals, as well as their concepts, are covered in NCERT Class 7 Mathematics Chapter 2 to help students improve their fundamental knowledge and analytical skills. Some benefits of NCERT Solutions for Class 7 Mathematics Chapter 2 provided by Extramarks are:

  • It incorporates answers to textbook questions for assisting students in solving textbook questions easily.
  • The solutions are prepared by subject-matter experts and experienced faculty, so they are error-free and reliable.
  • They provide exam-style answers so that students can learn how to solve each question properly in the exam and practise religiously to come out with flying colours. 

NCERT Solutions for Class 7 Mathematics

The NCERT Solutions Class 7 Mathematics Chapter 2 has been carefully curated by subject experts to ensure that the entire Mathematics syllabus is solved and has enough practice exercises. Students can easily plan their exam preparation and revision with the help of these competent resources to achieve the best-desired exam results. 

You will also find exercise solutions for Fractions and Decimals Class 7 here, which will allow you to search for answers to all questions which are not so easy and may require time and effort to solve. 

Mathematics can be a  rewarding subject if you know how to solve the given problems. Using the NCERT Class 7 Solutions, you can perform confidently in your exams. The Mathematics textbook for Class 7 has 15 chapters, each with problems and questions at the end of the chapter. The chapter-by-chapter NCERT Solutions for Class 7 Mathematics provided here can be used to answer questions from the NCERT textbook. 

Chapter 1 - Integers

Chapter 2 - Fractions and Decimals

Chapter 3 - Data Handling

Chapter 4 - Simple Equations

Chapter 5 - Lines and Angles

Chapter 6 - The Triangle and Its Properties

Chapter 7 - Congruence of Triangles

Chapter 8 - Comparing Quantities

Chapter 9 - Rational Numbers

Chapter 10 - Practical Geometry

Chapter 11 - Perimeter and Area

Chapter 12 - Algebraic Expressions

Chapter 13 - Exponents and Powers

Chapter 14 - Symmetry

Chapter 15 - Visualising Solid Shapes

NCERT Solutions for Class 7

Do you need guidance with your Class 7 homework or textbook questions? You've come to the right place! NCERT Solutions Class 7 will not only assist you with your homework but will also provide accurate and reliable answers to all questions in the NCERT textbook. 

Science, Mathematics, Social Science, Hindi, and English are the five major subjects in NCERT Class 7. NCERT Class 7 Solutions covers answers to all subject related questions of the NCERT textbook.

Q.1

Solve: (i) 2- 3 5 (ii) 4+ 7 8 (iii) 3 5 + 2 7 (iv) 9 11 4 15 (v) 7 10 + 2 5 + 3 2 (vi) 2 2 3 +3 1 2 (vii) 8 1 2 3 5 8 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGtbGaae4BaiaabYgacaqG2bGaaeyz aiaacQdaaeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeOmaiaab2cada WcaaqaaiaaiodaaeaacaaI1aaaaiaabccacaqGGaGaaeiiaiaabcca caqGGaGaaeiiaiaabccacaqGOaGaaeyAaiaabMgacaqGPaGaaeiiai aabsdacaqGRaWaaSaaaeaacaaI3aaabaGaaGioaaaacaqGGaGaaGjb VlaaysW7caqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGOa GaaeyAaiaabMgacaqGPbGaaeykaiaabccadaWcaaqaaiaaiodaaeaa caaI1aaaaiabgUcaRmaalaaabaGaaGOmaaqaaiaaiEdaaaGaaeiiai aabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGa aeikaiaabMgacaqG2bGaaeykaiaabccadaWcaaqaaiaaiMdaaeaaca aIXaGaaGymaaaacqGHsisldaWcaaqaaiaaisdaaeaacaaIXaGaaGyn aaaaaeaacaqGOaGaaeODaiaabMcacaqGGaWaaSaaaeaacaaI3aaaba GaaGymaiaaicdaaaGaae4kamaalaaabaGaaGOmaaqaaiaaiwdaaaGa ae4kamaalaaabaGaaG4maaqaaiaaikdaaaGaaeiiaiaabccacaqGGa GaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabcca caqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiai aabccacaqGOaGaaeODaiaabMgacaqGPaGaaeiiaiaabkdadaWcaaqa aiaaikdaaeaacaaIZaaaaiabgUcaRiaaiodadaWcaaqaaiaaigdaae aacaaIYaaaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeikaiaa bAhacaqGPbGaaeyAaiaabMcacaqGGaGaaeioamaalaaabaGaaGymaa qaaiaaikdaaaGaeyOeI0IaaG4mamaalaaabaGaaGynaaqaaiaaiIda aaaaaaa@A032@

Ans.

(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@

Q.2

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

Ans.

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?

411911211311511711811111611

Ans.

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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Q.4

A rectangular sheet of paper is 1212cmlong and1023cm wide. Find its perimeter.

Ans.

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@

Q.4

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

Ans.

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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Q.5

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?

Ans.

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@

Q.6

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?

Ans.

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.

Q.7

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

Ans.

We are given:Michael worked for 712hourVaibhavworkedfor34hourTherefore,In order to find the longer hour of work we have to compareboth 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.

Q.8

Which of the following drawing (a) to (d) show:(i)2×15             (ii)2×12(iii)3×23            (iv)3×14

Ans.

(i) 2×15=25(ii)2×12=1(iii)3×23=2(iv)3×14=34(a) Here, there are three pictures and each pictures show 23  part=23+23+23                =2+2+23                =63=2Hence, (iii) denotes picture (a).(b) Here, there are two pictures. Each pictures show 12part.So,12+12=1+12=22=1Hence, (ii) denotes picture (b).(c) Here, there are three pictures. Each pictures show14part.Therefore,14+14+14=1+1+14=34Hence, (iv) denotes picture (c).(d) Here, there are two pictures and each pictures denotes15 part.Therefore,15+15=1+15=25Hence, (i) denotes picture (d).Thus:(i) denotes picture (d)(ii) denotes picture (b)(iii) denotes picture (a)(iv) denotes picture (c)

Q.9

Some pictures a to c are given below.Tell which of them show(i)3×15=35     (ii)2×13=23    (iii)3×34=214

Ans.

In picture ( a ): There are two pictures in left hand side, each denotes 1 3 and picture in right hand side denotes 2 3 . Therefore, it denotes 1 3 ×2= 2 3 which denotes ( ii ). In picture ( b ): Here are three pictures in left hand side, each shows 3 4 . There are three pictures in right hand side in which each of first two denotes 1. Therefore, 3 4 ×3=2 1 4 Thus picture ( b ) denotes ( iii ). In picture ( c ): There are three pictures in left hand side and each of them denotes 1 5 . Hence, ( i ) denotes picture ( c ) ( ii ) denotes picture ( a ) ( iii ) denotes picture ( b ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGjbGaaeOBaiaabccacaqGWbGaaeyA aiaabogacaqG0bGaaeyDaiaabkhacaqGLbGaaeiiamaabmaabaGaae yyaaGaayjkaiaawMcaaiaacQdaaeaacaqGubGaaeiAaiaabwgacaqG YbGaaeyzaiaabccacaqGHbGaaeOCaiaabwgacaqGGaGaaeiDaiaabE hacaqGVbGaaeiiaiaabchacaqGPbGaae4yaiaabshacaqG1bGaaeOC aiaabwgacaqGZbGaaeiiaiaabMgacaqGUbGaaeiiaiaabYgacaqGLb GaaeOzaiaabshacaqGGaGaaeiAaiaabggacaqGUbGaaeizaiaabcca caqGZbGaaeyAaiaabsgacaqGLbGaaiilaiaabccacaqGLbGaaeyyai aabogacaqGObGaaeiiaiaabsgacaqGLbGaaeOBaiaab+gacaqG0bGa aeyzaiaabohadaWcaaqaaiaaigdaaeaacaaIZaaaaaqaaiaabggaca qGUbGaaeizaiaabccacaqGWbGaaeyAaiaabogacaqG0bGaaeyDaiaa bkhacaqGLbGaaeiiaiaabMgacaqGUbGaaeiiaiaabkhacaqGPbGaae 4zaiaabIgacaqG0bGaaeiiaiaabIgacaqGHbGaaeOBaiaabsgacaqG GaGaae4CaiaabMgacaqGKbGaaeyzaiaabccacaqGKbGaaeyzaiaab6 gacaqGVbGaaeiDaiaabwgacaqGZbGaaGjbVpaalaaabaGaaGOmaaqa aiaaiodaaaGaaiOlaaqaaiaabsfacaqGObGaaeyzaiaabkhacaqGLb GaaeOzaiaab+gacaqGYbGaaeyzaiaacYcacaqGGaGaaeyAaiaabsha caqGGaGaaeizaiaabwgacaqGUbGaae4BaiaabshacaqGLbGaae4Cam aalaaabaGaaGymaaqaaiaaiodaaaGaey41aqRaaGOmaiabg2da9maa laaabaGaaGOmaaqaaiaaiodaaaGaae4DaiaabIgacaqGPbGaae4yai aabIgacaqGGaGaaeizaiaabwgacaqGUbGaae4BaiaabshacaqGLbGa ae4CaiaabccadaqadaqaaiaabMgacaqGPbaacaGLOaGaayzkaaGaai OlaaqaaiaabMeacaqGUbGaaeiiaiaabchacaqGPbGaae4yaiaabsha caqG1bGaaeOCaiaabwgacaqGGaWaaeWaaeaacaqGIbaacaGLOaGaay zkaaGaaeOoaaqaaiaabIeacaqGLbGaaeOCaiaabwgacaqGGaGaaeyy aiaabkhacaqGLbGaaeiiaiaabshacaqGObGaaeOCaiaabwgacaqGLb GaaeiiaiaabchacaqGPbGaae4yaiaabshacaqG1bGaaeOCaiaabwga caqGZbGaaeiiaiaabMgacaqGUbGaaeiiaiaabYgacaqGLbGaaeOzai aabshacaqGGaGaaeiAaiaabggacaqGUbGaaeizaiaabccacaqGZbGa aeyAaiaabsgacaqGLbGaaiilaiaabccacaqGLbGaaeyyaiaabogaca qGObGaaeiiaiaabohacaqGObGaae4BaiaabEhacaqGZbWaaSaaaeaa caaIZaaabaGaaGinaaaacaGGUaaabaGaaeivaiaabIgacaqGLbGaae OCaiaabwgacaqGGaGaaeyyaiaabkhacaqGLbGaaeiiaiaabshacaqG ObGaaeOCaiaabwgacaqGLbGaaeiiaiaabchacaqGPbGaae4yaiaabs hacaqG1bGaaeOCaiaabwgacaqGZbGaaeiiaiaabMgacaqGUbGaaeii aiaabkhacaqGPbGaae4zaiaabIgacaqG0bGaaeiiaiaabIgacaqGHb GaaeOBaiaabsgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabcca caqGPbGaaeOBaiaabccacaqG3bGaaeiAaiaabMgacaqGJbGaaeiAai aabccacaqGLbGaaeyyaiaabogacaqGObaabaGaae4BaiaabAgacaqG GaGaaeOzaiaabMgacaqGYbGaae4CaiaabshacaqGGaGaaeiDaiaabE hacaqGVbGaaeiiaiaabsgacaqGLbGaaeOBaiaab+gacaqG0bGaaeyz aiaabohacaqGGaGaaeymaiaac6caaeaacaqGubGaaeiAaiaabwgaca qGYbGaaeyzaiaabAgacaqGVbGaaeOCaiaabwgacaGGSaWaaSaaaeaa caaIZaaabaGaaGinaaaacqGHxdaTcaaIZaGaeyypa0JaaGOmamaala aabaGaaGymaaqaaiaaisdaaaaabaGaaeivaiaabIgacaqG1bGaae4C aiaabccacaqGWbGaaeyAaiaabogacaqG0bGaaeyDaiaabkhacaqGLb GaaeiiamaabmaabaGaaeOyaaGaayjkaiaawMcaaiaabccacaqGKbGa aeyzaiaab6gacaqGVbGaaeiDaiaabwgacaqGZbGaaeiiamaabmaaba GaaeyAaiaabMgacaqGPbaacaGLOaGaayzkaaGaaiOlaaqaaiaabMea caqGUbGaaeiiaiaabchacaqGPbGaae4yaiaabshacaqG1bGaaeOCai aabwgacaqGGaWaaeWaaeaacaqGJbaacaGLOaGaayzkaaGaaiOoaaqa aiaabsfacaqGObGaaeyzaiaabkhacaqGLbGaaeiiaiaabggacaqGYb GaaeyzaiaabccacaqG0bGaaeiAaiaabkhacaqGLbGaaeyzaiaabcca caqGWbGaaeyAaiaabogacaqG0bGaaeyDaiaabkhacaqGLbGaae4Cai aabccacaqGPbGaaeOBaiaabccacaqGSbGaaeyzaiaabAgacaqG0bGa aeiiaiaabIgacaqGHbGaaeOBaiaabsgacaqGGaGaae4CaiaabMgaca qGKbGaaeyzaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaaeyzaiaa bggacaqGJbGaaeiAaaqaaiaabccacaqGVbGaaeOzaiaabccacaqG0b GaaeiAaiaabwgacaqGTbGaaeiiaiaabsgacaqGLbGaaeOBaiaab+ga caqG0bGaaeyzaiaabohadaWcaaqaaiaaigdaaeaacaaI1aaaaiaac6 caaeaacaqGibGaaeyzaiaab6gacaqGJbGaaeyzaiaacYcaaeaadaqj EaabaeqabaWaaeWaaeaacaqGPbaacaGLOaGaayzkaaGaaeiiaiaabs gacaqGLbGaaeOBaiaab+gacaqG0bGaaeyzaiaabohacaqGGaGaaeiC aiaabMgacaqGJbGaaeiDaiaabwhacaqGYbGaaeyzaiaabccadaqada qaaiaabogaaiaawIcacaGLPaaaaeaadaqadaqaaiaabMgacaqGPbaa caGLOaGaayzkaaGaaeiiaiaabsgacaqGLbGaaeOBaiaab+gacaqG0b GaaeyzaiaabohacaqGGaGaaeiCaiaabMgacaqGJbGaaeiDaiaabwha caqGYbGaaeyzaiaabccadaqadaqaaiaabggaaiaawIcacaGLPaaaae aadaqadaqaaiaabMgacaqGPbGaaeyAaaGaayjkaiaawMcaaiaabcca caqGKbGaaeyzaiaab6gacaqGVbGaaeiDaiaabwgacaqGZbGaaeiiai aabchacaqGPbGaae4yaiaabshacaqG1bGaaeOCaiaabwgacaqGGaWa aeWaaeaacaqGIbaacaGLOaGaayzkaaaaaaaaaaa@05C7@

Q.10

Multiply and reduce to lowest form:(i)7×35 (ii)4×13 (iii)2×67(iv)5×29 (v)23×4 (vi)52×6(vii)11×47 (viii)20×45 (ix)13×13(x)15×35

Ans.

(i)7×35=71×35=215=415(ii)4×13=41×13=43=113(iii)2×67=21×67=127=157(iv)5×29=51×29=109=119(v)23×4=23×41=83=223(vi)​ 52×6=5×62=302=15(vii)11×47=111×47=447=627(viii)20×47=201×45=805=16(ix)13×13=131×13=133=413(x)15×35=151×35=455=9

Q.11

Shade:(i)12 of the circles in box (a)(ii)23of the triangles in box (b)(iii)35of the squares in box (c)

Ans.

(i)Here, there are 12 circles in box (a).Therefore,12of12=12×12=6Hence 6 circles to be coloured in (a).(ii)Here, there are 9 triangles in box (b) and 23 of thetriangles are to be coloured.So,23of9=23×9            =183=6Thus, 6 triangles are needed to be coloured.(iii)Here, there are 15 rectangles in box (c) and35of themare needed to be coloured.So,35of ​15=35×15         =455=9Thus, 9 rectangles are needed to be coloured.

Q.12

Find:(a)12of (i) 24 (ii) 46(b)23of (i) 18 (ii) 27(c)34of (i) 16 (ii) 36(d)45of (i) 20 (ii) 35

Ans.

(a) (i) 1 2 of24= 1 2 ×24= 24 2 = 12 (ii) 1 2 of46= 1 2 ×46= 46 2 = 23 (b) (i) 2 3 of18= 2 3 ×18= 36 3 = 12 (ii) 2 3 of27= 2 3 ×27= 54 3 = 18 (c) (i) 3 4 of16= 3 4 ×16= 48 4 = 12 (ii) 3 4 of36= 3 4 ×36= 108 4 = 27 (d) (i) 4 5 of20= 4 5 ×20= 80 5 = 16 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaGGOaGaamyyaiaacMcaaeaacaGGOaGa amyAaiaacMcacaaMe8+aaSaaaeaacaaIXaaabaGaaGOmaaaacaaMe8 Uaam4BaiaadAgacaaMe8UaaGOmaiaaisdacqGH9aqpdaWcaaqaaiaa igdaaeaacaaIYaaaaiabgEna0kaaikdacaaI0aGaeyypa0ZaaSaaae aacaaIYaGaaGinaaqaaiaaikdaaaGaeyypa0ZaauIhaeaacaaIXaGa aGOmaaaaaeaacaGGOaGaamyAaiaadMgacaGGPaGaaGjbVpaalaaaba GaaGymaaqaaiaaikdaaaGaaGjbVlaad+gacaWGMbGaaGinaiaaiAda cqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiabgEna0kaaisdaca aI2aGaeyypa0ZaaSaaaeaacaaI0aGaaGOnaaqaaiaaikdaaaGaeyyp a0ZaauIhaeaacaaIYaGaaG4maaaaaeaacaGGOaGaamOyaiaacMcaae aacaGGOaGaamyAaiaacMcacaaMe8+aaSaaaeaacaaIYaaabaGaaG4m aaaacaaMe8Uaam4BaiaadAgacaaMe8UaaGymaiaaiIdacqGH9aqpda WcaaqaaiaaikdaaeaacaaIZaaaaiabgEna0kaaigdacaaI4aGaeyyp a0ZaaSaaaeaacaaIZaGaaGOnaaqaaiaaiodaaaGaeyypa0ZaauIhae aacaaIXaGaaGOmaaaaaeaacaGGOaGaamyAaiaadMgacaGGPaWaaSaa aeaacaaIYaaabaGaaG4maaaacaaMe8Uaam4BaiaadAgacaaMe8UaaG OmaiaaiEdacqGH9aqpdaWcaaqaaiaaikdaaeaacaaIZaaaaiabgEna 0kaaikdacaaI3aGaeyypa0ZaaSaaaeaacaaI1aGaaGinaaqaaiaaio daaaGaeyypa0ZaauIhaeaacaaIXaGaaGioaaaacaaMe8oabaGaaiik aiaadogacaGGPaaabaGaaiikaiaadMgacaGGPaGaaGjbVpaalaaaba GaaG4maaqaaiaaisdaaaGaaGjbVlaad+gacaWGMbGaaGjbVlaaigda caaI2aGaeyypa0ZaaSaaaeaacaaIZaaabaGaaGinaaaacqGHxdaTca aIXaGaaGOnaiabg2da9maalaaabaGaaGinaiaaiIdaaeaacaaI0aaa aiabg2da9maaL4babaGaaGymaiaaikdaaaaabaGaaiikaiaadMgaca WGPbGaaiykaiaaysW7daWcaaqaaiaaiodaaeaacaaI0aaaaiaaysW7 caWGVbGaamOzaiaaysW7caaIZaGaaGOnaiabg2da9maalaaabaGaaG 4maaqaaiaaisdaaaGaey41aqRaaG4maiaaiAdacqGH9aqpdaWcaaqa aiaaigdacaaIWaGaaGioaaqaaiaaisdaaaGaeyypa0ZaauIhaeaaca aIYaGaaG4naaaaaeaacaGGOaGaamizaiaacMcaaeaacaGGOaGaamyA aiaacMcacaaMe8+aaSaaaeaacaaI0aaabaGaaGynaaaacaaMe8Uaam 4BaiaadAgacaaMe8UaaGOmaiaaicdacqGH9aqpdaWcaaqaaiaaisda aeaacaaI1aaaaiabgEna0kaaikdacaaIWaGaeyypa0ZaaSaaaeaaca aI4aGaaGimaaqaaiaaiwdaaaGaeyypa0ZaauIhaeaacaaIXaGaaGOn aaaaaaaa@F094@

Q.13

Multiply and express as a mixed fraction:(a)3×515 (b)5×634 (c)7×214(d)4×613 (e)314×6 (f)325×8

Ans.

( a )3×5 1 5 =3× 26 5 = 78 5 = 15 3 5 ( b )5×6 3 4 =5× 27 4 = 135 4 = 33 3 4 ( c )7×2 1 4 =7× 9 4 = 63 4 = 15 3 4 ( d )4×6 1 3 =4× 19 3 = 76 3 = 25 1 3 ( e )3 1 4 ×6= 13 4 ×6= 78 7 = 19 1 2 ( f )3 2 5 ×8= 17 5 ×8= 136 5 = 27 1 5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caaMe8UaaGjbVlaaiodacqGHxdaTcaaI1aWaaSaaaeaacaaIXaaaba GaaGynaaaacqGH9aqpcaaIZaGaey41aq7aaSaaaeaacaaIYaGaaGOn aaqaaiaaiwdaaaGaeyypa0ZaaSaaaeaacaaI3aGaaGioaaqaaiaaiw daaaGaeyypa0ZaauIhaeaacaaIXaGaaGynamaalaaabaGaaG4maaqa aiaaiwdaaaaaaaqaamaabmaabaGaaeOyaaGaayjkaiaawMcaaiaays W7caaMe8UaaGynaiabgEna0kaaiAdadaWcaaqaaiaaiodaaeaacaaI 0aaaaiabg2da9iaaiwdacqGHxdaTdaWcaaqaaiaaikdacaaI3aaaba GaaGinaaaacqGH9aqpdaWcaaqaaiaaigdacaaIZaGaaGynaaqaaiaa isdaaaGaeyypa0ZaauIhaeaacaaIZaGaaG4mamaalaaabaGaaG4maa qaaiaaisdaaaaaaaqaamaabmaabaGaae4yaaGaayjkaiaawMcaaiaa ysW7caaMe8UaaG4naiabgEna0kaaikdadaWcaaqaaiaaigdaaeaaca aI0aaaaiabg2da9iaaiEdacqGHxdaTdaWcaaqaaiaaiMdaaeaacaaI 0aaaaiabg2da9maalaaabaGaaGOnaiaaiodaaeaacaaI0aaaaiabg2 da9maaL4babaGaaGymaiaaiwdadaWcaaqaaiaaiodaaeaacaaI0aaa aaaaaeaadaqadaqaaiaabsgaaiaawIcacaGLPaaacaaMe8UaaGjbVl aaisdacqGHxdaTcaaI2aWaaSaaaeaacaaIXaaabaGaaG4maaaacqGH 9aqpcaaI0aGaey41aq7aaSaaaeaacaaIXaGaaGyoaaqaaiaaiodaaa Gaeyypa0ZaaSaaaeaacaaI3aGaaGOnaaqaaiaaiodaaaGaeyypa0Za auIhaeaacaaIYaGaaGynamaalaaabaGaaGymaaqaaiaaiodaaaaaaa qaamaabmaabaGaaeyzaaGaayjkaiaawMcaaiaaysW7caaMe8UaaG4m amaalaaabaGaaGymaaqaaiaaisdaaaGaey41aqRaaGOnaiabg2da9m aalaaabaGaaGymaiaaiodaaeaacaaI0aaaaiabgEna0kaaiAdacqGH 9aqpdaWcaaqaaiaaiEdacaaI4aaabaGaaG4naaaacqGH9aqpdaqjEa qaaiaaigdacaaI5aWaaSaaaeaacaaIXaaabaGaaGOmaaaaaaaabaWa aeWaaeaacaqGMbaacaGLOaGaayzkaaGaaGjbVlaaysW7caaIZaWaaS aaaeaacaaIYaaabaGaaGynaaaacqGHxdaTcaaI4aGaeyypa0ZaaSaa aeaacaaIXaGaaG4naaqaaiaaiwdaaaGaey41aqRaaGioaiabg2da9m aalaaabaGaaGymaiaaiodacaaI2aaabaGaaGynaaaacqGH9aqpdaqj EaqaaiaaikdacaaI3aWaaSaaaeaacaaIXaaabaGaaGynaaaaaaaaaa a@D0C3@

Q.14

Find(a)12of (i) 234 (ii) 429(b) 58of (i) 356 (ii) 923

Ans.

( a ) ( i ) 1 2 of2 3 4 = 1 2 × 11 4 = 11 8 = 1 3 8 ( ii ) 1 2 of4 2 9 = 1 2 × 38 9 = 19 9 = 2 1 9 ( b ) ( i ) 5 8 of3 5 6 = 5 8 × 23 6 = 115 48 = 2 19 48 ( ii ) 5 8 of9 2 3 = 5 8 × 29 3 = 145 24 = 6 1 24 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa aeaadaqadaqaaiaabMgaaiaawIcacaGLPaaacaaMe8+aaSaaaeaaca aIXaaabaGaaGOmaaaacaaMe8Uaae4BaiaabAgacaaMe8UaaGOmamaa laaabaGaaG4maaqaaiaaisdaaaGaeyypa0ZaaSaaaeaacaaIXaaaba GaaGOmaaaacqGHxdaTdaWcaaqaaiaaigdacaaIXaaabaGaaGinaaaa cqGH9aqpdaWcaaqaaiaaigdacaaIXaaabaGaaGioaaaacaaMe8Uaey ypa0ZaauIhaeaacaaIXaWaaSaaaeaacaaIZaaabaGaaGioaaaaaaaa baWaaeWaaeaacaqGPbGaaeyAaaGaayjkaiaawMcaaiaaysW7daWcaa qaaiaaigdaaeaacaaIYaaaaiaaysW7caqGVbGaaeOzaiaaysW7caaI 0aWaaSaaaeaacaaIYaaabaGaaGyoaaaacqGH9aqpdaWcaaqaaiaaig daaeaacaaIYaaaaiabgEna0oaalaaabaGaaG4maiaaiIdaaeaacaaI 5aaaaiabg2da9maalaaabaGaaGymaiaaiMdaaeaacaaI5aaaaiabg2 da9maaL4babaGaaGOmamaalaaabaGaaGymaaqaaiaaiMdaaaaaaaqa amaabmaabaGaaeOyaaGaayjkaiaawMcaaaqaamaabmaabaGaaeyAaa GaayjkaiaawMcaaiaaysW7daWcaaqaaiaaiwdaaeaacaaI4aaaaiaa ysW7caqGVbGaaeOzaiaaysW7caaIZaWaaSaaaeaacaaI1aaabaGaaG OnaaaacqGH9aqpdaWcaaqaaiaaiwdaaeaacaaI4aaaaiabgEna0oaa laaabaGaaGOmaiaaiodaaeaacaaI2aaaaiabg2da9maalaaabaGaaG ymaiaaigdacaaI1aaabaGaaGinaiaaiIdaaaGaaGjbVlabg2da9maa L4babaGaaGOmamaalaaabaGaaGymaiaaiMdaaeaacaaI0aGaaGioaa aaaaaabaWaaeWaaeaacaqGPbGaaeyAaaGaayjkaiaawMcaaiaaysW7 daWcaaqaaiaaiwdaaeaacaaI4aaaaiaaysW7caqGVbGaaeOzaiaayg W7caaMe8UaaGyoamaalaaabaGaaGOmaaqaaiaaiodaaaGaeyypa0Za aSaaaeaacaaI1aaabaGaaGioaaaacqGHxdaTdaWcaaqaaiaaikdaca aI5aaabaGaaG4maaaacqGH9aqpdaWcaaqaaiaaigdacaaI0aGaaGyn aaqaaiaaikdacaaI0aaaaiabg2da9maaL4babaGaaGOnamaalaaaba GaaGymaaqaaiaaikdacaaI0aaaaaaaaaaa@B877@

Q.15

Vidya and Pratab went for a picnic. Their mothergave them a water bag that contained 5 liters ofwater. Vidya consumed 25 of the water. Pratapconsumed the remaining water.(i) How much water did Vidya drink?(ii) What fraction of totalquantity of water did Pratab drink?

Ans.

Vidya and Pratab went for a picnic. Their mothergave them a water bag that contained 5 liters ofwater. Vidya consumed 25 of the water. Pratapconsumed the remaining water.(i) How much water did Vidya drink?(ii) What fraction of totalquantity of water did Pratab drink?

Q.16

Find: (i) 1 4 of ( a ) 1 4 ( b ) 3 5 ( c ) 4 3 (ii) 1 7 of ( a ) 2 9 ( b ) 6 5 ( c ) 3 10 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaWHgbGaaCyAaiaah6gacaWHKbGaaiOo aaqaaiaacIcacaWGPbGaaiykaiaaysW7daWcaaqaaiaaigdaaeaaca aI0aaaaiaah+gacaWHMbGaaiiOaiaacckacaGGGcGaaCzcamaabmaa baGaaCyyaaGaayjkaiaawMcaamaalaaabaGaaGymaaqaaiaaisdaaa GaaCzcaiaaxMaadaqadaqaaiaahkgaaiaawIcacaGLPaaacaaMe8+a aSaaaeaacaaIZaaabaGaaGynaaaacaWLjaWaaeWaaeaacaWHJbaaca GLOaGaayzkaaGaaGjbVpaalaaabaGaaGinaaqaaiaaiodaaaaabaGa aiikaiaadMgacaWGPbGaaiykaiaaysW7daWcaaqaaiaaigdaaeaaca aI3aaaaiaaysW7caWHVbGaaCOzaiaacckacaGGGcGaaiiOamaabmaa baGaaCyyaaGaayjkaiaawMcaaiaaysW7daWcaaqaaiaaikdaaeaaca aI5aaaaiaaxMaadaqadaqaaiaahkgaaiaawIcacaGLPaaacaaMb8Ua aGjbVpaalaaabaGaaGOnaaqaaiaaiwdaaaGaaGjbVlaaxMaadaqada qaaiaahogaaiaawIcacaGLPaaacaaMb8+aaSaaaeaacaaIZaaabaGa aGymaiaaicdaaaaaaaa@80E1@

Ans.

( i ) ( a ) 1 4 of 1 4 = 1 4 × 1 4 = 1 16 ( b ) 1 4 of 3 5 = 1 4 × 3 5 = 3 20 ( c ) 1 4 of 4 3 = 1 4 × 4 3 = 1 3 (ii) (a) 1 7 of 2 9 = 1 7 × 2 9 = 2 63 (b) 1 7 of 6 5 = 1 7 × 6 5 = 6 35 (c) 1 7 of 3 10 = 1 7 × 3 10 = 3 70 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqaaeaadaqadaqaaiaahMgaaiaawIcacaGLPaaa aeaadaqadaqaaiaahggaaiaawIcacaGLPaaacaaMe8+aaSaaaeaaca aIXaaabaGaaGinaaaacaaMe8Uaae4BaiaabAgadaWcaaqaaiaaigda aeaacaaI0aaaaiabg2da9maalaaabaGaaGymaaqaaiaaisdaaaGaey 41aq7aaSaaaeaacaaIXaaabaGaaGinaaaacqGH9aqpdaqjEaqaamaa laaabaGaaGymaaqaaiaaigdacaaI2aaaaaaaaeaadaqadaqaaiaahk gaaiaawIcacaGLPaaacaaMe8+aaSaaaeaacaaIXaaabaGaaGinaaaa caaMe8Uaae4BaiaabAgacaaMe8+aaSaaaeaacaaIZaaabaGaaGynaa aacqGH9aqpdaWcaaqaaiaaigdaaeaacaaI0aaaaiabgEna0oaalaaa baGaaG4maaqaaiaaiwdaaaGaeyypa0ZaauIhaeaadaWcaaqaaiaaio daaeaacaaIYaGaaGimaaaaaaaabaWaaeWaaeaacaWHJbaacaGLOaGa ayzkaaGaaGjbVpaalaaabaGaaGymaaqaaiaaisdaaaGaaGjbVlaab+ gacaqGMbGaaGjbVpaalaaabaGaaGinaaqaaiaaiodaaaGaeyypa0Za aSaaaeaacaaIXaaabaWaaqIaaeaacaaI0aaaaaaacqGHxdaTdaWcaa qaamaaKiaabaGaaGinaaaaaeaacaaIZaaaaiabg2da9maaL4babaWa aSaaaeaacaaIXaaabaGaaG4maaaaaaaabaGaaiikaiaadMgacaWGPb GaaiykaaqaaiaacIcacaWGHbGaaiykaiaaysW7daWcaaqaaiaaigda aeaacaaI3aaaaiaaysW7caqGVbGaaeOzaiaaysW7daWcaaqaaiaaik daaeaacaaI5aaaaiabg2da9maalaaabaGaaGymaaqaaiaaiEdaaaGa ey41aq7aaSaaaeaacaaIYaaabaGaaGyoaaaacqGH9aqpdaqjEaqaam aalaaabaGaaGOmaaqaaiaaiAdacaaIZaaaaaaaaeaacaGGOaGaamOy aiaacMcacaaMe8+aaSaaaeaacaaIXaaabaGaaG4naaaacaaMe8Uaae 4BaiaabAgacaaMe8+aaSaaaeaacaaI2aaabaGaaGynaaaacqGH9aqp daWcaaqaaiaaigdaaeaacaaI3aaaaiabgEna0oaalaaabaGaaGOnaa qaaiaaiwdaaaGaeyypa0ZaauIhaeaadaWcaaqaaiaaiAdaaeaacaaI ZaGaaGynaaaaaaaabaGaaiikaiaadogacaGGPaGaaGjbVpaalaaaba GaaGymaaqaaiaaiEdaaaGaaGjbVlaab+gacaqGMbGaaGjbVpaalaaa baGaaG4maaqaaiaaigdacaaIWaaaaiabg2da9maalaaabaGaaGymaa qaaiaaiEdaaaGaey41aq7aaSaaaeaacaaIZaaabaGaaGymaiaaicda aaGaeyypa0ZaauIhaeaadaWcaaqaaiaaiodaaeaacaaI3aGaaGimaa aaaaaaaaa@C4D2@

Q.17

Multiply and reduce to lowest form (if possible).(i)23×223 (ii)27×79 (iii)38×64(iv)95×35 (v)13×158 (vi)112×310(vii)45×127

Ans.

(i)23×223=23×83=169(ii)27×79=27×79=29(iii)38×64=32×4×2×34=916(iv)95×35=2725(v)13×158=13×3×58=58(vi)112×310=3320(vii)45×127=4835

Q.18

For the fractions given below:(a) Multiply and reduce the product to lowestform (if possible)(b) Tell whether the fraction obtained is properor improper.(c) If the fraction obtained is improper thenconvert it into a mixed fraction. (i) 25×514 (ii) 625×79 (iii) 32×513 (iv) 56×237(v) 325×47 (vi) 235×3 (vii) 347×35

Ans.

(i)(a)25×514=25×214=25×212×2=2110(b) It is an improper fraction.(c)2110=2110(ii)(a)625×79=325×79=22445(b) It is an improper fraction.(c)22445=44445(iii)(a)32×513=32×163=32×2×83=81=8(b) It is whole number.(iv)(a) 56×237=56×177=8542(b) It is an improper fraction.(c)8542=2142(v)(a) ​325×47=175×47=6835(b) It is an improper fraction.(c)6835=13335(vi)(a)235×3=135×31=395(b)It is an improper fraction.(c)395=745(vii)(a)347×35=257×35=5×57×35=157(b) It is an improper fraction.(c)157=217

Q.19

Which is greater:(i)27of34 or 35of58(ii)12of67or23of37

Ans.

( i ) 2 7 of 3 4 or 3 5 of 5 8 2 7 × 3 4 or 3 5 × 5 8 = 2 7 × 3 2 ×2 or 3 5 × 5 8 = 3 14 or 3 8 Now, the LCM of 14 and 8 is 56. So, we get 3 14 = 3×4 14×4 = 12 56 or 3 8 = 3×7 8×7 = 21 56 Since, 21 > 12, so 3 8 > 3 14 Hence , 3 5 of 5 8 > 2 7 of 3 4 ( ii ) 1 2 of 6 7 or 2 3 of 3 7 1 2 of 6 7 or 2 3 of 3 7 = 1 2 × 2 ×3 7 or 2 3 × 3 7 = 3 7 or 2 7 Since, 3>2, so, 3 7 > 2 7 Hence, 1 2 of 6 7 > 2 3 of 3 7 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqaaeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caaMe8oabaWaaSaaaeaacaaIYaaabaGaaG4naaaacaaMe8Uaae4Bai aabAgacaaMe8+aaSaaaeaacaaIZaaabaGaaGinaaaacaaMe8Uaae4B aiaabkhacaaMe8+aaSaaaeaacaaIZaaabaGaaGynaaaacaaMe8Uaae 4BaiaabAgacaaMe8+aaSaaaeaacaaI1aaabaGaaGioaaaaaeaadaWc aaqaaiaaikdaaeaacaaI3aaaaiabgEna0oaalaaabaGaaG4maaqaai aaisdaaaGaaGjbVlaaysW7caWGVbGaamOCaiaaysW7daWcaaqaaiaa iodaaeaacaaI1aaaaiabgEna0oaalaaabaGaaGynaaqaaiaaiIdaaa aabaGaeyypa0ZaaSaaaeaaceaIYaGbaybaaeaacaaI3aaaaiabgEna 0oaalaaabaGaaG4maaqaaiqaikdagaGfaiabgEna0kaaikdaaaGaaG jbVlaab+gacaqGYbGaaGjbVpaalaaabaGaaG4maaqaaiqaiwdagaGf aaaacqGHxdaTdaWcaaqaaiqaiwdagaGfaaqaaiaaiIdaaaaabaGaey ypa0ZaaSaaaeaacaaIZaaabaGaaGymaiaaisdaaaGaaGjbVlaab+ga caqGYbGaaGjbVpaalaaabaGaaG4maaqaaiaaiIdaaaaabaGaaiiOai aab6eacaqGVbGaae4DaiaacYcacaqGGaGaaeiDaiaabIgacaqGLbGa aeiiaiaabYeacaqGdbGaaeytaiaabccacaqGVbGaaeOzaiaabccaca qGXaGaaeinaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaaeioaiaa bccacaqGPbGaae4CaiaabccacaqG1aGaaeOnaiaac6cacaqGGaGaae 4uaiaab+gacaGGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqG LbGaaeiDaaqaamaalaaabaGaaG4maaqaaiaaigdacaaI0aaaaiabg2 da9maalaaabaGaaG4maiabgEna0kaaisdaaeaacaaIXaGaaGinaiab gEna0kaaisdaaaGaeyypa0ZaaSaaaeaacaaIXaGaaGOmaaqaaiaaiw dacaaI2aaaaiaaysW7caqGVbGaaeOCaiaaysW7daWcaaqaaiaaioda aeaacaaI4aaaaiabg2da9maalaaabaGaaG4maiabgEna0kaaiEdaae aacaaI4aGaey41aqRaaG4naaaacqGH9aqpdaWcaaqaaiaaikdacaaI XaaabaGaaGynaiaaiAdaaaaabaGaae4uaiaabMgacaqGUbGaae4yai aabwgacaGGSaGaaeiiaiaabkdacaqGXaGaaeiiaiabg6da+iaabcca caqGXaGaaeOmaiaacYcacaqGGaGaae4Caiaab+gacaaMe8+aaSaaae aacaaIZaaabaGaaGioaaaacqGH+aGpdaWcaaqaaiaaiodaaeaacaaI XaGaaGinaaaaaeaacaqGibGaaeyzaiaab6gacaqGJbGaaeyzaiaacc kacaGGSaGaaGjbVpaaL4babaWaaSaaaeaacaaIZaaabaGaaGynaaaa caaMe8Uaae4BaiaabAgacaaMe8+aaSaaaeaacaaI1aaabaGaaGioaa aacqGH+aGpdaWcaaqaaiaaikdaaeaacaaI3aaaaiaaysW7caqGVbGa aeOzaiaaysW7daWcaaqaaiaaiodaaeaacaaI0aaaaaaaaeaadaqada qaaiaabMgacaqGPbaacaGLOaGaayzkaaaabaWaaSaaaeaacaaIXaaa baGaaGOmaaaacaaMe8Uaae4BaiaabAgacaaMe8+aaSaaaeaacaaI2a aabaGaaG4naaaacaaMe8Uaam4BaiaadkhacaaMb8UaaiiOaiaaccka daWcaaqaaiaaikdaaeaacaaIZaaaaiaaysW7caqGVbGaaeOzaiaays W7daWcaaqaaiaaiodaaeaacaaI3aaaaaqaamaalaaabaGaaGymaaqa aiaaikdaaaGaaGjbVlaab+gacaqGMbGaaGjbVpaalaaabaGaaGOnaa qaaiaaiEdaaaGaaGjbVlaad+gacaWGYbGaaGjbVpaalaaabaGaaGOm aaqaaiaaiodaaaGaaGjbVlaab+gacaqGMbGaaGjbVpaalaaabaGaaG 4maaqaaiaaiEdaaaaabaGaeyypa0ZaaSaaaeaacaaIXaaabaGabGOm ayaawaaaaiabgEna0oaalaaabaGabGOmayaawaGaey41aqRaaG4maa qaaiaaiEdaaaGaaGjbVlaab+gacaqGYbGaaGjbVpaalaaabaGaaGOm aaqaaiqaiodagaGfaaaacqGHxdaTdaWcaaqaaiqaiodagaGfaaqaai aaiEdaaaaabaGaeyypa0ZaaSaaaeaacaaIZaaabaGaaG4naaaacaaM e8Uaae4BaiaabkhacaaMe8+aaSaaaeaacaaIYaaabaGaaG4naaaaae aacaqGtbGaaeyAaiaab6gacaqGJbGaaeyzaiaacYcacaqGGaGaae4m aiabg6da+iaabkdacaGGSaGaaeiiaiaabohacaqGVbGaaiilamaala aabaGaaG4maaqaaiaaiEdaaaGaeyOpa4ZaaSaaaeaacaaIYaaabaGa aG4naaaaaeaacaqGibGaaeyzaiaab6gacaqGJbGaaeyzaiaacYcaca aMe8+aauIhaeaadaWcaaqaaiaaigdaaeaacaaIYaaaaiaaysW7caqG VbGaaeOzaiaaysW7daWcaaqaaiaaiAdaaeaacaaI3aaaaiabg6da+m aalaaabaGaaGOmaaqaaiaaiodaaaGaaGjbVlaab+gacaqGMbGaaGjb VpaalaaabaGaaG4maaqaaiaaiEdaaaaaaaaaaa@6CB6@

Q.20

Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 34 m.Find the distance between the first and the last sapling.

Ans.

Here, the distance between the two saplings is given as 3 4 m So, the distance between first and the last sapling is 3 4 m×3= 9 4 m = 2 1 4 m MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqaaeaacaqGibGaaeyzaiaabkhacaqGLbGaaiil aiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeizaiaabMgacaqGZb GaaeiDaiaabggacaqGUbGaae4yaiaabwgacaqGGaGaaeOyaiaabwga caqG0bGaae4DaiaabwgacaqGLbGaaeOBaiaabccacaqG0bGaaeiAai aabwgacaqGGaGaaeiDaiaabEhacaqGVbGaaeiiaiaabohacaqGHbGa aeiCaiaabYgacaqGPbGaaeOBaiaabEgacaqGZbGaaeiiaiaabMgaca qGZbGaaeiiaiaabEgacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaa bggacaqGZbGaaGjbVpaalaaabaGaaG4maaqaaiaaisdaaaGaaGjbVl aab2gaaeaacaqGtbGaae4BaiaacYcacaqGGaGaaeiDaiaabIgacaqG LbGaaeiiaiaabsgacaqGPbGaae4CaiaabshacaqGHbGaaeOBaiaabo gacaqGLbGaaeiiaiaabkgacaqGLbGaaeiDaiaabEhacaqGLbGaaeyz aiaab6gacaqGGaGaaeOzaiaabMgacaqGYbGaae4CaiaabshacaqGGa Gaaeyyaiaab6gacaqGKbGaaeiiaiaabshacaqGObGaaeyzaiaabcca caqGSbGaaeyyaiaabohacaqG0bGaaeiiaiaabohacaqGHbGaaeiCai aabYgacaqGPbGaaeOBaiaabEgacaqGGaGaaeyAaiaabohaaeaadaWc aaqaaiaaiodaaeaacaaI0aaaaiaaysW7caqGTbGaey41aqRaae4mai aab2dadaWcaaqaaiaaiMdaaeaacaaI0aaaaiaab2gaaeaacaaMe8Ua aGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaab2daca aMe8+aauIhaeaacaqGYaWaaSaaaeaacaaIXaaabaGaaGinaaaacaqG Tbaaaaaaaa@BE8E@

Q.21

Lipika reads a book for134hours every day. Shereads the entire book in 6 days. How many hoursin all were required by her to read the book?

Ans.

Lipika reads1 3 4 = 7 4 hours a day. So, Number of hours required by her to read the book in 6 days is 7 4 hour×6= 7 2× 2 hour×( 2 ×3 ) = 21 2 hour = 10 1 2 hour MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqaaeaacaqGmbGaaeyAaiaabchacaqGPbGaae4A aiaabggacaqGGaGaaeOCaiaabwgacaqGHbGaaeizaiaabohacaaMe8 UaaGymamaalaaabaGaaG4maaqaaiaaisdaaaGaeyypa0ZaaSaaaeaa caaI3aaabaGaaGinaaaacaGGGcGaaeiAaiaab+gacaqG1bGaaeOCai aabohacaqGGaGaaeyyaiaabccacaqGKbGaaeyyaiaabMhacaGGUaaa baGaae4uaiaab+gacaGGSaGaaeiiaiaab6eacaqG1bGaaeyBaiaabk gacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqGObGaae4B aiaabwhacaqGYbGaae4CaiaabccacaqGYbGaaeyzaiaabghacaqG1b GaaeyAaiaabkhacaqGLbGaaeizaiaabccacaqGIbGaaeyEaiaabcca caqGObGaaeyzaiaabkhacaqGGaGaaeiDaiaab+gacaqGGaGaaeOCai aabwgacaqGHbGaaeizaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGa aeOyaiaab+gacaqGVbGaae4AaaqaaiaabMgacaqGUbGaaeiiaiaabA dacaqGGaGaaeizaiaabggacaqG5bGaae4CaiaabccacaqGPbGaae4C aiaaygW7daWcaaqaaiaaiEdaaeaacaaI0aaaaiaabIgacaqGVbGaae yDaiaabkhacqGHxdaTcaaI2aGaeyypa0ZaaSaaaeaacaaI3aaabaGa aGOmaiabgEna0kqaikdagaGfaaaacaqGObGaae4BaiaabwhacaqGYb Gaey41aq7aaeWaaeaaceaIYaGbaybacqGHxdaTcaaIZaaacaGLOaGa ayzkaaaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaG jbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaM e8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaays W7caaMe8UaaGjbVlabg2da9maalaaabaGaaGOmaiaaigdaaeaacaaI YaaaaiaabIgacaqGVbGaaeyDaiaabkhaaeaacaaMe8UaaGjbVlaays W7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjb VlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8 UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaeypaiaaysW7 daqjEaqaaiaabgdacaqGWaWaaSaaaeaacaaIXaaabaGaaGOmaaaaca aMe8UaaeiAaiaab+gacaqG1bGaaeOCaaaaaaaa@0AEC@

Q.22

A car runs 16 km using 1 litre of petrol. Howmuch distance will it cover using234 litres of petrol?

Ans.

In 1 litre of petrol, car runs 16 km. So,in2 3 4 litres of petrol,distance covered by car is 2 3 4 ×16km = 11 4 ×( 4 ×4 )km = 44km MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaabaeaacaqGjbGaaeOBaiaabccacaqGXaGaaeii aiaabYgacaqGPbGaaeiDaiaabkhacaqGLbGaaeiiaiaab+gacaqGMb GaaeiiaiaabchacaqGLbGaaeiDaiaabkhacaqGVbGaaeiBaiaacYca caqGGaGaae4yaiaabggacaqGYbGaaeiiaiaabkhacaqG1bGaaeOBai aabohacaqGGaGaaeymaiaabAdacaqGGaGaae4Aaiaab2gacaGGUaaa baGaae4uaiaab+gacaGGSaGaaeyAaiaab6gacaaMe8UaaGOmamaala aabaGaaG4maaqaaiaaisdaaaGaaeiBaiaabMgacaqG0bGaaeOCaiaa bwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabchacaqGLbGaae iDaiaabkhacaqGVbGaaeiBaiaacYcacaqGKbGaaeyAaiaabohacaqG 0bGaaeyyaiaab6gacaqGJbGaaeyzaiaabccacaqGJbGaae4BaiaabA hacaqGLbGaaeOCaiaabwgacaqGKbGaaeiiaiaabkgacaqG5bGaaeii aiaabogacaqGHbGaaeOCaiaabccacaqGPbGaae4CaiaaysW7aeaaca aIYaWaaSaaaeaacaaIZaaabaGaaGinaaaacqGHxdaTcaaIXaGaaGOn aiaaysW7caqGRbGaaeyBaaqaaiabg2da9maalaaabaGaaGymaiaaig daaeaaceaI0aGbaybaaaGaey41aq7aaeWaaeaaceaI0aGbaybacqGH xdaTcaaI0aaacaGLOaGaayzkaaGaaGjbVlaabUgacaqGTbaabaGaey ypa0ZaauIhaeaacaqG0aGaaeinaiaaysW7caqGRbGaaeyBaaaaaaaa @A994@

Q.23

(a)(i) Provide the number in the box ​,such that23 ×=1030(ii)The simplest form of the number obtained inis____(b)(i)Provide the number in the box,such that 35×=2475(ii) The simplest form of the number obtained inis____

Ans.

(a)(i)23×=1030=2×53×10=23×510So,23×510=1030Therefore, the number inis510.(ii)The simplest form of510is510==12(b)(i)35×=2475=3×85×15=35×815So,35×815=2475Therefore, the number inis815.(ii)The simplest form of815is815

Q.24

Find:(i)12÷34 (ii)14÷56 (iii)73(iv)83 (v)3÷213 (vi)5÷347

Ans.

(i) 12÷34=12×43=483=16(ii) 14÷56=14×65=845=1645(iii) 8÷73=8×37=247=337(iv) 4÷83=4×383=32=112(v) 3÷213=3÷73=3×37=97=127(vi) 5÷347=5÷257=5×7257=75=125

Q.25

Find the reciprocal of each of the following fractions. Classify the reciprocals as properfractions, improper fractions and whole numbers.(i)37 (ii)58 (iii)97 (iv)​ 65(v)​ 127 (vi)18 (vii)111

Ans.

(i)37Reciprocal of 37is73 and it is an improper fraction.(ii)58Reciprocal of 58is85 and it is an improper fraction(iii) 97Reciprocal of 97is79 and it is an proper fraction(iv)​ 65Reciprocal of 65is56 and it is an proper fraction(v)127Reciprocal of 127is712 and it is an proper fraction(vi)18Reciprocal of 18is81 and it is an whole fraction(vii)111Reciprocal of 111is11 and it is a whole number.

Q.26

Find:(i)73÷2 (ii)49÷5 (iii)613÷7(iv)413÷3 (v)312÷4 (vi)437÷7

Ans.

(i) 7 3 ÷2= 7 3 × 1 2 = 7 6 = 1 1 6 (ii) 4 9 ÷5= 4 9 × 1 5 = 4 45 (iii) 6 13 ÷7= 6 13 × 1 7 = 6 91 (iv)4 1 3 ÷3= 13 3 × 1 3 = 13 9 = 1 4 9 (v)3 1 2 ÷4= 7 2 × 1 4 = 7 8 (vi)4 3 7 ÷7= 31 7 × 1 7 = 31 49 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaGGOaGaamyAaiaacMcacaaMe8+aaSaa aeaacaaI3aaabaGaaG4maaaacqGH3daUcaaIYaGaeyypa0ZaaSaaae aacaaI3aaabaGaaG4maaaacqGHxdaTdaWcaaqaaiaaigdaaeaacaaI Yaaaaiabg2da9maalaaabaGaaG4naaqaaiaaiAdaaaGaeyypa0Zaau IhaeaacaaIXaWaaSaaaeaacaaIXaaabaGaaGOnaaaaaaaabaGaaiik aiaadMgacaWGPbGaaiykaiaaysW7daWcaaqaaiaaisdaaeaacaaI5a aaaiabgEpa4kaaiwdacqGH9aqpdaWcaaqaaiaaisdaaeaacaaI5aaa aiabgEna0oaalaaabaGaaGymaaqaaiaaiwdaaaGaeyypa0ZaauIhae aadaWcaaqaaiaaisdaaeaacaaI0aGaaGynaaaaaaaabaGaaiikaiaa dMgacaWGPbGaamyAaiaacMcacaaMe8+aaSaaaeaacaaI2aaabaGaaG ymaiaaiodaaaGaey49aGRaaG4naiabg2da9maalaaabaGaaGOnaaqa aiaaigdacaaIZaaaaiabgEna0oaalaaabaGaaGymaaqaaiaaiEdaaa Gaeyypa0ZaauIhaeaadaWcaaqaaiaaiAdaaeaacaaI5aGaaGymaaaa aaaabaGaaiikaiaadMgacaWG2bGaaiykaiaaysW7caaI0aWaaSaaae aacaaIXaaabaGaaG4maaaacqGH3daUcaaIZaGaeyypa0ZaaSaaaeaa caaIXaGaaG4maaqaaiaaiodaaaGaey41aq7aaSaaaeaacaaIXaaaba GaaG4maaaacqGH9aqpdaWcaaqaaiaaigdacaaIZaaabaGaaGyoaaaa cqGH9aqpdaqjEaqaaiaaigdadaWcaaqaaiaaisdaaeaacaaI5aaaaa aaaeaacaGGOaGaamODaiaacMcacaaMe8UaaG4mamaalaaabaGaaGym aaqaaiaaikdaaaGaey49aGRaaGinaiabg2da9maalaaabaGaaG4naa qaaiaaikdaaaGaey41aq7aaSaaaeaacaaIXaaabaGaaGinaaaacqGH 9aqpdaqjEaqaamaalaaabaGaaG4naaqaaiaaiIdaaaaaaaqaaiaacI cacaWG2bGaamyAaiaacMcacaaMe8UaaGinamaalaaabaGaaG4maaqa aiaaiEdaaaGaey49aGRaaG4naiabg2da9maalaaabaGaaG4maiaaig daaeaacaaI3aaaaiabgEna0oaalaaabaGaaGymaaqaaiaaiEdaaaGa eyypa0ZaauIhaeaadaWcaaqaaiaaiodacaaIXaaabaGaaGinaiaaiM daaaaaaaaaaa@BA0D@

Q.27 

Find (i) 2 5 ÷ 1 2 (ii) 4 9 ÷ 2 3 (iii) 3 7 ÷ 8 7 (iv)2 1 3 ÷ 3 5 (v)3 1 2 ÷ 8 3 (vi) 2 5 ÷1 1 2 (vii)3 1 5 ÷1 2 3 (viii)2 1 5 ÷1 1 5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGgbGaaeyAaiaab6gacaqGKbaabaGa aiikaiaadMgacaGGPaGaaGjbVpaalaaabaGaaGOmaaqaaiaaiwdaaa Gaey49aG7aaSaaaeaacaaIXaaabaGaaGOmaaaacaWLjaaabaGaaiik aiaadMgacaWGPbGaaiykaiaaygW7caaMe8+aaSaaaeaacaaI0aaaba GaaGyoaaaacqGH3daUdaWcaaqaaiaaikdaaeaacaaIZaaaaiaaxMaa aeaacaGGOaGaamyAaiaadMgacaWGPbGaaiykaiaaysW7daWcaaqaai aaiodaaeaacaaI3aaaaiabgEpa4oaalaaabaGaaGioaaqaaiaaiEda aaaabaGaaiikaiaadMgacaWG2bGaaiykaiaaysW7caaIYaWaaSaaae aacaaIXaaabaGaaG4maaaacqGH3daUdaWcaaqaaiaaiodaaeaacaaI 1aaaaaqaaiaacIcacaWG2bGaaiykaiaaysW7caaIZaWaaSaaaeaaca aIXaaabaGaaGOmaaaacqGH3daUdaWcaaqaaiaaiIdaaeaacaaIZaaa aaqaaiaacIcacaWG2bGaamyAaiaacMcacaaMe8+aaSaaaeaacaaIYa aabaGaaGynaaaacqGH3daUcaaIXaWaaSaaaeaacaaIXaaabaGaaGOm aaaaaeaacaGGOaGaamODaiaadMgacaWGPbGaaiykaiaaysW7caaIZa WaaSaaaeaacaaIXaaabaGaaGynaaaacqGH3daUcaaIXaWaaSaaaeaa caaIYaaabaGaaG4maaaaaeaacaGGOaGaamODaiaadMgacaWGPbGaam yAaiaacMcacaaMe8UaaGOmamaalaaabaGaaGymaaqaaiaaiwdaaaGa ey49aGRaaGymamaalaaabaGaaGymaaqaaiaaiwdaaaaaaaa@9B80@

Ans.

(i) 2 5 ÷ 1 2 = 2 5 × 2 1 = 4 5 (ii) 4 9 ÷ 2 3 = 4 9 × 3 2 = ( 2× 2 ) ( 3× 3 ) × 3 2 = 2 3 (iii) 3 7 ÷ 8 7 = 3 7 × 7 8 = 3 8 (iv)2 1 3 ÷ 3 5 = 7 3 × 5 3 = 35 9 = 3 8 9 (v)3 1 2 ÷ 8 3 = 7 2 × 3 8 = 21 16 = 1 5 16 (vi) 2 5 ÷1 1 2 = 2 5 ÷ 3 2 = 2 5 × 2 3 = 4 15 (vii)3 1 5 ÷1 2 3 = 16 5 ÷ 5 3 = 16 5 × 3 5 = 48 25 = 1 23 25 (viii)2 1 5 ÷1 1 5 = 11 5 ÷ 6 5 = 11 5 × 5 6 = 11 6 = 1 5 6 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaGGOaGaamyAaiaacMcacaaMe8+aaSaa aeaacaaIYaaabaGaaGynaaaacqGH3daUdaWcaaqaaiaaigdaaeaaca aIYaaaaiabg2da9maalaaabaGaaGOmaaqaaiaaiwdaaaGaey41aq7a aSaaaeaacaaIYaaabaGaaGymaaaacqGH9aqpdaqjEaqaamaalaaaba GaaGinaaqaaiaaiwdaaaaaaaqaaiaacIcacaWGPbGaamyAaiaacMca caaMe8+aaSaaaeaacaaI0aaabaGaaGyoaaaacqGH3daUdaWcaaqaai aaikdaaeaacaaIZaaaaiabg2da9maalaaabaGaaGinaaqaaiaaiMda aaGaey41aq7aaSaaaeaacaaIZaaabaGaaGOmaaaacqGH9aqpdaWcaa qaamaabmaabaGaaGOmaiabgEna0kqaikdagaGfaaGaayjkaiaawMca aaqaamaabmaabaGaaG4maiabgEna0kqaiodagaGfaaGaayjkaiaawM caaaaacqGHxdaTdaWcaaqaaiqaiodagaGfaaqaaiqaikdagaGfaaaa cqGH9aqpdaqjEaqaamaalaaabaGaaGOmaaqaaiaaiodaaaaaaaqaai aacIcacaWGPbGaamyAaiaadMgacaGGPaGaaGjbVpaalaaabaGaaG4m aaqaaiaaiEdaaaGaey49aG7aaSaaaeaacaaI4aaabaGaaG4naaaacq GH9aqpdaWcaaqaaiaaiodaaeaaceaI3aGbaybaaaGaey41aq7aaSaa aeaaceaI3aGbaybaaeaacaaI4aaaaiabg2da9maaL4babaWaaSaaae aacaaIZaaabaGaaGioaaaaaaaabaGaaiikaiaadMgacaWG2bGaaiyk aiaaysW7caaIYaWaaSaaaeaacaaIXaaabaGaaG4maaaacqGH3daUda WcaaqaaiaaiodaaeaacaaI1aaaaiabg2da9maalaaabaGaaG4naaqa aiaaiodaaaGaey41aq7aaSaaaeaacaaI1aaabaGaaG4maaaacqGH9a qpdaWcaaqaaiaaiodacaaI1aaabaGaaGyoaaaacqGH9aqpdaqjEaqa aiaaiodadaWcaaqaaiaaiIdaaeaacaaI5aaaaaaaaeaacaGGOaGaam ODaiaacMcacaaMe8UaaG4mamaalaaabaGaaGymaaqaaiaaikdaaaGa ey49aG7aaSaaaeaacaaI4aaabaGaaG4maaaacqGH9aqpdaWcaaqaai aaiEdaaeaacaaIYaaaaiabgEna0oaalaaabaGaaG4maaqaaiaaiIda aaGaeyypa0ZaaSaaaeaacaaIYaGaaGymaaqaaiaaigdacaaI2aaaai abg2da9maaL4babaGaaGymamaalaaabaGaaGynaaqaaiaaigdacaaI 2aaaaaaaaeaacaGGOaGaamODaiaadMgacaGGPaGaaGjbVpaalaaaba GaaGOmaaqaaiaaiwdaaaGaey49aGRaaGymamaalaaabaGaaGymaaqa aiaaikdaaaGaeyypa0ZaaSaaaeaacaaIYaaabaGaaGynaaaacqGH3d aUdaWcaaqaaiaaiodaaeaacaaIYaaaaiabg2da9maalaaabaGaaGOm aaqaaiaaiwdaaaGaey41aq7aaSaaaeaacaaIYaaabaGaaG4maaaacq GH9aqpdaqjEaqaamaalaaabaGaaGinaaqaaiaaigdacaaI1aaaaaaa aeaacaGGOaGaamODaiaadMgacaWGPbGaaiykaiaaysW7caaIZaWaaS aaaeaacaaIXaaabaGaaGynaaaacqGH3daUcaaIXaWaaSaaaeaacaaI YaaabaGaaG4maaaacqGH9aqpdaWcaaqaaiaaigdacaaI2aaabaGaaG ynaaaacqGH3daUdaWcaaqaaiaaiwdaaeaacaaIZaaaaiabg2da9maa laaabaGaaGymaiaaiAdaaeaacaaI1aaaaiabgEna0oaalaaabaGaaG 4maaqaaiaaiwdaaaGaeyypa0ZaaSaaaeaacaaI0aGaaGioaaqaaiaa ikdacaaI1aaaaiabg2da9maaL4babaGaaGymamaalaaabaGaaGOmai aaiodaaeaacaaIYaGaaGynaaaaaaaabaGaaiikaiaadAhacaWGPbGa amyAaiaadMgacaGGPaGaaGjbVlaaikdadaWcaaqaaiaaigdaaeaaca aI1aaaaiabgEpa4kaaigdadaWcaaqaaiaaigdaaeaacaaI1aaaaiab g2da9maalaaabaGaaGymaiaaigdaaeaacaaI1aaaaiabgEpa4oaala aabaGaaGOnaaqaaiaaiwdaaaGaeyypa0ZaaSaaaeaacaaIXaGaaGym aaqaaiqaiwdagaGfaaaacqGHxdaTdaWcaaqaaiqaiwdagaGfaaqaai aaiAdaaaGaeyypa0ZaaSaaaeaacaaIXaGaaGymaaqaaiaaiAdaaaGa eyypa0ZaauIhaeaacaaIXaWaaSaaaeaacaaI1aaabaGaaGOnaaaaaa aaaaa@180D@

Q.28

Which is greater?(i) 0.5 or 0.05(ii) 0.7 or 0.5(iii) 7 or 0.7(iv) 1.37 or 1.49(v) 2.03 or 2.30(vi) 0.8 or 0.88

Ans.

( i ) 0.5 or 0.05 0.5>0.05 ( ii ) 0.7 or 0.5 0.7>0.5 ( iii ) 7 or 0.7 7>0.7 ( iv ) 1.37 or 1.49 1.49>1.37 ( v ) 2.03 or 2.30 2.30>2.03 (vi)0.8or0.88 0.88>0.8 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaGimaiaac6cacaqG1aGaaeiiaiaab+gacaqGYbGaaeiiai aaicdacaGGUaGaaGimaiaabwdaaeaadaqjEaqaaiaaicdacaGGUaGa aGynaiabg6da+iaaicdacaGGUaGaaGimaiaaiwdaaaaabaWaaeWaae aacaqGPbGaaeyAaaGaayjkaiaawMcaaiaabccacaaIWaGaaiOlaiaa bEdacaqGGaGaae4BaiaabkhacaqGGaGaaGimaiaac6cacaqG1aaaba WaauIhaeaacaaIWaGaaiOlaiaaiEdacqGH+aGpcaaIWaGaaiOlaiaa iwdaaaaabaWaaeWaaeaacaqGPbGaaeyAaiaabMgaaiaawIcacaGLPa aacaqGGaGaae4naiaabccacaqGVbGaaeOCaiaabccacaaIWaGaaiOl aiaabEdaaeaadaqjEaqaaiaaiEdacqGH+aGpcaaIWaGaaiOlaiaaiE daaaaabaWaaeWaaeaacaqGPbGaaeODaaGaayjkaiaawMcaaiaabcca caqGXaGaaiOlaiaabodacaqG3aGaaeiiaiaab+gacaqGYbGaaeiiai aabgdacaGGUaGaaeinaiaabMdaaeaadaqjEaqaaiaaigdacaGGUaGa aGinaiaaiMdacqGH+aGpcaaIXaGaaiOlaiaaiodacaaI3aaaaaqaam aabmaabaGaaeODaaGaayjkaiaawMcaaiaabccacaqGYaGaaiOlaiaa icdacaqGZaGaaeiiaiaab+gacaqGYbGaaeiiaiaabkdacaGGUaGaae 4maiaaicdaaeaadaqjEaqaaiaaikdacaGGUaGaaG4maiaaicdacqGH +aGpcaaIYaGaaiOlaiaaicdacaaIZaaaaaqaaiaacIcacaWG2bGaam yAaiaacMcacaaMe8UaaGimaiaac6cacaaI4aGaaGjbVlaab+gacaqG YbGaaGzaVlaaysW7caaIWaGaaiOlaiaaiIdacaaI4aaabaWaauIhae aacaaIWaGaaiOlaiaaiIdacaaI4aGaeyOpa4JaaGimaiaac6cacaaI 4aaaaaaaaa@AD65@

Q.29

Express as rupees using decimals:(i) 7 paise (ii) 7 rupees 7 paise (iii) 77 rupees 77 paise (iv) 50 paise(v) 235 paise

Ans.

(i) 7 paiseSince,1paise=1100So,7paise=7×1100= 7100= 0.07 (ii) 7rupees 7 paiseSince,1paise=1100So,7rupees7paise=₹ 7+(1100)=₹ 7+₹ 0.07=₹ 7.07(iii) 77 rupees 77 paise.Since,1paise=1100So, 77 rupees 77 paise=77+(77100)=₹ 77+0.77=77.77(iv) 50 paise.Since,1paise=1100So,50paise=(50100)=0.50(v) 235 paise.Since,1paise=1100So, 235paise=(235100)=2.35

Q.30

(i) Express 5 cm in metre and kilometre(ii) Express 35 mm in cm, m and km.

Ans.

(i) Since, 1 cm =1100m=11000kmSo, we get5 cm=5100m=0.05 mand5 cm =51000 km=0.005 km(ii) Since, 1 mm =110cm=11000m=1100000kmSo, we get35 mm =3510cm=3.5 m35 mm =351000 m=0.035 mand35​ mm=35100000 km=0.00035 km

Q.31

Express in kg:(i)  200g(ii)3470g(iii)4kg8g(iv)2598mg

Ans.

Since, 1 g=11000 kgSo, 200 g=2001000 kg=10010×100 kg=210 kg=0.2 kg(ii) 3470 g3470 g=34701000 kg=347×10100×10kg=347100kg=3.47 kg(iii) 4 kg 8 g4 kg 8 g=4 kg+81000kg=4 kg+0.008 kg=4.008 kg(iv) 2598 mgSince, 1 mg ​=11000000kgSo, 2598 mg=25981000000 kg =0.002598 kg

Q.32

Write the following decimal numbers in the expanded form:(i)   20.03(ii)2.03 (iii) 200.03 (iv) 2.034

Ans.

(i)20.03= 2×10+0×1+0× 1 10 +3× 1 100 (ii)2.02= 2×1+0× 1 10 +3× 1 100 (iii)200.03= 2×100+0×10+0×1+0× 1 10 +3× 1 100 (iv)2.034= 2×1+0× 1 10 +3× 1 100 +4× 1 1000 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaaMe8UaaeOm aiaabcdacaqGUaGaaeimaiaabodacaqG9aGaaGjbVpaaL4babaGaae OmaiaabEnacaqGXaGaaeimaiaabUcacaqGWaGaae41aiaabgdacaqG RaGaaeimaiaabEnadaWcaaqaaiaabgdaaeaacaqGXaGaaeimaaaaca qGRaGaae4maiaabEnadaWcaaqaaiaabgdaaeaacaqGXaGaaeimaiaa bcdaaaaaaaqaaiaabIcacaqGPbGaaeyAaiaabMcacaaMe8UaaeOmai aab6cacaqGWaGaaeOmaiaab2dacaaMe8+aauIhaeaacaqGYaGaae41 aiaabgdacaqGRaGaaeimaiaabEnadaWcaaqaaiaabgdaaeaacaqGXa GaaeimaaaacaqGRaGaae4maiaabEnadaWcaaqaaiaabgdaaeaacaqG XaGaaeimaiaabcdaaaaaaaqaaiaabIcacaqGPbGaaeyAaiaabMgaca qGPaGaaGjbVlaabkdacaqGWaGaaeimaiaab6cacaqGWaGaae4maiaa b2dacaaMe8+aauIhaeaacaqGYaGaae41aiaabgdacaqGWaGaaeimai aabUcacaqGWaGaae41aiaabgdacaqGWaGaae4kaiaabcdacaqGxdGa aeymaiaabUcacaqGWaGaae41amaalaaabaGaaeymaaqaaiaabgdaca qGWaaaaiaabUcacaqGZaGaae41amaalaaabaGaaeymaaqaaiaabgda caqGWaGaaeimaaaaaaaabaGaaeikaiaabMgacaqG2bGaaeykaiaays W7caqGYaGaaeOlaiaabcdacaqGZaGaaeinaiaab2dacaaMe8+aauIh aeaacaqGYaGaae41aiaabgdacaqGRaGaaeimaiaabEnadaWcaaqaai aabgdaaeaacaqGXaGaaeimaaaacaqGRaGaae4maiaabEnadaWcaaqa aiaabgdaaeaacaqGXaGaaeimaiaabcdaaaGaae4kaiaabsdacaqGxd WaaSaaaeaacaqGXaaabaGaaeymaiaabcdacaqGWaGaaeimaaaaaaaa aaa@B11D@

Q.33

Write the place value of 2 in the following decimal numbers:(i)2.56 (ii)21.37 (iii)10.25 (iv)9.42 (v)63.352

Ans.

(i)2.56 The place value of 2 in 2.56 = 2 Ones(ii) 21.37The place value of 2 in 21.37 = 2 Tens(iii) 10.25The place value of 2 in 10.25 = 2 Tenths(iv) 9.42The place value of 2 in 9.42 = 2 Hundredths(v) 63.352 The place value of 2 in 63.352 = 2 Thousandths

Q.34

Dinesh went from place A to place B and from there toplace C. A is 7.5 km from B and B is 12.7 km from C.A yub went from place A to place D and from there toplace C. D is 9.3 km from A and C is 11.8 km from D.Who travelled more and by how much?

Ans.

We are given that: Dinesh went from A to B and from B to C. Ayub went from A to D and from D to C. So, the distance from A to B =7.5 km Distance from B to C=12.7 km Distance from C to D=11.8 km Distance from A to D=9.3 km Distance covered by Dinesh=7.5+12.7 km=20.2 km Distance covered by Ayub = 9.3+11.8 km=21.1 km Hence, Ayun travelled 21.1 km and 0.9 km more than Dinesh. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGxbGaaeyzaiaabccacaqGHbGaaeOC aiaabwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGa GaaeiDaiaabIgacaqGHbGaaeiDaiaabQdaaeaacaqGebGaaeyAaiaa b6gacaqGLbGaae4CaiaabIgacaqGGaGaae4DaiaabwgacaqGUbGaae iDaiaabccacaqGMbGaaeOCaiaab+gacaqGTbGaaeiiaiaabgeacaqG GaGaaeiDaiaab+gacaqGGaGaaeOqaiaabccacaqGHbGaaeOBaiaabs gacaqGGaGaaeOzaiaabkhacaqGVbGaaeyBaiaabccacaqGcbGaaeii aiaabshacaqGVbGaaeiiaiaaboeacaqGUaaabaGaaeyqaiaabMhaca qG1bGaaeOyaiaabccacaqG3bGaaeyzaiaab6gacaqG0bGaaeiiaiaa bAgacaqGYbGaae4Baiaab2gacaqGGaGaaeyqaiaabccacaqG0bGaae 4BaiaabccacaqGebGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqG MbGaaeOCaiaab+gacaqGTbGaaeiiaiaabseacaqGGaGaaeiDaiaab+ gacaqGGaGaae4qaiaab6caaeaacaqGtbGaae4BaiaabYcacaqGGaGa aeiDaiaabIgacaqGLbGaaeiiaiaabsgacaqGPbGaae4Caiaabshaca qGHbGaaeOBaiaabogacaqGLbGaaeiiaiaabAgacaqGYbGaae4Baiaa b2gacaqGGaGaaeyqaiaabccacaaMb8UaaeiDaiaab+gacaqGGaGaae OqaiaabccacaqG9aGaae4naiaab6cacaqG1aGaaeiiaiaabUgacaqG TbaabaGaaeiraiaabMgacaqGZbGaaeiDaiaabggacaqGUbGaae4yai aabwgacaqGGaGaaeOzaiaabkhacaqGVbGaaeyBaiaabccacaqGcbGa aeiiaiaabshacaqGVbGaaeiiaiaaboeacaqG9aGaaeymaiaabkdaca qGUaGaae4naiaabccacaqGRbGaaeyBaaqaaiaabseacaqGPbGaae4C aiaabshacaqGHbGaaeOBaiaabogacaqGLbGaaeiiaiaabAgacaqGYb Gaae4Baiaab2gacaqGGaGaae4qaiaabccacaqG0bGaae4Baiaabcca caqGebGaaeypaiaabgdacaqGXaGaaeOlaiaabIdacaqGGaGaae4Aai aab2gaaeaacaqGebGaaeyAaiaabohacaqG0bGaaeyyaiaab6gacaqG JbGaaeyzaiaabccacaqGMbGaaeOCaiaab+gacaqGTbGaaeiiaiaabg eacaqGGaGaaeiDaiaab+gacaqGGaGaaeiraiaab2dacaqG5aGaaeOl aiaabodacaqGGaGaae4Aaiaab2gaaeaacaqGebGaaeyAaiaabohaca qG0bGaaeyyaiaab6gacaqGJbGaaeyzaiaabccacaqGJbGaae4Baiaa bAhacaqGLbGaaeOCaiaabwgacaqGKbGaaeiiaiaabkgacaqG5bGaae iiaiaabseacaqGPbGaaeOBaiaabwgacaqGZbGaaeiAaiaab2dacaqG 3aGaaeOlaiaabwdacaqGRaGaaeymaiaabkdacaqGUaGaae4naiaabc cacaqGRbGaaeyBaiaab2dacaqGYaGaaeimaiaab6cacaqGYaGaaeii aiaabUgacaqGTbaabaGaaeiraiaabMgacaqGZbGaaeiDaiaabggaca qGUbGaae4yaiaabwgacaqGGaGaae4yaiaab+gacaqG2bGaaeyzaiaa bkhacaqGLbGaaeizaiaabccacaqGIbGaaeyEaiaabccacaqGbbGaae yEaiaabwhacaqGIbGaaeiiaiaab2dacaqGGaGaaeyoaiaab6cacaqG ZaGaae4kaiaabgdacaqGXaGaaeOlaiaabIdacaqGGaGaae4Aaiaab2 gacaqG9aGaaeOmaiaabgdacaqGUaGaaeymaiaabccacaqGRbGaaeyB aaqaaiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaeilaiaabccaca qGbbGaaeyEaiaabwhacaqGUbGaaeiiaiaabshacaqGYbGaaeyyaiaa bAhacaqGLbGaaeiBaiaabYgacaqGLbGaaeizaiaabccacaqGYaGaae ymaiaab6cacaqGXaGaaeiiaiaabUgacaqGTbGaaeiiaiaabggacaqG UbGaaeizaiaabccacaqGWaGaaeOlaiaabMdacaqGGaGaae4Aaiaab2 gacaqGGaGaaeyBaiaab+gacaqGYbGaaeyzaiaabccacaqG0bGaaeiA aiaabggacaqGUbGaaeiiaiaabseacaqGPbGaaeOBaiaabwgacaqGZb GaaeiAaiaab6caaaaa@6D80@

Q.35

Shyama bought 5 kg 300 g apples and 3 kg 250 gmangoes. Sarala bought 4 kg 800 g oranges and4 kg 150 g bananas.Who bought more fruit?

Ans.

Apples bought by Shyama=5 kg 300 g=5 kg+3001000 kg=5 kg+ 0.3 kg=5.3 kgMangoes bought by Shyama= 3 kg 250 g= 3 kg+2501000 kg= 3 kg+ 0.25 kg=3.25 kgThus, total fruits bought by Shyama= 5.3 kg+3.25 kg= 8.55 kg Oranges bought by Sarla= 4 kg 800 g= 4 kg+8001000 kg= 4 kg+ 0.8 kg=4.8 kg Bananas bought by Sarla = 4 kg 150 g = 4 kg+ 150 1000 kg = 4 kg+ 0.15 kg = 4.15 kg Thus, total fruits bought by Sarla = 4.8 kg+4.15 kg = 8.95 kg Thus, Sarla bought more fruits. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGcbGaaeyyaiaab6gacaqGHbGaaeOB aiaabggacaqGZbGaaeiiaiaabkgacaqGVbGaaeyDaiaabEgacaqGOb GaaeiDaiaabccacaqGIbGaaeyEaiaabccacaqGtbGaaeyyaiaabkha caqGSbGaaeyyaaqaaiaab2dacaqGGaGaaeinaiaabccacaqGRbGaae 4zaiaabccacaqGXaGaaeynaiaabcdacaqGGaGaae4zaaqaaiaab2da caqGGaGaaeinaiaabccacaqGRbGaae4zaiaabUcadaWcaaqaaiaaig dacaaI1aGaaGimaaqaaiaaigdacaaIWaGaaGimaiaaicdaaaGaaeii aiaabUgacaqGNbaabaGaaeypaiaabccacaqG0aGaaeiiaiaabUgaca qGNbGaae4kaiaabccacaqGWaGaaeOlaiaabgdacaqG1aGaaeiiaiaa bUgacaqGNbaabaGaaeypaiaabccacaqG0aGaaeOlaiaabgdacaqG1a GaaeiiaiaabUgacaqGNbaabaGaaeivaiaabIgacaqG1bGaae4Caiaa bYcacaqGGaGaaeiDaiaab+gacaqG0bGaaeyyaiaabYgacaqGGaGaae OzaiaabkhacaqG1bGaaeyAaiaabshacaqGZbGaaeiiaiaabkgacaqG VbGaaeyDaiaabEgacaqGObGaaeiDaiaabccacaqGIbGaaeyEaiaabc cacaqGtbGaaeyyaiaabkhacaqGSbGaaeyyaaqaaiaab2dacaqGGaGa aeinaiaab6cacaqG4aGaaeiiaiaabUgacaqGNbGaae4kaiaabsdaca qGUaGaaeymaiaabwdacaqGGaGaae4AaiaabEgaaeaacaqG9aGaaeii amaaL4babaGaaeioaiaab6cacaqG5aGaaeynaiaabccacaqGRbGaae 4zaaaaaeaacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabccacaqG tbGaaeyyaiaabkhacaqGSbGaaeyyaiaabccacaqGIbGaae4Baiaabw hacaqGNbGaaeiAaiaabshacaqGGaGaaeyBaiaab+gacaqGYbGaaeyz aiaabccacaqGMbGaaeOCaiaabwhacaqGPbGaaeiDaiaabohacaqGUa aaaaa@C547@

Q.36

How much less is 28 km than 42.6 km?

Ans.

Since,42.6 km 28 km=14.6 kmHence, 28 km is 14.6 km less than 42.6 km.

Q.37

Find: ( i ) 0.2 × 6 ( ii ) 8 × 4.6 ( iii ) 2.71 × 5 ( iv ) 20.1 × 4 ( v ) 0.05 × 7 ( vi ) 211.02 × 4 ( vii ) 2 × 0.86 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWFGa Gaa8hiaiaa=bdacaWFUaGaa8Nmaiaa=bcacaWFxdGaa8hiaiaa=zda caWFGcGaa8hOaaqaamaabmaabaGaa8xAaiaa=LgaaiaawIcacaGLPa aacaWFGaGaa8hiaiaa=HdacaWFGaGaa831aiaa=bcacaWF0aGaa8Nl aiaa=zdaaeaadaqadaqaaiaa=LgacaWFPbGaa8xAaaGaayjkaiaawM caaiaa=bcacaWFYaGaa8Nlaiaa=DdacaWFXaGaa8hiaiaa=DnacaWF GaGaa8xnaaqaamaabmaabaGaa8xAaiaa=zhaaiaawIcacaGLPaaaca WFGaGaa8Nmaiaa=bdacaWFUaGaa8xmaiaa=bcacaWFxdGaa8hiaiaa =rdaaeaadaqadaqaaiaa=zhaaiaawIcacaGLPaaacaWFGaGaa8hmai aa=5cacaWFWaGaa8xnaiaa=bcacaWFxdGaa8hiaiaa=Ddaaeaadaqa daqaaiaa=zhacaWFPbaacaGLOaGaayzkaaGaa8hiaiaa=jdacaWFXa Gaa8xmaiaa=5cacaWFWaGaa8Nmaiaa=bcacaWFxdGaa8hiaiaa=rda aeaadaqadaqaaiaa=zhacaWFPbGaa8xAaaGaayjkaiaawMcaaiaa=b cacaWFYaGaa8hiaiaa=DnacaWFGaGaa8hmaiaa=5cacaWF4aGaa8Nn aaaaaa@8C92@

Ans.

( i ) 0.2 × 6 = 1.2 ( ii ) 8 × 4.6 = 36.8 ( iii ) 2.71 × 5= 13.55 ( iv ) 20.1 × 4= 80.4 ( v ) 0.05 × 7= 0.35 ( vi ) 211.02 × 4 = 844.08 ( vii ) 2 × 0.86= 1.72 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaGimaiaac6cacaqGYaGaaeiiaiabgEna0kaabccacaqG2a Gaaeiiaiaab2dacaqGGaWaauIhaeaacaaIXaGaaiOlaiaaikdaaaaa baGaaiiOamaabmaabaGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGa GaaeioaiaabccacqGHxdaTcaqGGaGaaeinaiaac6cacaqG2aGaaiiO aiaaysW7cqGH9aqpdaqjEaqaaiaaiodacaaI2aGaaiOlaiaaiIdaaa aabaGaaiiOamaabmaabaGaaeyAaiaabMgacaqGPbaacaGLOaGaayzk aaGaaeiiaiaabkdacaGGUaGaae4naiaabgdacaqGGaGaey41aqRaae iiaiaabwdacqGH9aqpdaqjEaqaaiaaigdacaaIZaGaaiOlaiaaiwda caaI1aaaaaqaaiaacckadaqadaqaaiaabMgacaqG2baacaGLOaGaay zkaaGaaeiiaiaabkdacaaIWaGaaiOlaiaabgdacaqGGaGaey41aqRa aeiiaiaabsdacaaMe8UaaeypaiaabccadaqjEaqaaiaabIdacaqGWa GaaeOlaiaabsdaaaaabaWaaeWaaeaacaqG2baacaGLOaGaayzkaaGa aeiiaiaaicdacaGGUaGaaGimaiaabwdacaqGGaGaey41aqRaaeiiai aabEdacaaMe8UaaeypaiaabccadaqjEaqaaiaabcdacaqGUaGaae4m aiaabwdaaaaabaWaaeWaaeaacaqG2bGaaeyAaaGaayjkaiaawMcaai aabccacaqGYaGaaeymaiaabgdacaGGUaGaaGimaiaabkdacaqGGaGa ey41aqRaaeiiaiaabsdacaaMe8UaaiiOaiabg2da9maaL4babaGaaG ioaiaaisdacaaI0aGaaiOlaiaaicdacaaI4aaaaaqaaiaacckadaqa daqaaiaabAhacaqGPbGaaeyAaaGaayjkaiaawMcaaiaabccacaqGYa GaaeiiaiabgEna0kaabccacaaIWaGaaiOlaiaabIdacaqG2aGaaGjb Vlaab2dacaqGGaWaauIhaeaacaqGXaGaaeOlaiaabEdacaqGYaaaaa aaaa@B98D@

Q.38

Find the are a of rectangle whose length is 5.7cmand breadth is 3 cm.

Ans.

The area of a rectangle is given by length×breadth.So, the area of given rectangle would be 5.7 cm×3 cm= 17.1 cm2

Q.39

Find: ( i ) 1.3 × 10 ( ii ) 36.8 × 10 ( iii ) 153.7 × 10 ( iv ) 168.07 × 10 ( v ) 31.1 × 100 ( vi ) 156.1 × 100 ( vii ) 3.62 × 100 ( viii ) 43.07 × 100 ( ix ) 0.5 × 10 ( x ) 0.08 × 10 ( xi ) 0.9 × 100 (xii ) 0.03 × 1000 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWFXa Gaa8Nlaiaa=ndacaWFGaGaa831aiaa=bcacaWFXaGaa8hmaiaa=bka caWFGcGaa8hOaiaa=bkacaWFGcGaa8hOaiaa=bkacaWFGcGaaCzcam aabmaabaGaa8xAaiaa=LgaaiaawIcacaGLPaaacaWFGaGaa83maiaa =zdacaWFUaGaa8hoaiaa=bcacaWFxdGaa8hiaiaa=fdacaWFWaGaa8 hOaiaa=bkacaWFGcGaa8hOaiaa=bkacaWFGcGaa8hOaiaa=bkacaWF GcGaa8hOaiaa=bkacaWFGcGaaCzcamaabmaabaGaa8xAaiaa=Lgaca WFPbaacaGLOaGaayzkaaGaa8hiaiaa=fdacaWF1aGaa83maiaa=5ca caWF3aGaa8hiaiaa=DnacaWFGaGaa8xmaiaa=bdaaeaacaWFGcWaae WaaeaacaWFPbGaa8NDaaGaayjkaiaawMcaaiaa=bkacaWFXaGaa8Nn aiaa=HdacaWFUaGaa8hmaiaa=DdacaWFGaGaa831aiaa=bcacaWFXa Gaa8hmaiaaxMaadaqadaqaaiaa=zhaaiaawIcacaGLPaaacaWFGaGa a83maiaa=fdacaWFUaGaa8xmaiaa=bcacaWFxdGaa8hiaiaa=fdaca WFWaGaa8hmaiaa=bkacaWFGcGaa8hOaiaa=bkacaWFGcGaa8hOaiaa =bkacaWFGcGaa8hOaiaa=bkacaWFGcGaaCzcamaabmaabaGaa8NDai aa=LgaaiaawIcacaGLPaaacaWFGaGaa8xmaiaa=vdacaWF2aGaa8Nl aiaa=fdacaWFGaGaa831aiaa=bcacaWFXaGaa8hmaiaa=bdaaeaada qadaqaaiaa=zhacaWFPbGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWF ZaGaa8Nlaiaa=zdacaWFYaGaa8hiaiaa=DnacaWFGaGaa8xmaiaa=b dacaWFWaGaa8hOaiaa=bkacaWFGcGaa8hOamaabmaabaGaa8NDaiaa =LgacaWFPbGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWF0aGaa83mai aa=5cacaWFWaGaa83naiaa=bcacaWFxdGaa8hiaiaa=fdacaWFWaGa a8hmaiaaxMaacaaMe8UaaGjbVlaaysW7caaMe8UaaGjbVpaabmaaba Gaa8xAaiaa=HhaaiaawIcacaGLPaaacaWFGaGaa8hmaiaa=5cacaWF 1aGaa8hiaiaa=DnacaWFGaGaa8xmaiaa=bdaaeaacaWFGcWaaeWaae aacaWF4baacaGLOaGaayzkaaGaa8hiaiaa=bdacaWFUaGaa8hmaiaa =HdacaWFGaGaa831aiaa=bcacaWFXaGaa8hmaiaa=bkacaWFGcGaa8 hOaiaa=bkacaWFGcGaa8hOaiaaysW7daqadaqaaiaa=HhacaWFPbaa caGLOaGaayzkaaGaa8hiaiaa=bdacaWFUaGaa8xoaiaa=bcacaWFxd Gaa8hiaiaa=fdacaWFWaGaa8hmaiaa=bkacaWFGcGaa8hOaiaa=bka caWFGcGaa8hOaiaa=bkacaWFGcGaa8hOaiaaysW7caaMe8UaaGjbVp aabmaabaGaa8hEaiaa=LgacaWFPbaacaGLOaGaayzkaaGaa8hiaiaa =bdacaWFUaGaa8hmaiaa=ndacaWFGaGaa831aiaa=bcacaWFXaGaa8 hmaiaa=bdacaWFWaaaaaa@17CB@

Ans.

( i ) 1.3 × 10= 13 ( ii ) 36.8 × 10 = 368 ( iii ) 153.7 × 10= 1537 ( iv ) 168.07 × 10= 1680.7 ( v ) 31.1 × 100= 3110 ( vi ) 156.1 × 100= 15610 ( vii ) 3.62 × 100= 362 ( viii ) 43.07 × 100= 4307 ( ix ) 0.5 × 10= 5 ( x ) 0.08 × 10 = 0.8 ( xi ) 0.9 × 100= 90 ( xii ) 0.03 × 1000= 30 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaeymaiaac6cacaqGZaGaaeiiaiabgEna0kaabccacaqGXa GaaGimaiabg2da9maaL4babaGaaGymaiaaiodaaaGaaiiOaaqaamaa bmaabaGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaae4maiaabA dacaGGUaGaaeioaiaabccacqGHxdaTcaqGGaGaaeymaiaaicdacaGG GcGaeyypa0ZaauIhaeaacaaIZaGaaGOnaiaaiIdaaaaabaWaaeWaae aacaqGPbGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaaeymaiaa bwdacaqGZaGaaiOlaiaabEdacaqGGaGaey41aqRaaeiiaiaabgdaca aIWaGaeyypa0ZaauIhaeaacaaIXaGaaGynaiaaiodacaaI3aaaaaqa aiaacckadaqadaqaaiaabMgacaqG2baacaGLOaGaayzkaaGaaiiOai aabgdacaqG2aGaaeioaiaac6cacaaIWaGaae4naiaabccacqGHxdaT caqGGaGaaeymaiaaicdacqGH9aqpdaqjEaqaaiaaigdacaaI2aGaaG ioaiaaicdacaGGUaGaaG4naaaaaeaadaqadaqaaiaabAhaaiaawIca caGLPaaacaqGGaGaae4maiaabgdacaGGUaGaaeymaiaabccacqGHxd aTcaqGGaGaaeymaiaaicdacaaIWaGaeyypa0ZaauIhaeaacaaIZaGa aGymaiaaigdacaaIWaaaaaqaamaabmaabaGaaeODaiaabMgaaiaawI cacaGLPaaacaqGGaGaaeymaiaabwdacaqG2aGaaiOlaiaabgdacaqG GaGaey41aqRaaeiiaiaabgdacaaIWaGaaGimaiabg2da9maaL4baba GaaGymaiaaiwdacaaI2aGaaGymaiaaicdaaaaabaWaaeWaaeaacaqG 2bGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaae4maiaac6caca qG2aGaaeOmaiaabccacqGHxdaTcaqGGaGaaeymaiaaicdacaaIWaGa eyypa0ZaauIhaeaacaaIZaGaaGOnaiaaikdaaaaabaGaaiiOamaabm aabaGaaeODaiaabMgacaqGPbGaaeyAaaGaayjkaiaawMcaaiaabcca caqG0aGaae4maiaac6cacaaIWaGaae4naiaabccacqGHxdaTcaqGGa GaaeymaiaaicdacaaIWaGaeyypa0ZaauIhaeaacaaI0aGaaG4maiaa icdacaaI3aaaaaqaamaabmaabaGaaeyAaiaabIhaaiaawIcacaGLPa aacaqGGaGaaGimaiaac6cacaqG1aGaaeiiaiabgEna0kaabccacaqG XaGaaGimaiabg2da9maaL4babaGaaGynaaaaaeaacaGGGcWaaeWaae aacaqG4baacaGLOaGaayzkaaGaaeiiaiaaicdacaGGUaGaaGimaiaa bIdacaqGGaGaey41aqRaaeiiaiaabgdacaaIWaGaaiiOaiabg2da9m aaL4babaGaaGimaiaac6cacaaI4aaaaaqaamaabmaabaGaaeiEaiaa bMgaaiaawIcacaGLPaaacaqGGaGaaGimaiaac6cacaqG5aGaaeiiai abgEna0kaabccacaqGXaGaaGimaiaaicdacqGH9aqpdaqjEaqaaiaa iMdacaaIWaaaaaqaamaabmaabaGaaeiEaiaabMgacaqGPbaacaGLOa GaayzkaaGaaeiiaiaaicdacaGGUaGaaGimaiaabodacaqGGaGaey41 aqRaaeiiaiaabgdacaaIWaGaaGimaiaaicdacqGH9aqpdaqjEaqaai aaiodacaaIWaaaaaaaaa@05CA@

Q.40

A twowheeler covers a distance of 55.3 km in one litre of petrol.How much distance will it cover in 10 litresof petrol?

Ans.

Distance covered by a two-wheeler in 1 litres= 55.3 kmSo, distance covered by a two-wheeler in 10 litres=55.3 km ×10= 553 km

Q.41

Find: ( i ) 2.5 × 0.3 ( ii ) 0.1 × 51.7 ( iii ) 0.2 × 316.8 ( iv ) 1.3 × 3.1 ( v ) 0.5 × 0.05 ( vi ) 11.2 × 0.15 ( vii ) 1.07 × 0.02 ( viii )10.05 × 1.05 ( ix ) 101.01 × 0.01 ( x ) 100.01 × 1.1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWFYa Gaa8Nlaiaa=vdacaWFGaGaa831aiaa=bcacaWFWaGaa8Nlaiaa=nda caWLjaGaaCzcaiaa=bkadaqadaqaaiaa=LgacaWFPbaacaGLOaGaay zkaaGaa8hiaiaa=bdacaWFUaGaa8xmaiaa=bcacaWFxdGaa8hiaiaa =vdacaWFXaGaa8Nlaiaa=DdacaWFGcGaa8hOaiaa=bkacaWFGcGaaC zcaiaa=bkadaqadaqaaiaa=LgacaWFPbGaa8xAaaGaayjkaiaawMca aiaa=bcacaWFWaGaa8Nlaiaa=jdacaWFGaGaa831aiaa=bcacaWFZa Gaa8xmaiaa=zdacaWFUaGaa8hoaiaa=bkacaWFGcaabaWaaeWaaeaa caWFPbGaa8NDaaGaayjkaiaawMcaaiaa=bcacaWFXaGaa8Nlaiaa=n dacaWFGaGaa831aiaa=bcacaWFZaGaa8Nlaiaa=fdacaWLjaGaaCzc amaabmaabaGaa8NDaaGaayjkaiaawMcaaiaa=bcacaWFWaGaa8Nlai aa=vdacaWFGaGaa831aiaa=bcacaWFWaGaa8Nlaiaa=bdacaWF1aGa a8hOaiaa=bkacaWFGcGaaCzcaiaa=bkadaqadaqaaiaa=zhacaWFPb aacaGLOaGaayzkaaGaa8hiaiaa=fdacaWFXaGaa8Nlaiaa=jdacaWF GaGaa831aiaa=bcacaWFWaGaa8Nlaiaa=fdacaWF1aaabaGaa8hOam aabmaabaGaa8NDaiaa=LgacaWFPbaacaGLOaGaayzkaaGaa8hiaiaa =fdacaWFUaGaa8hmaiaa=DdacaWFGaGaa831aiaa=bcacaWFWaGaa8 Nlaiaa=bdacaWFYaGaa8hOaiaa=bkadaqadaqaaiaa=zhacaWFPbGa a8xAaiaa=LgaaiaawIcacaGLPaaacaWFXaGaa8hmaiaa=5cacaWFWa Gaa8xnaiaa=bcacaWFxdGaa8hiaiaa=fdacaWFUaGaa8hmaiaa=vda caWFGcGaaCzcamaabmaabaGaa8xAaiaa=HhaaiaawIcacaGLPaaaca WFGaGaa8xmaiaa=bdacaWFXaGaa8Nlaiaa=bdacaWFXaGaa8hiaiaa =DnacaWFGaGaa8hmaiaa=5cacaWFWaGaa8xmaaqaamaabmaabaGaa8 hEaaGaayjkaiaawMcaaiaa=bcacaWFXaGaa8hmaiaa=bdacaWFUaGa a8hmaiaa=fdacaWFGaGaa831aiaa=bcacaWFXaGaa8Nlaiaa=fdaaa aa@D0D5@

Ans.

( i ) 2.5 × 0.3 = 0.75 ( ii ) 0.1 × 51.7 = 5.17 ( iii ) 0.2 × 316.8 = 63.36 ( iv ) 1.3 × 3.1= 4.03 ( v ) 0.5 × 0.05 = 0.025 ( vi ) 11.2 × 0.15= 1.68 ( vii ) 1.07 × 0.02= 0.0214 ( viii )10.05 × 1.05= 10.5525 ( ix ) 101.01 × 0.01= 1.0101 ( x ) 100.01 × 1.1= 110.011 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaeOmaiaac6cacaqG1aGaaeiiaiabgEna0kaabccacaaIWa GaaiOlaiaabodacaqGGaGaaeypaiaabccadaqjEaqaaiaabcdacaqG UaGaae4naiaabwdaaaaabaGaaiiOamaabmaabaGaaeyAaiaabMgaai aawIcacaGLPaaacaqGGaGaaGimaiaac6cacaqGXaGaaeiiaiabgEna 0kaabccacaqG1aGaaeymaiaac6cacaqG3aGaaiiOaiabg2da9maaL4 babaGaaGynaiaac6cacaaIXaGaaG4naaaaaeaacaGGGcWaaeWaaeaa caqGPbGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaaGimaiaac6 cacaqGYaGaaeiiaiabgEna0kaabccacaqGZaGaaeymaiaabAdacaGG UaGaaeioaiaacckacaGGGcGaeyypa0ZaauIhaeaacaaI2aGaaG4mai aac6cacaaIZaGaaGOnaaaaaeaadaqadaqaaiaabMgacaqG2baacaGL OaGaayzkaaGaaeiiaiaabgdacaGGUaGaae4maiaabccacqGHxdaTca qGGaGaae4maiaac6cacaqGXaGaaeypaiaabccadaqjEaqaaiaabsda caqGUaGaaeimaiaabodaaaaabaWaaeWaaeaacaqG2baacaGLOaGaay zkaaGaaeiiaiaaicdacaGGUaGaaeynaiaabccacqGHxdaTcaqGGaGa aGimaiaac6cacaaIWaGaaeynaiaacckacqGH9aqpdaqjEaqaaiaaic dacaGGUaGaaGimaiaaikdacaaI1aaaaaqaaiaacckadaqadaqaaiaa bAhacaqGPbaacaGLOaGaayzkaaGaaeiiaiaabgdacaqGXaGaaiOlai aabkdacaqGGaGaey41aqRaaeiiaiaaicdacaGGUaGaaeymaiaabwda caqG9aGaaeiiamaaL4babaGaaeymaiaab6cacaqG2aGaaeioaaaaae aacaGGGcWaaeWaaeaacaqG2bGaaeyAaiaabMgaaiaawIcacaGLPaaa caqGGaGaaeymaiaac6cacaaIWaGaae4naiaabccacqGHxdaTcaqGGa GaaGimaiaac6cacaaIWaGaaeOmaiaab2dacaqGGaWaauIhaeaacaqG WaGaaeOlaiaabcdacaqGYaGaaeymaiaabsdaaaaabaWaaeWaaeaaca qG2bGaaeyAaiaabMgacaqGPbaacaGLOaGaayzkaaGaaeymaiaaicda caGGUaGaaGimaiaabwdacaqGGaGaey41aqRaaeiiaiaabgdacaGGUa GaaGimaiaabwdacaqG9aWaauIhaeaacaqGXaGaaeimaiaab6cacaqG 1aGaaeynaiaabkdacaqG1aaaaaqaamaabmaabaGaaeyAaiaabIhaai aawIcacaGLPaaacaqGGaGaaeymaiaaicdacaqGXaGaaiOlaiaaicda caqGXaGaaeiiaiabgEna0kaabccacaaIWaGaaiOlaiaaicdacaqGXa GaaeypamaaL4babaGaaeymaiaab6cacaqGWaGaaeymaiaabcdacaqG XaaaaaqaamaabmaabaGaaeiEaaGaayjkaiaawMcaaiaabccacaqGXa GaaGimaiaaicdacaGGUaGaaGimaiaabgdacaqGGaGaey41aqRaaeii aiaabgdacaGGUaGaaeymaiaab2dadaqjEaqaaiaabgdacaqGXaGaae imaiaab6cacaqGWaGaaeymaiaabgdaaaaaaaa@FB8F@

Q.42

Find: ( i ) 0.4÷2 ( ii ) 0.35÷5 ( iii ) 2.48÷4 ( iv ) 65.4÷6 ( v ) 651.2÷4 ( vi ) 14.49÷7 ( vii ) 3.96÷4 ( viii )0.80÷5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWFWa Gaa8Nlaiaa=rdacaWF3dGaa8Nmaiaa=bkacaWFGcGaa8hOaiaa=bka caWFGcGaaCzcamaabmaabaGaa8xAaiaa=LgaaiaawIcacaGLPaaaca WFGaGaa8hmaiaa=5cacaWFZaGaa8xnaiaa=DpacaWF1aGaa8hOaiaa xMaadaqadaqaaiaa=LgacaWFPbGaa8xAaaGaayjkaiaawMcaaiaa=b cacaWFYaGaa8Nlaiaa=rdacaWF4aGaa839aiaa=rdacaWLjaWaaeWa aeaacaWFPbGaa8NDaaGaayjkaiaawMcaaiaa=bcacaWF2aGaa8xnai aa=5cacaWF0aGaa839aiaa=zdaaeaadaqadaqaaiaa=zhaaiaawIca caGLPaaacaWFGaGaa8Nnaiaa=vdacaWFXaGaa8Nlaiaa=jdacaWF3d Gaa8hnaiaaxMaadaqadaqaaiaa=zhacaWFPbaacaGLOaGaayzkaaGa a8hiaiaa=fdacaWF0aGaa8Nlaiaa=rdacaWF5aGaa839aiaa=Ddaca WFGcGaa8hOamaabmaabaGaa8NDaiaa=LgacaWFPbaacaGLOaGaayzk aaGaa8hiaiaa=ndacaWFUaGaa8xoaiaa=zdacaWF3dGaa8hnaiaaxM aadaqadaqaaiaa=zhacaWFPbGaa8xAaiaa=LgaaiaawIcacaGLPaaa caWFWaGaa8Nlaiaa=HdacaWFWaGaa839aiaa=vdaaaaa@97B8@

Ans.

( i ) 0.4÷2 = 0.2 ( ii ) 0.35÷5 = 0.07 ( iii ) 2.48÷4= 0.62 ( iv ) 65.4÷6= 10.9 ( v ) 651.2÷4= 162.8 ( vi ) 14.49÷7 = 2.07 ( vii ) 3.96÷4= 0.99 ( viii )0.80÷5= 0.16 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaGimaiaac6cacaqG0aGaey49aGRaaeOmaiaacckacqGH9a qpdaqjEaqaaiaaicdacaGGUaGaaGOmaaaaaeaacaGGGcWaaeWaaeaa caqGPbGaaeyAaaGaayjkaiaawMcaaiaabccacaaIWaGaaiOlaiaabo dacaqG1aGaey49aGRaaeynaiaacckacqGH9aqpdaqjEaqaaiaaicda caGGUaGaaGimaiaaiEdaaaaabaWaaeWaaeaacaqGPbGaaeyAaiaabM gaaiaawIcacaGLPaaacaqGGaGaaeOmaiaac6cacaqG0aGaaeioaiab gEpa4kaabsdacaqG9aGaaeiiamaaL4babaGaaeimaiaab6cacaqG2a GaaeOmaaaaaeaadaqadaqaaiaabMgacaqG2baacaGLOaGaayzkaaGa aeiiaiaabAdacaqG1aGaaiOlaiaabsdacqGH3daUcaqG2aGaaeypai aabccadaqjEaqaaiaabgdacaqGWaGaaeOlaiaabMdaaaaabaWaaeWa aeaacaqG2baacaGLOaGaayzkaaGaaeiiaiaabAdacaqG1aGaaeymai aac6cacaqGYaGaey49aGRaaeinaiaab2dacaqGGaWaauIhaeaacaqG XaGaaeOnaiaabkdacaqGUaGaaeioaaaaaeaadaqadaqaaiaabAhaca qGPbaacaGLOaGaayzkaaGaaeiiaiaabgdacaqG0aGaaiOlaiaabsda caqG5aGaey49aGRaae4naiaacckacqGH9aqpdaqjEaqaaiaaikdaca GGUaGaaGimaiaaiEdaaaaabaGaaiiOamaabmaabaGaaeODaiaabMga caqGPbaacaGLOaGaayzkaaGaaeiiaiaabodacaGGUaGaaeyoaiaabA dacqGH3daUcaqG0aGaaeypaiaabccadaqjEaqaaiaabcdacaqGUaGa aeyoaiaabMdaaaaabaWaaeWaaeaacaqG2bGaaeyAaiaabMgacaqGPb aacaGLOaGaayzkaaGaaGimaiaac6cacaqG4aGaaGimaiabgEpa4kaa bwdacaqG9aGaaeiiamaaL4babaGaaeimaiaab6cacaqGXaGaaeOnaa aaaaaa@B6C7@

Q.43

Find: ( i ) 4.8÷10 ( ii ) 52.5÷10 ( iii ) 0.7÷10 ( iv ) 33.1÷10 ( v ) 272.23÷10 ( vi ) 0.56÷10 ( vii )3.97÷10 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWF0a Gaa8Nlaiaa=HdacaWF3dGaa8xmaiaa=bdacaWFGcGaa8hOaiaa=bka caWFGcWaaeWaaeaacaWFPbGaa8xAaaGaayjkaiaawMcaaiaa=bcaca WF1aGaa8Nmaiaa=5cacaWF1aGaa839aiaa=fdacaWFWaGaa8hOaiaa =bkacaWFGcGaa8hOaiaa=bkadaqadaqaaiaa=LgacaWFPbGaa8xAaa GaayjkaiaawMcaaiaa=bcacaWFWaGaa8Nlaiaa=DdacaWF3dGaa8xm aiaa=bdacaWLjaGaa8hOamaabmaabaGaa8xAaiaa=zhaaiaawIcaca GLPaaacaWFGaGaa83maiaa=ndacaWFUaGaa8xmaiaa=DpacaWFXaGa a8hmaaqaamaabmaabaGaa8NDaaGaayjkaiaawMcaaiaa=bcacaWFYa Gaa83naiaa=jdacaWFUaGaa8Nmaiaa=ndacaWF3dGaa8xmaiaa=bda caWFGcGaa8hOaiaa=bkadaqadaqaaiaa=zhacaWFPbaacaGLOaGaay zkaaGaa8hiaiaa=bdacaWFUaGaa8xnaiaa=zdacaWF3dGaa8xmaiaa =bdacaWFGcGaa8hOaiaa=bkacaWFGcWaaeWaaeaacaWF2bGaa8xAai aa=LgaaiaawIcacaGLPaaacaWFZaGaa8Nlaiaa=LdacaWF3aGaa839 aiaa=fdacaWFWaaaaaa@9888@

Ans.

( i ) 4.8÷10= 0.48 ( ii ) 52.5÷10 = 5.25 ( iii ) 0.7÷10= 0.07 ( iv ) 33.1÷10= 3.31 ( v ) 272.23÷10= 27.223 ( vi ) 0.56÷10= 0.056 ( vii )3.97÷10= 0.397 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaeinaiaac6cacaqG4aGaey49aGRaaeymaiaaicdacqGH9a qpdaqjEaqaaiaaicdacaGGUaGaaGinaiaaiIdaaaaabaGaaiiOamaa bmaabaGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaaeynaiaabk dacaGGUaGaaeynaiabgEpa4kaabgdacaaIWaGaaiiOaiabg2da9maa L4babaGaaGynaiaac6cacaaIYaGaaGynaaaaaeaacaGGGcWaaeWaae aacaqGPbGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaaGimaiaa c6cacaqG3aGaey49aGRaaeymaiaaicdacqGH9aqpdaqjEaqaaiaaic dacaGGUaGaaGimaiaaiEdaaaaabaWaaeWaaeaacaqGPbGaaeODaaGa ayjkaiaawMcaaiaabccacaqGZaGaae4maiaac6cacaqGXaGaey49aG RaaeymaiaaicdacqGH9aqpdaqjEaqaaiaaiodacaGGUaGaaG4maiaa igdaaaaabaWaaeWaaeaacaqG2baacaGLOaGaayzkaaGaaeiiaiaabk dacaqG3aGaaeOmaiaac6cacaqGYaGaae4maiabgEpa4kaabgdacaaI WaGaeyypa0ZaauIhaeaacaaIYaGaaG4naiaac6cacaaIYaGaaGOmai aaiodaaaaabaGaaiiOamaabmaabaGaaeODaiaabMgaaiaawIcacaGL PaaacaqGGaGaaGimaiaac6cacaqG1aGaaeOnaiabgEpa4kaabgdaca aIWaGaeyypa0ZaauIhaeaacaaIWaGaaiOlaiaaicdacaaI1aGaaGOn aaaaaeaacaGGGcWaaeWaaeaacaqG2bGaaeyAaiaabMgaaiaawIcaca GLPaaacaqGZaGaaiOlaiaabMdacaqG3aGaey49aGRaaeymaiaaicda cqGH9aqpdaqjEaqaaiaaicdacaGGUaGaaG4maiaaiMdacaaI3aaaaa aaaa@ACB1@

Q.44

Find: ( i ) 2.7÷100 ( ii ) 0.3÷100 ( iii ) 0.78÷100 ( iv ) 432.6÷100 ( v ) 23.6÷10 0 (vi)98.53÷100 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWFYa Gaa8Nlaiaa=DdacaWF3dGaa8xmaiaa=bdacaWFWaGaa8hOaiaa=bka caWFGcGaa8hOaiaaxMaadaqadaqaaiaa=LgacaWFPbaacaGLOaGaay zkaaGaa8hiaiaa=bdacaWFUaGaa83maiaa=DpacaWFXaGaa8hmaiaa =bdacaWFGcGaa8hOaiaa=bkacaWFGcGaaCzcaiaaxMaacaWFGcWaae WaaeaacaWFPbGaa8xAaiaa=LgaaiaawIcacaGLPaaacaWFGaGaa8hm aiaa=5cacaWF3aGaa8hoaiaa=DpacaWFXaGaa8hmaiaa=bdaaeaada qadaqaaiaa=LgacaWF2baacaGLOaGaayzkaaGaa8hiaiaa=rdacaWF ZaGaa8Nmaiaa=5cacaWF2aGaa839aiaa=fdacaWFWaGaa8hmaiaa=b kacaWFGcGaaCzcaiaa=bkadaqadaqaaiaa=zhaaiaawIcacaGLPaaa caWFGaGaa8Nmaiaa=ndacaWFUaGaa8Nnaiaa=DpacaWFXaGaa8hmai aa=bkacaWFWaGaaCzcaiaaxMaacaWFOaGaa8NDaiaa=LgacaWFPaGa aGjbVlaa=LdacaWF4aGaa8Nlaiaa=vdacaWFZaGaa839aiaa=fdaca WFWaGaa8hmaaaaaa@9213@

Ans.

( i ) 2.7÷100= 0.027 ( ii ) 0.3÷100 = 0.003 ( iii ) 0.78÷100= 0.0078 ( iv ) 432.6÷100 = 4.326 ( v ) 23.6÷10 0= 0.236 (vi)98.53÷100= 0.9853 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaeOmaiaac6cacaqG3aGaey49aGRaaeymaiaabcdacaaIWa Gaeyypa0ZaauIhaeaacaaIWaGaaiOlaiaaicdacaaIYaGaaG4naaaa aeaadaqadaqaaiaabMgacaqGPbaacaGLOaGaayzkaaGaaeiiaiaaic dacaGGUaGaaG4maiabgEpa4kaabgdacaqGWaGaaGimaiaacckacqGH 9aqpdaqjEaqaaiaaicdacaGGUaGaaGimaiaaicdacaaIZaGaaiiOaa aaaeaacaGGGcWaaeWaaeaacaqGPbGaaeyAaiaabMgaaiaawIcacaGL PaaacaqGGaGaaGimaiaac6cacaqG3aGaaeioaiabgEpa4kaabgdaca aIWaGaaGimaiabg2da9maaL4babaGaaGimaiaac6cacaaIWaGaaGim aiaaiEdacaaI4aaaaaqaamaabmaabaGaaeyAaiaabAhaaiaawIcaca GLPaaacaqGGaGaaGinaiaaiodacaaIYaGaaiOlaiaaiAdacqGH3daU caqGXaGaaeimaiaaicdacaGGGcGaeyypa0ZaauIhaeaacaaI0aGaai OlaiaaiodacaaIYaGaaGOnaaaaaeaadaqadaqaaiaabAhaaiaawIca caGLPaaacaqGGaGaaGOmaiaaiodacaGGUaGaaGOnaiabgEpa4kaabg dacaaIWaGaaiiOaiaaicdacqGH9aqpdaqjEaqaaiaaicdacaGGUaGa aGOmaiaaiodacaaI2aaaaaqaaiaacIcacaWG2bGaamyAaiaacMcaca aMe8UaaGyoaiaaiIdacaGGUaGaaGynaiaaiodacqGH3daUcaqGXaGa aeimaiaaicdacqGH9aqpdaqjEaqaaiaaicdacaGGUaGaaGyoaiaaiI dacaaI1aGaaG4maaaaaaaa@A5E3@

Q.45

Find: ( i ) 7.9÷1000 ( ii ) 26.3÷1000 ( iii ) 38.53÷1000 ( iv ) 128.9÷1000 ( v ) 0.5÷1000 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWF3a Gaa8Nlaiaa=LdacaWF3dGaa8xmaiaa=bdacaWFWaGaa8hmaiaa=bka caWFGcGaa8hOaiaa=bkacaWLjaWaaeWaaeaacaWFPbGaa8xAaaGaay jkaiaawMcaaiaa=bcacaWFYaGaa8Nnaiaa=5cacaWFZaGaa839aiaa =fdacaWFWaGaa8hmaiaa=bdacaWFGcGaa8hOaiaa=bkacaWFGcGaaC zcaiaaxMaacaWFGcWaaeWaaeaacaWFPbGaa8xAaiaa=LgaaiaawIca caGLPaaacaWFGaGaa83maiaa=HdacaWFUaGaa8xnaiaa=ndacaWF3d Gaa8xmaiaa=bdacaWFWaGaa8hmaaqaamaabmaabaGaa8xAaiaa=zha aiaawIcacaGLPaaacaWFGaGaa8xmaiaa=jdacaWF4aGaa8Nlaiaa=L dacaWF3dGaa8xmaiaa=bdacaWFWaGaa8hmaiaa=bkacaWFGcGaaCzc aiaa=bkadaqadaqaaiaa=zhaaiaawIcacaGLPaaacaWFGaGaa8hmai aa=5cacaWF1aGaa839aiaa=fdacaWFWaGaa8hmaiaa=bdacaWLjaGa aCzcaaaaaa@8951@

Ans.

( i ) 7.9÷1000= 0.0079 ( ii ) 26.3÷1000 = 0.0263 ( iii ) 38.53÷1000= 0.03853 ( iv ) 128.9÷1000 = 0.1289 ( v ) 0.5÷1000= 0.0005 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaG4naiaac6cacaaI5aGaey49aGRaaeymaiaabcdacaaIWa GaaGimaiabg2da9maaL4babaGaaGimaiaac6cacaaIWaGaaGimaiaa iEdacaaI5aaaaaqaamaabmaabaGaaeyAaiaabMgaaiaawIcacaGLPa aacaqGGaGaaGOmaiaaiAdacaGGUaGaaG4maiabgEpa4kaabgdacaqG WaGaaeimaiaaicdacaGGGcGaeyypa0ZaauIhaeaacaaIWaGaaiOlai aaicdacaaIYaGaaGOnaiaaiodaaaaabaGaaiiOamaabmaabaGaaeyA aiaabMgacaqGPbaacaGLOaGaayzkaaGaaeiiaiaaiodacaaI4aGaai OlaiaabwdacaqGZaGaey49aGRaaeymaiaaicdacaaIWaGaaGimaiab g2da9maaL4babaGaaGimaiaac6cacaaIWaGaaG4maiaaiIdacaaI1a GaaG4maaaaaeaadaqadaqaaiaabMgacaqG2baacaGLOaGaayzkaaGa aeiiaiaaigdacaaIYaGaaGioaiaac6cacaaI5aGaey49aGRaaeymai aabcdacaqGWaGaaGimaiaacckacqGH9aqpdaqjEaqaaiaaicdacaGG UaGaaGymaiaaikdacaaI4aGaaGyoaaaaaeaadaqadaqaaiaabAhaai aawIcacaGLPaaacaqGGaGaaGimaiaac6cacaaI1aGaey49aGRaaeym aiaabcdacaaIWaGaaGimaiabg2da9maaL4babaGaaGimaiaac6caca aIWaGaaGimaiaaicdacaaI1aaaaaaaaa@991D@

Q.46

Find: ( i ) 7÷3.5 ( ii ) 36÷0.2 ( iii ) 3.25÷0.5 ( iv ) 30.94÷0.7 ( v ) 0.5÷0.25 ( vi ) 7.75÷0.25 ( vii ) 76.5÷0.15 ( viii )37.8÷14 (ix) 2.73÷1.3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGa a8NoaaqaamaabmaabaGaa8xAaaGaayjkaiaawMcaaiaa=bcacaWF3a Gaa839aiaa=ndacaWFUaGaa8xnaiaa=bkacaWFGcGaa8hOaiaa=bka caWFGcGaaCzcamaabmaabaGaa8xAaiaa=LgaaiaawIcacaGLPaaaca WFGaGaa83maiaa=zdacaWF3dGaa8hmaiaa=5cacaWFYaGaa8hOaiaa xMaadaqadaqaaiaa=LgacaWFPbGaa8xAaaGaayjkaiaawMcaaiaa=b cacaWFZaGaa8Nlaiaa=jdacaWF1aGaa839aiaa=bdacaWFUaGaa8xn aiaaxMaadaqadaqaaiaa=LgacaWF2baacaGLOaGaayzkaaGaa8hiai aa=ndacaWFWaGaa8Nlaiaa=LdacaWF0aGaa839aiaa=bdacaWFUaGa a83naaqaamaabmaabaGaa8NDaaGaayjkaiaawMcaaiaa=bcacaWFWa Gaa8Nlaiaa=vdacaWF3dGaa8hmaiaa=5cacaWFYaGaa8xnaiaaxMaa caWLjaGaaCzcamaabmaabaGaa8NDaiaa=LgaaiaawIcacaGLPaaaca WFGaGaa83naiaa=5cacaWF3aGaa8xnaiaa=DpacaWFWaGaa8Nlaiaa =jdacaWF1aGaa8hOaiaa=bkadaqadaqaaiaa=zhacaWFPbGaa8xAaa GaayjkaiaawMcaaiaa=bcacaWF3aGaa8Nnaiaa=5cacaWF1aGaa839 aiaa=bdacaWFUaGaa8xmaiaa=vdaaeaadaqadaqaaiaa=zhacaWFPb Gaa8xAaiaa=LgaaiaawIcacaGLPaaacaWFZaGaa83naiaa=5cacaWF 4aGaa839aiaa=fdacaWF0aGaaCzcaiaaxMaacaWLjaGaa8hkaiaa=L gacaWF4bGaa8xkaiaa=bcacaWFYaGaa8Nlaiaa=DdacaWFZaGaa839 aiaa=fdacaWFUaGaa83maaaaaa@ACC2@

Ans.

( i ) 7÷3.5= 2 ( ii ) 36÷0.2= 180 ( iii ) 3.25÷0.5= 6.5 ( iv ) 30.94÷0.7= 44.2 ( v ) 0.5÷0.25=2 ( vi ) 7.75÷0.25= 31 ( vii ) 76.5÷0.15= 510 ( viii )37.8÷14= 2.7 (ix) 2.73÷1.3= 2.1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaae4naiabgEpa4kaabodacaqGUaGaaeynaiaab2dacaqGGa WaauIhaeaacaqGYaaaaaqaamaabmaabaGaaeyAaiaabMgaaiaawIca caGLPaaacaqGGaGaae4maiaabAdacqGH3daUcaqGWaGaaeOlaiaabk dacaqG9aWaauIhaeaacaqGXaGaaeioaiaabcdaaaaabaWaaeWaaeaa caqGPbGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaaG4maiaac6 cacaaIYaGaaGynaiabgEpa4kaaicdacaGGUaGaaeynaiaab2dacaqG GaWaauIhaeaacaqG2aGaaeOlaiaabwdaaaaabaWaaeWaaeaacaqGPb GaaeODaaGaayjkaiaawMcaaiaabccacaaIZaGaaGimaiaac6cacaaI 5aGaaGinaiabgEpa4kaabcdacaqGUaGaae4naiaab2dadaqjEaqaai aabsdacaqG0aGaaeOlaiaabkdaaaaabaWaaeWaaeaacaqG2baacaGL OaGaayzkaaGaaeiiaiaaicdacaGGUaGaaGynaiabgEpa4kaabcdaca qGUaGaaeOmaiaabwdacaqG9aGaaeOmaaqaamaabmaabaGaaeODaiaa bMgaaiaawIcacaGLPaaacaqGGaGaaG4naiaac6cacaaI3aGaaGynai abgEpa4kaabcdacaqGUaGaaeOmaiaabwdacaqG9aWaauIhaeaacaqG ZaGaaeymaaaaaeaacaGGGcWaaeWaaeaacaqG2bGaaeyAaiaabMgaai aawIcacaGLPaaacaqGGaGaaG4naiaaiAdacaGGUaGaaGynaiabgEpa 4kaabcdacaqGUaGaaeymaiaabwdacaqG9aGaaeiiamaaL4babaGaae ynaiaabgdacaqGWaaaaaqaamaabmaabaGaaeODaiaabMgacaqGPbGa aeyAaaGaayjkaiaawMcaaiaaiodacaaI3aGaaiOlaiaaiIdacqGH3d aUcaqGXaGaaeinaiaab2dadaqjEaqaaiaabkdacaqGUaGaae4naaaa aeaacaqGOaGaaeyAaiaabIhacaqGPaGaaeiiaiaabkdacaqGUaGaae 4naiaabodacqGH3daUcaaIXaGaaiOlaiaaiodacqGH9aqpdaqjEaqa aiaaikdacaGGUaGaaGymaaaaaaaa@BE8C@

Q.47

A vehicle covers a distance of 43.2 km in 2.4 litres of petrol.How much distance will it cover in one litre of petrol?

Ans.

Distance coverd by a vehicle in 2.4 litres = 43.2 kmSo, the distance vehicle will cover in 1 litre of petrol= 43.22.4km=18 km

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The main topics covered in Fractions and Decimals Class 7 are- addition and subtraction of decimals and fractions, multiplication of decimals and fractions, multiplication of fractions with whole numbers, multiplication of fractions with fractions, multiplication of decimal numbers by 10, 100, and 1000, division of fractions and decimals, division of whole numbers by fractions, division of a fraction by a whole number, reciprocal of fraction, division of a fraction by another fraction, and division of decimals by 10, 100, 1000. The chapter is well-structured, covering all the important sub-topics to help students prepare for their final exams.

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