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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

by aglasem
August 23, 2021
in 7th Class
Reading Time: 1 min read
0
NCERT Solutions

NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals – Here are all the NCERT solutions for Class 7 Maths Chapter 2. This solution contains questions, answers, images, explanations of the complete chapter 2 titled Fractions and Decimals of Maths taught in class 7. If you are a student of class 7 who is using NCERT Textbook to study Maths, then you must come across chapter 2 Fractions and Decimals. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals in one place.

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NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals

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Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 7 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 2 Fractions and Decimals , Maths, Class 7.

Class 7
Subject Maths
Book Mathematics
Chapter Number 2
Chapter Name

Fractions and Decimals

NCERT Solutions Class 7 Maths chapter 2 Fractions and Decimals

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Class 7, Maths chapter 2, Fractions and Decimals solutions are given below in PDF format. You can view them online or download PDF file for future use.

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Question & Answer

Q.1: Solve: 
(i) \( 2 - \frac { 3 } { 5 } \) 
(ii) \( 4 + \frac { 7 } { 8 } \) 
(iii) \( \frac { 3 } { 5 } + \frac { 2 } { 7 } \) 
(iv) \( \frac { 9 } { 11 } - \frac { 4 } { 15 } \) 
(v) \( \frac { 7 } { 10 } + \frac { 2 } { 5 } + \frac { 3 } { 2 } \) 
(vi) \( 2 \frac { 2 } { 3 } + 3 \frac { 1 } { 2 } \) 
(vii) \( 8 \frac { 1 } { 2 } - 3 \frac { 5 } { 8 } \)

Ans : (i) \( 2 - \frac { 3 } { 5 } = \frac { 2 \times 5 } { 5 } - \frac { 3 } { 5 } = \frac { 10 - 3 } { 5 } = \frac { 7 } { 5 } \) (ii) \( 4 + \frac { 7 } { 8 } = \frac { 4 \times 8 } { 8 } + \frac { 7 } { 8 } = \frac { ( 4 \times 8 ) + 7 } { 8 } = \frac { 39 } { 8 } = 4 \frac { 7 } { 8 } \) (iii) \( \frac { 3 } { 5 } + \frac { 2 } { 7 } = \frac { 3 \times 7 } { 5 \times 7 } + \frac { 2 \times 5 } { 7 \times 5 } = \frac { 21 + 10 } { 35 } = \frac { 31 } { 35 } \) (iv) \( \frac { 9 } { 11 } - \frac { 4 } { 15 } = \frac { 9 \times 15 } { 11 \times 15 } - \frac { 4 \times 11 } { 15 \times 11 } = \frac { 135 - 44 } { 165 } = \frac { 91 } { 165 } \) (v) \( \frac { 7 } { 10 } + \frac { 2 } { 5 } + \frac { 3 } { 2 } = \frac { 7 } { 10 } + \frac { 2 \times 2 } { 5 \times 2 } + \frac { 3 \times 5 } { 2 \times 5 } = \frac { 7 + 4 + 15 } { 10 } = \frac { 26 } { 10 } = \frac { 13 } { 5 } = 2 \frac { 3 } { 5 } \) (vi) \( 2 \frac { 2 } { 3 } + 3 \frac { 1 } { 2 } = \frac { 8 } { 3 } + \frac { 7 } { 2 } = \frac { 8 \times 2 } { 3 \times 2 } + \frac { 7 \times 3 } { 2 \times 3 } = \frac { 16 + 21 } { 6 } = \frac { 37 } { 6 } = 6 \frac { 1 } { 6 } \) (vii) \( 8 \frac { 1 } { 2 } - 3 \frac { 5 } { 8 } = \frac { 17 } { 2 } - \frac { 29 } { 8 } = \frac { 17 \times 4 } { 2 \times 4 } - \frac { 29 } { 8 } = \frac { 68 - 29 } { 8 } = \frac { 39 } { 8 } = 4 \frac { 7 } { 8 } \)

Q.2: Arrange the following in descending order:
(i) \( \frac { 2 } { 9 } , \frac { 2 } { 3 } , \frac { 8 } { 21 } \) 
(ii) \( \frac { 1 } { 5 } , \frac { 3 } { 7 } , \frac { 7 } { 10 } \)

Ans : (i) Changing them to like fractions, we obtain \( \begin{array} { l } { \frac { 2 } { 9 } = \frac { 2 \times 7 } { 9 \times 7 } = \frac { 14 } { 63 } } \\ { \frac { 2 } { 3 } = \frac { 2 \times 21 } { 3 \times 21 } = \frac { 42 } { 63 } } \\ { \frac { 8 } { 21 } = \frac { 8 \times 3 } { 21 \times 3 } = \frac { 24 } { 63 } } \end{array} \) Since \( 42 > 24 > 14 \) , \( \therefore \frac { 2 } { 3 } > \frac { 8 } { 21 } > \frac { 2 } { 9 } \) (ii) Changing them to like fractions, we obtain \( \begin{array} { l } { \frac { 1 } { 5 } = \frac { 1 \times 14 } { 5 \times 14 } = \frac { 14 } { 70 } } \\ { \frac { 3 } { 7 } = \frac { 3 \times 10 } { 7 \times 10 } = \frac { 30 } { 70 } } \\ { \frac { 7 } { 10 } = \frac { 7 \times 7 } { 10 \times 7 } = \frac { 49 } { 70 } } \end{array} \) As \( 49 > 30 > 14 \) \( \therefore \frac { 7 } { 10 } > \frac { 3 } { 7 } > \frac { 1 } { 5 } \)

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

Ans : Along the first row, sum = \( \frac { 4 } { 11 } + \frac { 9 } { 11 } + \frac { 2 } { 11 } = \frac { 15 } { 11 } \) Along the second row, sum = \( \frac { 3 } { 11 } + \frac { 5 } { 11 } + \frac { 7 } { 11 } = \frac { 15 } { 11 } \) Along the third row, sum = \( \frac { 8 } { 11 } + \frac { 1 } { 11 } + \frac { 6 } { 11 } = \frac { 15 } { 11 } \) Along the first column, sum = \( \frac { 4 } { 11 } + \frac { 3 } { 11 } + \frac { 8 } { 11 } = \frac { 15 } { 11 } \) Along the second column, sum = \( \frac { 9 } { 11 } + \frac { 5 } { 11 } + \frac { 1 } { 11 } = \frac { 15 } { 11 } \) Along the third column, sum = \( \frac { 2 } { 11 } + \frac { 7 } { 11 } + \frac { 6 } { 11 } = \frac { 15 } { 11 } \) Along the first diagonal, sum = \( \frac { 2 } { 11 } + \frac { 5 } { 11 } + \frac { 8 } { 11 } = \frac { 15 } { 11 } \) Since the sum of the numbers in each row, in each column, and along the diagonals is the same, it is a magic square.

Q.4: A rectangular sheet of paper is \( 12 \frac { 1 } { 2 } \) cm long and \( 10 \frac { 2 } { 3 } \) cm wide. Find its perimeter.

Ans : Length = \( 12 \frac { 1 } { 2 } \mathrm { cm } = \frac { 25 } { 2 } \mathrm { cm } \) Breadth = \( 10 \frac { 2 } { 3 } \mathrm { cm } = \frac { 32 } { 3 } \mathrm { cm } \) Perimeter = \( 2 \times ( \text { Length } + \text { Breadth } ) \) \( = 2 \times \left[ \frac { 25 } { 2 } + \frac { 32 } { 3 } \right] = 2 \times \left[ \frac { ( 25 \times 3 ) + ( 32 \times 2 ) } { 6 } \right] = 2 \times \left[ \frac { 75 + 64 } { 6 } \right] \) \( = 2 \times \frac { 139 } { 6 } = \frac { 139 } { 3 } = 46 \frac { 1 } { 3 } \mathrm { cm } \)

Q.5: Find the perimeters of (i) \( \triangle \mathrm { ABE } \) (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Ans : (i) Perimeter of \( \triangle \mathrm { ABE } \) = AB + BE+ EA \( = \left( \frac { 5 } { 2 } + 2 \frac { 3 } { 4 } + 3 \frac { 3 } { 5 } \right) = \left( \frac { 5 } { 2 } + \frac { 11 } { 4 } + \frac { 18 } { 5 } \right) \) \( = \left( \frac { 5 \times 10 } { 2 \times 10 } + \frac { 11 \times 5 } { 4 \times 5 } + \frac { 18 \times 4 } { 5 \times 4 } \right) \) \( = \frac { 50 + 55 + 72 } { 20 } = \frac { 177 } { 20 } = 8 \frac { 17 } { 20 } \mathrm { cm } \) (ii) Perimeter of rectangle = 2 ( Length + Breadth ) Perimeter of rectangle = \( 2 \left[ \frac { 11 } { 4 } + \frac { 7 } { 6 } \right] \) \( 2 \left[ \frac { 11 \times 3 } { 4 \times 3 } + \frac { 7 \times 2 } { 6 \times 2 } \right] = 2 \left[ \frac { 33 + 14 } { 12 } \right] \) \( 2 \times \frac { 47 } { 12 } = \frac { 47 } { 6 } = 7 \frac { 5 } { 6 } \mathrm { cm } \) Perimeter of \( \triangle \mathrm { ABE } \) = \( \frac { 177 } { 20 } \mathrm { cm } \) Changing them to like fractions, we obtain \( \frac { 177 } { 20 } = \frac { 177 \times 3 } { 20 \times 3 } = \frac { 531 } { 60 } \) \( \frac { 43 } { 6 } = \frac { 43 \times 10 } { 6 \times 10 } = \frac { 430 } { 60 } \) As \( 531 > 430 \) \( \therefore \frac { 177 } { 20 } > \frac { 43 } { 6 } \) Perimeter ( \( \triangle \mathrm { ABE } \) ) > Perimeter (BCDE)

NCERT / CBSE Book for Class 7 Maths

You can download the NCERT Book for Class 7 Maths in PDF format for free. Otherwise you can also buy it easily online.

  • Click here for NCERT Book for Class 7 Maths
  • Click here to buy NCERT Book for Class 7 Maths

All NCERT Solutions Class 7

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You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

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