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.
NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals
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| Class | 7 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 2 |
| Chapter Name |
Fractions and Decimals |
NCERT Solutions Class 7 Maths chapter 2 Fractions and Decimals
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.
Fractions and Decimals
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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)
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