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
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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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