NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT Solutions for Class 7 Maths Chapter 2

Fractions and Decimals for Class 7 is an important topic covered in Class 7 Maths Chapter 2. This chapter helps students understand how fractions and decimals are used in daily life, such as while measuring quantities, handling money, or comparing values. Fractions and decimals class 7 focuses on different types of fractions, conversion between fractions and decimals, and operations on them.

In class 7 maths chapter 2, students learn how to add, subtract, multiply, and divide fractions and decimals using easy methods and clear rules. These concepts are explained step by step, helping students avoid common mistakes. Practising regularly builds accuracy and confidence in calculations.

Class 7 maths chapter 2 solutions are designed to help students understand the correct approach to solving textbook problems. The class 7 maths chapter 2 NCERT solutions provide clear explanations and solved examples strictly based on the latest NCERT syllabus. They are useful for homework, exams, and quick revision.

The fractions and decimals class 7 questions answers section includes a variety of problems to test understanding and improve problem-solving skills. With fractions and decimals class 7 NCERT solutions, students can check their answers, learn proper methods, and strengthen their basics. Overall, mastering this chapter prepares students for advanced maths topics in higher classes.

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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

Below we have provided NCERT Solutions for Class 7 Maths Chapter 2 for students to help them understand the NCERT Solutions for Class 7 Maths Chapter 2 chapter better and to score good marks in their examination.

1. Solve:

(i) 2 – (3/5)

Solution:-

For subtraction of two unlike fractions, first change them to like fractions. LCM of 1, 5 = 5 Now, let us change each of the given fractions into an equivalent fraction having 5 as the denominator. = [(2/1) × (5/5)] = (10/5) = [(3/5) × (1/1)] = (3/5) Now, = (10/5) – (3/5) = [(10 – 3)/5] = (7/5)

(ii) 4 + (7/8)

Solution:-

For addition of two unlike fractions, first change them to like fractions. LCM of 1, 8 = 8 Now, let us change each of the given fractions into an equivalent fraction having 8 as the denominator. = [(4/1) × (8/8)] = (32/8) = [(7/8) × (1/1)] = (7/8) Now, = (32/8) + (7/8) = [(32 + 7)/8] = (39/8) =

(iii) (3/5) + (2/7)

Solution:-

For addition of two unlike fractions, first change them to like fractions. LCM of 5, 7 = 35 Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator. = [(3/5) × (7/7)] = (21/35) = [(2/7) × (5/5)] = (10/35) Now, = (21/35) + (10/35) = [(21 + 10)/35] = (31/35)

(iv) (9/11) – (4/15)

Solution:-

For subtraction of two unlike fractions, first change them to like fractions. LCM of 11, 15 = 165 Now, let us change each of the given fractions into an equivalent fraction having 165 as the denominator. = [(9/11) × (15/15)] = (135/165) = [(4/15) × (11/11)] = (44/165) Now, = (135/165) – (44/165) = [(135 – 44)/165] = (91/165)

(v) (7/10) + (2/5) + (3/2)

Solution:-

For addition of two unlike fractions, first change them to like fractions. LCM of 10, 5, 2 = 10 Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator. = [(7/10) × (1/1)] = (7/10) = [(2/5) × (2/2)] = (4/10) = [(3/2) × (5/5)] = (15/10) Now, = (7/10) + (4/10) + (15/10) = [(7 + 4 + 15)/10] = (26/10) = (13/5) =

(vi)

Solution:-

First convert mixed fraction into improper fraction, == 8/3 = 3 ½ = 7/2 For addition of two unlike fractions, first change them to like fractions. LCM of 3, 2 = 6 Now, let us change each of the given fractions into an equivalent fraction having 6 as the denominator. = [(8/3) × (2/2)] = (16/6) = [(7/2) × (3/3)] = (21/6) Now, = (16/6) + (21/6) = [(16 + 21)/6] = (37/6) =

(vii)

Solution:-

First convert mixed fraction into improper fraction, = 8 ½ = 17/2 == 29/8 For subtraction of two unlike fractions, first change them to like fractions. LCM of 2, 8 = 8 Now, let us change each of the given fractions into an equivalent fraction having 35 as the denominator. = [(17/2) × (4/4)] = (68/8) = [(29/8) × (1/1)] = (29/8) Now, = (68/8) – (29/8) = [(68 – 29)/8] = (39/8) =

2. Arrange the following in descending order:

(i) 2/9, 2/3, 8/21

Solution:-

LCM of 9, 3, 21 = 63 Now, let us change each of the given fractions into an equivalent fraction having 63 as the denominator. [(2/9) × (7/7)] = (14/63) [(2/3) × (21/21)] = (42/63) [(8/21) × (3/3)] = (24/63) Clearly, (42/63) > (24/63) > (14/63) Hence, (2/3) > (8/21) > (2/9) Hence, the given fractions in descending order are (2/3), (8/21), (2/9)

(ii) 1/5, 3/7, 7/10

Solution:-

LCM of 5, 7, 10 = 70 Now, let us change each of the given fractions into an equivalent fraction having 70 as the denominator. [(1/5) × (14/14)] = (14/70) [(3/7) × (10/10)] = (30/70) [(7/10) × (7/7)] = (49/70) Clearly, (49/70) > (30/70) > (14/70) Hence, (7/10) > (3/7) > (1/5) Hence, the given fractions in descending order are (7/10), (3/7), (1/5)

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?

Solution:-

Sum along the first row = (4/11) + (9/11) + (2/11) = (15/11) Sum along the second row = (3/11) + (5/11) + (7/11) = (15/11) Sum along the third row = (8/11) + (1/11) + (6/11) = (15/11) Sum along the first column = (4/11) + (3/11) + (8/11) = (15/11) Sum along the second column = (9/11) + (5/11) + (1/11) = (15/11) Sum along the third column = (2/11) + (7/11) + (6/11) = (15/11) Sum along the first diagonal = (4/11) + (5/11) + (6/11) = (15/11) Sum along the second diagonal = (2/11) + (5/11) + (8/11) = (15/11) Yes. The sum of the numbers in each row, in each column and along the diagonals is the same, so it is a magic square.

4. A rectangular sheet of paper is 12 ½ cm long andcm wide. Find its perimeter.

Solution:-

From the question, it is given that, Length = 12 ½ cm = 25/2 cm Breadth =cm = 32/3 cm We know that, Perimeter of the rectangle = 2 × (length + breadth) = 2 × [(25/2) + (32/3)] = 2 × {[(25 × 3) + (32 × 2)]/6} = 2 × [(75 + 64)/6] = 2 × [139/6] = 139/3 cm Hence, the perimeter of the sheet of paper iscm

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

Solution:-

From the fig, AB = (5/2) cm AE == 18/5 cm BE == 11/4 cm ED = 7/6 cm (i) We know that, Perimeter of the triangle = Sum of all sides Then, Perimeter of triangle ABE = AB + BE + EA = (5/2) + (11/4) + (18/5) The LCM of 2, 4, 5 = 20 Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator. = {[(5/2) × (10/10)] + [(11/4) × (5/5)] + [(18/5) × (4/4)]} = (50/20) + (55/20) + (72/20) = (50 + 55 + 72)/20 = 177/20 =cm (ii) Now, we have to find the perimeter of the rectangle, We know that, Perimeter of the rectangle = 2 × (length + breadth) Then, Perimeter of rectangle BCDE = 2 × (BE + ED) = 2 × [(11/4) + (7/6)] The LCM of 4, 6 = 12 Now, let us change each of the given fractions into an equivalent fraction having 20 as the denominator = 2 × {[(11/4) × (3/3)] + [(7/6) × (2/2)]} = 2 × [(33/12) + (14/12)] = 2 × [(33 + 14)/12] = 2 × (47/12) = 47/6 =Finally, we have to find which one is having a greater perimeter. Perimeter of triangle ABE = (177/20) Perimeter of rectangle BCDE = (47/6) The two perimeters are in the form of unlike fractions. Changing perimeters into like fractions we have, (177/20) = (177/20) × (3/3) = 531/60 (43/6) = (43/6) × (10/10) = 430/60 Clearly, (531/60) > (430/60) Hence, (177/20) > (43/6) ∴ Perimeter of Triangle ABE > Perimeter of Rectangle (BCDE)

6. Salil wants to put a picture in a frame. The picture iscm wide. To fit in the frame the picture cannot be more thancm wide. How much should the picture be trimmed?

Solution:-

From the question, it is given that, Picture having a width of == 38/5 cm Frame having a width of == 73/10 cm ∴ The picture should be trimmed by = [(38/5) – (73/10)] The LCM of 5, 10 = 10 Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator. = [(38/5) × (2/2)] – [(73/10) × (1/1)] = (76/10) – (73/10) = (76 – 73)/10 = 3/10 cm Thus, the picture should be trimmed by (3/10) cm

7. Ritu ate (3/5) part of an apple and the remaining apple was eaten by her brother Somu. What part of the apple did Somu eat? Who had the larger share? By how much?

Solution:-

From the question, it is given that, Part of the apple eaten by Ritu is = (3/5) Part of the apple eaten by Somu is = 1 – Part of the apple eaten by Ritu = 1 – (3/5) The LCM of 1, 5 = 5 Now, let us change each of the given fractions into an equivalent fraction having 10 as the denominator. = [(1/1) × (5/5)] – [(3/5) × (1/1)] = (5/5) – (3/5) = (5 – 3)/5 = 2/5 ∴ Part of the apple eaten by Somu is (2/5) So, (3/5) > (2/5) hence, Ritu ate larger size of the apple. Now, the difference between the 32 shares = (3/5) – (2/5) = (3 – 2)/5 = 1/5 Thus, Ritu’s share is larger than the share of Somu by (1/5)

8. Michael finished colouring a picture in (7/12) hour. Vaibhav finished colouring the same picture in (3/4) hour. Who worked longer? By what fraction was it longer?

Solution:-

From the question, it is given that, Time taken by the Michael to colour the picture is = (7/12) Time taken by the Vaibhav to colour the picture is = (3/4) The LCM of 12, 4 = 12 Now, let us change each of the given fraction into an equivalent fraction having 12 as the denominator. (7/12) = (7/12) × (1/1) = 7/12 (3/4) = (3/4) × (3/3) = 9/12 Clearly, (7/12) < (9/12) Hence, (7/12) < (3/4) Thus, Vaibhav worked for longer time. So, Vaibhav worked longer time by = (3/4) – (7/12) = (9/12) – (7/12) = (9 – 7)/12 = (2/12) = (1/6) of an hour.

Exercise 2.2 Page: 36

1. Which of the drawings (a) to (d) show:

(i) 2 × (1/5) (ii) 2 × ½ (iii) 3 × (2/3) (iv) 3 × ¼

Solution:-

(i) 2 × (1/5) represents the addition of 2 figures, each represents 1 shaded part out of the given 5 equal parts. ∴ 2 × (1/5) is represented by fig (d). (ii) 2 × ½ represents the addition of 2 figures, each represents 1 shaded part out of the given 2 equal parts. ∴ 2 × ½ is represented by fig (b). (iii) 3 × (2/3) represents the addition of 3 figures, each represents 2 shaded parts out of the given 3 equal parts. ∴ 3 × (2/3) is represented by fig (a). (iii) 3 × ¼ represents the addition of 3 figures, each represents 1 shaded part out of the given 4 equal parts. ∴ 3 × ¼ is represented by fig (c).

2. Some pictures (a) to (c) are given below. Tell which of them show:

(i) 3 × (1/5) = (3/5) (ii) 2 × (1/3) = (2/3) (iii) 3 × (3/4) = 2 ¼

Solution:-

(i) 3 × (1/5) represents the addition of 3 figures, each represents 1 shaded part out of the given 5 equal parts and (3/5) represents 3 shaded parts out of 5 equal parts. ∴ 3 × (1/5) = (3/5) is represented by fig (c). (ii) 2 × (1/3) represents the addition of 2 figures, each represents 1 shaded part out of the given 3 equal parts and (2/3) represents 2 shaded parts out of 3 equal parts. ∴ 2 × (1/3) = (2/3) is represented by fig (a). (iii) 3 × (3/4) represents the addition of 3 figures, each represents 3 shaded parts out of the given 4 equal parts and 2 ¼ represents 2 fully and 1 figure having 1 part as shaded out of 4 equal parts. ∴ 3 × (3/4) = 2 ¼ is represented by fig (b).

3. Multiply and reduce to lowest form and convert into a mixed fraction:

(i) 7 × (3/5)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (7/1) × (3/5) = (7 × 3)/ (1 × 5) = (21/5) =

(ii) 4 × (1/3)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (4/1) × (1/3) = (4 × 1)/ (1 × 3) = (4/3) =

(iii) 2 × (6/7)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (2/1) × (6/7) = (2 × 6)/ (1 × 7) = (12/7) =

(iv) 5 × (2/9)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (5/1) × (2/9) = (5 × 2)/ (1 × 9) = (10/9) =

(v) (2/3) × 4

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (2/3) × (4/1) = (2 × 4)/ (3 × 1) = (8/3) =

(vi) (5/2) × 6

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (5/2) × (6/1) = (5 × 6)/ (2 × 1) = (30/2) = 15

(vii) 11 × (4/7)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (11/1) × (4/7) = (11 × 4)/ (1 × 7) = (44/7) =

(viii) 20 × (4/5)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (20/1) × (4/5) = (20 × 4)/ (1 × 5) = (80/5) = 16

(ix) 13 × (1/3)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (13/1) × (1/3) = (13 × 1)/ (1 × 3) = (13/3) =

(x) 15 × (3/5)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (15/1) × (3/5) = (15 × 3)/ (1 × 5) = (45/5) = 9 = (3/20) (c) (4/3) We have, = ¼ × (4/3) By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = ¼ × (4/3) = (1 × 4)/ (4 × 3) = (4/12) = 1/3

(ii) 1/7 of (a) 2/9 (b) 6/5 (c) 3/10

Solution:-

(a) 2/9 We have, = (1/7) × (2/9) By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (1/7) × (2/9) = (1 × 2)/ (7 × 9) = (2/63) (b) 6/5 We have, = (1/7) × (6/5) By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (1/7) × (6/5) = (1 × 6)/ (7 × 5) = (6/35) (c) 3/10 We have, = (1/7) × (3/10) By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (1/7) × (3/10) = (1 × 3)/ (7 × 10) = (3/70)

2. Multiply and reduce to lowest form (if possible):

(i) (2/3) ×

Solution:-

First convert the given mixed fraction into improper fraction. == 8/3 Now, = (2/3) × (8/3) By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (2 × 8)/ (3 × 3) = (16/9) =

(ii) (2/7) × (7/9)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (2 × 7)/ (7 × 9) = (2 × 1)/ (1 × 9) = (2/9)

(iii) (3/8) × (6/4)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (3 × 6)/ (8 × 4) = (3 × 3)/ (4 × 4) = (9/16)

(iv) (9/5) × (3/5)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (9 × 3)/ (5 × 5) = (27/25) =

(v) (1/3) × (15/8)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (1 × 15)/ (3 × 8) = (1 × 5)/ (1 × 8) = (5/8)

(vi) (11/2) × (3/10)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (11 × 3)/ (2 × 10) = (33/20) =

(vii) (4/5) × (12/7)

Solution:-

By the rule Multiplication of fraction, Product of fraction = (product of numerator)/ (product of denominator) Then, = (4 × 12)/ (5 × 7) = (48/35) =

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Decimals are just another way to express fractions with denominators like 10, 100, or 1000. Understanding place value is the secret to mastering decimals. As you move to the right of the decimal point, the values become tenths, hundredths, and thousandths. When you multiply decimals, you ignore the decimal point initially and then count the total decimal places in the original numbers to place the point in your final answer.

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