NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

NCERT Solutions For Class 7 Maths Chapter 2 PDF Free Download

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals are given here. Students can either practice online or download these files and practice different types of questions related to this chapter and achieve maximum marks in their examinations. Students have learnt addition and subtraction of fractions and addition and subtraction of decimals in earlier classes. In this class, students are going to learn about multiplication and division of fraction and multiplication and division of decimals. Subject experts prepare these NCERT Solutions for Class 7 Maths for fractions and decimals to help students for their exam preparations and are available for free in downloadable PDF format.

Chapter 2 – Fractions and Decimals contains 7 exercises, and the NCERT Solutions Class 7 available on this page provides solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.

  • Addition and Subtraction of Fractions
  • Multiplication of Fraction
  • Multiplication of a Fraction by a Whole Number
  • Multiplication of a Fraction by a Fraction
  • Division of Fraction
  • Division of Whole Number by a Fraction
  • Reciprocal of Fraction
  • Division of a fraction by a Whole Number
  • Division of Fraction by Another Fraction
  • Multiplication of Decimal Numbers
  • Multiplication of Decimal Numbers by 10, 100 and 1000
  • Division of Decimal Numbers
  • Division of Decimals by 10, 100 and 1000
  • Division of a Decimal Number by a Whole Number
  • Division of a Decimal Number by Another Decimal Number

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 2 Fractions and Decimals

 

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Access answers to Maths NCERT Solutions For Class 7 Chapter 2 – Fractions and Decimals

Exercise 2.1 Page: 31

1. Solve:

(i) 2 – (3/5)

Solution:-

For subtraction of two unlike fractions, first change them to the like fractions.

LCM of 1, 5 = 5

Now, let us change each of the given fraction 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 the like fractions.

LCM of 1, 8 = 8

Now, let us change each of the given fraction 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 1

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

Solution:-

For addition of two unlike fractions, first change them to the like fractions.

LCM of 5, 7 = 35

Now, let us change each of the given fraction 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 the like fractions.

LCM of 11, 15 = 165

Now, let us change each of the given fraction 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 the like fractions.

LCM of 10, 5, 2 = 10

Now, let us change each of the given fraction 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 2

(vi) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 3

Solution:-

First convert mixed fraction into improper fraction,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 4= 8/3

= 3 ½ = 7/2

For addition of two unlike fractions, first change them to the like fractions.

LCM of 3, 2 = 6

Now, let us change each of the given fraction 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 5

(vii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 6

Solution:-

First convert mixed fraction into improper fraction,

= 8 ½ = 17/2

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 7= 29/8

For Subtraction of two unlike fractions, first change them to the like fractions.

LCM of 2, 8 = 8

Now, let us change each of the given fraction 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 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 fraction 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 fraction 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?

4/11

9/11

2/11

3/11

5/11

7/11

8/11

1/11

6/11

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 and NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 9 cm wide. Find its perimeter.

Solution:-

From the question, it is given that,

Length = 12 ½ cm = 25/2 cm

Breadth =
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 10cm = 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 is
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 11cm

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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 12

Solution:-

From the fig,

AB = (5/2) cm

AE =NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 13= 18/5 cm

BE =NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 14= 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 fraction 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

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 15cm

(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 fraction 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

= NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 16

Finally, we have find which one is having greater perimeter.

Perimeter of triangle ABE = (177/20)

Perimeter of rectangle BCDE = (47/6)

The two perimeters are in the form of unlike fraction.

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 is NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 17 cm wide. To fit in the frame the picture cannot be more than NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 18 cm wide. How much should the picture be trimmed?

Solution:-

From the question, it is given that,

Picture having a width of =NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 19= 38/5 cm

Frame having a width of =NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 20= 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 fraction 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. How much 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 apple eaten by Ritu is = (3/5)

Part of apple eaten by Somu is = 1 – Part of apple eaten by Ritu

= 1 – (3/5)

The LCM of 1, 5 = 5

Now, let us change each of the given fraction 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 apple eaten by Somu is (2/5)

So, (3/5) > (2/5) hence, Ritu ate larger size of apple.

Now, the difference between the 32 shares = (3/5) – (2/5)

= (3 – 2)/5

= 1/5

Thus, Ritu’s share is larger than 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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 21

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 part 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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 22

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 23

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 part 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 24

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 25

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 26

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 27

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 28

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 29

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 30

(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

4. Shade:

(i) ½ of the circles in box (a) (b) 2/3 of the triangles in box (b)

(iii) 3/5 of the squares in the box (c)

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 31

Solution:-

(i) From the question,

We may observe that there are 12 circles in the given box. So, we have to shade ½ of the circles in the box.

∴ 12 × ½ = 12/2

= 6

So we have to shade any 6 circles in the box.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 32

(ii) From the question,

We may observe that there are 9 triangles in the given box. So, we have to shade 2/3 of the triangles in the box.

∴ 9 × (2/3) = 18/3

= 6

So we have to shade any 6 triangles in the box.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 33

(iii) From the question,

We may observe that there are 15 squares in the given box. So, we have to shade 3/5 of the squares in the box.

∴ 15 × (3/5) = 45/5

= 9

So we have to shade any 9 squares in the box.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 34

5. Find:

(a) ½ of (i) 24 (ii) 46

Solution:-

(i) 24

We have,

= ½ × 24

= 24/2

= 12

(ii) 46

We have,

= ½ × 46

= 46/2

= 23

(b) 2/3 of (i) 18 (ii) 27

Solution:-

(i) 18

We have,

= 2/3 × 18

= 2 × 6

= 12

(ii) 27

We have,

= 2/3 × 27

= 2 × 9

= 18

(c) ¾ of (i) 16 (ii) 36

Solution:-

(i) 16

We have,

= ¾ × 16

= 3 × 4

= 12

(ii) 36

We have

= ¾ × 36

= 3 × 9

= 27

(d) 4/5 of (i) 20 (ii) 35

Solution:-

(i) 20

We have,

= 4/5 × 20

= 4 × 4

= 16

(ii) 35

We have,

= 4/5 × 35

= 4 × 7

= 28

6. Multiply and express as a mixed fraction:

(a) 3 × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 35

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 36= 26/5

Now,

= 3 × (26/5)

= 78/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 37

(b) 5 × 6 ¾

Solution:-

First convert the given mixed fraction into improper fraction.

= 6 ¾ = 27/4

Now,

= 5 × (27/4)

= 135/4

= 33 ¾

(c) 7 × 2 ¼

Solution:-

First convert the given mixed fraction into improper fraction.

= 2 ¼ = 9/4

Now,

= 7 × (9/4)

= 63/4

= 15 ¾

(d) 4 × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 38

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 39= 19/3

Now,

= 4 × (19/3)

= 76/3

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 40

(e) 3 ¼ × 6

Solution:-

First convert the given mixed fraction into improper fraction.

= 3 ¼ = 13/4

Now,

= (13/4) × 6

= (13/2) × 3

= 39/2

= 19 ½

(f) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 41 × 8

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 42= 17/5

Now,

= (17/5) × 8

= 136/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 43

7. Find:

(a) ½ of (i) 2 ¾ (ii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 44

Solution:-

(i) 2 ¾

First convert the given mixed fraction into improper fraction.

= 2 ¾ = 11/4

Now,

= ½ × 11/4

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ½ × (11/4)

= (1 × 11)/ (2 × 4)

= (11/8)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 45

(ii)
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 46

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 47= 38/9

Now,

= ½ × (38/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ½ × (38/9)

= (1 × 38)/ (2 × 9)

= (38/18)

= 19/9

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 48

(b) 5/8 of (i) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 49 (ii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 50

Solution:-

(i)
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 51

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 52= 23/6

Now,

= (5/8) × (23/6)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) × (23/6)

= (5 × 23)/ (8 × 6)

= (115/48)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 53

(ii)
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 54

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 55= 29/3

Now,

= (5/8) × (29/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5/8) × (29/3)

= (5 × 29)/ (8 × 3)

= (145/24)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 56

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 liters water. Vidya consumed 2/5 of the water. Pratap consumed the remaining water.

(i) How much water did Vidya drink?

(ii) What fraction of the total quantity of water did Pratap drink?

Solution:-

(i) From the question, it is given that,

Amount of water in the water bottle = 5 liters

Amount of water consumed by Vidya = 2/5 of 5 liters

= (2/5) × 5

= 2 liters

So, the total amount of water drank by Vidya is 2 liters

(ii) From the question, it is given that,

Amount of water in the water bottle = 5 liters

Then,

Amount of water consumed by Pratap = (1 – water consumed by Vidya)

= (1 – (2/5))

= (5-2)/5

= 3/5

∴ Total amount of water consumed by Pratap = 3/5 of 5 liters

= (3/5) × 5

= 3 liters

So, the total amount of water drank by Pratap is 3 liters


Exercise 2.3 Page: 41

1. Find:

(i) ¼ of (a) ¼ (b) 3/5 (c) 4/3

Solution:-

(a) ¼

We have,

= ¼ × ¼

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × ¼

= (1 × 1)/ (4 × 4)

= (1/16)

(b) 3/5

We have,

= ¼ × (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= ¼ × (3/5)

= (1 × 3)/ (4 × 5)

= (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) × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 57

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 58= 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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 59

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 60

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 61

(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)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 62

3. Multiply the following fractions:

(i) (2/5) × 5 ¼

Solution:-

First convert the given mixed fraction into improper fraction.

= 5 ¼ = 21/4

Now,

= (2/5) × (21/4)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2 × 21)/ (5 × 4)

= (1 × 21)/ (5 × 2)

= (21/10)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 63

(ii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 64 × (7/9)

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 65= 32/5

Now,

= (32/5) × (7/9)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (32 × 7)/ (5 × 9)

= (224/45)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 66

(iii) (3/2) × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 67

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 68= 16/3

Now,

= (3/2) × (16/3)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (3 × 16)/ (2 × 3)

= (1 × 8)/ (1 × 1)

= 8

(iv) (5/6) × NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 69

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 70= 17/7

Now,

= (5/6) × (17/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (5 × 17)/ (6 × 7)

= (85/42)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 71

(v) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 72 × (4/7)

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 73= 17/5

Now,

= (17/5) × (4/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (17 × 4)/ (5 × 7)

= (68/35)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 74

(vi)NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 75 × 3

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 76= 13/5

Now,

= (13/5) × (3/1)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (13 × 3)/ (5 × 1)

= (39/5)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 77

(vi) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 78 × (3/5)

Solution:-

First convert the given mixed fraction into improper fraction.

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 79= 25/7

Now,

= (25/7) × (3/5)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (25 × 3)/ (7 × 5)

= (5 × 3)/ (7 × 1)

= (15/7)

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 80

4. Which is greater:

(i) (2/7) of (3/4) or (3/5) of (5/8)

Solution:-

We have,

= (2/7) × (3/4) and (3/5) × (5/8)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (2/7) × (3/4)

= (2 × 3)/ (7 × 4)

= (1 × 3)/ (7 × 2)

= (3/14) … [i]

And,

= (3/5) × (5/8)

= (3 × 5)/ (5 × 8)

= (3 × 1)/ (1 × 8)

= (3/8) … [ii]

Now, convert [i] and [ii] into like fractions,

LCM of 14 and 8 is 56

Now, let us change each of the given fraction into an equivalent fraction having 56 as the denominator.

[(3/14) × (4/4)] = (12/56)

[(3/8) × (7/7)] = (21/56)

Clearly,

(12/56) < (21/56)

Hence,

(3/14) < (3/8)

(ii) (1/2) of (6/7) or (2/3) of (3/7)

Solution:-

We have,

= (1/2) × (6/7) and (2/3) × (3/7)

By the rule Multiplication of fraction,

Product of fraction = (product of numerator)/ (product of denominator)

Then,

= (1/2) × (6/7)

= (1 × 6)/ (2 × 7)

= (1 × 3)/ (1 × 7)

= (3/7) … [i]

And,

= (2/3) × (3/7)

= (2 × 3)/ (3 × 7)

= (2 × 1)/ (1 × 7)

= (2/7) … [ii]

By comparing [i] and [ii],

Clearly,

(3/7) < (2/7)

5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling.

Solution:-

From the question, it is given that,

The distance between two adjacent saplings = ¾ m

Number of saplings planted by Saili in a row = 4

Then, number of gap in saplings = ¾ × 4

= 3

∴The distance between the first and the last saplings = 3 × ¾

= (9/4) m

= 2 ¼ m

Hence, the distance between the first and the last saplings is 2 ¼ m.

6. Lipika reads a book for 1 ¾ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Solution:-

From the question, it is given that,

Lipika reads the book for = 1 ¾ hours every day = 7/4 hours

Number of days she took to read the entire book = 6 days

∴Total number of hours required by her to complete the book = (7/4) × 6

= (7/2) × 3

= 21/2

= 10 ½ hours

Hence, the total number of hours required by her to complete the book is 10 ½ hours.

7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 ¾ litres of petrol.

Solution:-

From the question, it is given that,

The total number of distance travelled by a car in 1 liter of petrol = 16 km

Then,

Total quantity of petrol = 2 ¾ liter = 11/4 liters

Total number of distance travelled by car in 11/4 liters of petrol = (11/4) × 16

= 11 × 4

= 44 km

∴Total number of distance travelled by car in 11/4 liters of petrol is 44 km.

8. (a) (i) provide the number in the box [ ], such that (2/3) × [ ] = (10/30)

Solution:-

Let the required number be x,

Then,

= (2/3) × (x) = (10/30)

By cross multiplication,

= x = (10/30) × (3/2)

= x = (10 × 3) / (30 × 2)

= x = (5 × 1) / (10 × 1)

= x = 5/10

∴The required number in the box is (5/20)

(ii) The simplest form of the number obtained in [ ] is

Solution:-

The number in the box is 5/10

Then,

The simplest form of 5/10 is ½

(b) (i) provide the number in the box [ ], such that (3/5) × [ ] = (24/75)

Solution:-

Let the required number be x,

Then,

= (3/5) × (x) = (24/75)

By cross multiplication,

= x = (24/75) × (5/3)

= x = (24 × 5) / (75 × 3)

= x = (8 × 1) / (15 × 1)

= x = 8/15

∴The required number in the box is (8/15)

(ii) The simplest form of the number obtained in [ ] is

Solution:-

The number in the box is 8/15

Then,

The simplest form of 8/15 is 8/15


Exercise 2.4 Page: 46

1. Find:

(i) 12 ÷ ¾

Solution:-

We have,

= 12 × reciprocal of ¾

= 12 × (4/3)

= 4 × 4

= 16

(ii) 14 ÷ (5/6)

Solution:-

We have,

= 14 × reciprocal of (5/6)

= 14 × (6/5)

= 84/5

(iii) 8 ÷ (7/3)

Solution:-

We have,

= 8 × reciprocal of (7/3)

= 8 × (3/7)

= (24/7)

(iv) 4 ÷ (8/3)

Solution:-

We have,

= 4 × reciprocal of (8/3)

= 4 × (3/8)

= 1 × (3/2)

= 3/2

(v) 3 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 81

Solution:-

While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 82= 7/3

Then,

= 3 ÷ (7/3)

= 3 × reciprocal of (7/3)

= 3 × (3/7)

= 9/7

(vi) 5 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 83

Solution:-

While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 84= 25/7

Then,

= 5 ÷ (25/7)

= 5 × reciprocal of (25/7)

= 5 × (7/25)

= 1 × (7/5)

= 7/5

2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

(i) 3/7

Solution:-

Reciprocal of (3/7) is (7/3) [∵ ((3/7) × (7/3)) = 1]

So, it is an improper fraction.

Improper fraction is that fraction in which numerator is greater than its denominator.

(ii) 5/8

Solution:-

Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]

So, it is an improper fraction.

Improper fraction is that fraction in which numerator is greater than its denominator.

(iii) 9/7

Solution:-

Reciprocal of (9/7) is (7/9) [∵ ((9/7) × (7/9)) = 1]

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(iv) 6/5

Solution:-

Reciprocal of (6/5) is (5/6) [∵ ((6/5) × (5/6)) = 1]

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(v) 12/7

Solution:-

Reciprocal of (12/7) is (7/12) [∵ ((12/7) × (7/12)) = 1]

So, it is a proper fraction.

A proper fraction is that fraction in which denominator is greater than the numerator of the fraction.

(vi) 1/8

Solution:-

Reciprocal of (1/8) is (8/1) or 8 [∵ ((1/8) × (8/1)) = 1]

So, it is a whole number.

Whole numbers are collection of all positive integers including 0.

(vii) 1/11

Solution:-

Reciprocal of (1/11) is (11/1) or 11 [∵ ((1/11) × (11/1)) = 1]

So, it is a whole number.

Whole numbers are collection of all positive integers including 0.

3. Find:

(i) (7/3) ÷ 2

Solution:-

We have,

= (7/3) × reciprocal of 2

= (7/3) × (1/2)

= (7 × 1) / (3 × 2)

= 7/6

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 85

(ii) (4/9) ÷ 5

Solution:-

We have,

= (4/9) × reciprocal of 5

= (4/9) × (1/5)

= (4 × 1) / (9 × 5)

= 4/45

(iii) (6/13) ÷ 7

Solution:-

We have,

= (6/13) × reciprocal of 7

= (6/13) × (1/7)

= (6 × 1) / (13 × 7)

= 6/91

(iv) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 86 ÷ 3

Solution:-

First covert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 87= 13/3

Then,

= (13/3) × reciprocal of 3

= (13/3) × (1/3)

= (13 × 1) / (3 × 3)

= 13/9

(iv) 3 ½ ÷ 4

Solution:-

First covert the mixed fraction into improper fraction.

We have,

= 3 ½ = 7/2

Then,

= (7/2) × reciprocal of 4

= (7/2) × (1/4)

= (7 × 1) / (2 × 4)

= 7/8

(iv) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 88 ÷ 7

Solution:-

First covert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 89= 31/7

Then,

= (31/7) × reciprocal of 7

= (31/7) × (1/7)

= (31 × 1) / (7 × 7)

= 31/49

4. Find:

(i) (2/5) ÷ (½)

Solution:-

We have,

= (2/5) × reciprocal of ½

= (2/5) × (2/1)

= (2 × 2) / (5 × 1)

= 4/5

(ii) (4/9) ÷ (2/3)

Solution:-

We have,

= (4/9) × reciprocal of (2/3)

= (4/9) × (3/2)

= (4 × 3) / (9 × 2)

= (2 × 1) / (3 × 1)

= 2/3

(iii) (3/7) ÷ (8/7)

Solution:-

We have,

= (3/7) × reciprocal of (8/7)

= (3/7) × (7/8)

= (3 × 7) / (7 × 8)

= (3 × 1) / (1 × 8)

= 3/8

(iv) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 90 ÷ (3/5)

Solution:-

First covert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 91= 7/3

Then,

= (7/3) × reciprocal of (3/5)

= (7/3) × (5/3)

= (7 × 5) / (3 × 3)

= 35/9

(v) 3 ½ ÷ (8/3)

Solution:-

First covert the mixed fraction into improper fraction.

We have,

= 3 ½ = 7/2

Then,

= (7/2) × reciprocal of (8/3)

= (7/2) × (3/8)

= (7 × 3) / (2 × 8)

= 21/16

(vi) (2/5) ÷ 1 ½

Solution:-

First covert the mixed fraction into improper fraction.

We have,

= 1 ½ = 3/2

Then,

= (2/5) × reciprocal of (3/2)

= (2/5) × (2/3)

= (2 × 2) / (5 × 3)

= 4/15

(vii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 92 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 93

Solution:-

First covert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 94= 16/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 95= 5/3

Then,

= (16/5) × reciprocal of (5/3)

= (16/5) × (3/5)

= (16 × 3) / (5 × 5)

= 48/25

(viii) NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 96 ÷ NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 97

Solution:-

First covert the mixed fraction into improper fraction.

We have,

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 98= 11/5

=NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 99= 6/5

Then,

= (11/5) × reciprocal of (6/5)

= (11/5) × (5/6)

= (11 × 5) / (5 × 6)

= (11 × 1) / (1 × 6)

= 11/6


Exercise 2.5 Page: 47

1. Which is grater?

(i) 0.5 or 0.05

Solution:-

By comparing whole number, 0 = 0

By comparing the tenths place digit, 5 > 0

∴ 0.5 > 0.05

(ii) 0.7 or 0.5

Solution:-

By comparing whole number, 0 = 0

By comparing the tenths place digit, 7 > 5

∴ 0.7 > 0.5

(iii) 7 or 0.7

Solution:-

By comparing whole number, 7 > 0

∴ 7 > 0.7

(iv) 1.37 or 1.49

Solution:-

By comparing whole number, 1 = 1

By comparing the tenths place digit, 3 < 4

∴ 1.37 < 1.49

(v) 2.03 or 2.30

Solution:-

By comparing whole number, 2 = 2

By comparing the tenths place digit, 0 < 3

∴ 2.03 < 2.30

(vi) 0.8 or 0.88

Solution:-

By comparing whole number, 0 = 0

By comparing the tenths place digit, 8 = 8

By comparing the hundredths place digit, 0 < 8

∴ 0.8 < 0.88

2. Express as rupees as decimals:

(i) 7 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 paise = ₹ (7/100)

= ₹ 0.07

(ii) 7 rupees 7 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 rupees 7 paise = ₹ 7 + ₹ (7/100)

= ₹ 7 + ₹ 0.07

= ₹ 7.07

(iii) 77 rupees 77 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 77 rupees 77 paise = ₹ 77 + ₹ (77/100)

= ₹ 77 + ₹ 0.77

= ₹ 7.77

(iv) 50 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 50 paise = ₹ (50/100)

= ₹ 0.50

(v) 235 paise

Solution:-

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 235 paise = ₹ (235/100)

= ₹ 2.35

3. (i) Express 5 cm in meter and kilometer

Solution:-

We know that,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 5 cm = (5/100)

= 0.05 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.05 m = (0.05/1000)

= 0. 00005 km

(i) Express 35 mm in cm, m and km

Solution:-

We know that,

= 1 cm = 10 mm

Then,

= 1 mm = (1/10) cm

= 35 mm = (35/10) cm

= 3.5 cm

And,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 3.5 cm = (3.5/100) m

= (35/1000) m

= 0.035 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.035 m = (0.035/1000)

= 0. 000035 km

4. Express in kg:

(i) 200 g

Solution:-

We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 200 g = (200/1000) kg

= (2/10)

= 0.2 kg

(ii) 3470 g

Solution:-

We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 3470 g = (3470/1000) kg

= (3470/100)

= 3.470 kg

(ii) 4 kg 8 g

Solution:-

We know that,

= 1 kg = 1000 g

Then,

= 1 g = (1/1000) kg

= 4 kg 8 g = 4 kg + (8/1000) kg

= 4 kg + 0.008

= 4.008 kg

5. Write the following decimal numbers in the expanded form:

(i) 20.03

Solution:-

We have,

20.03 = (2 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(ii) 2.03

Solution:-

We have,

2.03 = (2 × 1) + (0 × (1/10)) + (3 × (1/100))

(iii) 200.03

Solution:-

We have,

200.03 = (2 × 100) + (0 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(iv) 2.034

Solution:-

We have,

2.03 = (2 × 1) + (0 × (1/10)) + (3 × (1/100)) + (4 × (1/1000))

6. Write the place value of 2 in the following decimal numbers:

(i) 2.56

Solution:-

From the question, we observe that,

The place value of 2 in 2.56 is ones

(ii) 21.37

Solution:-

From the question, we observe that,

The place value of 2 in 21.37 is tens

(iii) 10.25

Solution:-

From the question, we observe that,

The place value of 2 in 10.25 is tenths.

(iv) 9.42

Solution:-

From the question, we observe that,

The place value of 2 in 9.42 is hundredth.

(v) 63.352

Solution:-

From the question, we observe that,

The place value of 2 in 63.352 is thousandth.

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

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals image 100

Solution:-

From the question, it is given that,

Distance travelled by Dinesh = AB + BC

= 7.5 + 12.7

= 20.2 km

∴Dinesh travelled 20.2 km

Distance travelled by Ayub = AD + DC

= 9.3 + 11.8

= 21.1 km

∴Ayub travelled 21.1km

Clearly, Ayub travelled more distance by = (21.1 – 20.2)

= 0.9 km

∴Ayub travelled 0.9 km more than Dinesh.

8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?

Solution:-

From the question, it is given that,

Fruits bought by Shyama = 5 kg 300 g

= 5 kg + (300/1000) kg

= 5 kg + 0.3 kg

= 5.3 kg

Fruits bought by Sarala = 4 kg 800 g + 4 kg 150 g

= (4 + (800/1000)) + (4 + (150/1000))

= (4 + 0.8) kg + (4 + .150) kg

= 4.8 kg + 4.150kg

= 8.950 kg

So, Sarala bought more fruits.

9. How much less is 28 km than 42.6 km?

Solution:-

Now, we have to find the difference of 42.6 km and 28 km

42.6

-28.0

14.6

∴ 14.6 km less is 28 km than 42.6 km.


Exercise 2.6 Page: 52

Find:

(i) 0.2 × 6

Solution:-

We have,

= (2/10) × 6

= (12/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 1.2

(ii) 8 × 4.6

Solution:-

We have,

= (8) × (46/10)

= (368/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 36.8

(iii) 2.71 × 5

Solution:-

We have,

= (271/100) × 5

= (1355/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 13.55

(iv) 20.1 × 4

Solution:-

We have,

= (201/10) × 4

= (804/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 80.4

(v) 0.05 × 7

Solution:-

We have,

= (5/100) × 7

= (35/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.35

(vi) 211.02 × 4

Solution:-

We have,

= (21102/100) × 4

= (84408/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 844.08

(vii) 2 × 0.86

Solution:-

We have,

= (2) × (86/100)

= (172/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 1.72

2. Find the area of rectangle whose length is 5.7cm and breadth is 3 cm.

Solution:-

From the question, it is given that,

Length of the rectangle = 5.7 cm

Breadth of the rectangle = 3 cm

Then,

Area of the rectangle = length × Breadth

= 5.7 × 3

= 17.1 cm2

3. Find:

(i) 1.3 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 1.3 × 10 = 13

(ii) 36.8 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 36.8 × 10 = 368

(iii) 153.7 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 153.7 × 10 = 1537

(iv) 168.07 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 168.07 × 10 = 1680.7

(v) 31.1 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 31.1 × 100 = 3110

(vi) 156.1 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 156.1 × 100 = 15610

(vii) 3.62 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 3.62 × 100 = 362

(viii) 43.07 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 43.07 × 100 = 4307

(ix) 0.5 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.5 × 10 = 5

(x) 0.08 × 10

Solution:-

On multiplying a decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 0.08 × 10 = 0.8

(xi) 0.9 × 100

Solution:-

On multiplying a decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 0.9 × 100 = 90

(xii) 0.03 × 1000

Solution:-

On multiplying a decimal by 1000, the decimal point is shifted to the right by three places.

We have,

= 0.03 × 1000 = 30

4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?

Solution:-

From the question, it is given that,

Distance covered by two-wheeler in 1L of petrol = 55.3 km

Then,

Distance covered by two wheeler in 10L of petrol = (10 × 55.3)

= 553 km

∴Two-wheeler covers a distance in 10L of petrol is 553 km.

5. Find:

(i) 2.5 × 0.3

Solution:-

We have,

= (25/10) × (3/10)

= (75/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.75

(ii) 0.1 × 51.7

Solution:-

We have,

= (1/10) × (517/10)

= (517/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 5.17

(iii) 0.2 × 316.8

Solution:-

We have,

= (2/10) × (3168/10)

= (6336/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 63.36

(iv) 1.3 × 3.1

Solution:-

We have,

= (13/10) × (31/10)

= (403/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 4.03

(v) 0.5 × 0.05

Solution:-

We have,

= (5/10) × (5/100)

= (25/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 0.025

(vi) 11.2 × 0.15

Solution:-

We have,

= (112/10) × (15/100)

= (1680/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 1.680

(vii) 1.07 × 0.02

Solution:-

We have,

= (107/100) × (2/100)

= (214/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 0.0214

(viii) 10.05 × 1.05

Solution:-

We have,

= (1005/100) × (105/100)

= (105525/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 10.5525

(ix) 101.01 × 0.01

Solution:-

We have,

= (10101/100) × (1/100)

= (10101/10000)

On dividing a decimal by 10000, the decimal point is shifted to the left by four places.

Then,

= 1.0101

(x) 100.01 × 1.1

Solution:-

We have,

= (10001/100) × (11/10)

= (110011/1000)

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

Then,

= 110.011


Exercise 2.7 Page: 55

1. Find:

(i) 0.4 ÷ 2

Solution:-

We have,

= (4/10) ÷ 2

Then,

= (4/10) × (1/2)

= (2/10) × (1/1)

= (2/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 0.2

(ii) 0.35 ÷ 5

Solution:-

We have,

= (35/100) ÷ 5

Then,

= (35/100) × (1/5)

= (7/100) × (1/1)

= (7/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.07

(iii) 2.48 ÷ 4

Solution:-

We have,

= (248/100) ÷ 4

Then,

= (248/100) × (1/4)

= (62/100) × (1/1)

= (62/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.62

(iv) 65.4 ÷ 6

Solution:-

We have,

= (654/10) ÷ 6

Then,

= (654/10) × (1/6)

= (109/10) × (1/1)

= (109/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 10.9

(v) 651.2 ÷ 4

Solution:-

We have,

= (6512/10) ÷ 4

Then,

= (6512/10) × (1/4)

= (1628/10) × (1/1)

= (1628/10)

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

Then,

= 162.8

(vi) 14.49 ÷ 7

Solution:-

We have,

= (1449/100) ÷ 7

Then,

= (1449/100) × (1/7)

= (207/100) × (1/1)

= (207/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 2.07

(vii) 3.96 ÷ 4

Solution:-

We have,

= (396/100) ÷ 4

Then,

= (396/100) × (1/4)

= (99/100) × (1/1)

= (99/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.99

(viii) 0.80 ÷ 5

Solution:-

We have,

= (80/100) ÷ 5

Then,

= (80/100) × (1/5)

= (16/100) × (1/1)

= (16/100)

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.16

2. Find:

(i) 4.8 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 4.8 ÷ 10

= (4.8/10)

= 0.48

(ii) 52.5 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 52.5 ÷ 10

= (52.5/10)

= 5.25

(iii) 0.7 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.7 ÷ 10

= (0.7/10)

= 0.07

(iv) 33.1 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 33.1 ÷ 10

= (33.1/10)

= 3.31

(v) 272.23 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 272.23 ÷ 10

= (272.23/10)

= 27.223

(vi) 0.56 ÷ 10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 0.56 ÷ 10

= (0.56/10)

= 0.056

(vii) 3.97 ÷10

Solution:-

On dividing a decimal by 10, the decimal point is shifted to the left by one place.

We have,

= 3.97 ÷ 10

= (3.97/10)

= 0.397

3. Find:

(i) 2.7 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 2.7 ÷ 100

= (2.7/100)

= 0.027

(ii) 0.3 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.3 ÷ 100

= (0.3/100)

= 0.003

(iii) 0.78 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 0.78 ÷ 100

= (0.78/100)

= 0.0078

(iv) 432.6 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 432.6 ÷ 100

= (432.6/100)

= 4.326

(v) 23.6 ÷100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 23.6 ÷ 100

= (23.6/100)

= 0.236

(vi) 98.53 ÷ 100

Solution:-

On dividing a decimal by 100, the decimal point is shifted to the left by two places.

We have,

= 98.53 ÷ 100

= (98.53/100)

= 0.9853

4. Find:

(i) 7.9 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 7.9 ÷ 1000

= (7.9/1000)

= 0.0079

(ii) 26.3 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 26.3 ÷ 1000

= (26.3/1000)

= 0.0263

(iii) 38.53 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 38.53 ÷ 1000

= (38.53/1000)

= 0.03853

(iv) 128.9 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 128.9 ÷ 1000

= (128.9/1000)

= 0.1289

(v) 0.5 ÷ 1000

Solution:-

On dividing a decimal by 1000, the decimal point is shifted to the left by three places.

We have,

= 0.5 ÷ 1000

= (0.5/1000)

= 0.0005

5. Find:

(i) 7 ÷ 3.5

Solution:-

We have,

= 7 ÷ (35/10)

= 7 × (10/35)

= 1 × (10/5)

= 2

(ii) 36 ÷ 0.2

Solution:-

We have,

= 36 ÷ (2/10)

= 36 × (10/2)

= 18 × 10

= 180

(iii) 3.25 ÷ 0.5

Solution:-

We have,

= (325/100) ÷ (5/10)

= (325/100) × (10/5)

= (325 × 10)/ (100 × 5)

= (65 × 1)/ (10 × 1)

= 65/10

= 6.5

(iv) 30.94 ÷ 0.7

Solution:-

We have,

= (3094/100) ÷ (7/10)

= (3094/100) × (10/7)

= (3094 × 10)/ (100 × 7)

= (442 × 1)/ (10 × 1)

= 442/10

= 44.2

(v) 0.5 ÷ 0.25

Solution:-

We have,

= (5/10) ÷ (25/100)

= (5/10) × (100/25)

= (5 × 100)/ (10 × 25)

= (1 × 10)/ (1 × 5)

= 10/5

= 2

(vi) 7.75 ÷ 0.25

Solution:-

We have,

= (775/100) ÷ (25/100)

= (775/100) × (100/25)

= (775 × 100)/ (100 × 25)

= (155 × 1)/ (1 × 5)

= (31 × 1)/ (1 × 1)

= 31

(vii) 76.5 ÷ 0.15

Solution:-

We have,

= (765/10) ÷ (15/100)

= (765/10) × (100/15)

= (765 × 100)/ (10 × 15)

= (51 × 10)/ (1 × 1)

= 510

(viii) 37.8 ÷ 1.4

Solution:-

We have,

= (378/10) ÷ (14/10)

= (378/10) × (10/14)

= (378 × 10)/ (10 × 14)

= (27 × 1)/ (1 × 1)

= 27

(ix) 2.73 ÷ 1.3

Solution:-

We have,

= (273/100) ÷ (13/10)

= (273/100) × (10/13)

= (273 × 10)/ (100 × 13)

= (21 × 1)/ (10 × 1)

= 21/10

= 2.1

6. 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?

Solution:-

From the question, it is given that,

Total distance covered by vehicle in 2.4 litres of petrol = 43.2 km

Then,

Distance covered in 1 litre of petrol = 43.2 ÷ 2.4

= (432/10) ÷ (24/10)

= (432/10) × (10/24)

= (432 × 10)/ (10 × 24)

= (36 × 1)/ (1 × 2)

= (18 × 1)/ (1 × 1)

= 18 km

∴Total distance covered in 1 liter of petrol is 18 km.


Also, access other exercises of NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals

Exercise 2.1 Solutions

Exercise 2.2 Solutions

Exercise 2.3 Solutions

Exercise 2.4 Solutions

Exercise 2.5 Solutions

Exercise 2.6 Solutions

Exercise 2.7 Solutions

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