41
Class- VII-CBSE-Mathematics Fractions and Decimals Practice more on Fractions and Decimals Page - 1 www.embibe.com CBSE NCERT Solutions for Class 7 Mathematics Chapter 2 Back of Chapter Questions Exercise: 2.1 1. Solve: (i) 2 3 5 (ii) 4+ 7 8 (iii) 3 5 + 2 7 (iv) 9 11 4 15 (v) 7 10 + 2 5 + 3 2 (vi) 2 2 3 +3 1 2 (vii) 8 1 2 3 5 8 Solution: (i) 2 3 5 2 1 × 5 5 = 10 5 (since the LCM of the denominators 1 and 5 is 5) 10 5 3 5 = 10 3 5 = 7 5 =1 2 5 (7÷5 gives quotient 1 and remainder 2) (ii) 4+ 7 8 4 1 × 8 8 = 32 8 (since the LCM of the denominators 1 and 8 is 8) 32 8 + 7 8 = 32 + 7 8 = 39 8 =4 7 8 (39 ÷ 8 gives quotient 4 and remainder 7)

CBSE NCERT Solutions for Class 7 Mathematics …...Class- VII-CBSE-Mathematics Fractions and Decimals Practice more on Fractions and Decimals Page - 1 CBSE NCERT Solutions for Class

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Class- VII-CBSE-Mathematics Fractions and Decimals

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CBSE NCERT Solutions for Class 7 Mathematics Chapter 2 Back of Chapter Questions

Exercise: 2.1

1. Solve:

(i) 2 − 35

(ii) 4 + 78

(iii) 3 5

+ 27

(iv) 911− 4

15

(v) 710

+ 25

+ 32

(vi) 2 23

+ 3 12

(vii) 8 12− 3 5

8

Solution:

(i) 2 − 35

21

× 55

= 105

(since the LCM of the denominators 1 and 5 is 5)

105−

35

=10 − 3

5

=75

= 1 25 (7÷5 gives quotient 1 and remainder 2)

(ii) 4 + 78

41

× 88

= 328

(since the LCM of the denominators 1 and 8 is 8)

328

+78

=32 + 7

8

=398

= 4 78 (39 ÷ 8 gives quotient 4 and remainder 7)

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(iii) 3 5

+ 27

35

× 77

= 2135

and 27

× 55

= 1035

(since the LCM of the denominators 5 and 7 is 35)

2135

+1035

=21 + 10

35

=3135

(iv) 911− 4

15

911

× 1515

= 135165

and 415

× 1111

= 44165

(since the LCM of the denominators 11 and 15 is 165)

135165

−44

165=

135 − 44165

=91

165

(v) 710

+ 25

+ 32

= 7+4+1510

(since the LCM of the denominators 10, 5 and 2 is 10)

=2610

=135

= 2 35

(13 ÷ 5 gives quotient 2 and remainder 3)

(vi) 2 23

+ 3 12

= 83

+ 72

= 16+216

(since the LCM of the denominators 3 and 2 is 6)

=376

= 6 16 (37 ÷ 6 gives quotient 6 and remainder 1)

(vii) 8 12− 3 5

8= 17

2− 29

8

= 68− 298

(since the LCM of the denominators 2 and 8 is 8)

=398

= 4 78 (39 ÷ 8 gives quotient 4 and remainder 7)

2. Arrange the following in descending order:

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(i) 29 , 23 , 821

(ii) 15 , 37 , 710

Solution:

(i) 29 , 23 , 821

The above fractions are unlike fractions (i.e., the denominators are not the same)

The LCM of the denominators 9, 3 and 21 is 63

29

×77

=1463

23

×2121

=4263

821

×33

=2463

∴ the descending order of the fractions is 4263

> 2463

> 1463

i.e., 23 > 8

21 > 2

9

(ii) 15 , 37 , 710

The above fractions are unlike fractions (i.e., the denominators are not the same)

The LCM of the denominators 5, 7 and 10 is 70

15

×1414

=1470

37

×1010

=3070

710

×77

=4970

∴ the descending order of the fractions is 4970

> 3070

> 1470

i.e., 710

> 37 > 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? Along the first row

4

11+

911

+2

11=

1511

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411

9

11

211

311

5

11

711

8

11

111

6

11

Solution:

Sum of the remaining rows:

3

11+

511

+7

11=

1511

8

11+

111

+6

11=

1511

Sum of the three columns

4

11+

311

+8

11=

1511

9

11+

511

+1

11=

1511

2

11+

711

+6

11=

1511

Sum of the diagonals

4

11+

511

+6

11=

1511

2

11+

511

+8

11=

1511

The sum of the fractions row wise, column wise and diagonal wise is the same i.e., 15

11

Hence, it is a magic square

4. A rectangular sheet of paper is 12 12 cm long and 10 2

3 cm wide. Find its perimeter.

Solution:

Length, l = 12 12

= 252

cm Breadth, b = 10 23

= 323

cm

Perimeter of the sheet = 2 (l + b)

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= 2 �252

+323�

= 2 �75+646

� (since the LCM of the denominators 2 and 3 is 6)

= 2 ×139

6

=139

3

= 46 13

cm (139 ÷ 3 gives quotient 46 and remainder 1)

Therefore, the perimeter of the rectangular sheet is 46 13 cm

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

Solution:

Given that AB = 52 cm, BE = 2 3

4= 11

4 cm AE = 3 3

5= 18

5 cm

Perimeter of a triangle = sum of the three sides

Perimeter of the Δ ABE = AB + BE + AE

=52

+114

+185

= 50+55+7220

(since the LCM of the denominators 2, 4 and 5 is 20)

=17720

= 8 1720

cm (177 ÷ 20 gives quotient 8 and remainder 17)

Therefore, the perimeter of the Δ ABE is 8 1720

cm

(ii) Length (BE = CD) l = 2 34

= 114

cm Breadth (BC = ED) b = 76 cm

Perimeter of a rectangle BCDE = 2 ( l + b)

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= 2 �114

+76�

= 2 �33+1412

� (since the LCM of the denominators 4 and 6 is 12)

= 2 ×4712

=476

= 7 56 cm (47 ÷ 6 gives quotient 7 and remainder 5)

Now, 8 1720

cm > 7 56

cm

Therefore, the perimeter of the triangle ABE is greater than the rectangle BCDE is 7 5

6 cm

6. Salil wants to put a picture in a frame. The picture is 7 35 cm wide. To fit in the

frame the picture cannot be more than 7 310

cm wide. How much should the picture be trimmed?

Solution:

The size of the picture = 7 35

= 385

cm

The size of the picture that will fit in the frame = 7 310

= 7310

cm

The picture should be trimmed by = 385− 73

10

= 76 − 7310

(since the LCM of the denominators 5 and 10 is 10)

= 310

cm

7. Ritu ate 35 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:

Part of the apple eaten by Ritu = 35

Part of the apple eaten by Somu = 1 − 35

= 5 − 35

(since the LCM of the denominators 1 and 5 is 5)

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=25

⇒ 35

> 25

∴ Ritu has eaten the larger share by=35− 2

5

=15

8. Michael finished colouring a picture in 712

hour. Vaibhav finished colouring the

same picture in 34 hour. Who worked longer? By what fraction was it longer?

Solution:

Time taken by Michael to colour the picture = 712

hour

Time taken by Vaibhav to colour the picture = 34 hour

712

× 11

= 712

(since the LCM of the denominators 12 and 4 is 12)

34

× 33

= 912

(since the LCM of the denominators 12 and 4 is 12)

912

> 712

i.e., 34 > 7

12

∴ Vaibhav has taken longer time than Michael

⇒ 9

12−

712

=2

12=

16

Therefore, Vaibhav has taken 16 hour longer than Michael to colour the picture

Exercise: 2.2

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

(i) 2 × 15

(ii) 2 × 12

(iii) 3 × 23

(iv) 3 × 14

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(a)

(b)

(c)

(d)

Solution:

2 ×15

= (d) ⇒ 2 ×15

=15

+15

2 ×12

= (b) ⇒ 2 ×12

=12

+12

3 ×23

= (a) ⇒ 3 ×23

=23

+23

+23

3 ×14

= (c) ⇒ 14

+14

+14

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

(i) 3 × 15

= 35

(ii) 2 × 13

= 23

(iii) 3 × 34

= 2 14

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(a)

(b)

(c)

Solution:

3 ×15

=35

= (c) ⇒ 3 ×15

=15

+15

+15

2 ×13

=23

= (a) ⇒ 2 ×13

=13

+13

3 ×34

= 214

= (b) ⇒ 3 ×34

= 214

=34

+34

+34

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

(i) 7 × 35

(ii) 4 × 13

(iii) 2 × 67

(iv) 5 × 29

(v) 23

× 4

(vi) 52

× 6

(vii) 11 × 47

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(viii) 20 × 45

(ix) 13 × 13

(x) 15 × 35

Solution:

7 ×35

=215

= 415

4 ×13

=43

= 113

2 ×67

=127

= 157

5 ×29

=109

= 119

23

× 4 =83

= 223

52

× 6 = 15

11 ×47

=447

= 627

20 ×45

= 16

13 ×13

=133

= 413

15 ×35

= 9

4. Shade: (i) 12 of the circles in box (a) (ii) 2

3 of the triangles in box (b) (iii) 3

5 of the

squares in box (c)

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Solution:

Shade 12 of the circles in box (a)

Shade 23 of the triangles in box (b)

Shade 35 of the squares in box (c)

5. Find

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(a) 12 of (i) 24 (ii) 46

(b) 23 of (i) 18 (ii) 27

(c) 34 of (i) 16(ii) 36

(d) 45 of (i) 20 (ii) 35

Solution:

(i) 12 of 24 = 1

2× 24

= 12

(ii) 12 of 46 = 1

2× 46

= 23

(i) 23 of 18 = 2

3× 18

= 12

(ii) 23 of 27 = 2

3× 27

= 18

(i) 34 of 16 = 3

4× 16

= 12

(ii) 34 of 36 = 3

4× 36

= 27

(i) 45 of 20 = 4

5× 20

= 16

(ii) 45 of 35 = 4

5× 35

=28

6. Multiply and express as a mixed fraction:

(a) 3 × 5 15

(b) 5 × 6 34

(c) 7 × 2 14

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(d) 4 × 6 13

(e) 3 14

× 6

(f) 3 25

× 8

Solution:

(a) 3 × 5 15

= 3 × 265

=785

= 15 35(since 78 ÷ 5 gives 15 as quotient and 3 as remainder)

(b) 5 × 6 34

= 5 × 274

=135

4

= 33 34(since 135 ÷ 4 gives 33 as quotient and 3 as remainder)

(c) 7 × 2 14

= 7 × 94

=634

= 15 34(since 63 ÷ 4 gives 15 as quotient and 3 as remainder)

(d) 4 × 6 13

= 4 × 193

=763

=25 13(since 76 ÷3 gives 25 as quotient and 1 as remainder)

(e) 3 14×6=13

4×6

=392

= 19 12(since 39 ÷ 2 gives 19 as quotient and 1 as remainder)

(f) 3 25×8=17

5×8

=136

5

= 27 15(since 136 ÷ 5 gives 27 as quotient and 1 as remainder)

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7. Find:

(a) 12of (i) 2 3

4 (ii) 4 2

9

(b) 58of (i) 3 5

6 (ii) 9 2

3

Solution:

(a) (i) 12 of 2 3

4= 1

2× 11

4

=118

= 1 38(since 11 ÷ 8 gives 1 as quotient and 3 as remainder)

(ii) 12 of 4 2

9= 1

2× 38

9

=199

= 2 19(since 19 ÷ 9 gives 2 as quotient and 1 as remainder)

(b) (i) 58 of 3 5

6= 5

8× 23

6

=11548

= 2 1948

(since 115 ÷ 48 gives 2 as quotient and 19 as remainder)

(ii) 58 of 9 2

3= 5

8× 29

3

=14524

= 6 124

(since 145 ÷ 24 gives 6 as quotient and 1 as remainder)

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of 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:

Amount of water in the bottle = 5 litres

water consumed by Vidya = 25

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Therefore, amount of water Vidya drank = 25

× 5

= 2 litres

Fraction of water in the bottle = 1 (whole)

water consumed by Vidya = 25

Therefore, fraction of water consumed by Pratap = 1 − 25

= 5 − 25

(LCM of 1 and 5 is 5)

=35

Exercise: 2.3

1. Find: 12 of (a) 1

4 (b) 3

5 (c) 4

3

17 of (a) 2

9 (b) 6

5 (c) 3

10

Solution:

(a) 12 of 1

4= 1

2× 1

4

=18

(b) 12 of 3

5= 1

2× 3

5

=3

10

(c) 12 of 4

3= 1

2× 4

3

=23

(a) 17 of 2

9= 1

7× 2

9

=2

63

(b) 17 of 6

5= 1

7× 6

5

=6

35

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(c) 17 of 3

10= 1

7× 3

10

=3

70

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

(i) 23

× 2 23

(ii) 27

× 79

(iii) 38

× 64

(iv) 95

× 35

(v) 13

× 158

(vi) 112

× 310

(vii) 45

× 127

Solution:

(i) 23

× 2 23

= 23

× 83

=169

= 179

(ii) 27

× 79

= 29

(iii) 38

× 64

= 916

(iv) 95

× 35

= 2725

= 12

25

(v) 13

× 158

= 58

(vi) 112

× 310

= 3320

= 11320

(vii) 45

× 127

= 4835

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= 11335

3. Multiply the following fractions:

(i) 25

× 5 14

(ii) 6 25

× 79

(iii) 32

× 5 13

(iv) 56

× 2 37

(v) 3 25

× 47

(vi) 2 35

× 3

(vii) 3 47

× 35

Solution:

(i) 25

× 5 14

= 25

× 214

=2110

= 21

10

(ii) 6 25

× 79

= 325

× 79

=22445

= 44445

(iii) 32

× 5 13

= 32

× 163

= 8

(iv) 56

× 2 37

= 56

× 177

=8542

= 21

42

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(v) 3 25

× 47

= 175

× 47

=6835

= 13335

(vi) 2 35

× 3 = 135

× 3

=395

= 745

(vii) 3 47

× 35

= 257

× 35

=157

= 217

4. Which is greater: 27 of 3

4 or 3

5 of 5

8

12 of 6

7 or 2

3 of 3

7

Solution: 27 of 3

4 or 3

5 of 5

8

27 of 3

4= 2

7× 3

4

=3

14

35 of 5

8= 3

5× 5

8

=38

⇒ 38

> 314

(when the numerator is equal, the fraction with smaller denominator is greater)

Therefore, 38 is greater than 3

14

12 of 6

7 or 2

3 of 3

7

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12 of 6

7= 1

2× 6

7

=37

23 of 3

7= 2

3× 3

7

=27

⇒37

>27

Therefore, 37 is greater than 2

7

5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is 3

4 m. Find the distance between the first and the last sapling.

Solution:

No of saplings in a row in the garden = 4

The distance between two adjacent saplings = 34

m

Therefore, the distance between the first and the last sapling = 3 × 34

=94

= 214

m

6. Lipika reads a book for 1 34 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:

Time spent every day by Lipika in reading the book = 1 34

= 74 hours

No of days she takes to finish the entire book = 6 days

Therefore, total hours required by her to read the book = 74

× 6

=212

= 1012

hours

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7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 34 litres of petrol?

Solution:

The distance covered by the car using 1 litre of petrol = 16 kms

Therefore, distance covered using 2 34 litres = 2 3

4× 16

=114

× 16

= 44 kms

(a) (i) Provide the number in the box � , such that 23

�= 1030

(ii) The simplest form of the number obtained in � is _____.

(b) (i) Provide the number in the box � , such that 35

�= 2475

(ii) The simplest form of the number obtained in � is _____.

Solution:

(a) (i) 23

�=1030

=1030

×32

=5

10

(ii) The simplest form of the number = 12

(b) (i) 35

�= 2475

=2475

× 53

=8

15

(ii) The simplest form of the number = 815

Exercise: 2.4

1. Find:

(i) 12 ÷ 34

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(ii) 14 ÷ 56

(iii) 8 ÷ 73

(iv) 4 ÷ 83

(v) 3 ÷ 2 13

(vi) 5 ÷ 3 47

Solution:

(i) 12 ÷ 34

= 12 × 43

= 16

(ii) 14 ÷ 56

= 14 × 65

=845

= 1645

(iii) 8 ÷ 73

= 8 × 37

=247

= 337

(iv) 4 ÷ 83

= 4 × 38

=32

= 112

(v) 3 ÷ 2 13

= 3 × 37

=97

= 127

(vi) 5 ÷ 3 47

= 5 × 725

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=75

= 125

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

(i) 37

(ii) 58

(iii) 97

(iv) 65

(v) 127

(vi) 18

(vii) 111

Solution:

(i) 37

Reciprocal of 37 is 7

3 and 7

3 is an improper fraction

(ii) 58

Reciprocal of 58 is 8

5 and 8

5 is an improper fraction

(iii) 97

Reciprocal of 97 is 7

9 and 7

9 is a proper fraction

(iv) 65

Reciprocal of 65 is 5

6 and 5

6 is a proper fraction

(v) 127

Reciprocal of 127

is 712

and 712

is a proper fraction

(vi) 18

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Reciprocal of 18 is 8

1=8 and 8 is a whole number

(vii) 111

Reciprocal of 111

is 111

=11 and 11 is a whole fraction

3. Find:

(i) 73

÷ 2

(ii) 49

÷ 5

(iii) 613

÷ 7

(iv) 4 13

÷ 3

(v) 3 12

÷ 4

(vi) 4 37

÷ 7

Solution:

(i) 73

÷ 2 = 73

× 12

=76

(ii) 49

÷ 5 = 49

× 15

=4

45

(iii) 613

÷ 7 = 613

× 17

=6

91

(iv) 4 13

÷ 3 = 133

÷ 3

=133

×13

=139

(v) 3 12

÷ 4 = 72

÷ 4

=72

×14

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=78

(vi) 4 37

÷ 7 = 317

÷ 7

=317

×17

=3149

4. Find:

(i) 25

÷ 12

(ii) 49

÷ 23

(iii) 37

÷ 87

(iv) 2 13

÷ 35

(v) 3 12

÷ 83

(vi) 25

÷ 1 12

(vii) 3 15

÷ 1 23

(viii) 2 15

÷ 1 15

Solution:

(i) 25

÷ 12

= 25

× 21

=45

(ii) 49

÷ 23

= 49

× 32

=23

(iii) 37

÷ 87

= 37

× 78

=38

(iv) 2 13

÷ 35

= 73

÷ 35

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=73

×53

=359

= 3 89

(v) 3 12

÷ 83

= 72

÷ 83

=72

×38

=2116

= 15

16

(vi) 25

÷ 1 12

= 25

÷ 32

=25

×23

=4

15

(vii) 3 15

÷ 1 23

= 165

÷ 53

=165

×35

=4825

= 12325

(viii) 2 15

÷ 1 15

= 115

÷ 65

=115

×56

=116

= 156

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Exercise: 2.5

1. Which is greater?

(i) 0.5 or 0.05

(ii) 0.7 or 0.5

(iii) 7 or 0.7

(iv) 1.37 or 1.49

(v) 2.03 or 2.30

(vi) 0.8 or 0.88

Solution:

(i) 0.5 or 0.05

0.5 =5

10

0.05 =5

100

⇒ 510

> 5100

(when the numerator is equal, the fraction with the smaller denominator is greater)

Therefore, 0.5 is greater than 0.05

(ii) 0.7 or 0.5

0.7 =7

10

0.5 =5

10

⇒ 710

> 510

[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]

Therefore, 0.7 is greater than 0.5

(iii) 7 or 0.7

7 =71

0.7 =7

10

⇒ 71

> 710

(when the numerator is equal, the fraction with the smaller denominator is greater)

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Therefore, 7 is greater than 0.7

(iv) 1.37 or 1.49

1. 37 =137100

1.49 =149100

⇒ 149100

> 137100

[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]

Therefore, 1.49 is greater than 1.37

(v) 2.03 or 2.30

2.03 =203100

2.30 =230100

⇒ 230100

> 203100

[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]

Therefore, 2.30 is greater than 2.03

(vi) 0.8 or 0.88

0.8 =80

100

0.88 =88

100

⇒ 80100

> 88100

[when the denominator is equal (like fractions), the fraction with the greater numerator is greater]

Therefore, 0.88 is greater than 0.8

2. Express as rupees using decimals:

(i) 7 paise

(ii) 7 rupees 7 paise

(iii) 77 rupees 77 paise

(iv) 50 paise

(v) 235 paise

Solution:

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(i) 7 paise

Rs. 1 = 100 paise

7 paise = 7100

= Rs 0.07

(ii) 7 rupees 7 paise

Rs. 1 = 100 paise

7 rupees 7 paise = 7 + 7100

= 7 + 0.07

= Rs 7.07

(iii) 77 rupees 77 paise

Rs. 1 = 100 paise

77 rupees 77 paise = 77 + 77100

= 77 + 0.77

= Rs 77.77

(iv) 50 paise

Rs. 1 = 100 paise

50 paise = 50100

= Rs 0.50

(v) 235 paise

Rs. 1 = 100 paise

235 paise = 235100

= Rs 2.35

(i) Express 5 cm in metre and kilometer

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

Solution:

3. Express 5 cm in metre and kilometer

Solution:

100cm = 1m

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∴ 5cm =5

100

= 0.05 m

100cm = 1m

1000m = 1km

⇒ 100000cm = 1km

100000cm = 1 km

∴ 5cm =5

100000

= 0.00005 km

4. Express 35 mm in cm, m and km

Solution:

10mm = 1cm

∴ 35mm =3510

= 3.5 cm

10mm = 1cm

100cm = 1m

⇒ 1000mm = 1m

∴ 35mm =35

1000

= 0.035 m

10mm = 1cm

100cm = 1m

⇒ 1000000mm = 1km

∴ 35mm =35

1000000

= 0.000035 km

5. Express in kg:

(i) 200 g

(ii) 3470 g

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(iii) 4 kg 8 g

Solution:

200 g

1000 g = 1kg

200 g =200

1000

= 0.2Kg

3470 g

1000 g = 1kg

3470g =34701000

= 3.47 Kg

4 kg 8 g

1000 g = 1kg

4 kg 8 g = 4 kg +8

1000

= 4 kg + 0.008

= 4.008 Kg

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

(i) 20.03

(ii) 2.03

(iii) 200.03

(iv) 2.034

Solution:

20.03 = (2 × 10) + (0 × 1) + �0 ×1

10� + �3 ×

1100

2.03 = (2 × 10) + �0 ×1

10� + �3 ×

1100

200.03 = (2 × 100) + (0 × 10) + (0 × 1) + �0 ×1

10� + �3 ×

1100

2.034 = (2 × 1) + �0 ×1

10� + �3 ×

1100

� + �4 ×1

1000�

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

(i) 2.56

(ii) 21.37

(iii) 10.25

(iv) 9.42

(v) 63.352

Solution:

(i) 2.56

Place value of 2 is ones

(ii) 21.37

Place value of 2 is tens

(iii) 10.25

Place value of 2 is one tenths

(iv) 9.42

Place value of 2 is one hundredths

(v) 63.352

Place value of 2 is one thousandths

8. 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?

Solution:

Distance from A to B = 7.5 km

Distance from B to C = 12.7 km

Distance travelled by Dinesh = A to B + B to C

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= 7.5 + 12.7

= 20.2 km

Distance from A to D = 9.3 km

Distance from D to C = 11.8 km

Distance travelled by Ayub = A to D + D to C

= 9.3 + 11.8

= 21.1 km

Difference between the distance travelled = 21.1 − 20.2

= 0.9 km

Therefore, Ayub travelled 0.9 kms more than Dinesh

9. 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:

Quantity of fruits bought by Shyama:

Apples = 5.300 Kg

Mangoes = 3.250 kg

Total fruits bought by Shyama = 8.550 kg

Quantity of fruits bought by Sarala:

Oranges = 4.800 Kg

Bananas = 4.150 kg

Total fruits bought by Shyama = 8.950 kg

Difference in the quantity = 8.950 – 8.550

= 0.4 kg

Therefore, Sarala has bought 0.4 kg or 400 g more fruits than Shyama

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

Solution:

42.6 km – 28 Km = 14.6 Km

Therefore, 28 km is less than 42.6 km by 14.6 km

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Exercise: 2.6

1. Find:

(i) 0.2 × 6

(ii) 8 × 4.6

(iii) 2.71 × 5

(iv) 20.1 × 4

(v) 0.05 × 7

(vi) 211.02 × 4

(vii) 2 × 0.86

Solution:

(i) 0.2 × 6 = 1.2

(ii) 8 × 4.6 = 36.8

(iii) 2.71 × 5 = 13.55

(iv) 20.1 × 4 = 80.4

(v) 0.05 × 7 = 0.35

(vi) 211.02 × 4 = 844.08

(vii) 2 × 0.86 = 1.72

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

Solution:

Length of the rectangle, l = 5.7cm

Breadth of the rectangle, b = 3cm

∴ Area of the rectangle = l × b

= 5.7 × 3

= 17.1cm2

2. Find:

(i) 1.3 × 10

(ii) 36.8 × 10

(iii) 153.7 × 10

(iv) 168.07 × 10

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(v) 31.1 × 100

(vi) 156.1 × 100

(vii) 3.62 × 100

(viii) 43.07 × 100

(ix) 0.5 × 10

(x) 0.08 × 10

(xi) 0.9 × 100

(xii) 0.03 × 1000

Solution:

(i) 1.3 × 10 = 13

(ii) 36.8 × 10 = 368

(iii) 153.7 × 10 = 1537

(iv) 168.07 × 10 = 1680.7

(v) 31.1 × 100 = 311

(vi) 156.1 × 100 = 15610

(vii) 3.62 × 100 = 362

(viii) 43.07 × 100 = 4307

(ix) 0.5 × 10 = 5

(x) 0.08 × 10 = 0.8

(xi) 0.9 × 100 = 90

(xii) 0.03 × 1000 = 30

3. 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:

Distance covered in one litre of petrol = 55.3 km

∴ Distance covered in 10 litres = 55.3 × 10

= 553 km

4. Find:

(i) 2.5 × 0.3

(ii) 0.1 × 51.7

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(iii) 0.2 × 316.8

(iv) 1.3 × 3.1

(v) 0.5 × 0.05

(vi) 11.2 × 0.15

(vii) 1.07 × 0.02

(viii) 10.05 × 1.05

(ix) 101.01 × 0.01

(x) 100.01 × 1.1

Solution:

(i) 2.5 × 0.3 = 0.75

(ii) 0.1 × 51.7 = 5.17

(iii) 0.2 × 316.8 = 63.36

(iv) 1.3 × 3.1 = 4.03

(v) 0.5 × 0.05 = 0.025

(vi) 11.2 × 0.15 = 1.68

(vii) 1.07 × 0.02 = 0.0214

(viii) 10.05 × 1.05 = 10.5525

(ix) 101.01 × 0.01 = 1.0101

(x) 100.01 × 1.1 = 110.011

Exercise: 2.7

1. Find:

(i) 0.4 ÷ 2

(ii) 0.35 ÷ 5

(iii) 2.48 ÷ 4

(iv) 65.4 ÷ 6

(v) 651.2 ÷ 4

(vi) 14.49 ÷ 7

(vii) 3.96 ÷ 4

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(viii) 0.80 ÷ 5

Solution:

(i) 0.4 ÷ 2 = 410

÷ 2

=4

10×

12

=2

10

= 0.2

(ii) 0.35 ÷ 5 = 35100

÷ 5

=35

100×

15

=7

100

= 0.07

(iii) 2.48 ÷ 4 = 248100

÷ 4

=248100

×14

=62

100

= 0.62

(iii) 65.4 ÷ 6 = 65410

÷ 6

=65410

×16

=10910

= 10.9

(iv) 651.2 ÷ 4 = 651210

÷ 4

=6512

10×

14

=1628

10

= 162.8

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(v) 14.49 ÷ 7 = 1449100

÷ 7

=1449100

×17

=207100

= 2.07

(vi) 3.96 ÷ 4 = 396100

÷ 4

=396100

×14

=99

100

= 0.99

(vii) 0.80 ÷ 5 = 80100

÷ 5

=80

100×

15

=16

100

= 0.16

2. Find:

(i) 4.8 ÷ 10

(ii) 52.5 ÷ 10

(iii) 0.7 ÷ 10

(iv) 33.1 ÷ 10

(v) 272.23 ÷ 10

(vi) 0.56 ÷ 10

(vii) 3.97 ÷ 10

Solution:

(i) 4.8 ÷ 10 = 0.48

(ii) 52.5 ÷ 10 = 5.25

(iii) 0.7 ÷ 10 = 0.07

(iv) 33.1 ÷ 10 = 3.31

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(v) 272.23 ÷ 10 = 27.223

(vi) 0.56 ÷ 10 = 0.056

(vii) 3.97 ÷ 10 = 0.397

3. Find:

(i) 2.7 ÷ 100

(ii) 0.3 ÷ 100

(iii) 0.78 ÷ 100

(iv) 432.6 ÷ 100

(v) 23.6 ÷ 100

(vi) 98.53 ÷ 100

Solution:

(i) 2.7 ÷ 100 = 0.027

(ii) 0.3 ÷ 100 = 0.003

(iii) 0.78 ÷ 100 = 0.0078

(iv) 432.6 ÷ 100 = 4.326

(v) 23.6 ÷ 100 = 0.236

(vi) 98.53 ÷ 100 = 0.9853

4. Find:

(i) 7.9 ÷ 1000

(ii) 26.3 ÷ 1000

(iii) 38.53 ÷ 1000

(iv) 128.9 ÷ 1000

(v) 0.5 ÷ 1000

Solution:

(i) 7.9 ÷ 1000 = 0.0079

(ii) 26.3 ÷ 1000 = 0.0263

(iii) 38.53 ÷ 1000 = 0.03853

(iv) 128.9 ÷ 1000 = 0.1289

(v) 0.5 ÷ 1000 = 0.0005

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5. Find:

(i) 7 ÷ 3.5

(ii) 36 ÷ 0.2

(iii) 3.25 ÷ 0.5

(iv) 30.94 ÷ 0.7

(v) 0.5 ÷ 0.25

(vi) 7.75 ÷ 0.25

(vii) 76.5 ÷ 0.15

(viii) 37.8 ÷ 1.4

(ix) 2.73 ÷ 1.3

Solution:

(i) 7 ÷ 3.5 = 7 ÷ 3510

= 7 ×1035

= 2

(ii) 36 ÷ 0.2 = 36 ÷ 210

= 36 ×102

= 180

(iii) 3.25 ÷ 0.5 = 325100

÷ 510

=325100

× 105

=6510

= 6.5

(iv) 30.94 ÷ 0.7 = 3094100

÷ 710

=3094100

× 107

=44210

= 44.2

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(v) 0.5 ÷ 0.25 = 510

÷ 25100

=5

10×

10025

=2010

= 2

(vi) 7.75 ÷ 0.25 = 775100

÷ 25100

=775100

× 10025

=77525

= 31

(vii) 76.5 ÷ 0.15 = 76510

÷ 15100

=76510

× 10015

= 255 × 2

= 510

(viii) 37.8 ÷ 1.4 = 37810

÷ 1410

=37810

× 1014

=37814

= 27

(ix) 2.73 ÷ 1.3 = 273100

÷ 1310

=273100

× 1013

=2110

= 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:

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Distance covered by the vehicle = 43.2 km

Quantity of petrol used = 2.4 litres

Therefore, distance that can be covered in 1 litre = 43.22.4

× 1010

=43224

= 18 kms