RS Aggarwal Solutions for Class 7 Maths Exercise 3D Chapter 3 Decimals

RS Aggarwal Solutions for Class 7 Maths Exercise 3D Chapter 3 Decimals in simple PDF are given here. This exercise of RS Aggarwal Solutions for Class 7 Chapter 3 contains questions related to the Division of Decimals by 10, 100 and 1000, dividing a decimal by a whole number. On dividing a decimal by 10, 100 and 1000 the decimal point is shifted to the left by one, two and three places respectively. Students are suggested to try solving the questions from RS Aggarwal Solutions for Class 7 Maths Chapter 3 Decimals.

Download the PDF of RS Aggarwal Solutions For Class 7 Maths Chapter 3 Decimals – Exercise 3D

 

rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d
rs aggarwal solution class 7 maths chapter ex 3d

 

Access answers to Maths RS Aggarwal Solutions for Class 7 Chapter 3 – Decimals Exercise 3D

1. Divide:

(i) 131.6 by 10

Solution:-

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

We have,

= 131.6 ÷ 10

= (131.6/10)

= 13.16

(ii) 32.56 by 10

Solution:-

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

We have,

= 32.56 ÷ 10

= (32.56/10)

= 3.256

(iii) 4.38 by 10

Solution:-

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

We have,

= 4.38 ÷ 10

= (4.38/10)

= 0.438

(iv) 0.34 by 10

Solution:-

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

We have,

= 0.34 ÷ 10

= (0.34/10)

= 0.034

(v) 0.08 by 10

Solution:-

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

We have,

= 0.08 ÷ 10

= (0.08/10)

= 0.008

(vi) 0.062 by 10

Solution:-

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

We have,

= 0.062 ÷ 10

= (0.062/10)

= 0.0062

2. Divide:

(i) 137.2 by 100

Solution:-

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

We have,

= 137.2 ÷ 100

= (137.2/100)

= 1.372

(ii) 23.4 by 100

Solution:-

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

We have,

= 23.4 ÷ 100

= (23.4/100)

= 0.234

(iii) 4.7 by 100

Solution:-

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

We have,

= 4.7 ÷ 100

= (4.7/100)

= 0.047

(iv) 0.3 by 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

(v) 0.58 by 100

Solution:-

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

We have,

= 0.58 ÷ 100

= (0.58/100)

= 0.0058

(vi) 0.02 by 100

Solution:-

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

We have,

= 0.02 ÷ 100

= (0.02/100)

= 0.0002

3. Divide:

(i) 1286.5 by 1000

Solution:-

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

We have,

= 1286.5 ÷ 1000

= (1286.5/1000)

= 1.2865

(ii) 354.16 by 1000

Solution:-

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

We have,

= 354.16 ÷ 1000

= (354.16/1000)

= 0.35416

(iii) 38.9 by 1000

Solution:-

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

We have,

= 38.9 ÷ 1000

= (38.9/1000)

= 0.0389

(iv) 4.6 by 1000

Solution:-

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

We have,

= 4.6 ÷ 1000

= (4.6/1000)

= 0.0046

(v) 0.8 by 1000

Solution:-

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

We have,

= 0.8 ÷ 1000

= (0.8/1000)

= 0.0008

(vi) 2 by 1000

Solution:-

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

We have,

= 2 ÷ 1000

= (2/1000)

= 0.0002

4. Divide:-

(i) 12 by 8

Solution:-

The above question can be written as, 12 ÷ 8

Then,

= (12/8) … [÷2]

= (3/2)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 20

∴12 ÷ 8 = 1.5

(ii) 63 by 15

Solution:-

The above question can be written as, 63 ÷ 15

Then,

= (63/15) … [÷3]

= (21/5)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 21

∴63 ÷ 15 = 1.5

(iii) 47 by 20

Solution:-

The above question can be written as, 47 ÷ 20

Then,

= (47/20)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 22

∴47 ÷ 20 = 2.35

(iv) 101 by 25

Solution:-

The above question can be written as, 101 ÷ 25

Then,

= (101/25)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 23

∴101 ÷ 25 = 4.04

(v) 31 by 40

Solution:-

The above question can be written as, 31 ÷ 40

Then,

= (31/40)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 24

∴31 ÷ 40= 0.775

(vi) 11 by 16

Solution:-

The above question can be written as, 11 ÷ 16

Then,

= (31/40)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 25

∴11 ÷ 16=0.06875

5. Divide:

(i) 43.2 by 6

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

43.2÷6

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 26

∴43.2÷6 = 7.2

(ii) 60.48 by 12

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

60.48÷12

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 27

∴60.48÷12 = 5.04

(iii) 117.6 by 21

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

117.6÷21

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 28

∴117.6÷21=5.6

(iv) 217.44 by 18

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

217.44÷18

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 29

∴217.44÷18=12.08

(v) 2.575 by 25

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

2.575÷25

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 30

∴2.575÷25= 0.103

(vi) 6.08 by 8

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

6.08÷6

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 31

∴6.08÷6=0.76

(vii) 0.765 by 9

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

0.765÷9

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 32

∴0.765÷9=0.085

(viii) 0.768 by16

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

0.768÷16

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 33

∴0.769÷16=0.048

(ix) 0.175 by 25

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

0.175÷25

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 34

∴0.175÷25= 0.007

(x) 0.3322 by 11

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

0.3322÷11

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 35

∴0.3322÷11= 0.0302

(xi) 2.13 by 15

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

2.13÷15

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 36

∴2.13÷15 = 0.14

(xii) 6.54 by 12

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

6.54÷12

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 37

∴6.54÷12=0.545

(xiii) 5.52 by 16

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

5.52÷16

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 38

∴5.52÷16=0.345

(xiv) 1.001 by 14

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

1.001÷14

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 39

∴1.001÷14=0.0715

(xv) 0.477 by 18

Solution:-

Perform the division by considering the dividend a whole number.

When the division of whole-part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers.

We have,

0.477÷18

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 40

∴0.477÷18=0.0265

6. Divide:

(i) 16.46 ÷ 20

Solution:

The above question can be written as,

⇒ (16.46/20)

Multiply by 100 for both numerator and denominator, then we get,

= [(16.46 × 100)/ (20×100)]

= (1646/2000)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 41

∴ (1646/2000) = 0.823

(ii) 403.8 ÷ 30

Solution:

The above question can be written as,

⇒ (403.8/30)

Multiply by 10 for both numerator and denominator, then we get,

= [(403.8 × 10)/ (30×10)]

= (4038/300)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 42

∴ (4038/300) = 13.46

(iii) 19.2 ÷ 80

Solution:

The above question can be written as,

⇒ (19.2/80)

Multiply by 10 for both numerator and denominator, then we get,

= [(19.2 × 10)/ (80×10)]

= (192/800)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 43

∴ (192/800) = 0.24

(iv) 156.8 ÷ 200

Solution:

The above question can be written as,

⇒ (156.8/200)

Multiply by 10 for both numerator and denominator, then we get,

= [(156.8 × 10)/ (200×10)]

= (1568/2000)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 44

∴ (1568/2000) = 0.784

(v) 12.8 ÷ 500

Solution:

The above question can be written as,

⇒ (12.8/500)

Multiply by 10 for both numerator and denominator, then we get,

= [(12.8 × 10)/ (500×10)]

= (128/5000)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 45

∴ (128/5000) = 0.0256

(vi) 18.08 ÷ 400

Solution:

The above question can be written as,

⇒ (18.08/400)

Multiply by 100 for both numerator and denominator, then we get,

= [(18.08 × 100)/ (400×100)]

= (1808/40000)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 46

∴ (1808/40000) = 0.0256

7. Divide:

(i) 3.28 by 0.8

Solution:-

The above question can be written as,

⇒ (3.28/0.8)

Multiply by 10 for both numerator and denominator, then we get,

= [(3.28 × 10)/ (0.8×10)]

= (32.8/8)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 47

∴ (32.8/8)=4.1

(ii) 0.288 by 0.9

Solution:-

The above question can be written as,

⇒ (0.288/0.9)

Multiply by 10 for both numerator and denominator, then we get,

= [(0.288 × 10)/ (0.9×10)]

= (2.88/9)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 48

∴ (2.88/9)=0.32

(iii) 25.395 by 1.5

Solution:-

The above question can be written as,

⇒ (25.395/1.5)

Multiply by 10 for both numerator and denominator, then we get,

= [(25.395 × 10)/ (1.5×10)]

= (253.95/15)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 49

∴ (253.95/15)=16.93

(iv) 2.0484 by 0.18

Solution:-

The above question can be written as,

⇒ (2.0484/0.18)

Multiply by 100 for both numerator and denominator, then we get,

= [(2.0484× 100)/ (0.18×100]

= (204.84/18)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 50

∴ (204.84/18) = 11.38

(v) 0.228 by 0.38

Solution:-

The above question can be written as,

⇒ (0.228/0.38)

Multiply by 100 for both numerator and denominator, then we get,

= [(0.228× 100)/ (0.38×100]

= (22.8/38)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 51

∴ (22.8/38) = 0.6

(vi) 0.8085 by 0.35

Solution:-

The above question can be written as,

⇒ (0.8085/0.35)

Multiply by 100 for both numerator and denominator, then we get,

= [(0.8085× 100)/ (0.35×100]

= (80.85/35)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 52

∴ (80.85/35) = 2.31

(vii) 21.976 by 1.64

Solution:-

The above question can be written as,

⇒ (21.976/1.64)

Multiply by 100 for both numerator and denominator, then we get,

= [(21.976× 100)/ (1.64×100]

= (2197.6/164)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 53

∴ (2197.6/164) = 13.4

(viii) 11.04 by 1.6

Solution:-

The above question can be written as,

⇒ (11.04/1.6)

Multiply by 10 for both numerator and denominator, then we get,

= [(11.04× 10)/ (1.6×10]

= (110.4/16)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 54

∴ (110.4/16) = 6.9

(ix) 6.612 by 11.6

Solution:-

The above question can be written as,

⇒ (6.612/11.6)

Multiply by 10 for both numerator and denominator, then we get,

= [(6.612× 10)/ (11.6×10]

= (66.12/116)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 55

∴ (66.12/116) = 0.57

(x) 0.076 by 0.19

Solution:-

The above question can be written as,

⇒ (0.076/0.19)

Multiply by 100 for both numerator and denominator, then we get,

= [(0.076× 100)/ (0.19×100]

= (7.6/19)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 56

∴ (7.6/19) = 0.4

(xi) 148 by 0.074

Solution:-

The above question can be written as,

⇒ (148/0.074)

Multiply by 1000 for both numerator and denominator, then we get,

= [(148× 1000)/ (0.074×1000]

= (148000/74)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 57

∴ (148000/74) = 2000

(xii) 16.578 by 5.4

Solution:-

The above question can be written as,

⇒ (16.578/5.4)

Multiply by 10 for both numerator and denominator, then we get,

= [(16.578× 10)/ (5.4×10]

= (165.78/54)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 58

∴ (165.78/54) = 3.07

(xiii) 28 by 0.56

Solution:-

The above question can be written as,

⇒ (28/0.56)

Multiply by 100 for both numerator and denominator, then we get,

= [(28× 100)/ (0.56×100]

= (2800/56)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 59

∴ (2800/56) = 50

(xiv) 204 by 0.17

Solution:-

The above question can be written as,

⇒ (204/0.17)

Multiply by 100 for both numerator and denominator, then we get,

= [(204× 100)/ (0.17×100]

= (20400/17)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 60

∴ (20400/17) = 1200

(xv) 3 by 80

Solution:-

The above question can be written as,

⇒ (3/80)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 61

∴ (3/80) = 0.0375

8. The total cost of 24 chairs is ₹ 9255.60. Find the cost of each chair.

Solution:-

The number of chairs= 24

The cost of 24 chairs is = ₹ 9255.60

Then the cost of each chair = (9255.60/24)

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 62

∴ the cost of each chair is ₹ 385.65

 


Access other exercises of RS Aggarwal Solutions For Class 7 Chapter 3 – Decimals

Exercise 3A Solutions

Exercise 3B Solutions

Exercise 3C Solutions

Exercise 3E Solutions