Selina Solutions Concise Maths Class 7 Chapter 4: Decimal Fractions (Decimals)

Selina Solutions Concise Maths Class 7 Chapter 4 Decimal Fractions (Decimals) help students understand the basic concepts covered in this chapter. Students can solve the exercise wise problems by using the solutions designed by faculty at BYJU’S. The solutions are created, keeping in mind the understanding capacity of students. Selina Solutions Concise Maths Class 7 Chapter 4 Decimal Fractions (Decimals), PDF links are given here.

Chapter 4 explains concepts like converting decimal fraction into a fraction and vice versa, decimal places and various mathematical operations on them. Students can refer to the PDF of solutions and solve textbook problems, without any time constraints.

Selina Solutions Concise Maths Class 7 Chapter 4: Decimal Fractions (Decimals) Download PDF

 

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Exercises of Selina Solutions Concise Maths Class 7 Chapter 4 – Decimal Fractions (Decimals)

Exercise 4A Solutions

Exercise 4B Solutions

Exercise 4C Solutions

Exercise 4D Solutions

Exercise 4E Solutions

Exercise 4F Solutions

Access Selina Solutions Concise Maths Class 7 Chapter 4: Decimal Fractions (Decimals)

Exercise 4A page: 57

1. Convert the following into fractions in their lowest terms:

(i) 3.75

(ii) 0.5

(iii) 2.04

(iv) 0.65

(v) 2.405

(vi) 0.085

(vii) 8.025

Solution:

(i) 3.75

We can write it as

= 375/ 100

Here the HCF of 375 and 100 is 25

= (375 ÷ 25)/ (100 ÷ 25)

So we get

= 15/4

(ii) 0.5

We can write it as

= 5/10

= 1/2

(iii) 2.04

We can write it as

= 204/100

Here the HCF of 204 and 100 is 4

= (204 ÷ 4)/ (100 ÷ 4)

So we get

= 51/25

(iv) 0.65

We can write it as

= 65/100

Here the HCF of 65 and 100 is 5

= (65 ÷ 5)/ (100 ÷ 5)

So we get

= 13/20

(v) 2.405

We can write it as

= 2405/1000

Here the HCF of 2405 and 1000 is 5

= (2405 ÷ 5)/ (1000 ÷ 5)

So we get

= 481/200

(vi) 0.085

We can write it as

= 85/1000

Here the HCF of 85 and 1000 is 5

= (85 ÷ 5)/ (1000 ÷ 5)

So we get

= 17/200

(vii) 8.025

We can write it as

= 8025/1000

Here the HCF of 8025 and 1000 is 25

= (8025 ÷ 25)/ (1000 ÷ 25)

So we get

= 321/ 40

2. Convert into decimal fractions:

(i) 2 4/5

(ii) 79/100

(iii) 37/10,000

(iv) 7543/104

(v) 3/4

(vi) 9 3/5

(vii) 8 5/8

(viii) 5 7/8

Solution:

(i) 2 4/5

We can write it as

= 14/5 × 2/2

So we get

= 28/10

= 2.8

(ii) 79/100

We can write it as

= 79/100

= 0.79

(iii) 37/10,000

We can write it as

= 37/10,000

= 0.0037

(iv) 7543/104

We can write it as

= 7543/10000

= 0.7543

(v) 3/4

Selina Solutions Concise Maths Class 7 Chapter 4 Image 1

By division we get

= 0.75

(vi) 9 3/5

We can write it as

= 48/5

By division

Selina Solutions Concise Maths Class 7 Chapter 4 Image 2

So we get

= 9.6

(vii) 8 5/8

Selina Solutions Concise Maths Class 7 Chapter 4 Image 3

By division we get

= 8.625

(viii) 5 7/8

Selina Solutions Concise Maths Class 7 Chapter 4 Image 4

By division we get

= 5.875

3. Write the number of decimal places in:

(i) 0.4762

(ii) 7.00349

(iii) 8235.403

(iv) 35.4

(v) 2.608

(vi) 0.000879

Solution:

(i) 0.4762

There are four decimal places.

(ii) 7.00349

There are five decimal places.

(iii) 8235.403

There are three decimal places.

(iv) 35.4

There is one decimal place.

(v) 2.608

There are three decimal places.

(vi) 0.000879

There are six decimal places.

4. Write the following decimals as word statements:

(i) 0.4, 0.9, 0.1

(ii) 1.9, 4.4, 7.5

(iii) 0.02, 0.56, 13.06

(iv) 0.005, 0.207, 111.519

(v) 0.8, 0.08, 0.008, 0.0008

(vi) 256.1, 10.22, 0.634

Solution:

(i) 0.4 = zero-point-four

0.9 = zero-point-nine

0.1 = zero-point-one

(ii) 1.9 = one-point-nine

4.4 = four-point-four

7.5 = seven-point-five

(iii) 0.02 = zero-point-zero-two

0.56 = zero-point-five-six

13.06 = thirteen-point-zero-six

(iv) 0.005 = zero-point-zero-zero-five

0.207 = zero-point-two-zero-seven

111.519 = one hundred eleven-point-five-one-nine

(v) 0.8 = zero-point-eight

0.08 = zero-point-zero-eight

0.008 = zero-point-zero-zero-eight

0.0008 = zero-point-zero-zero-zero-eight

(vi) 256.1 = two hundred fifty six-point-one

10.22 = ten-point-two-two

0.634 = zero-point-six-three-four

5. Convert the given fractions into like fractions:

(i) 0.5, 3.62, 43.981 and 232.0037

(ii) 215.78, 33.0006, 530.3 and 0.03569

Solution:

(i) 0.5, 3.62, 43.981 and 232.0037

The greatest decimal places is 4

So we get

0.5 = 0.5000

3.62 = 3.6200

43.981 = 43.9810

232.0037 = 232.0037

(ii) 215.78, 33.0006, 530.3 and 0.03569

The greatest decimal places is 5

215.78 = 215.78000

33.0006 = 33.00060

530.3 = 530.30000

0.03569 = 0.03569

Exercise 4B page: 59

1. Add:

(i) 0.5 and 0.37

(ii) 3.8 and 8.7

(iii) 0.02, 0.008 and 0.309

(iv) 0. 4136, 0. 3195 and 0.52

(v) 9.25, 3.4 and 6.666

(vi) 3.007, 0.587 and 18.341

(vii) 0.2, 0.02 and 2.0002

(viii) 6. 08, 60.8, 0.608 and 0.0608

(ix) 29.03, 0.0003, 0.3 and 7.2

(x) 3.4, 2.025, 9.36 and 3.6221

Solution:

(i) 0.5 and 0.37

Selina Solutions Concise Maths Class 7 Chapter 4 Image 5

So we get

0.5 + 0.37 = 0.87

(ii) 3.8 and 8.7

Selina Solutions Concise Maths Class 7 Chapter 4 Image 6

So we get

3.8 + 8.7 = 12.5

(iii) 0.02, 0.008 and 0.309

Selina Solutions Concise Maths Class 7 Chapter 4 Image 7

So we get

0.02 + 0.008 + 0.309 = 0.337

(iv) 0.4136, 0.3195 and 0.52



Selina Solutions Concise Maths Class 7 Chapter 4 Image 8

So we get

0.4136 + 0.3195 + 0.52 = 1.2531

(v) 9.25, 3.4 and 6.666

Selina Solutions Concise Maths Class 7 Chapter 4 Image 9

So we get

9.25 + 3.4 + 6.666 = 19.316

(vi) 3.007, 0.587 and 18.341

Selina Solutions Concise Maths Class 7 Chapter 4 Image 10

So we get

3.007 + 0.587 + 18.341 = 21.935

(vii) 0.2, 0.02 and 2.0002



Selina Solutions Concise Maths Class 7 Chapter 4 Image 11

So we get

0.2 + 0.02 + 2.0002 = 2.2202

(viii) 6.08, 60.8, 0.608 and 0.0608

Selina Solutions Concise Maths Class 7 Chapter 4 Image 12

So we get

6.08 + 60.8 + 0.608 + 0.0608 = 67.5488

(ix) 29.03, 0.0003, 0.3 and 7.2

Selina Solutions Concise Maths Class 7 Chapter 4 Image 13

So we get

29.03 + 0.0003 + 0.3 + 7.2 = 36.5303

(x) 3.4, 2.025, 9.36 and 3.6221

Selina Solutions Concise Maths Class 7 Chapter 4 Image 14

So we get

3.4 + 2.025 + 9.36 + 3.6221 = 18.4071

2. Subtract the first number from the second:
(i) 5.4, 9.8
(ii) 0.16, 4.3
(iii) 0.82, 8.6
(v) 2.237, 9.425
(vi) 41 .03, 59.46
(vii) 3.92. 26.86
(viii) 4.73, 8.5
(ix) 12.63, 36.2
(x) 0.845, 3.71

Solution:

(i) 5.4, 9.8
It can be written as

9.8 – 5.4 = 4.4

Selina Solutions Concise Maths Class 7 Chapter 4 Image 15

(ii) 0.16, 4.3
It can be written as

4.3 – 0.16 = 4.14

Selina Solutions Concise Maths Class 7 Chapter 4 Image 16

(iii) 0.82, 8.6
It can be written as

8.6 – 0.82 = 7.78

Selina Solutions Concise Maths Class 7 Chapter 4 Image 17

(v) 2.237, 9.425
It can be written as

9.425 – 2.237 = 7.188

Selina Solutions Concise Maths Class 7 Chapter 4 Image 18

(vi) 41 .03, 59.46
It can be written as

59.46 – 41.03 = 18.43

Selina Solutions Concise Maths Class 7 Chapter 4 Image 19

(vii) 3.92, 26.86
It can be written as

26.86 – 3.92 = 22.94

Selina Solutions Concise Maths Class 7 Chapter 4 Image 20

(viii) 4.73, 8.5
It can be written as

8.5 – 4.73 = 3.77

Selina Solutions Concise Maths Class 7 Chapter 4 Image 21

(ix) 12.63, 36.2
It can be written as

36.2 – 12.63 = 23.57

Selina Solutions Concise Maths Class 7 Chapter 4 Image 22

(x) 0.845, 3.71

It can be written as

3.71 – 0.845 = 2.865

Selina Solutions Concise Maths Class 7 Chapter 4 Image 23

3. Simplify:
(i) 28.796 -13.42 – 2.555
(ii) 93.354 – 62.82 – 13.045
(iii) 36 – 18.59 – 3.2
(iv) 86 + 16.95 – 3.0042
(v) 32.8 – 13 – 10.725 +3.517
(vi) 4000 – 30.51 – 753.101 – 69.43
(vii) 0.1835 + 163.2005 – 25.9 – 100
(viii) 38.00 – 30 + 200.200 – 0.230
(ix) 555.555 + 55.555 – 5.55 – 0.555

Solution:

(i) 28.796 -13.42 – 2.555
It can be written as

= 28.796 – (13.42 + 2.555)

On further calculation

= 28.796 – 15.975

= 12.821

Selina Solutions Concise Maths Class 7 Chapter 4 Image 24

(ii) 93.354 – 62.82 – 13.045
It can be written as

= 93.354 – (62.82 + 13.045)

On further calculation

= 93.354 – 75.865

= 17.489

Selina Solutions Concise Maths Class 7 Chapter 4 Image 25

(iii) 36 – 18.59 – 3.2
It can be written as

= 36 – (18.59 + 3.2)

On further calculation

= 36 – 21.79

= 14.21

Selina Solutions Concise Maths Class 7 Chapter 4 Image 26

(iv) 86 + 16.95 – 3.0042
It can be written as

= 102.95 – 3.0042

On further calculation

= 99.9458

Selina Solutions Concise Maths Class 7 Chapter 4 Image 27

(v) 32.8 – 13 – 10.725 +3.517
It can be written as

= (32.8 + 3.517) – (13 + 10.725)

On further calculation

= 36.317 – 23.725

= 12.592

Selina Solutions Concise Maths Class 7 Chapter 4 Image 28

(vi) 4000 – 30.51 – 753.101 – 69.43
It can be written as

= 4000 – (30.51 + 753.101 + 69.43)

On further calculation

= 4000 – 853.041

= 3146.959

Selina Solutions Concise Maths Class 7 Chapter 4 Image 29

(vii) 0.1835 + 163.2005 – 25.9 – 100
It can be written as

= (0.1835 + 163.2005) – (25.9 + 100)

On further calculation

= 163.2840 – 125.9

= 37.484

Selina Solutions Concise Maths Class 7 Chapter 4 Image 30

(viii) 38.00 – 30 + 200.200 – 0.230
It can be written as

= (38.00 + 200.200) – (30 + 0.230)

On further calculation

= 238.200 – 30.230

= 207.970

= 207.97

Selina Solutions Concise Maths Class 7 Chapter 4 Image 31

(ix) 555.555 + 55.555 – 5.55 – 0.555

It can be written as

= (555.555 + 55.555) – (5.55 + 0.555)

On further calculation

= 611.110 – 6.105

= 605.005

Selina Solutions Concise Maths Class 7 Chapter 4 Image 32

4. Find the difference between 6.85 and 0.685.

Solution:

The difference between 6.85 and 0.685 = 6.85 – 0.685

= 6.165

Selina Solutions Concise Maths Class 7 Chapter 4 Image 33

5. Take out the sum of 19.38 and 56.025, then subtract it from 200.111.

Solution:

We know that the sum of 19.38 and 56.025 can be written as

19.38 + 56.025 = 75.405

Selina Solutions Concise Maths Class 7 Chapter 4 Image 34

We can write it as

Difference between 200.111 and 75.405

200.111 – 75.405 = 124.706

Selina Solutions Concise Maths Class 7 Chapter 4 Image 35

6. Add 13.95 and 1.003, and from the result, subtract the sum of 2.794 and 6.2.

Solution:

We know that addition of 13.95 and 1.003 can be written as

13.95 + 1.003 = 14.953

Selina Solutions Concise Maths Class 7 Chapter 4 Image 36

Similarly the sum of 2.794 and 6.2 can be written as

2.794 + 6.2 = 8.994

Selina Solutions Concise Maths Class 7 Chapter 4 Image 37

Here the difference between 14.953 and 8.994

14.953 – 8.994 = 5.959

Selina Solutions Concise Maths Class 7 Chapter 4 Image 38

7. What should be added to 39.587 to give 80.375?

Solution:

It is given that

Sum of two numbers = 80.375

One number = 39.587

So the other number = 80.375 – 39.587 = 40.788

Selina Solutions Concise Maths Class 7 Chapter 4 Image 39

8. What should be subtracted from 100 to give 19.29?

Solution:

It is given that

Sum of two numbers = 100

One number = 19.29

So the other number = 100 – 19.29 = 80.71

Selina Solutions Concise Maths Class 7 Chapter 4 Image 40

9. What is the excess of 584.29 over 213.95?

Solution:

It is given that

Sum of two numbers = 584.29

One number = 213.95

So the other number = 584.29 – 213.95 = 370.34

Selina Solutions Concise Maths Class 7 Chapter 4 Image 41

10. Evaluate:

(i) (5.4 – 0.8) + (2.97 -1.462)

(ii) (6.25 + 0.36) – (17.2 – 8.97)

(iii) 9.004 + (3 -2.462)

(iv) 879.4 – (87.94 – 8 .794)

Solution:

(i) (5.4 – 0.8) + (2.97 -1.462)
It can be written as

= 4.6 + 1.508

On further calculation

= 6.108

Selina Solutions Concise Maths Class 7 Chapter 4 Image 42

(ii) (6.25 + 0.36) – (17.2 – 8.97)
It can be written as

= 6.61 – 8.23

On further calculation

= – 1.62

Selina Solutions Concise Maths Class 7 Chapter 4 Image 43

(iii) 9.004 + (3 – 2.462)
It can be written as

= 9.004 + 0.538

On further calculation

= 9.542

Selina Solutions Concise Maths Class 7 Chapter 4 Image 44

(iv) 879.4 – (87.94 – 8 .794)

It can be written as

= 879.4 – 79.146

On further calculation

= 800.254

Selina Solutions Concise Maths Class 7 Chapter 4 Image 45

11. What is the excess of 75 over 48.29?

Solution:

We know that the excess of 75 over 48.29 can be written as

Selina Solutions Concise Maths Class 7 Chapter 4 Image 46

Hence, the excess of 75 over 48.29 is 26.71.

12. If A = 237.98 and B = 83.47.

Find:

(i) A – B

(ii) B – A.

Solution:

(i) A – B

It is given that A = 237.98 and B = 83.47

Substituting the values

A – B = 237.98 – 83.47

A – B = 154.51

Selina Solutions Concise Maths Class 7 Chapter 4 Image 47

(ii) B – A

It is given that A = 237.98 and B = 83.47

Substituting the values

B – A = 83.47 – 237.98

B – A = – 154.51

Selina Solutions Concise Maths Class 7 Chapter 4 Image 48

13. The cost of one kg of sugar increases from 28.47 to 32.65. Find the increase in cost.

Solution:

Cost of sugar = 28.47

Cost of sugar is raised = 32.65

Increase in the cost of sugar = 32.65 – 28.47 = 4.18

Selina Solutions Concise Maths Class 7 Chapter 4 Image 49

Exercise 4C page: 61

1. Multiply:

(i) 0.87 by 10

(ii) 2.948 by 100

(iii) 6.4 by 1000

(iv) 5.8 by 4

(v) 16.32 by 28

(vi) 5. 037 by 8

(vi) 4.6 by 2.1

(viii) 0.568 by 6.4

Solution:

(i) 0.87 by 10
It can be written as

0.87 × 10 = 8.7

(ii) 2.948 by 100
It can be written as

2.948 × 100 = 294.8

(iii) 6.4 by 1000
It can be written as

6.4 × 1000 = 6400

(iv) 5.8 by 4
It can be written as

5.8 × 4 = 23.2

Selina Solutions Concise Maths Class 7 Chapter 4 Image 50

(v) 16.32 by 28
It can be written as

16.32 × 28 = 456.96

Selina Solutions Concise Maths Class 7 Chapter 4 Image 51

(vi) 5.037 by 8
It can be written as

5.037 × 8 = 40.296

Selina Solutions Concise Maths Class 7 Chapter 4 Image 52

(vi) 4.6 by 2.1
It can be written as

4.6 × 2.1 = 9.66

Selina Solutions Concise Maths Class 7 Chapter 4 Image 53

(viii) 0.568 by 6.4

It can be written as

0.568 × 6.4 = 3.6352

Selina Solutions Concise Maths Class 7 Chapter 4 Image 54

2. Multiply each number by 10, 100, 1000:

(i) 0.5

(ii) 0.112

(iii) 4.8

(iv) 0.0359

(v) 16.27

(vi) 234.8

Solution:

(i) 0.5
It can be written as

0.5 × 10 = 5

0.5 × 100 = 50

0.5 × 1000 = 500

(ii) 0.112
It can be written as

0.112 × 10 = 1.12

0.112 × 100 = 11.2

0.112 × 1000 = 112

(iii) 4.8
It can be written as

4.8 × 10 = 48

4.8 × 100 = 480

4.8 × 1000 = 4800

(iv) 0.0359
It can be written as

0.0359 × 10 = 0.359

0.0359 × 100 = 3.59

0.0359 × 1000 = 35.9

(v) 16.27
It can be written as

16.27 × 10 = 162.7

16.27 × 100 = 1627

16.27 × 1000 = 16270

(vi) 234.8

It can be written as

234.8 × 10 = 2348

234.8 × 100 = 23480

234.8 × 1000 = 234800

3. Evaluate:

(i) 5.897 x 2.3

(ii) 0.894 x 87

(iii) 0.01 x 0.001

(iv) 0.84 x 2.2 x 4

(v) 4.75 x 0.08 x 3

(vi) 2.4 x 3.5 x 4.8

(vii) 0.8 x 1.2 x 0.25

(viii) 0.3 x 0.03 x 0.003

Solution:

(i) 5.897 x 2.3
We know that

5.897 x 2.3 = 13.5631

Selina Solutions Concise Maths Class 7 Chapter 4 Image 55

(ii) 0.894 x 87
We know that

0.894 x 87 = 77.778

Selina Solutions Concise Maths Class 7 Chapter 4 Image 56

(iii) 0.01 x 0.001
We know that

0.01 x 0.001 = 0.00001

(iv) 0.84 x 2.2 x 4
It can be written as

= 0.84 x 8.8

= 7.392

Selina Solutions Concise Maths Class 7 Chapter 4 Image 57

(v) 4.75 x 0.08 x 3
It can be written as

= 4.75 x 0.24

= 1.1400

= 1.14

Selina Solutions Concise Maths Class 7 Chapter 4 Image 58

(vi) 2.4 x 3.5 x 4.8
It can be written as

= 8.40 x 4.8

= 8.4 x 4.8

We get

= 40.32

Selina Solutions Concise Maths Class 7 Chapter 4 Image 59

(vii) 0.8 x 1.2 x 0.25
It can be written as

= 0.96 x 0.25
= 0.2400

= 0.24

Selina Solutions Concise Maths Class 7 Chapter 4 Image 60

(viii) 0.3 x 0.03 x 0.003

It can be written as

= 0.009 x 0.003

= 0.000027

4. Divide:

(i) 54.9 by 10

(ii) 7.8 by 100

(iii) 324.76 by 1000

(iv) 12.8 by 4

(v) 27.918 by 9

(vi) 4.672 by 8

(vii) 4.32 by 1.2

(viii) 7.644 by 1.4

(ix) 4.8432 by 0.08

Solution:

(i) 54.9 by 10
It can be written as

54.9 ÷ 10 = 5.49

(ii) 7.8 by 100
It can be written as

7.8 ÷ 100 = 0.078

(iii) 324.76 by 1000
It can be written as

324.76 ÷ 1000 = 0.32476

(iv) 12.8 by 4
It can be written as

12.8 ÷ 4 = 3.2

(v) 27.918 by 9
It can be written as

27.918 ÷ 9 = 3.102

(vi) 4.672 by 8

It can be written as

4.672 ÷ 8 = 0.584



Selina Solutions Concise Maths Class 7 Chapter 4 Image 61

(vii) 4.32 by 1.2
It can be written as

4.32 ÷ 1.2

Multiplying by 100

432 ÷ 120 = 3.6

Selina Solutions Concise Maths Class 7 Chapter 4 Image 62

(viii) 7.644 by 1.4
It can be written as

7.644 ÷ 1.4

Multiplying by 1000

7644 ÷ 1400 = 5.46

Selina Solutions Concise Maths Class 7 Chapter 4 Image 63

(ix) 4.8432 by 0.08

It can be written as

4.8432 ÷ 0.08

So we get

48432 ÷ 800 = 60.54

Selina Solutions Concise Maths Class 7 Chapter 4 Image 64

5. Divide each of the given numbers by 10, 100, 1000 and 10000

(i) 2.1

(ii) 8.64

(iii) 5-01

(iv) 0.0906

(v) 0.125

(vi) 111.11

Solution:

(i) 2.1
It can be written as

2.1 ÷ 10 = 0.21

2.1 ÷ 100 = 0.021

2.1 ÷ 1000 = 0.0021

2.1 ÷ 10000 = 0.00021

(ii) 8.64
It can be written as

8.64 ÷ 10 = 0.864

8.64 ÷ 100 = 0.0864

8.64 ÷ 1000 = 0.00864

8.64 ÷ 10000 = 0.000864

(iii) 5.01
It can be written as

5.01 ÷ 10 = 0.501

5.01 ÷ 100 = 0.0501

5.01 ÷ 1000 = 0.00501

5.01 ÷ 10000 = 0.000501

(iv) 0.0906
It can be written as

0.0906 ÷ 10 = 0.00906

0.0906 ÷ 100 = 0.000906

0.0906 ÷ 1000 = 0.0000906

0.0906 ÷ 10000 = 0.00000906

(v) 0.125
It can be written as

0.125 ÷ 10 = 0.0125

0.125 ÷ 100 = 0.00125

0.125 ÷ 1000 = 0.000125

0.125 ÷ 10000 = 0.0000125

(vi) 111.11

It can be written as

111.11 ÷ 10 = 11.111

111.11 ÷ 100 = 1.1111

111.11 ÷ 1000 = 0.11111

111.11 ÷ 10000 = 0.011111

6. Evaluate :

(i) 9.75 ÷ 5

(ii) 4.4064 ÷ 4

(iii) 27.69 ÷ 30

(iv) 19.25 ÷ 25

(v) 20.64 ÷ 16

(vi) 3.204 ÷ 9

(vii) 0.125 ÷ 25

(viii) 0.14616 ÷ 72

(ix) 0.6227 ÷ 1300

(x) 257.894 ÷ 0.169

(xi) 6.3 ÷ (0.3)²

Solution:

(i) 9.75 ÷ 5
We get

9.75 ÷ 5 = 1.95

Selina Solutions Concise Maths Class 7 Chapter 4 Image 65

(ii) 4.4064 ÷ 4
We get

4.4064 ÷ 4 = 1.016

(iii) 27.69 ÷ 30
We get

27.69 ÷ 30 = 0.923

Selina Solutions Concise Maths Class 7 Chapter 4 Image 66

(iv) 19.25 ÷ 25
We get

19.25 ÷ 25 = 0.77

Selina Solutions Concise Maths Class 7 Chapter 4 Image 67

(v) 20.64 ÷ 16
We get

20.64 ÷ 16 = 1.29

Selina Solutions Concise Maths Class 7 Chapter 4 Image 68

(vi) 3.204 ÷ 9
We get

3.204 ÷ 9 = 0.356

Selina Solutions Concise Maths Class 7 Chapter 4 Image 69

(vii) 0.125 ÷ 25
We get

0.125 ÷ 25 = 0.005

Selina Solutions Concise Maths Class 7 Chapter 4 Image 70

(viii) 0.14616 ÷ 72
We get

0.14616 ÷ 72 = 0.00203

Selina Solutions Concise Maths Class 7 Chapter 4 Image 71

(ix) 0.6227 ÷ 1300
We get

0.6227 ÷ 1300 = 0.000479

Selina Solutions Concise Maths Class 7 Chapter 4 Image 72

(x) 257.894 ÷ 0.169
Multiplying by 1000

257894 ÷ 169 = 1526

Selina Solutions Concise Maths Class 7 Chapter 4 Image 73

(xi) 6.3 ÷ (0.3)²

We can write it as

= 6.3 ÷ (0.3 × 0.3)

By further calculation

= 6.3 ÷ 0.09

Multiply both sides by 100

= 630 ÷ 9 = 70

7. Evaluate:

(i) 4.3 x 0.52 x 0.3

(ii) 3.2 x 2.5 x 0.7

(iii) 0.8 x 1.5 x 0.6

(iv) 0.3 x 0.3 x 0.3

(v) 1.2 x 1.2 x 0.4

(vi) 0.4 x 0.04 x 0.004

(vii) 0.5 x 0.6 x 0.7

(viii) 0.5 x 0.06 x 0.007

Solution:

(i) 4.3 x 0.52 x 0.3
We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 74

Here the sum of decimal places = 1 + 2 + 1 = 4

So we get

4.3 x 0.52 x 0.3 = 0.6708

(ii) 3.2 x 2.5 x 0.7
We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 75

Here the sum of decimal places = 1 + 1 + 1 = 3

So we get

3.2 x 2.5 x 0.7 = 5.600 or 5.6

(iii) 0.8 x 1.5 x 0.6
We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 76

Here the sum of decimal places = 1 + 1 + 1 = 3

So we get

0.8 x 1.5 x 0.6 = 0.720 or 0.72

(iv) 0.3 x 0.3 x 0.3
We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 77

Here the sum of decimal places = 1 + 1 + 1 = 3

So we get

0.3 x 0.3 x 0.3 = 0.027

(v) 1.2 x 1.2 x 0.4
We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 78

Here the sum of decimal places = 1 + 1 + 1 = 3

So we get

1.2 x 1.2 x 0.4 = 0.576

(vi) 0.4 x 0.04 x 0.004
We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 79

Here the sum of decimal places = 1 + 2 + 3 = 6

So we get

0.4 x 0.04 x 0.004 = 0.000064

(vii) 0.5 x 0.6 x 0.7

We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 80

Here the sum of decimal places = 1 + 1 + 1 = 3

So we get

0.5 x 0.6 x 0.7 = 0.210 or 0.21

(viii) 0.5 x 0.06 x 0.007

We know that

Selina Solutions Concise Maths Class 7 Chapter 4 Image 81

Here the sum of decimal places = 1 + 2 + 3 = 5

So we get

0.5 x 0.06 x 0.007 = 0.00021

8. Evaluate:

(i) (0.9)²

(ii) (0.6)² x 0.5

(iii) 0.3 x (0.5)²

(iv) (0.4)³

(v) (0.2)3 x 5

(vi) (0.2)3 x 0.05

Solution:

(i) (0.9)²
It can be written as

0.9 x 0.9 = 0.81

Here the sum of decimal places is 1 + 1 = 2

(ii) (0.6)² x 0.5
It can be written as

= 0.6 x 0.6 x 0.5

On further calculation

= 0.36 x 0.5

= 0.180 or 0.18

Here the sum of decimal places is 1 + 1 + 1 = 3

(iii) 0.3 x (0.5)²
It can be written as

= 0.3 x 0.5 x 0.5

On further calculation

= 0.3 x 0.25

= 0.075

Here the sum of decimal places is 1 + 1 + 1 = 3

(iv) (0.4)³
It can be written as

= 0.4 x 0.4 x 0.4

On further calculation

= 0.16 x 0.4

= 0.064

Here the sum of decimal places is 1 + 1 + 1 = 3

(v) (0.2)3 x 5
It can be written as

= 0.2 x 0.2 x 0.2 x 5

On further calculation

= 0.008 x 5

= 0.40 or 0.4

Here the sum of decimal places is 1 + 1 + 1 = 3

(vi) (0.2)3 x 0.05

It can be written as

= 0.2 x 0.2 x 0.2 x 0.05

On further calculation

= 0.008 x 0.05

= 0.00040

Here the sum of decimal places is 1 + 1 + 1 + 1 + 1 = 5

9. Find the cost of 36.75 kg wheat at the rate of ₹12.80 per kg.

Solution:

It is given that

Weight of wheat = 36.75 kg

Cost of wheat per kg = ₹12.80

So the cost of 36.75 kg wheat = 36.75 x 12.80 = ₹470.40

Selina Solutions Concise Maths Class 7 Chapter 4 Image 82

10. The cost of a pen is ₹56.15. Find the cost of 16 such pens.

Solution:

It is given that

Cost of a pen = ₹56.15

So the cost of 16 such pens = 16 x 56.15 = ₹898.40

Selina Solutions Concise Maths Class 7 Chapter 4 Image 83

11. Evaluate:

(i) 0.0072 ÷ 0.06

(ii) 0.621 ÷ 0.3

(iii) 0.0532 ÷ 0.005

(iv) 0.01162 ÷ 0.14

(v) (7.5 x 40.4) ÷ 25

(vi) 2.1 ÷ (0.1 x 0.1)

Solution:

(i) 0.0072 ÷ 0.06
Multiplying both numerator and denominator by 100

= (0.0072 x 100)/ (0.06 x 100)

On further calculation

= 0.72/6

= 0.12

(ii) 0.621 ÷ 0.3
Multiplying both numerator and denominator by 10

= (0.621 x 10)/ (0.3 x 10)

On further calculation

= 6.21/3

= 2.07

(iii) 0.0532 ÷ 0.005
Multiplying both numerator and denominator by 1000

= (0.0532 x 1000)/ (0.005 x 1000)

On further calculation

= 53.2/5

= 10.64

(iv) 0.01162 ÷ 0.14
Multiplying both numerator and denominator by 100

= (0.01162 x 100)/ (0.14 x 100)

On further calculation

= 1.162/14

= 0.083

(v) (7.5 x 40.4) ÷ 25
It can be written as

= 303/25

= 12.12

(vi) 2.1 ÷ (0.1 x 0.1)

Multiplying both numerator and denominator by 100

= (2.1 x 100)/ (0.01 x 100)

On further calculation

= 210/1

= 210

12. Fifteen identical articles weigh 31.50 kg. Find the weight of each article.

Solution:

It is given that

Total weight of 15 identical articles = 31.50 kg

So the weight of each article = 31.50 – 15 = 2.1 kg

Hence, the weight of each article is 2.1 kg.

13. The product of two numbers is 211.2. If one of these two numbers is 16.5, find the other number.

Solution:

It is given that

Product of two numbers = 211.2

One of the two numbers = 16.5

So the other number = 211.2 ÷ 16.5

On further calculation

= (211.2 x 10)/ (16.5 x 10)

So we get

= 2112/165

= 12.8

14. One dozen identical articles cost ₹45.96. Find the cost of each article.

Solution:

It is given that

Cost of one dozen articles = ₹45.96

We know that one dozen = 12

So the cost of one article = 45.96 ÷ 12 = ₹3.83

Exercise 4D page: 65

1. Find whether the given division forms a terminating decimal or a non-terminating decimal:

(i) 3 ÷ 8

(ii) 8 ÷ 3

(iii) 6÷ 5

(iv) 5 ÷ 6

(v) 12.5 ÷ 4

(vi) 23 ÷ 0.7

(vii) 42 ÷ 9

(viii) 0.56÷ 0.11

Solution:

(i) 3 ÷ 8
We know that

3 ÷ 8 = 0.375

Therefore, it is terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 84

(ii) 8 ÷ 3
We know that

8 ÷ 3 = 2.666

Therefore, it is a non-terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 85

(iii) 6 ÷ 5
We know that

6 ÷ 5 = 1.2

Therefore, it is terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 86

(iv) 5 ÷ 6
We know that

5 ÷ 6 = 0.8333

Therefore, it is non-terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 87

(v) 12.5 ÷ 4
We know that

12.5 ÷ 4 = 3.125

Therefore, it is terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 88

(vi) 23 ÷ 0.7
Multiplying by 10 we get

230 ÷ 7 = 32.8571428

Therefore, it is non-terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 89

(vii) 42 ÷ 9
We know that

42 ÷ 9 = 4.666

Therefore, it is non-terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 90

(viii) 0.56 ÷ 0.11

Multiplying by 100

56 ÷ 11 = 5.0909

Therefore, it is non-terminating decimal.

Selina Solutions Concise Maths Class 7 Chapter 4 Image 91

2. Express as recurring decimals:

(i) 1 1/3

(ii) 10/11

(iii) 5/6

(iv) 2/13

(v) 1/9

(vi) 17/90

(vii) 5/18

(viii) 7/12

Solution:

(i) 1 1/3

It can be written as

1 1/3 = 4/3
Selina Solutions Concise Maths Class 7 Chapter 4 Image 92

Selina Solutions Concise Maths Class 7 Chapter 4 Image 93

(ii) 10/11

It can be written as

10/11 = 0.909090… Selina Solutions Concise Maths Class 7 Chapter 4 Image 94

Selina Solutions Concise Maths Class 7 Chapter 4 Image 95

(iii) 5/6

It can be written as

5/6 = 0.8333…. Selina Solutions Concise Maths Class 7 Chapter 4 Image 96

Selina Solutions Concise Maths Class 7 Chapter 4 Image 97

(iv) 2/13

It can be written as

2/13 = 0.153846153846 Selina Solutions Concise Maths Class 7 Chapter 4 Image 98

Selina Solutions Concise Maths Class 7 Chapter 4 Image 99

(v) 1/9

It can be written as

1/9 = 0.1111 …. Selina Solutions Concise Maths Class 7 Chapter 4 Image 100

Selina Solutions Concise Maths Class 7 Chapter 4 Image 101

(vi) 17/90

It can be written as

17/90 = 0.1888 Selina Solutions Concise Maths Class 7 Chapter 4 Image 102

Selina Solutions Concise Maths Class 7 Chapter 4 Image 103

(vii) 5/18

It can be written as

5/18 = 0.2777 … Selina Solutions Concise Maths Class 7 Chapter 4 Image 104

Selina Solutions Concise Maths Class 7 Chapter 4 Image 105

(viii) 7/12

It can be written as

7/12 = 0.58333…. Selina Solutions Concise Maths Class 7 Chapter 4 Image 106

Selina Solutions Concise Maths Class 7 Chapter 4 Image 107

3. Convert into vulgar fraction:

Selina Solutions Concise Maths Class 7 Chapter 4 Image 108

Solution:

Selina Solutions Concise Maths Class 7 Chapter 4 Image 109

It can be written as

= 3/9

So we get

= (3 – 0)/ 9

= 3/9

= 1/3

Selina Solutions Concise Maths Class 7 Chapter 4 Image 110

It can be written as

= 8/9

So we get

= (8 – 0)/ 9

= 8/9

Selina Solutions Concise Maths Class 7 Chapter 4 Image 111

It can be written as

= 44/9

So we get

= (44 – 4)/ 9

= 40/9

= 4 4/9

Selina Solutions Concise Maths Class 7 Chapter 4 Image 112

It can be written as

= 237/9

So we get

= (237 – 23)/ 9

= 214/9

= 23 7/9

4. Convert into vulgar fraction:

Selina Solutions Concise Maths Class 7 Chapter 4 Image 113

Solution:

Selina Solutions Concise Maths Class 7 Chapter 4 Image 114

It can be written as

= 35/99

So we get

= (35 – 0)/ 99

= 35/99

Selina Solutions Concise Maths Class 7 Chapter 4 Image 115

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 4 Image 116

So we get

= 2 + (23 – 0)/ 99

On further calculation

= 2 + 23/99

= 2 33/99

Selina Solutions Concise Maths Class 7 Chapter 4 Image 117

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 4 Image 118

So we get

= 1 + (28 – 0)/ 99

On further calculation

= 1 + 28/99

= 1 28/99

Selina Solutions Concise Maths Class 7 Chapter 4 Image 119

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 4 Image 120

So we get

= 5 + (234 – 0)/ 999

On further calculation

= 5 234/999

5. Convert into vulgar fraction:

Selina Solutions Concise Maths Class 7 Chapter 4 Image 121

Solution:

Selina Solutions Concise Maths Class 7 Chapter 4 Image 122

It can be written as

= (37 – 3)/ 90

So we get

= 34/90

= 17/45

Selina Solutions Concise Maths Class 7 Chapter 4 Image 123

It can be written as

= (245 – 2)/ 990

So we get

= 243/990

On further calculation

= 81/330

= 27/110

Selina Solutions Concise Maths Class 7 Chapter 4 Image 124

It can be written as

= (685 – 68)/ 900

So we get

= 617/ 900

Selina Solutions Concise Maths Class 7 Chapter 4 Image 125

It can be written as

= (442 – 4)/ 990

So we get

= 438/ 990

= 219/ 495

Exercise 4E PAGE: 67

1. Round off:

(i) 0.07, 0.112, 3.59, 9.489 to the nearest tenths.

(ii) 0.627, 100.479, 0.065 and 0.024 to the nearest hundredths.

(iii) 4.83, 0.86, 451.943 and 9.08 to the nearest whole number.

Solution:

(i) We know that

0.07 = 0.1

0.112 = 0.1

3.59 = 3.6

9.489 = 9.5

(ii) 0.627 = 0.63

100.479 = 100.48

0.065 = 0.07

0.024 = 0.02

(iii) 4.83 = 5

0.86 = 1

451.943 = 452

9.08 = 9

2. Simplify, and write your answers correct to the nearest hundredths:

(i) 18.35 × 1.2

(ii) 62.89 × 0.02

Solution:

(i) 18.35 × 1.2 = 22.02

Selina Solutions Concise Maths Class 7 Chapter 4 Image 126

(ii) 62.89 × 0.02 = 1.2578 = 1.26

Selina Solutions Concise Maths Class 7 Chapter 4 Image 127

3. Write the number of significant figures (digits) in:

(i) 35.06

(ii) 0.35

(iii) 7.0068

(iv) 19.0

(v) 0.0062

(vi) 4.2 × 0.6

(vii) 0.08 × 25

(viii) 3.6 ÷ 0.12

Solution:

(i) The number of significant figures in 35.06 is 4.

(ii) The number of significant figures in 0.35 is 2.

(iii) The number of significant figures in 7.0068 is 5.

(iv) The number of significant figures in 19.0 is 3.

(v) The number of significant figures in 0.0062 is 2.

(vi) The number of significant figures in 4.2 × 0.6 = 2.52 is 3.

(vii) The number of significant figures in 0.08 × 25 = 2.00 = 2 is 1.

(viii) The number of significant figures in 3.6 ÷ 0.12 or 360 ÷ 12 = 30 is 2.

4. Write:

(i) 35.869, 0.008426, 4.952 and 382.7 correct to three significant figures.

(ii) 60.974, 2.8753, 0.001789 and 400.04 correct to four significant figures.

(iii) 14.29462, 19.2, 46356.82 and 69 correct to five significant figures.

Solution:

(i) Here by correcting to three significant figures.

35.869 = 35.9

0.008426 = 0.00843

4.952 = 4.95

382.7 = 383

(ii) Here by correcting to four significant figures

60.974 = 60.97

2.8753 = 2.875

0.001789 = 0.001789

400.04 = 400.0

(iii) Here by correcting to five significant figures

14.29462 = 14.295

19.2 = 19.200

46356.82 = 46357

69 = 69.000

Exercise 4F page: 67

1. The weight of an object is 3.06 kg. Find the total weight of 48 similar objects.

Solution:

It is given that

Weight of an object = 3.06 kg

So the weight of 48 objects = 3.06 × 48 = 146.88 kg

Selina Solutions Concise Maths Class 7 Chapter 4 Image 128

2. Find the cost of 17.5 m cloth at the rate of ₹ 112.50 per metre.

Solution:

It is given that

Cost of cloth per metre = ₹ 112.50

So the cost of 17.5 m cloth = ₹ 112.50 × 17.5

On further calculation

= ₹ 1968.750

= ₹ 1968.75

Selina Solutions Concise Maths Class 7 Chapter 4 Image 129

3. One kilogramme of oil costs ₹ 73.40. Find the cost of 9.75 kilogramme of the oil.

Solution:

It is given that

Cost of 1 kg oil = ₹ 73.40

So the cost of 9.75 kg oil = ₹ 73.40 × 9.75

On further calculation

= ₹ 715.6500

= ₹ 715.65

Selina Solutions Concise Maths Class 7 Chapter 4 Image 130

4. Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.

Solution:

It is given that

Weight of 8 identical objects = 51.2 kg

So the weight of 1 object = 51.2 ÷ 8 = 6.4 kg

5. 18.5 m of cloth costs ₹ 666. Find the cost of 3.8 m cloth.

Solution:

It is given that

Cost of 18.5 m cloth = ₹ 666

So the cost of 1m cloth = ₹ 666 ÷ 18.5 and cost of 3.8 m cloth

We can write it as

= (666 ÷ 18.5) × 3.8

Multiplying by 10

= (6660 ÷ 185) × 3.8

= 36 × 3.8

So we get

= ₹ 136.80

Selina Solutions Concise Maths Class 7 Chapter 4 Image 131

6. Find the value of:

(i) 0.5 of ₹ 7.60 + 1.62 of ₹ 30

(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg

(iii) 6.25 of 8.4 – 4.7 of 3.24

(iv) 0.98 of 235 – 0.09 of 3.2

Solution:

(i) 0.5 of ₹ 7.60 + 1.62 of ₹ 30

It can be written as

= ₹ 3.80 + ₹ 48.60

So we get

= ₹ 52.40

Selina Solutions Concise Maths Class 7 Chapter 4 Image 132

(ii) 2.3 of 7.3 kg + 0.9 of 0.48 kg

It can be written as

= 16.79 kg + 0.432 kg

So we get

= 17.222 kg

Selina Solutions Concise Maths Class 7 Chapter 4 Image 133

(iii) 6.25 of 8.4 – 4.7 of 3.24

It can be written as

= 52.500 – 15.228

So we get

= 37.272

Selina Solutions Concise Maths Class 7 Chapter 4 Image 134

(iv) 0.98 of 235 – 0.09 of 3.2

It can be written as

= 230.30 – 0.288

So we get

= 230.012

Selina Solutions Concise Maths Class 7 Chapter 4 Image 135

7. Evaluate:

(i) 5.6 – 1.5 of 3.4

(ii) 4.8 ÷ 0.04 of 5

(iii) 0.72 of 80 ÷ 0.2

(iv) 0.72 ÷ 80 of 0.2

(v) 6.45 ÷ (3.9 – 1.75)

(vi) 0.12 of (0.104 – 0.02) + 0.36 × 0.5

Solution:

(i) 5.6 – 1.5 of 3.4

It can be written as

= 5.6 – 5.1

So we get

= 0.5

Selina Solutions Concise Maths Class 7 Chapter 4 Image 136

(ii) 4.8 ÷ 0.04 of 5

It can be written as

= 4.8 ÷ 0.20

Multiplying by 10

= 48 ÷ 2

= 24

(iii) 0.72 of 80 ÷ 0.2

It can be written as

= 57.60 ÷ 0.2

Multiplying by 10

= 576 ÷ 2

= 288

(iv) 0.72 ÷ 80 of 0.2

It can be written as

= 0.72 ÷ 16.0

Multiplying by 100

= 72 ÷ 1600

= 0.045

Selina Solutions Concise Maths Class 7 Chapter 4 Image 137

(v) 6.45 ÷ (3.9 – 1.75)

It can be written as

= 6.45 ÷ 2.15

Multiplying by 100

= 645 ÷ 215

= 3

Selina Solutions Concise Maths Class 7 Chapter 4 Image 138

(vi) 0.12 of (0.104 – 0.02) + 0.36 × 0.5

It can be written as

= 0.12 of 0.084 + 0.36 × 0.5

So we get

= 0.01008 + 0.180

= 0.19008

Selina Solutions Concise Maths Class 7 Chapter 4 Image 139

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