RS Aggarwal Solutions for Class 7 Maths Chapter 3 Decimals

RS Aggarwal Solutions for Class 7 Maths Chapter 3 – Decimals, download the pdf given below. If students want to score higher in Maths, it requires the right amount of practice for every topic. More marks can be achieved by referring to RS Aggarwal Class 7 Solutions. Problems are solved step by step by BYJU’S experts with neat explanations.

Students obtain more knowledge by referring to RS Aggarwal Solutions for Class 7 Maths Chapter 3. Download pdf of Class 7 Chapter 3 in their respective links.

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

 

Also, access RS Aggarwal Solutions for Class 7 Chapter 3 Exercises

Exercise 3A

Exercise 3B

Exercise 3C

Exercise 3D

Exercise 3E

Exercise 3A

Convert each of the following into a fraction in its simplest from:

(i) .8

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= .8 = (8/10)

Now reduce the above fraction to the simplest form.

= (8/10) … [÷ by 2]

= (4/5)

(ii) .75

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= .75 = (75/100)

Now reduce the above fraction to the simplest form.

= (75/100) … [÷ by 5]

= (3/4)

(iii) .06

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= .06 = (6/100)

Now reduce the above fraction to the simplest form.

= (6/100) … [÷ by 2]

= (3/50)

(iv) .285

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= .285 = (285/1000)

Now reduce the above fraction to the simplest form.

= (285/1000) … [÷ by 5]

= (5/200)

2. Convert each of the following as a mixed fraction:

(i) 5.6

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= 5.6 = (56/10)

Now reduce the above fraction to the simplest form.

= (56/10) … [÷ by 2]

= (28/5)

= [5(3/5)]

(ii) 12.25

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= 12.25 = (1225/100)

Now reduce the above fraction to the simplest form.

= (1225/100) … [÷ by 5]

= (49/4)

= [12(1/4)]

(iii) 6.004

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= 6.004 = (6004/1000)

Now reduce the above fraction to the simplest form.

= (6004/1000) … [÷ by 2]

= (1501/250)

= [6(1/250)]

(iv) 4.625

Solution:-

Now write the given decimal without the decimal point as the numerator of the fraction.

In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.

Then we have,

= 4.625 = (4625/1000)

Now reduce the above fraction to the simplest form.

= (4625/1000) … [÷ by 125]

= (37/8)

= [4(5/8)]

3. Convert each of the following into a decimals:

(i) (47/10)

Solution:-

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

rs aggarwal class 7 maths chapter 3 - 1

(ii) (156/100)

Solution:-

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 1

∴ (156/100) = 1.56

(iii) (2516/100)

Solution:-

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 2

∴ (2516/100) = 25.16

(iv) (3524/1000)

Solution:-

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

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

∴ (3524/1000) = 3.524

(v) (25/8)

Solution:-

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 4

∴ (25/8) = 3.125

(vi) [3(2/5)]

Solution:-

Convert mixed fraction into improper fraction,

= [3(2/5)] = (17/5)

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 5

∴ (17/5) = 3.4

(vii) [2(2/25)]

Solution:-

Convert mixed fraction into improper fraction,

= [2(2/25)] = (52/25)

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 6

∴ (52/25) = 2.08

(viii) (17/20)

Solution:-

Dived the numerator by the denominator till a nonzero remainder is obtained.

Put a decimal point in the dividend as well as in the quotient.

Put a zero on the decimal point in the dividend as well as on the right of the remainder.

Divide again just as we do in whole numbers.

Repeat the steps 3 and 4, till the remainder is zero.

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

∴ (17/20) = 0.85

4. Convert each of the following into like decimals:

(i) 6.5, 16.03, 0.274, 119.4

Solution:-

Decimals having the same number of decimal place are called like decimals.

By converting, we have,

6.500, 16.030, 0.274, 119.400

(ii) 3.5, 0.67, 15.6, 4

Solution:-

Decimals having the same number of decimal place are called like decimals.

By converting, we have

3.50, 0.67, 15.60, 4.00

5. Fill in each of the place holders with the correct symbol > or <.

(i) 78.23 ……. 69.85

Solution:-

By comparing whole numbers 78 > 69

∴ 78.23>69.85

(ii) 3.406……3.46

Solution:-

By comparing whole number, 3 = 3

By comparing the tenths place digit, 4 = 4

By comparing the hundredths place digit, 0 < 6

∴ 3.406 < 3.46

(iii) 5.68……5.86

Solution:-

By comparing whole number, 5 = 5

By comparing the tenths place digit, 6 < 8

∴ 5.68 < 5.86

(iv) 14.05……14.005

Solution:-

By comparing whole number, 14 = 14

By comparing the tenths place digit, 0 = 0

By comparing the hundredths place digit, 5 > 0

∴ 14.05 > 14.005

(v) 1.85……1.80

Solution:-

By comparing whole number, 1 = 1

By comparing the tenths place digit, 8 = 8

By comparing the hundredths place digit, 5 > 0

∴ 1.85 > 1.80

(vi) 0.98……1.07

Solution:-

By comparing whole number, 0 < 1

∴ 0.98 < 1.07


Exercise 3B

Add:

1. 16, 8.7, 0.94, 6.8 and 7.77

Solution:-

First convert the given decimals into like decimals.

We get,

16.00, 8.70, 0.94, 6.80, 7.77

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

rs aggarwal class 7 maths chapter 3 - 2

Hence, the sum of the given decimals is 40.21

2. 18.6, 206.37, 8.008, 26.4 and 6.9

Solution:-

First convert the given decimals into like decimals.

We get,

18.600, 206.370, 8.008, 26.400, 6.900

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

rs aggarwal class 7 maths chapter 3 - 3

Hence, the sum of the given decimals is 266.278

3. 63.5, 9.7, 0.8, 26.66 and 12.17

Solution:-

First convert the given decimals into like decimals.

We get,

63.50, 9.70, 0.80, 26.66 and 12.17

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

rs aggarwal class 7 maths chapter 3 - 4

Hence, the sum of the given decimals is 112.83

4. 17.4, 86,39, 9.435, 8.8 and 0.06

Solution:-

First convert the given decimals into like decimals.

We get,

17.400, 86,390, 9.435, 8.800 and 0.060

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

rs aggarwal class 7 maths chapter 3 - 5

Hence, the sum of the given decimals is 122.085

5. 26.9, 19.74, 231.769 and 0.048

Solution:-

First convert the given decimals into like decimals.

We get,

26.900, 19.740, 231.769 and 0.048

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

26.900

19.740

234.769

0.048

______________

278.457

______________

Hence, the sum of the given decimals is 278.457

6. 23.8, 8.94, 0.078 and 214.6

Solution:-

First convert the given decimals into like decimals.

We get,

23.800, 8.940, 0.078 and 214.600

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

23.800

8.940

0.078

214.600

________________

247.418

________________

Hence, the sum of the given decimals is 247.418

7. 6.606, 66.6, 666, 0.066, 0.66

Solution:-

First convert the given decimals into like decimals.

We get,

6.606, 66.600, 666.000, 0.066, 0.660

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

6.606

66.600

666.000

0.066

0.660

_______________

739.932

_______________

Hence, the sum of the given decimals is 739.932

8. 9.09, 0.909, 99.9, 9.99, 0.099

Solution:-

First convert the given decimals into like decimals.

We get,

9.090, 0.909, 99.900, 9.990, 0.099

Write the addends on under the other in column form, keeping the decimal points of all the addends in the same column and the digits of the same place in the same column.

Writing these decimals in column from and adding, we get:

9.090

0.909

99.900

9.990

0.099

____________

119.988

_____________

Subtract:

9. 14.79 from 72.43

Solution:-

Writing them into column form with the larger one at the top and subtracting, we get:

72.79

-14.79

____________

57.64

____________

Hence (14.79 – 72.43) = 57.64

10. 36.74 from 52.6

Solution:-

Converting the decimals into like decimals, we get 36.74 and 52.60

Writing them into column form with the larger one at the top and subtracting, we get:

52.60

-36.74

_____________

15.86

_____________

Hence (36.74 – 52.6) = 15.86

11. 13.876 from 22

Solution:-

Converting the decimals into like decimals, we get 13.876 from 22.000

Writing them into column form with the larger one at the top and subtracting, we get:

22.000

-13.876

_____________

8.124

_____________

Hence (13.876-22) = 8.124


Exercise 3C

1. Find the product:

(i) 73.92 × 10

Solution:-

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

We have,

73.92 × 10 = 739.2

(ii) 7.54 × 10

Solution:-

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

We have,

7.54 × 10 = 75.4

(iii) 84.003 × 10

Solution:-

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

We have,

84.003 × 10 = 840.03

(iv) 0.83 × 10

Solution:-

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

We have,

0.83 × 10 = 8.3

(v) 0.7 × 10

Solution:-

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

We have,

0.7 × 10 = 7

(vi) 0.032 × 10

Solution:-

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

We have,

0.032 × 10 = 0.32

2. Find the product:

(i) 2.397 × 100

Solution:-

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

We have,

2.397 × 100 = 239.7

(ii) 6.83 × 100

Solution:-

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

We have,

6.83 × 100 = 683

(iii) 2.9 × 100

Solution:-

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

We have,

2.9 × 100 = 290

(iv) 0.08 × 100

Solution:-

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

We have,

0.08 × 100 = 8

(v) 0.6 × 100

Solution:-

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

We have,

0.6 × 100 = 60

(vi) 0.06 × 100

Solution:-

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

We have,

0.003 × 100 = 0.3

3. Find the product:

(i) 6.7314 × 1000

Solution:-

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

We have,

6.7314 × 1000 = 6731.4

(ii) 0.182 × 1000

Solution:-

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

We have,

0.182 × 1000 = 182

(iii) 0.076 × 1000

Solution:-

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

We have,

0.076 × 1000 = 76

(iv) 6.25 × 1000

Solution:-

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

We have,

6.25 × 1000 = 6250

(v) 4.8 × 1000

Solution:-

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

We have,

4.8 × 1000 = 4800

(vi) 0.06 × 1000

Solution:-

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

We have,

0.06 × 1000 = 60

4. Find the product:

(i) 5.4 × 16

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

54 × 16 = 864

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 5.4 × 16 = 8.64

(ii) 3.65 × 19

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

365 × 19 = 6935

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 3.65 × 19 = 69.35

(iii) 0.854 × 12

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

854 × 12 = 10248

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 0.854 × 12 = 10.248

(iv) 36.73 × 48

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

3673 × 48 = 176304

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 36.73 × 48 = 1763.04

(v) 4.125 × 86

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

4125 × 86 = 354750

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 4.125 × 86 = 35.4750

(vi) 104.06 × 75

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

10406 × 75 = 780450

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 104.06 × 75 = 780.450

(vii) 6.032 × 124

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

6.032 × 124 = 747968

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 6.032 × 124 = 747.968

(viii) 0.0146 × 69

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

146 × 69 = 10074

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 0.0146 × 69 = 1.0074

(ix) 0.00125 × 327

Solution:-

Multiply the decimal without the decimal point by the given whole number.

We have,

125 × 327 = 40875

Mark the decimal point in the product to have as many places of decimal as are there are in the given decimal.

∴ 0.00125 × 327 = 0.40875

5. Find the product :-

(i) 7.6 × 2.4

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 8

∴ 76 × 24 = 1824

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 2

∴ 7.6 × 2.4 = 18.24

(ii) 3.45 × 6.3

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 9

∴ 345 × 63 = 21735

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 3

∴ 3.45 × 6.3 = 21.735

(iii) 0.54 × 0.27

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 10

∴ 54 × 27 = 1458

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 4

∴ 0.54 × 0.27 = 0.1458

(iv) 0.568 × 4.9

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 11

∴ 568 × 49 = 27832

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 4

∴ 0.568 × 4.9 = 2.7832

(v) 6.54 × 0.09

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 12

∴ 654 × 9 = 5886

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 4

∴ 6.54 × 0.09 =0.5886

(vi) 3.87 × 1.25

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 13

∴ 387 × 125 = 48375

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 4

∴ 3.87 × 1.25 = 4.8375

(vii) 0.06 × 0.38

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 14

∴ 6 × 38 = 228

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 4

∴ 0.06 × 0.38 =0.0228

(viii) 0.623 × 0.75

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 15

∴ 623 × 75 = 46725

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 5

∴ 0.623 × 0.75 = 0.46725

(ix) 0.014 × 0.46

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 16

∴ 14 × 46 = 644

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 5

∴ 0.014 × 0.46 = 0.00644

(x) 54.5 × 1.76

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 17

∴ 545 × 176 = 95920

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 3

∴ 54.5 × 1.76 = 95.92

(xi) 0.045 × 2.4

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 18

∴ 45 × 24 = 1080

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 4

∴ 0.045 × 2.4 = 0.108

(xii) 1.245 × 6.4

Solution:-

Multiply the two decimal without the decimal point just like whole numbers.

RS Aggarwal Solutions for Class 7 Mathematics chapter 3 Decimals Image 19

∴ 1245 × 64 = 79680

Mark the decimal point in the product in such a way that the number of decimal places in the product is equal to the sum of the decimal place in the given decimals.

Sum of decimal places in the given decimals = 4

∴ 1.245 × 6.4 = 7.9680

6. Find the product:

(i) 13 × 1.3 × 0.13

Solution:-

First we find the product of 13 × 13 × 13

Now,

=13 × 13 = 169

= 169 × 13= 2197

Sum of decimal places in the given decimals = 3

So, the product must contain 3 places of decimals.

∴ 13 × 1.3 × 0.13 = 2.197

(ii) 2.4 × 1.5 × 2.5

Solution:-

First we find the product of 24 × 15 × 25

Now,

=24 × 15 = 360

= 360 × 25= 9000

Sum of decimal places in the given decimals = 3

So, the product must contain 3 places of decimals.

∴ 2.4 × 1.5 × 2.5 = 9

(iii) 0.8 × 3.5 × 0.05

Solution:-

First we find the product of 8 × 35 × 5

Now,

=8 × 35 = 280

= 280 × 5= 1400

Sum of decimal places in the given decimals = 4

So, the product must contain 3 places of decimals.

∴ 0.8 × 3.5 × 0.05=0.14

(iv) 0.2 × 0.02 × 0.002

Solution:-

First we find the product of 2 × 2 × 2

Now,

=2 × 2 = 4

= 4 × 2= 8

Sum of decimal places in the given decimals = 6

So, the product must contain 3 places of decimals.

∴ 0.2 × 0.02 × 0.002= 0.000008

(v) 11.1 × 1.1 × 0.11

Solution:-

First we find the product of 111 × 11 × 11

Now,

=111 × 11 = 1221

= 1221 × 11= 13431

Sum of decimal places in the given decimals = 4

So, the product must contain 3 places of decimals.

∴ 11.1 × 1.1 × 0.11= 1.3431

(vi) 2.1 × 0.21 × 0.021

Solution:-

First we find the product of 21 × 21 × 21

Now,

=21 × 21 = 441

= 441 × 21= 9261

Sum of decimal places in the given decimals = 6

So, the product must contain 3 places of decimals.

∴ 2.1 × 0.21 × 0.021= 0.009261

7. Evaluate:

(i) (1.2)2

Solution:-

The above question can be written as, (1.2) × (1.2)

First we find the product of,

= (12) × (12) = 144

Sum of decimal places in the given decimals = 2

So, the product must contain 2 places of decimals

∴ (1.2)2 = 1.44

(ii) (0.7)2

Solution:-

The above question can be written as, (0.7) × (0.7)

First we find the product of,

= (7) × (7) = 49

Sum of decimal places in the given decimals = 2

So, the product must contain 2 places of decimals

∴ (0.7)2 = 0.49

(iii) (0.04)2

Solution:-

The above question can be written as, (0.04) × (0.04)

First we find the product of,

= (4) × (4) = 16

Sum of decimal places in the given decimals = 4

So, the product must contain 2 places of decimals

∴ (0.04)2 = 0.0016

(iv) (0.11)2

Solution:-

The above question can be written as, (0.11) × (0.11)

First we find the product of,

= (11) × (11) = 121

Sum of decimal places in the given decimals = 4

So, the product must contain 2 places of decimals

∴ (0.11)2 = 0.0121

8. Evaluate:

(i) (0.3)3

Solution:-

The above question can be written as, (0.3) × (0.3) × (0.3)

First we find the product of,

= (3) × (3) = 9

= (9) × (3) = 27

Sum of decimal places in the given decimals = 3

So, the product must contain 2 places of decimals

∴ (0.3)3 = 0.027

(ii) (0.05)3

Solution:-

The above question can be written as, (0.05) × (0.05) × (0.05)

First we find the product of,

= (5) × (5) = 25

= (25) × (5) = 125

Sum of decimal places in the given decimals = 6

So, the product must contain 2 places of decimals

∴ (0.05)3 = 0.000125

(iii) (1.5)3

Solution:-

The above question can be written as, (1.5) × (1.5) × (1.5)

First we find the product of,

= (15) × (15) = 225

= (225) × (15) = 3375

Sum of decimal places in the given decimals = 3

So, the product must contain 2 places of decimals

∴ (1.5)3 = 3.375

9. A bus can cover 62.5 km in one hour. How much distance can it cover in 18 hours?

Solution:-

A bus can cover distance in one hour is = 62.5 km

Total distance it covers in 18 hours = (62.5 × 18)

First we find the product of,

= (625 × 18) = 11250

The product contain 1 places of decimals

∴ Total distance covered by bus in 18 hours is 1125


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


Exercise 3E

Mark against the correct answer in each of the following:

1. 0.06=?

a)3/5 b)(3/50) c)(3/500) d)none of these

Solution:-

a. (3/50)

Because,

=0.06

= (6/100) … [÷ 2]

= (3/50)

2. 1.04 =?

a) [1(1/50)] b)[1(2/5)] c)[1(1/25)] d)none of these

Solution:-

C) [1(1/25)]

Because,

=1.04

= (104/100)

= (26/25) … [÷ 4]

= [1(1/25)]

3. [2(2/25)] = ?

a)2.8 b)2.08 c)2.008 d)none of these

Solution:-

b) 2.08

Because,

= [2(2/25)]

= (52/25) … [÷ 25]

= 2.08

4. 6cm =?

a)0.006km b)0.0006km c)0.00006km d)none of these

Solution:-

C) 0.00006km

WKT,

=1m = 100 cm

Then,

= 6cm

= (6/100) m

= 0.06m

WKT,

= 1km = 1000m

Then,

= 0.06m

= (0.06/1000)

= (0.00006) km

5. 70 g =?

a)0.7kg b)0.07kg c)0.007kg d)none of these

Solution:-

B) 0.07kg

WKT,

= 1 kg = 1000g

= 70 g

= (70/1000)

= 0.07 kg

6. 5kg 6g =?

a)5.0006kg b) 5.06kg c)5.006kg d) 5.6kg

Solution:-

C) 5.006kg

WKT,

= 1 kg = 1000g

= 5kg = 5000g

= 5000 + 6

= 5006g

= (5006/1000)

= 5.006kg

7. 2 km 5 m =?

a)2.5km b)2.05km c)2.005km d)2.0005km

Solution:-

c) 2.005km

WKT,

= 1 km = 1000m

= 2 km = 2000m

= 2000 + 5

= 2005m

= (2005/1000)

8. (1.007-0.7) =?

a)1 b)0.37 c)0.307 d)none of these

Solution:-

C) 0.307

First convert the given decimals into like decimals

= (1.007 – 0.700)

= 0.307

9. What should be subtracted from 0.1 to get 0.03?

a)0.7 b) 0.07 c)0.007 d)none of these

Solution:-

C) 0.07

Let us assume the missing number be x,

Then,

= 0.1 – x = 0.03

By sending – x from left hand side to the right hand side it become x and 0.03 from right hand side to the left hand side it become -0.03. We get,

= 0.1 – 0.03 = x

= 0.07

10. What should be added to 3.07 to get 3.5?

0.57 b)0.34 c)0.43 d)0.02

Solution:-

C) 0.43

Let us assume the missing number be x,

Then,

= 3.07 + x = 3.5

By sending 3.07 from left hand side to the right hand side it become -3.07. We get,

= 3.5 – 3.07 = x

= 0.43


 

RS Aggarwal Solutions for Class 7 Maths Chapter 3 – Decimals

Chapter 3 – Decimals contains 5 exercises, and also RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercise. Now, let us have a look at some of the important concepts discussed in this chapter.

  • Method of Converting a Decimal into a Fraction
  • Converting a Fraction into Decimal
  • Addition and Subtraction of Decimals
  • Addition of Decimals
  • Subtraction of Decimals
  • Multiplication of Decimals
  • Multiplication of a Decimal by 10, 100, 1000, etc.
  • Multiplication of a Decimal by a Whole Number
  • Multiplication of a Decimal by a Decimal
  • Division of Decimals
  • Dividing a Decimal By a Whole Number

Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 3 – Decimals

RS Aggarwal Solutions for Class 7 Maths Chapter 3 – Decimals, is any number from the base-ten number system. In this Chapter, we will be focusing on the numbers that contain one or more digits to the right of the point. The decimal point helps in separating the ones place from the tenths place in the number. The solutions are solved in such a way that students will understand clearly, and they will never get confused.