Class 7 Exam  >  Class 7 Notes  >  Mathematics (Maths) Class 7  >  RD Sharma Solutions: Decimals (Exercise 3.3)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download

Q.1. Divide:

(i) 142.45 by 10

(ii) 54.25 by 10

(iii) 3.45 by 10

(iv) 0.57 by 10

(v) 0.043 by 10

(vi) 0.004 by 10

Ans: 

(i)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 14.245

(ii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 5.425

(iii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.345

(iv)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.057

(v)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.0043

(vi)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.0004


Q.2. Divide:

(i) 459.5 by 100

(ii) 74.3 by 100

(iii) 5.8 by 100

(iv) 0.7 by 100

(v) 0.48 by 100

(vi) 0.03 by 100

Ans: 

(i)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 4.595 

(ii)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.743

(iii)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.058

(iv)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.007

(v)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.0048

(vi)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.0003


Q.3. Divide:

(i) 235.41 by 1000

(ii) 29.5 by 1000

(iii) 3.8 by 1000

(iv) 0.7 by 1000

Ans: 

(i)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.23451

(ii)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.0295 

(iii)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.0038 

(iv)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.0007


Q.4. Divide:

(i) 0.45 by 9

(ii) 217.44 by 18

(iii) 319.2 by 2.28

(iv) 40.32 by 9.6

(v) 0.765 by 0.9

(vi) 0.768 by 1.6

Ans: 

(i)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.05

(ii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 12.08

(iii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 140

(iv)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 4.2

(v)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.85 

(vi)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.48


Q.5. Divide:

(i) 16.64 by 20

(ii) 0.192 by 12

(iii) 163.44 by 24

(iv) 403.2 by 96

(v) 16.344 by 12

(vi) 31.92 by 228

Ans: 

(i)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.832

(ii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.016

(iii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 6.81

(iv)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 4.2

(v)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 1.362

(vi)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.14


Q.6. Divide:

(i) 15.68 by 20

(ii) 164.6 by 200

(iii) 403.80 by 30

Ans: 

(i)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.784 

(ii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.823 

(iii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 13.46


Q.7. Divide:

(i) 76 by 0.019

(ii) 88 by 0.08

(iii) 148 by 0.074

(iv) 7 by 0.014

Ans: 

(i)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 4000

(ii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 1100 

(iii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 2000

(iv)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 500


Q.8. Divide:

(i) 20 by 50

(ii) 8 by 100

(iii) 72 by 576

(iv) 144 by 15

Ans: 

(i)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.4

(ii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.08 

(iii)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.125

(iv)

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 9.6


Q.9. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it travel in 1 litre of petrol?

Ans: Distance travelled by the vehicle in 2.4 litres of petrol = 43.2 km

∴ Distance travelled in 1 litre of petrol will be:

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 18

Thus, the vehicle will travel 18 km in 1 litre of petrol.


Q.10. The total weight of some bags of wheat is 1743 kg. If each bag weights 49.8 kg, how many bags are there?

Ans: Let the number of bags of wheat be x.

Given:

Weight of x bags = 1,743 kg

Weight of one bag = 49.8 kg

Therefore, 49.8 × x = 1,743

x = Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 35

Thus, there are 35 bags of wheat.


Q.11. Shikha cuts 50 m of cloth into pieces of 1.25 m each. How many pieces does she get?

Ans: Length of cloth = 50 m

Length of one piece of cloth = 1.25 m

Let the number of pieces be x.

∴ Number of pieces of cloth × Length of each piece of cloth = Total length of cloth

x ×1.25 = 50

x = Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 40

Thus, she gets 40 pieces of 1.25 m cloth from a total of 50 m.


Q.12. Each side of a rectangular polygon is 2.5 cm in length. The perimeter of the polygon is 12.5 cm. How many sides does the polygon have? 

Ans: Length of each side of the polygon  = 2.5 cm

Perimeter of the polygon = 12.5 cm

Let the number of sides of the polygon be z.

∴ 2.5 × z = 12.5 ( Perimeter = Length of one side multiplied by the number of sides)

or, z =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 5 

Thus, the polygon has 5 sides.


Q.13. The product of two decimals is 42.987. If one of them is 12.46, find the other.

Ans: One decimal is 12.46.

Let the other decimal be x.

According to the question:

12.46 × x = 42.987

 x =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 3.45

Thus, the other decimal is 3.45.


Q.14. The weight of 34 bags of sugar is 3483.3 kg. If all bags weigh equally, find the weight of each bag.

Ans: Number of bags = 34

Weight of 34 bags = 3483.3 kg

∴ Weight of one bag =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 102.45

Thus, weight of each bag = 102.45 kg 


Q.15. How many buckets of equal capacity can be filled from 586.5 litres of water, if each bucket has capacity of 8.5 litres? 

Ans: Capacity of one bucket = 8.5 litres

Total water in all buckets = 586.5 litres

Let the number of buckets with equal capacity be x.

According to the question:

8.5 × x = 586.5

x =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 69

Thus, there are 69 buckets.


Objective Type Questions


Q.1. Mark the correct alternative in each of the following:

When 0.48 is written in the simplest form of its terms, the sum of its numerator and denominator is

(a) 148                                         

(b) 74                                         

(c) 37                                         

(d) 147  

Ans: Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7(HCF of 48 and 100=4)

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Here,

Numerator = 12

Denominator = 25

∴ Sum of the numerator and denominator = 12 + 25 = 37

Hence, the correct answer is option (c).


Q.2. Mark the correct alternative in each of the following:

The improper fraction Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7 in decimal form is

(a) 2.4                                        

(b) 2.04                                        

(c) 2.004                                         

(d) None of these

Ans: The given fraction isDecimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7. Now,

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7 

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 2 + 0.04

= 2.04 

Hence, the correct answer is option (b).


Q.3. Mark the correct alternative in each of the following:

4 + 4.4 + 44.4 + 4.04 + 444 =

(a) 500.88                                         

(b) 577.2                                         

(c) 495.22                                         

(d) 472.88

Ans: Converting the decimals to like decimals, we have 4.00, 4.40, 44.40, 4.04 and 444.00.

Now,

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

∴ 4 + 4.4 + 44.4 + 4.04 + 444 = 500.84

Disclaimer: None of the options given in the question matches with the answer.


Q.4. Mark the correct alternative in each of the following:

1.04 as an improper fraction is

(a)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7                                       

(b)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7                                 

(c)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7                                

(d)Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Ans: The given decimal is 1.04.

1.04

= 1 + 0.04

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Hence, the correct answer is option (c).


Q.5. Mark the correct alternative in each of the following:

If 24.125 = 24 + Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7, then A + B + C is

(a) 3                                         

(b) 6                                         

(c) 13                                         

(d) 8 

Ans: 

24.125 = 24 + 0.125 

= 24+0.1 + 0.02 + 0.005

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Comparing this with the given expression, we get

A = 1, B = 2 and C = 5

∴ A + B + C = 1 + 2 + 5 = 8

Hence, the correct answer is option (d).


Q.6. Mark the correct alternative in each of the following:

0.002 × 0.5 =

(a) 0.0001                                         

(b) 0.001                                         

(c) 0.01                                        

(d) 1 

Ans: 0.002 × 0.5

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.001 

Hence, the correct answer is option (b).


Q.7. Mark the correct alternative in each of the following:

3 × 0.3 × 0.03 × 0.003 × 30 =  

(a) 0.0000243                     

(b) 0.000243                     

(c) 0.00243                     

(d) 0.243 

Ans: 3× 0.3 × 0.03 × 0.003 × 30

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 0.00243                  (Decimal point is shifted to left by 5 places)

Hence, the correct answer is option (c).


Q.8. Mark the correct alternative in each of the following:

0.012 × 0.15 =  

(a) 0.8                           

(b) 0.08                          

(c) 0.008                         

(d) 0.0018

Ans: We have,

12 × 15 = 180

It can be seen that the sum of the decimals in the given decimals is 3 + 2 = 5.

So, the product must contain 5 places of decimals.

∴ 0.012 × 0.15 = 0.00180 = 0.0018

Hence, the correct answer is option (d).


Q.9. Mark the correct alternative in each of the following:

75.57 ÷ 0.01 =

(a) 7557                                         

(b) 0.7557                                        

(c) 755.7                                         

(d) 7.557

Ans: 75.57 ÷ 0.01

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

(Multiply numerator and denominator by 100 to convert the divisor into whole number)

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=7557

Hence, the correct answer is option (a).


Q.10. Mark the correct alternative in each of the following:

What should be subtracted from 0.1 to get 0.06?

(a) 0.4                                         

(b) 0.04                                        

(c) 0.004                                         

(d) None of these

Ans: The decimal which should be subtracted from 0.1 to get 0.06 can be obtained by subtracting 0.06 from 0.1.

Converting the given decimals into like decimals, we have 0.10 and 0.06.

Now,

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

∴ Required decimal = 0.1 − 0.06 = 0.04

Hence, the correct answer is option (b).


Q.11. Mark the correct alternative in each of the following:

What should be added to 5.09 to get 5.5?

(a) 0.41                                         

b) 0.59                                         

(c) 0.49                                         

(d) 0.95

Ans: The decimal number which should be added to 5.09 to get 5.5 is obtained by subtracting 5.09 from 5.5.

Converting the given decimals to like decimals, we have 5.09 and 5.50.

Now,

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

∴ Required decimal = 5.50 − 5.09 = 0.41

Thus, 0.41 must be added to 5.09 to get 5.5.

Hence, the correct answer is option (a).


Q.12. Mark the correct alternative in each of the following:

0.3 × 0.3 × 0.3 =

(a) 2.7                                         

(b) 0.27                                         

(c) 0.027                                         

(d) None of these

Ans: We have,

3 × 3 × 3 = 27

The sum of the decimal places in the given decimals is 1 + 1 + 1 = 3.

So, the product must contain 3 places of decimals.

∴ 0.3 × 0.3 × 0.3 = 0.027

Hence, the correct answer is (c).


Q.13. Mark the correct alternative in each of the following:

0.25 × 0.8 =

(a) 0.02                                        

(b) 0.2                                        

(c) 0.002                                         

(d) 2

Ans: In order to find the product, we first multiply 8 by 25.

We have, 25 × 8 = 200

Now, 0.25 has 2 decimal places and 0.8 has 1 decimal place.

The sum of the decimal places is 2 + 1 = 3.

So, the product must contain 3 places of decimals.

∴ 0.25 × 0.8 = 0.200 = 0.2

Hence, the correct answer is option (b).


Q.14. Mark the correct alternative in each of the following:

5 kg 5 g written in decimal notation is

(a) 5.5 kg                             

(b) 5.05 kg                             

(c) 5.005 kg                             

(d) 5.0005 kg

Disclaimer: The units are missing from the options given in the book.

Ans: We know that,

1 g =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Now,

5 kg 5 g = 5 kg + 5 g

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 5 kg + 0.005 kg

= 5.005 kg

∴ 5 kg 5 g = 5.005 kg

Hence, the correct answer is option (c).


Q.15. Mark the correct alternative in each of the following:

0.012 ÷ 1.5 = ?

(a) 0.8                                         

(b) 0.08                                         

(c) 0.008                                         

(d) None of these 

Ans: 0.012 ÷ 1.5

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

(Multiply the numerator and denominator by 10 to convert the divisor into a whole number)

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

∴ 0.012 ÷ 1.5 = 0.008

Hence, the correct answer is option (c).


Q.16. Mark the correct alternative in each of the following:

0.02 × 0.05 =

(a) 0.1                                        

(b) 0.01                                        

(c) 0.001                                         

(d) 0.0001

Ans: In order to find the product, we first multiply 2 by 5.

We have, 2 × 5 = 10

Now, 0.02 has 2 decimal places and 0.05 has 2 decimal places.

The sum of the decimal places is 2 + 2 = 4.

So, the product must contain 4 places of decimals.

∴ 0.02 × 0.05 = 0.0010 = 0.001

Hence, the correct answer is option (c).


Q.17. Mark the correct alternative in each of the following:

5 km 5 m = ?

(a) 5.5 km                        

(b) 5.05 km                         

(c) 5.005 km                         

(d) 5.0005 km

Ans: We know that,

1 m =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Now,

5 km 5 m = 5 km + 5 m

=Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

= 5 km + 0.005 km

= 5.005 km

∴ 5 km 5 m = 5.005 km

Hence, the correct answer is option (c).


Q.18. Mark the correct alternative in each of the following:

The value of 2.2 × 0.2 × 0.001 is

(a) 4.2                                         

(b) 0.00044                                        

(c) 4.4                                         

(d) None of these

Ans: In order to find the product, we first multiply 22 by 2.

We have, 22 × 2 = 44

Now, 2.2 has 1 decimal place, 0.2 has 1 decimal place and 0.001 has 3 decimal places.

The sum of the decimal places is 1 + 1 + 3 = 5.

So, the product must contain 5 places of decimals.

∴ 2.2 × 0.2 × 0.001 = 0.00044

Thus, the value of 2.2 × 0.2 × 0.001 is 0.00044.

Hence, the correct answer is option (b).


Q.19. Mark the correct alternative in each of the following:

If 14 × 4 = 56, then 0.014 × 4 =

(a) 0.56                                         

(b) 5.6                                         

(c) 0.056                                         

(d) None of these

Ans: It is given that,

14 × 4 = 56

Now, 0.014 has 3 decimal places.

So, the required product must contain 3 places of decimals.

∴ 0.014 × 4 = 0.056

Hence, the correct answer is option (c).


Q.20. Mark the correct alternative in each of the following:

8 ml is equal to

(a) 0.8 l                                         

(b) 0.08 l                                         

(c) 0.008 l                                         

(d) None of these

Ans: We know that,

1 mL =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

∴ 8 mL =Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7

Hence, the correct answer is option (c).

The document Decimals (Exercise 3.3) RD Sharma Solutions | Mathematics (Maths) Class 7 is a part of the Class 7 Course Mathematics (Maths) Class 7.
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FAQs on Decimals (Exercise 3.3) RD Sharma Solutions - Mathematics (Maths) Class 7

1. What is the importance of learning decimals in Class 7 mathematics?
Ans. Learning decimals in Class 7 mathematics is important as it helps students understand and work with numbers that are less than one. Decimals are used in various real-life scenarios such as money, measurements, and data analysis. It is also a foundation for more advanced mathematical concepts in higher grades.
2. How can I convert a decimal into a fraction?
Ans. To convert a decimal into a fraction, you can use the following steps: 1. Identify the place value of the last digit in the decimal. 2. Write the decimal without the decimal point. 3. Write the digits after the decimal point as the numerator. 4. Write the place value as the denominator. 5. Simplify the fraction, if possible. For example, to convert 0.75 into a fraction: 1. The place value of the last digit 5 is hundredths. 2. Write 75 without the decimal point. 3. The numerator is 75. 4. The denominator is 100. 5. Simplifying the fraction gives 3/4.
3. How can I compare decimals?
Ans. To compare decimals, you can follow these steps: 1. If the whole part of the decimals is different, compare the whole parts. The decimal with the greater whole part is greater. 2. If the whole part is the same, compare the tenths place. The decimal with the greater digit in the tenths place is greater. 3. Continue comparing each subsequent place value until the digits are different or all the digits have been compared. For example, to compare 0.75 and 0.9: 1. The whole part is 0 for both decimals. 2. The digit in the tenths place for 0.75 is 7, and for 0.9, it is 9. Therefore, 0.9 is greater.
4. How do I add and subtract decimals?
Ans. To add and subtract decimals, follow these steps: 1. Write the decimals one below the other, aligning the decimal points. 2. If necessary, add zeros to the right of the decimal point to make the number of decimal places equal. 3. Add or subtract the decimals as usual, starting from the rightmost place value. 4. Carry over or borrow as necessary. 5. Write the decimal point in the answer directly below the decimal points in the given decimals. For example, to add 1.25 and 0.75: 1.25 + 0.75 ------- 2.00 To subtract 1.25 from 2.00: 2.00 - 1.25 ------- 0.75
5. How can I multiply and divide decimals?
Ans. To multiply and divide decimals, follow these steps: 1. Multiply or divide the decimals as you would with whole numbers, ignoring the decimal point. 2. Count the total number of decimal places in the given decimals. 3. Place the decimal point in the product or quotient by counting from the right, starting with the rightmost digit. For example, to multiply 2.5 by 0.6: 2.5 x 0.6 ------ 15 The given decimals have one decimal place each. Therefore, the product should have a total of one decimal place. So, the answer is 15.0. To divide 2.5 by 0.6: 2.5 ÷ 0.6 = 4.1666... The given decimals have one decimal place each. Therefore, the quotient should have a total of one decimal place. So, the answer is approximately 4.2.
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