The exercise-wise solutions are depicted in an interactive manner to help students with their exam preparation. RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals Exercise 3.2 are provided here. Students can download this PDF from the provided links. The subject experts at BYJU’S provide students with PDFs of solutions to solve the problems according to the CBSE syllabus. The solutions for exercise-wise problems are prepared after vast research conducted on each topic. The step-wise solutions prepared by subject experts help students to solve problems effortlessly.
By practising the RD Sharma Solutions for Class 7, students will be able to grasp the concepts correctly. Conceptual knowledge is crucial as some of the topics are continued in higher classes.
RD Sharma Solutions for Class 7 Maths Chapter 3 – Decimals Exercise 3.2
Access answers to Maths RD Sharma Solutions for Class 7 Chapter 3 – Decimals Exercise 3.2
1. Find the product:
(i) 4.74 × 10
(ii) 0.45 × 10
(iii) 0.0215 × 10
(iv) 0.0054 × 10
Solution:
(i) Given 4.74 × 10
Here we have to do normal multiplication with shifting the decimal point by one place to the right
Therefore 4.74 × 10 = 47.4
(ii) Given 0.45 × 10
Here we have to do normal multiplication with shifting the decimal point by one place to the right
Therefore 0.45 × 10 = 4.5
(iii) Given 0.0215 × 10
Here we have to do normal multiplication with shifting the decimal point by one place to the right
Therefore 0.0215 × 10 = 0.215
(iv) Given 0.0054 × 10
Here we have to do normal multiplication with shifting the decimal point by one place to the right
Therefore 0.0054 × 10 = 0.054
2. Find the product:
(i) 35.853 × 100
(ii) 42.5 × 100
(iii) 12.075 × 100
(iv) 100 × 0.005
Solution:
(i) Given 35.853 × 100
Here we have to do normal multiplication with shifting the decimal point by two places to the right
Therefore 35.853 × 100 = 3585.3
(ii) Given 42.5 × 100
Here we have to do normal multiplication with shifting the decimal point by two places to the right
Therefore 42.5 × 100 = 4250
(iii) Given 12.075 × 100
Here we have to do normal multiplication with shifting the decimal point by two places to the right
Therefore 12.075 × 100 = 1207.50
(iv) Given 100 × 0.005
Here we have to do normal multiplication with shifting the decimal point by two places to the right
Therefore 100 × 0.005 = 0.5
3. Find the product:
(i) 2.506 × 1000
(ii) 20.708 × 1000
(iii) 0.0529 × 1000
(iv) 1000 × 0.1
Solution:
(i) Given 2.506 × 1000
Here we have to do normal multiplication with shifting the decimal point by three places to the right
Therefore 2.506 × 1000 = 2506
(ii) Given 20.708 × 1000
Here we have to do normal multiplication with shifting the decimal point by three places to the right
Therefore 20.708 × 1000 = 20708
(iii) Given 0.0529 × 1000
Here we have to do normal multiplication with shifting the decimal point by three places to the right
Therefore 0.0529 × 1000 = 52.9
(iv) Given 1000 × 0.1
Here we have to do normal multiplication with shifting the decimal point by three places to the right
Therefore 1000 × 0.1 = 100
4. Find the product:
(i) 3.14 × 17
(ii) 0.745 × 12
(iii) 28.73 × 47
(iv) 0.0415 × 59
Solution:
(i) Given 3.14 × 17
First multiply as usual without looking at the decimal point
3.14 × 17 = 578
Now mark the decimal point in the product to have one place of decimal as there in the given decimal
3.14 × 17 = 57.8
(ii) Given 0.745 × 12
First multiply as usual without looking at the decimal point
0.745 × 12 = 894
Now mark the decimal point in the product to have three places of decimal as there in the given decimal
0.745 × 12 = 8.94
(iii) Given 28.73 × 47
First multiply as usual without looking at the decimal point
28.73 × 47 = 135031
Now mark the decimal point in the product to have two places of decimal as there in the given decimal
28.73 × 47 = 1350.31
(iv) Given 0.0415 × 59
First multiply as usual without looking at the decimal point
0.0415 × 59 = 24485
Now mark the decimal point in the product to have four places of decimal as there in the given decimal
0.0415 × 59 = 2.4485
5. Find:
(i) 1.07 × 0.02
(ii) 211.9 × 1.13
(iii) 10.05 × 1.05
(iv) 13.01 × 5.01
Solution:
(i) Given 1.07 × 0.02
First multiply as usual without looking at the decimal point
1.07 × 0.02 = 00214
Sum of the decimals is 4
Now mark the decimal point in the product to have four places of decimal as there in the given decimal
1.07 × 0.02 = 0.0214
(ii) Given 211.9 × 1.13
First multiply as usual without looking at the decimal point
211.9 × 1.13 = 239447
Sum of the decimals is 3
Now mark the decimal point in the product to have three places of decimal as there in the given decimal
211.9 × 1.13 = 239.447
(iii) Given 10.05 × 1.05
First multiply as usual without looking at the decimal point
10.05 × 1.05 = 105525
Sum of the decimals is 4
Now mark the decimal point in the product to have four places of decimal as there in the given decimal
10.05 × 1.05 = 10.5525
(iv) Given 13.01 × 5.01
First multiply as usual without looking at the decimal point
13.01 × 5.01 = 651801
Sum of the decimals is 4
Now mark the decimal point in the product to have four places of decimal as there in the given decimal
13.01 × 5.01 = 65.1801
6. Find the area of a rectangle whose length is 5.5m and breadth is 3.4m.
Solution:
Given length of rectangle = 5.5m
Breadth of rectangle = 3.4 m
Area of rectangle = length × breadth
= 5.5 × 3.4
= 18.7 m2
7. If the cost of a book is Rs 25.57, find the cost 0f 24 such books.
Solution:
Given cost of a book is Rs 25.57
Cost of 24 books = 25.57 × 24
= Rs 618.00
8. A car covers a distance of 14.75km in one litre of petrol. How much distance it will cover in 15.5 litres of petrol?
Solution:
Given that distance covered by car in 1 litre of petrol = 14.75 km
Distance covered by car in 15.5 litres of petrol = 14.75 × 15.5
= 228.625 km
9. One kg of rice costs Rs 42.65. What will be the cost of 18.25 kg of rice?
Solution:
Given cost of 1kg of rice = 42.65
Cost of 18.25kg of rice = 42.65 × 18.25
= Rs 778.3625
10. One metre of cloth costs Rs 152.50. What is the cost of 10.75 metres of cloth?
Solution:
Given that cost of 1m cloth = Rs 152.50
Cost of 10.75 m of cloth = 152.50 × 10.75
= Rs 1639.375
RD Sharma Solutions for Class 7 Maths Chapter – 3 Decimals Exercise 3.2
RD Sharma Solutions for Class 7 Maths Chapter 3 Decimals Exercise 3.2 have problems which are based on the multiplication of decimals. Some of the topics focused on prior to this exercise are listed below.
- Multiplication of a decimal by 10, 100 and 1000
- Multiplication of a decimal by a whole number
- Multiplication of decimal by another decimal
Students are suggested to try solving the questions from RD Sharma book of Class 7 and then refer to these solutions to know the best way of approaching the different questions.
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