Posted by RBSE Solutions for Class 6 Maths Chapter 8: Decimals offers students a thorough understanding of decimal concepts, ensuring clarity in solving problems. This chapter, part of the RBSE curriculum, covers crucial decimal-related topics and helps Class 6 students master fundamental math skills. With step-by-step solutions, students can easily practice and enhance their knowledge.
Key Concepts Covered in Chapter 8: Decimals
- Understanding Decimals
- Decimal Fractions
- Adding and Subtracting Decimals
- Comparing Decimals
- Word Problems on Decimals
This chapter lays the foundation for working with decimals, which are essential for real-life applications and advanced math topics.
Detailed RBSE Class 6 Solutions for Chapter 8: Decimals
Below are the solutions for Class 6 Maths Chapter 8. These solutions are created to help students understand each concept in detail and solve problems effectively. Each question is solved with detailed steps, allowing students to develop problem-solving skills and improve accuracy.
NCERT Solutions for Class 6 Chapter 8: Decimals Exercise 8.1
Table of Contents
1. Write the following numbers in the given table.
| Hundreds | Tens | Ones | Tenths |
| (100) | (10) | (1) | (1 / 10) |
Solutions:
| Rows | Hundreds | Tens | Ones | Tenths |
| a | 0 | 3 | 1 | 2 |
| b | 1 | 1 | 0 | 4 |
2. Write the following decimals in the place value table.
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9
Solutions:
| Hundreds | Tens | Ones | Tenths | |
| 19.4 | 0 | 1 | 9 | 4 |
| 0.3 | 0 | 0 | 0 | 3 |
| 10.6 | 0 | 1 | 0 | 6 |
| 205.9 | 2 | 0 | 5 | 9 |
3. Write each of the following as decimals:
(a) Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Solutions:
(a) The decimal form of Seven-tenths is 7 / 10 = 0.7
(b) The decimal form of two tens and nine tenths is 20 + 9 / 10 = 20.9
(c) The decimal form of fourteen point six is 14.6
(d) The decimal form of one hundred and two ones is 100 + 2 = 102.0
(e) The decimal form of six hundred point eight is 600.8
4. Write each of the following as decimals:
(a) 5 / 10
(b) 3 + 7 / 10
(c) 200 + 60 + 5 + 1 / 10
(d) 70 + 8 / 10
(e) 88 / 10
(f)
(g) 3 / 2
(h) 2 / 5
(i) 12 / 5
(j)
(k)
Solutions:
(a) 5 / 10 = 0.5
(b) 3 + 7 / 10 = 3 + 0.7
= 3.7
(c) 200 + 60 + 5 + 1 / 10 = 265 + 0.1
= 265.1
(d) 70 + 8 / 10 = 70 + 0.8
= 70.8
(e) 88 / 10 = 80 / 10 + 8 / 10
= 8 + 0.8
= 8.8
(f)
(g) 3 / 2 = (2 + 1) / 2
= 2 / 2 + 1 / 2
= 1 + 0.5
= 1.5
(h) 2 / 5 = 0.4
(i) 12 / 5 = (10 + 2) / 5
= 10 / 5 + 2 / 5
= 2 + 0.4
= 2.4
(j)
(k)
5. Write the following decimals as fractions. Reduce the fraction to lowest form.
(a) 0.6
(b) 2.5
(c) 1.0
(d) 3.8
(e) 13.7
(f) 21.2
(g) 6.4
Solutions:
(a) 0.6 = 6 / 10
= 3 / 5
(b) 2.5 = 25 / 10
= 5 / 2
(c) 1.0 = 1
= 1
(d) 3.8 = 38 / 10
= 19 / 5
(e) 13. 7 = 137 / 10
(f) 21.2 = 212 / 10
= 106 / 5
(g) 6.4 = 64 / 10
= 32 / 5
6. Express the following as cm using decimals.
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2 mm
(e) 162 mm
(f) 83 mm
Solutions:
We know that
1 cm = 10 mm
1 mm = 1 / 10 cm
(a) 2 mm = 2 / 10 cm
= 0.2 cm
(b) 30 mm = 30 / 10 cm
= 3.0 cm
(c) 116 mm = 116 / 10 cm
= 11.6 cm
(d) 4 cm 2 mm = [(4 + 2 / 10)] cm
= 4.2 cm
(e) 162 mm = 162 / 10 cm
= 16.2 cm
(f) 83 mm = 83 / 10 cm
= 8.3 cm
7. Between which two whole numbers on the number line are the given numbers lie?
Which of these whole numbers is nearer the number?
(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Solutions:
(a) 0.8 lies between 0 and 1
0.8 is nearer to 1
(b) 5.1 lies between 5 and 6
5.1 is nearer to 5
(c) 2.6 lies between 2 and 3
2.6 is nearer to 3
(d) 6.4 lies between 6 and 7
6.4 is nearer to 6
(e) 9.1 lies between 9 and 10
9.1 is nearer to 9
(f) 4.9 lies between 4 and 5
4.9 is nearer to 5
8. Show the following numbers on the number line.
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Solutions:
(a) 0.2 lies between the points 0 and 1 on the number line. The space between 0 and 1 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 0.2 is the second point between 0 and 1
(b) 1.9 lies between the points 1 and 2 on the number line. The space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.9 is the ninth point between 1 and 2
(c) 1.1 lies between the points 1 and 2 on the number line such that the space between 1 and 2 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 1.1 is the first point between 1 and 2
(d) 2.5 lies between the points 2 and 3 on the number line such that the space between 2 and 3 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 2.5 is the fifth point between 2 and 3
9. Write the decimal number represented by the points A, B, C, and D on the given number line.
Solutions:
(a) Point A represents 0.8 cm on the given number line.
(b) Point B represents 1.3 cm on the given number line
(c) Point C represents 2.2 cm on the given number line
(d) Point D represents 2.9 cm on the given number line
10. (a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Solutions:
(a) The length of Ramesh notebook is 9 cm 5 mm
The length in cm is [(9 + 5 / 10)] cm
= 9.5 cm
(b) The length of a gram plant is 65 mm
Hence, the length in cm is 65 / 10
= 6.5 cm
NCERT Solutions for Class 6 Maths Chapter 8: Decimals Exercise 8.1 (old 8.3)
1. Which is greater?
(a) 0.3 or 0.4
(b) 0.07 or 0.02
(c) 3 or 0.8
(d) 0.5 or 0.05
(e) 1.23 or 1.20
(f) 0.099 or 0.19
(g) 1.5 or 1.50
(h) 1.431 or 1.490
(i) 3.3 or 3.300
(j) 5.64 or 5.603
Solutions:
(a) 0.3 or 0.4
The whole parts for both numbers are the same. We know that the tenth part of 0.4 is greater than that of 0.3.
∴ 0.4 > 0.3.
(b) 0.07 or 0.02
Both the numbers have the same parts up to the tenth place, but the hundredth part of 0.07 is greater than that of 0.02.
∴ 0.07 > 0.02.
(c) 3 or 0.8
The whole part of 3 is greater than that of 0.8.
∴ 3 > 0.8.
(d) 0.5 or 0.05
The whole parts for both numbers are the same. But, the tenth part of 0.5 is greater than that of 0.05.
∴ 0.5 > 0.05.
(e) 1.23 or 1.20
Here, both the numbers have the same parts up to the tenth place. The hundredth part of 1.23 is greater than that of 1.20.
∴ 1.23 > 1.20.
(f) 0.099 or 0.19
The whole parts for both numbers are the same. But, the tenth part of 0.19 is greater than that of 0.099.
∴ 0.099 < 0.19.
(g) 1.5 or 1.50
We can find that both numbers have the same parts up to the tenth place. Here, 1.5 have no digit at the hundredth place. It represents that this digit is 0, which is equal to the digit at the hundredth place of 1.50.
∴ Both these numbers are equal.
(h) 1.431 or 1.490
Here, both the numbers have the same parts up to the tenth place, but the hundredth part of 1.490 is greater than that of 1.431.
∴ 1.431 < 1.490.
(i) 3.3 or 3.300
Here, both numbers have the same parts up to the tenth place. There are no digits at the hundredth and thousandth place of 3.3. It represents that these numbers are 0, which is equal to the digits at the hundredth and thousandth place of 3.300.
∴ Both these numbers are equal.
(j) 5.64 or 5.603
Here, both numbers have the same parts up to the tenth place, but the hundredth part of 5.64 is greater than that of 5.603.
∴ 5.64 > 5.603.
2. Make five more examples and find the greater number from them.
Solutions:
Five more examples are given below:
(a) 32.55 or 32.5
The whole parts for both numbers are the same. The tenth part is also equal, but the hundredth part of 32.55 is greater than that of 32.5.
Hence, 32.55 > 32.5.
(b) 1 or 0.99
The whole part of 1 is greater than that of 0.99.
∴ 1 > 0.99.
(c) 1.09 or 1.093
Here, both numbers have the same parts up to the hundredth part. But the thousandth part of 1.093 is greater than that of 1.09.
∴ 1.093 > 1.09.
(d) 2 or 1.99
The whole part of 2 is greater than that of 1.99.
∴ 2 > 1.99.
(e) 2.08 or 2.085
Here, both numbers have the same parts up to the hundredth. But the thousandth part of 2.085 is greater than that of 2.08.
∴ 2.085 > 2.08
NCERT Solutions for Class 6 Chapter 8: Decimals Exercise 8.2 (old 8.4)
1. Express as rupees using decimals:
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Solutions:
We know that there are 100 paise in 1 rupee
(a) 5 paise = 5 / 100 rupees
= Rupees 0.05
(b) 75 paise = 75 / 100 rupees
= Rupees 0.75
(c) 20 paise = 20 / 100 rupees
= Rupees 0.20
(d) 50 rupees 90 paise = [(50 + 90 / 100)] rupees
= Rupees 50.90
(e) 725 paise = 725 / 100 rupees
= Rupees 7.25
2. Express as metres using decimals:
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7 cm
(e) 419 cm
Solutions:
We know that there are 100 cm in 1 metre
(a) 15 cm = 15 / 100 m
= 0.15 m
(b) 6 cm = 6 / 100 m
= 0.06 m
(c) 2 m 45 cm = [(2 + 45 / 100)] m
= 2.45 m
(d) 9 m 7 cm = [(9 + 7 / 100)] m
= 9.07 m
(e) 419 cm = 419 / 100 m
= 4.19 m
3. Express as cm using decimals:
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Solutions:
We know that there are 10 mm in 1 cm
(a) 5 mm = 5 / 10 cm
= 0.5 cm
(b) 60 mm = 60 / 10 cm
= 6.0 cm
(c) 164 mm = 164 / 10 cm
= 16.4 cm
(d) 9 cm 8 mm = [(9 + 8 / 10)] cm
= 9.8 cm
(e) 93 mm = 93 / 10 cm
= 9.3 cm
4. Express as km using decimals:
(a) 8 m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Solutions:
We know that there are 1000 metres in 1 km
(a) 8 m = 8 / 1000 km
= 0.008 km
(b) 88 m = 88 / 1000 km
= 0.088 km
(c) 8888 m = 8888 / 1000 km
= 8.888 km
(d) 70 km 5 m = [(70 + 5 / 1000)] km
= 70.005 km
5. Express as kg using decimals:
(a) 2 g
(b) 100 g
(c) 3750 g
(d) 5 kg 8 g
(e) 26 kg 50 g
Solutions:
We know that there are 1000 grams in 1 kg
(a) 2 g = 2 / 1000 kg
= 0.002 kg
(b) 100 g = 100 / 1000 kg
= 0.1 kg
(c) 3750 g = 3750 / 1000 kg
= 3.750 kg
(d) 5 kg 8 g = [(5 + 8 / 1000)] kg
= 5.008 kg
(e) 26 kg 50 g = [(26 + 50 / 1000)] kg
= 26.050 kg
NCERT Solutions for Class 6 Chapter 8: Decimals Exercise 8.3 (old 8.5)
1. Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
Solutions:
(a) Sum of 0.007 + 8.5 + 30.08
0.007
8.500
+ 30.080
__________
38.587
__________
(b) Sum of 15 + 0.632 + 13.8
15.000
0.632
+ 13.800
_________
29.432
__________
(c) Sum of 27.076 + 0.55 + 0.004
27.076
0.550
+ 0.004
_____________
27.630
______________
(d) Sum of 25.65 + 9.005 + 3.7
25.650
9.005
+ 3.700
__________
38.355
___________
(e) Sum of 0.75 + 10.425 + 2
0.750
10.425
+ 2.000
_________
13.175
__________
(f) Sum of 280.69 + 25.2 + 38
280.69
25.20
+ 38.00
__________
343.89
___________
2. Rashid spent ₹ 35.75 for Maths book and ₹ 32.60 for Science book. Find the total amount spent by Rashid.
Solutions:
Cost of Maths book = ₹ 35.75
Cost of Science book = ₹ 32.60
Total amount spent by Rashid is
35.75
+ 32.60
__________
68.35
___________
∴ Total amount of money spent by Rashid is ₹ 68.35
3. Radhika’s mother gave her ₹ 10.50 and her father gave her ₹ 15.80, find the total amount given to Radhika by the parents.
Solutions:
Amount given by Radhika’s mother = ₹ 10.50
Amount given by Radhika’s father = ₹ 15.80
Total amount given by her parents
10.50
+ 15.80
__________
26.30
___________
∴ Total amount of money given by Radhika’s parents is ₹ 26.30
4. Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Solutions:
Cloth of shirt = 3 m 20 cm
Cloth of trouser = 2 m 5 cm
Total length of cloth is
3.20
+ 2.05
________
5.25
_________
∴ Total length of cloth bought by Nasreen is 5.25 m
5. Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solutions:
Distance walked by Naresh in the morning = 2 km 35 m
= [(2 + 35 /1000)] km
= 2.035 km
Distance walked by him in the evening = 1 km 7 m
= [(1 + 7 / 1000)] km
= 1.007 km
Total distance walked by Naresh is
2.035
+ 1.007
_______
3.042
________
∴ Total distance walked by Naresh is 3.042 km
6. Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Solutions:
Distance travelled by bus = 15 km 268 m
= [(15 + 268 / 1000)] km
= 15.268 km
Distance travelled by car = 7 km 7 m
= [(7 + 7 / 1000)] km
= 7.007 km
Distance walked by Sunita = 500 m
= 500 / 1000
= 0.500 km
Total distance of school from her residence is
15.268
7.007
+ 0.500
________
22.775
________
∴ Total distance of the school from her residence is 22.775 km
7. Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Solutions:
Weight of rice = 5 kg 400 g
= [(5 + 400 / 1000)] kg
= 5.400 kg
Weight of sugar = 2 kg 20 g
= [(2 + 20 / 1000)] kg
= 2.020 kg
Weight of flour = 10 kg 850 g
= [(10 + 850 / 1000)] kg
= 10.850 kg
Total weight of his purchases is
5.400
2.020
+ 10.850
___________
18.270
____________
∴ Total weight of his purchases is 18.270 kg
NCERT Solutions for Class 6 Chapter 8: Decimals Exercise 8.4 (old 8.6)
1. Subtract:
(a) ₹ 18.25 from ₹ 20.75
(b) 202.54 m from 250 m
(c) ₹ 5.36 from ₹ 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Solutions:
(a) ₹ 20.75 – ₹ 18.75
20.75
– 18.25
__________
2.50
___________
₹ 2.50
(b) 250 m – 202.54 m
250.00
– 202.54
___________
47.46
____________
47.46 m
(c) ₹ 8.40 – ₹ 5.36
8.40
– 5.36
_________
3.04
_________
₹ 3.04
(d) 5.206 km – 2.051 km
5.206
– 2.051
__________
3.155
__________
3.155 km
(e) 2.107 kg – 0.314 kg
2.107
– 0.314
_________
1.793
__________
1.793 kg
2. Find the value of:
(a) 9.756 – 6.28
(b) 21.05 – 15.27
(c) 18.5 – 6.79
(d) 11.6 – 9.847
Solutions:
(a) 9.756
– 6.280
_________
3.476
_________
(b) 21.05
– 15.27
___________
5.78
____________
(c) 18.50
– 6.79
___________
11.71
___________
(d) 11.600
– 9.847
____________
1.753
____________
3. Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Solutions:
The money given to the shopkeeper = ₹ 50.00
The price of the book = ₹ 35.65
The money that Raju will get back from the shopkeeper will be the difference between these two.
∴ The money left with Raju is
50.00
– 35.65
___________
14.35
___________
Hence, the money left with Raju is ₹ 14.35
4. Rani had ₹ 18.50. She bought one ice cream for ₹ 11.75. How much money does she have now?
Solutions:
The money with Rani = ₹ 18.50
The price of an ice cream = ₹ 11.75
Now, the money left with Rani will be the difference between these two.
Hence, the money left with her is
18.50
– 11.75
__________
6.75
___________
∴ The money left with Rani is ₹ 6.75
5. Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Solutions:
The length of the cloth = 20 m 5 cm
= 20.05 m
The length of the cloth to make a curtain = 4 m 50 cm
= 4.50 m
The length of the cloth left with Tina will be the difference between these two.
Thus, the length of the cloth left with her is
20.05
– 4.50
________
15.55
________
∴ The length of the remaining cloth left with Tina is 15.55 m
6. Namita travels 20 km 50 m every day. Out of this, she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Solutions:
Total distance travelled by Namita = 20 km 50 m
= 20.050 km
Distance travelled by bus = 10 km 200 m
= 10.200 km
Distance travelled by auto = Total distance travelled – Distance travelled by bus
∴ Distance to be travelled by auto is
20.050
– 10.200
________
9.850
________
∴ Namita travelled 9.850 km by auto.
7. Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes, and the rest is potatoes. What is the weight of the potatoes?
Solutions:
Total weight of vegetables Aakash bought = 10.000 kg
Weight of onions = 3 kg 500 g
= 3.500 kg
Weight of tomatoes = 2 kg 75 g
= 2.075 kg
Weight of potatoes = Total weight of vegetables bought – (Weight of onions + Weight of tomatoes)
= 10.000 – (3.500 + 2.075)
3.500
+ 2.075
________
5.575
________
10.000
– 5.575
_________
4.425
_________
∴ 4.425 kg is the weight of the potatoes.
- Rbse Solutions for Class 6 Chapter 1: Knowing Our Numbers
- Rbse Solutions for Class 6 Chapter 2: Whole Numbers
- Rbse Solutions For Class 6 Maths Chapter 3 Playing with Numbers
- Rbse Solutions for Class 6 Chapter 4: Basic Geometrical Ideas
- Rbse Solutions for Class 6 Chapter 5: Understanding Elementary Shapes
- Rbse Solutions for Class 6 Chapter 6: Integers
- Rbse Solutions to Class 6 Chapter 7: Fractions
- RBSE Solutions for Class 6 Maths Chapter 8: Decimals | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 9: Data Handling | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 10: Mensuration | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 11: Algebra | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 12: Ratio and Proportion | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 13: Symmetry | Updated for 2024-25
- RBSE Solutions for Class 6 Maths Chapter 14: Practical Geometry | Updated for 2024-25
Benefits of RBSE Class 6 Chapter 8 Solutions
- Clear Understanding: Step-by-step solutions for each problem help students understand complex topics easily.
- Practice for Exams: Focused practice on decimals is useful for exam preparation.
- Real-Life Application: Applying decimals in real-life scenarios like money, measurements, and calculations.
FAQs on RBSE Solutions for Class 6 Maths Chapter 8: Decimals
Q1: What is the importance of learning decimals in Class 6?
A1: Decimals are foundational for advanced mathematical concepts and real-life applications, such as handling money, measurements, and precise calculations.
Q2: How does Chapter 8 of Class 6 Maths help students?
A2: It covers the basics of decimals, including addition, subtraction, and comparisons, which are essential skills in both academics and everyday life.
Q3: Where can I find detailed solutions for Chapter 8 Decimals?
A3: Detailed solutions are available on rbsesolution.in, helping students with easy-to-understand, step-by-step explanations.
Q4: Why is practice important for mastering decimals?
A4: Practicing decimals improves accuracy and confidence, especially in real-life situations involving decimal-based calculations.
Q5: How can RBSE Solutions help in exam preparation?
A5: They provide a thorough understanding of each topic, making it easier for students to tackle similar questions in exams confidently.
Conclusion
The RBSE Solutions for Class 6 Maths Chapter 8: Decimals provides an essential resource for students looking to excel in mathematics. The solutions not only help with exam preparation but also enhance real-world skills related to decimal calculations. For students, practicing these solutions will boost their confidence and mastery in decimal-related topics, laying a strong foundation for advanced math studies.